Drorbn - User contributions [en]

0708-1300/Class notes for Thursday, January 31

2008-04-25T16:23:58Z

Megan:

{{0708-1300/Navigation}}

==Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

[[Image:0708-1300-January31p1.jpg|400px]]
[[Image:0708-1300-January31p2.jpg|400px]]
[[Image:0708-1300-January31p3.jpg|400px]]
[[Image:0708-1300-January31p4.jpg|400px]]

Template:0708-1300/Navigation

2008-04-25T16:22:50Z

Megan:

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|-
|colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]] [[0708-1300/Class Photo|Add your name / see who's in!]]
|- align=center
!#
!Week of...
!Links
|-
|colspan=3 align=center|'''Fall Semester'''
|- align=left
|align=center|1
|Sep 10
|[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]
|- align=left
|align=center|2
|Sep 17
|[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]
|- align=left
|align=center|3
|Sep 24
|[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]
|- align=left
|align=center|4
|Oct 1
|[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]
|- align=left
|align=center|5
|Oct 8
|Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]
|- align=left
|align=center|6
|Oct 15
|[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]
|- align=left
|align=center|7
|Oct 22
|[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]
|- align=left
|align=center|8
|Oct 29
|[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]
|- align=left
|align=center|9
|Nov 5
|[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]
|- align=left
|align=center|10
|Nov 12
|[[0708-1300/Class notes for Tuesday, November 13|Tue]], <strike>[[0708-1300/Class notes for Thursday, November 15|Thu]]</strike>
|- align=left
|align=center|11
|Nov 19
|[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Class notes for Thursday, November 22|Thu]], [[0708-1300/Homework Assignment 5|HW5]]
|- align=left
|align=center|12
|Nov 26
|[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]
|- align=left
|align=center|13
|Dec 3
|[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Class notes for Thursday, December 6|Thu]], [[0708-1300/Homework Assignment 6|HW6]]
|-
|colspan=3 align=center|'''Spring Semester'''
|- align=left
|align=center|14
|Jan 7
|[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]
|- align=left
|align=center|15
|Jan 14
|[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]
|- align=left
|align=center|16
|Jan 21
|[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]], [[0708-1300/Homework Assignment 8|HW8]]
|- align=left
|align=center|17
|Jan 28
|[[0708-1300/Class notes for Tuesday, January 29|Tue]], [[0708-1300/Class notes for Thursday, January 31|Thu]]
|- align=left
|align=center|18
|Feb 4
|[[0708-1300/Class notes for Tuesday, February 5|Tue]]
|- align=left
|align=center|19
|Feb 11
|[[0708-1300/Term Exam 2|TE2]], [[0708-1300/Class notes for Tuesday, February 12|Tue]], [[0708-1300/Homework Assignment 9|HW9]], [[0708-1300/Class notes for Thursday, February 14|Thu]], Feb 17: last chance to drop class
|- align=left
|align=center|R
|Feb 18
|Reading week
|- align=left
|align=center|20
|Feb 25
|[[0708-1300/Class notes for Tuesday, February 26|Tue]], [[0708-1300/Class notes for Thursday, February 28|Thu]], [[0708-1300/Homework Assignment 10|HW10]]
|- align=left
|align=center|21
|Mar 3
|[[0708-1300/Class notes for Tuesday, March 4|Tue]], [[0708-1300/Class notes for Thursday, March 6|Thu]]
|- align=left
|align=center|22
|Mar 10
|[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Class notes for Thursday, March 13|Thu]], [[0708-1300/Homework Assignment 11|HW11]]
|- align=left
|align=center|23
|Mar 17
|[[0708-1300/Class notes for Tuesday, March 18|Tue]], [[0708-1300/Class notes for Thursday, March 20|Thu]]
|- align=left
|align=center|24
|Mar 24
|[[0708-1300/Class notes for Tuesday, March 25|Tue]], [[0708-1300/Homework Assignment 12|HW12]], [[0708-1300/Class notes for Thursday, March 27|Thu]]
|- align=left
|align=center|25
|Mar 31
|[[0708-1300/Democracy At Last|Referendum]],[[0708-1300/Class notes for Tuesday, April 1|Tue]], [[0708-1300/Class notes for Thursday, April 3|Thu]]
|- align=left
|align=center|26
|Apr 7
|[[0708-1300/Class notes for Tuesday, April 8|Tue]], [[0708-1300/Class notes for Thursday April 10|Thu]],
|- align=left
|align=center|R
|Apr 14
|[[0708-1300/The Final Exam|Office hours]]
|- align=left
|align=center|R
|Apr 21
|[[0708-1300/The Final Exam|Office hours]]
|- align=left
|align=center|F
|Apr 28
|[[0708-1300/The Final Exam|Office hours]], [[0708-1300/The Final Exam|Final]] (Fri, May 2)
|- align=center
|colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]
|- align=center
|colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]
|}
</div></div>
|}
<div align=center style="color: red; font-size: 150%; display: none;">Announcements go here</div>

File:0708-1300-January31p4.jpg

2008-04-25T16:21:28Z

Megan: January 31 notes, page 4 of 4

January 31 notes, page 4 of 4

File:0708-1300-January31p3.jpg

2008-04-25T16:20:40Z

Megan: January 31 notes, page 3 of 4

January 31 notes, page 3 of 4

File:0708-1300-January31p2.jpg

2008-04-25T16:19:35Z

Megan: January 31 notes, page 2 of 4

January 31 notes, page 2 of 4

File:0708-1300-January31p1.jpg

2008-04-25T16:18:36Z

Megan: January 31 notes, page 1 of 4

January 31 notes, page 1 of 4

0708-1300/Class notes for Thursday, November 22

2008-04-25T16:17:26Z

Megan:

File:0708-1300-November22p3.jpg

2008-04-25T16:15:09Z

Megan: November 22 notes, page 3 of 3

November 22 notes, page 3 of 3

File:0708-1300-November22p2.jpg

2008-04-25T16:13:45Z

Megan: November 22 notes, page 2 of 3

November 22 notes, page 2 of 3

File:0708-1300-November22p1.jpg

2008-04-25T16:12:56Z

Megan: November 22 notes, page 1 of 3

November 22 notes, page 1 of 3

Template:0708-1300/Navigation

2008-04-25T16:08:25Z

Megan:

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|-
|colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]] [[0708-1300/Class Photo|Add your name / see who's in!]]
|- align=center
!#
!Week of...
!Links
|-
|colspan=3 align=center|'''Fall Semester'''
|- align=left
|align=center|1
|Sep 10
|[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]
|- align=left
|align=center|2
|Sep 17
|[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]
|- align=left
|align=center|3
|Sep 24
|[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]
|- align=left
|align=center|4
|Oct 1
|[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]
|- align=left
|align=center|5
|Oct 8
|Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]
|- align=left
|align=center|6
|Oct 15
|[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]
|- align=left
|align=center|7
|Oct 22
|[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]
|- align=left
|align=center|8
|Oct 29
|[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]
|- align=left
|align=center|9
|Nov 5
|[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]
|- align=left
|align=center|10
|Nov 12
|[[0708-1300/Class notes for Tuesday, November 13|Tue]], <strike>[[0708-1300/Class notes for Thursday, November 15|Thu]]</strike>
|- align=left
|align=center|11
|Nov 19
|[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Class notes for Thursday, November 22|Thu]], [[0708-1300/Homework Assignment 5|HW5]]
|- align=left
|align=center|12
|Nov 26
|[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]
|- align=left
|align=center|13
|Dec 3
|[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Class notes for Thursday, December 6|Thu]], [[0708-1300/Homework Assignment 6|HW6]]
|-
|colspan=3 align=center|'''Spring Semester'''
|- align=left
|align=center|14
|Jan 7
|[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]
|- align=left
|align=center|15
|Jan 14
|[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]
|- align=left
|align=center|16
|Jan 21
|[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]], [[0708-1300/Homework Assignment 8|HW8]]
|- align=left
|align=center|17
|Jan 28
|[[0708-1300/Class notes for Tuesday, January 29|Tue]]
|- align=left
|align=center|18
|Feb 4
|[[0708-1300/Class notes for Tuesday, February 5|Tue]]
|- align=left
|align=center|19
|Feb 11
|[[0708-1300/Term Exam 2|TE2]], [[0708-1300/Class notes for Tuesday, February 12|Tue]], [[0708-1300/Homework Assignment 9|HW9]], [[0708-1300/Class notes for Thursday, February 14|Thu]], Feb 17: last chance to drop class
|- align=left
|align=center|R
|Feb 18
|Reading week
|- align=left
|align=center|20
|Feb 25
|[[0708-1300/Class notes for Tuesday, February 26|Tue]], [[0708-1300/Class notes for Thursday, February 28|Thu]], [[0708-1300/Homework Assignment 10|HW10]]
|- align=left
|align=center|21
|Mar 3
|[[0708-1300/Class notes for Tuesday, March 4|Tue]], [[0708-1300/Class notes for Thursday, March 6|Thu]]
|- align=left
|align=center|22
|Mar 10
|[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Class notes for Thursday, March 13|Thu]], [[0708-1300/Homework Assignment 11|HW11]]
|- align=left
|align=center|23
|Mar 17
|[[0708-1300/Class notes for Tuesday, March 18|Tue]], [[0708-1300/Class notes for Thursday, March 20|Thu]]
|- align=left
|align=center|24
|Mar 24
|[[0708-1300/Class notes for Tuesday, March 25|Tue]], [[0708-1300/Homework Assignment 12|HW12]], [[0708-1300/Class notes for Thursday, March 27|Thu]]
|- align=left
|align=center|25
|Mar 31
|[[0708-1300/Democracy At Last|Referendum]],[[0708-1300/Class notes for Tuesday, April 1|Tue]], [[0708-1300/Class notes for Thursday, April 3|Thu]]
|- align=left
|align=center|26
|Apr 7
|[[0708-1300/Class notes for Tuesday, April 8|Tue]], [[0708-1300/Class notes for Thursday April 10|Thu]],
|- align=left
|align=center|R
|Apr 14
|[[0708-1300/The Final Exam|Office hours]]
|- align=left
|align=center|R
|Apr 21
|[[0708-1300/The Final Exam|Office hours]]
|- align=left
|align=center|F
|Apr 28
|[[0708-1300/The Final Exam|Office hours]], [[0708-1300/The Final Exam|Final]] (Fri, May 2)
|- align=center
|colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]
|- align=center
|colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]
|}
</div></div>
|}
<div align=center style="color: red; font-size: 150%; display: none;">Announcements go here</div>

0708-1300/Class notes for Thursday April 10

2008-04-25T16:06:03Z

Megan:

File:0708-1300-April10p2.jpg

2008-04-25T16:05:16Z

Megan: April 10 notes, page 2 of 2

April 10 notes, page 2 of 2

File:0708-1300-April10p1.jpg

2008-04-25T16:04:55Z

Megan: April 10 notes, page 1 of 2

April 10 notes, page 1 of 2

0708-1300/Class notes for Tuesday, April 8

2008-04-25T16:03:28Z

Megan:

{{0708-1300/Navigation}}

==Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

[[Image:0708-1300-april8p1.jpg|400px]]
[[Image:0708-1300-April8p2.jpg|400px]]
[[Image:0708-1300-April8p3.jpg|400px]]
[[Image:0708-1300-April8p4.jpg|400px]]
[[Image:0708-1300-April8p5.jpg|400px]]
[[Image:0708-1300-April8p6.jpg|400px]]

0708-1300/Class notes for Tuesday, April 8

2008-04-25T16:02:00Z

Megan:

==Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

[[Image:0708-1300-april8p1.jpg|400px]]
[[Image:0708-1300-April8p2.jpg|400px]]
[[Image:0708-1300-April8p3.jpg|400px]]
[[Image:0708-1300-April8p4.jpg|400px]]
[[Image:0708-1300-April8p5.jpg|400px]]
[[Image:0708-1300-April8p6.jpg|400px]]

File:0708-1300-April8p6.jpg

2008-04-25T15:58:36Z

Megan: April 8 notes, page 6 of 6

April 8 notes, page 6 of 6

File:0708-1300-April8p5.jpg

2008-04-25T15:58:18Z

Megan: April 8 notes, page 5 of 6

April 8 notes, page 5 of 6

File:0708-1300-April8p4.jpg

2008-04-25T15:58:00Z

Megan: April 8 notes, page 4 of 6

April 8 notes, page 4 of 6

File:0708-1300-April8p3.jpg

2008-04-25T15:57:40Z

Megan: April 8 notes, page 3 of 6

April 8 notes, page 3 of 6

File:0708-1300-April8p2.jpg

2008-04-25T15:57:15Z

Megan: April 8 notes, page 2 of 6

April 8 notes, page 2 of 6

File:0708-1300-april8p1.jpg

2008-04-25T15:56:10Z

Megan: April 8 notes, page 1 of 6

April 8 notes, page 1 of 6

Template:0708-1300/Navigation

2008-04-25T15:39:37Z

Megan:

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|-
|colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]] [[0708-1300/Class Photo|Add your name / see who's in!]]
|- align=center
!#
!Week of...
!Links
|-
|colspan=3 align=center|'''Fall Semester'''
|- align=left
|align=center|1
|Sep 10
|[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]
|- align=left
|align=center|2
|Sep 17
|[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]
|- align=left
|align=center|3
|Sep 24
|[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]
|- align=left
|align=center|4
|Oct 1
|[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]
|- align=left
|align=center|5
|Oct 8
|Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]
|- align=left
|align=center|6
|Oct 15
|[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]
|- align=left
|align=center|7
|Oct 22
|[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]
|- align=left
|align=center|8
|Oct 29
|[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]
|- align=left
|align=center|9
|Nov 5
|[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]
|- align=left
|align=center|10
|Nov 12
|[[0708-1300/Class notes for Tuesday, November 13|Tue]], <strike>[[0708-1300/Class notes for Thursday, November 15|Thu]]</strike>
|- align=left
|align=center|11
|Nov 19
|[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Homework Assignment 5|HW5]]
|- align=left
|align=center|12
|Nov 26
|[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]
|- align=left
|align=center|13
|Dec 3
|[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Class notes for Thursday, December 6|Thu]], [[0708-1300/Homework Assignment 6|HW6]]
|-
|colspan=3 align=center|'''Spring Semester'''
|- align=left
|align=center|14
|Jan 7
|[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]
|- align=left
|align=center|15
|Jan 14
|[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]
|- align=left
|align=center|16
|Jan 21
|[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]], [[0708-1300/Homework Assignment 8|HW8]]
|- align=left
|align=center|17
|Jan 28
|[[0708-1300/Class notes for Tuesday, January 29|Tue]]
|- align=left
|align=center|18
|Feb 4
|[[0708-1300/Class notes for Tuesday, February 5|Tue]]
|- align=left
|align=center|19
|Feb 11
|[[0708-1300/Term Exam 2|TE2]], [[0708-1300/Class notes for Tuesday, February 12|Tue]], [[0708-1300/Homework Assignment 9|HW9]], [[0708-1300/Class notes for Thursday, February 14|Thu]], Feb 17: last chance to drop class
|- align=left
|align=center|R
|Feb 18
|Reading week
|- align=left
|align=center|20
|Feb 25
|[[0708-1300/Class notes for Tuesday, February 26|Tue]], [[0708-1300/Class notes for Thursday, February 28|Thu]], [[0708-1300/Homework Assignment 10|HW10]]
|- align=left
|align=center|21
|Mar 3
|[[0708-1300/Class notes for Tuesday, March 4|Tue]], [[0708-1300/Class notes for Thursday, March 6|Thu]]
|- align=left
|align=center|22
|Mar 10
|[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Class notes for Thursday, March 13|Thu]], [[0708-1300/Homework Assignment 11|HW11]]
|- align=left
|align=center|23
|Mar 17
|[[0708-1300/Class notes for Tuesday, March 18|Tue]], [[0708-1300/Class notes for Thursday, March 20|Thu]]
|- align=left
|align=center|24
|Mar 24
|[[0708-1300/Class notes for Tuesday, March 25|Tue]], [[0708-1300/Homework Assignment 12|HW12]], [[0708-1300/Class notes for Thursday, March 27|Thu]]
|- align=left
|align=center|25
|Mar 31
|[[0708-1300/Democracy At Last|Referendum]],[[0708-1300/Class notes for Tuesday, April 1|Tue]], [[0708-1300/Class notes for Thursday, April 3|Thu]]
|- align=left
|align=center|26
|Apr 7
|[[0708-1300/Class notes for Tuesday, April 8|Tue]], [[0708-1300/Class notes for Thursday April 10|Thu]],
|- align=left
|align=center|R
|Apr 14
|[[0708-1300/The Final Exam|Office hours]]
|- align=left
|align=center|R
|Apr 21
|[[0708-1300/The Final Exam|Office hours]]
|- align=left
|align=center|F
|Apr 28
|[[0708-1300/The Final Exam|Office hours]], [[0708-1300/The Final Exam|Final]] (Fri, May 2)
|- align=center
|colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]
|- align=center
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0708-1300/Class notes for Tuesday, January 15

2008-02-04T23:34:38Z

Megan: /* First Hour */

{{0708-1300/Navigation}}

==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

We begin by reformulating our previous lemma into a more general form:

'''Lemma 1'''

Let <math>p:(X,x_0)\rightarrow(B,b_0)</math> be a covering map. Then every family of paths <math>\gamma:Y\times I\rightarrow B </math> such that <math>\forall y\in Y\ \gamma(y,0)=b_0</math> has a unique lift <math>\tilde{\gamma}:Y\times I\rightarrow X</math> such that

0) <math>\forall y\ \tilde{\gamma}(y,0) = x_0</math>

1) <math>\gamma = p\circ\tilde{\gamma}</math>

Claim 1: <math>Ind[\gamma]</math> is well defined, hence <math>\pi_1(S^1)\cong\mathbb{Z}</math>
<math>(\gamma:I\rightarrow S, \gamma(0)=1, ind \gamma = \tilde{\gamma})</math>

''Proof of Claim 1''

We construct the homotopy H between two such gamma's which, schematically, is a square with <math>\gamma_1</math> along the bottom, <math>\gamma_0</math> along the top and along the side the parameter Y where the homotopy is just horizontal lines between <math>\gamma_1</math> and <math>\gamma_0.</math>

Then, the lemma implies the existence of a homotopy <math>\tilde{H}</math> which is schematically a square with <math>\tilde{\gamma_1}</math> along the bottom, <math>\tilde{\gamma_0}</math> along the top, x=0 on the left and side and the right hand side taking values in <math>p^{-1}(1)</math>

''Proof of Lemma 1''

Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each <math>y\in Y</math> we can get a <math>\tilde{\gamma}</math>

We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about <math>y_0</math>.

Now by a "good" open set in B, we mean that the preimage under p CAN be written as a product.

Hence, we choose a neighborhood about the first interval in <math>\mathbb{R}</math> extending from <math>y_0</math> (see proof of one path case for explanation) and this gets mapped to a "good" open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the "good" open set in B.

Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.

'''Applications of <math>\pi_1(S^1)\cong\mathbb{Z}</math>'''

1) We get again the proof of non existence of a retract <math>r:D^2\rightarrow S^1</math>

Indeed, assume we DID have such a retract. We would have the following commuting diagram.

<math>\begin{matrix}
D^2&\rightarrow^{r}&S^1\\
\uparrow &\nearrow_{I}& \\
S^1&&\\
\end{matrix}</math>

Applying the functor <math>\pi_1</math> would yield the diagram

<math>\begin{matrix}
\{0\}&\rightarrow^{r}&\mathbb{Z}\\
\uparrow &\nearrow_{I}& \\
\mathbb{Z}&&\\
\end{matrix}</math>

But clearly this does not exist, as the only map from <math>\{0\}</math> is the trivial one.

Recall the non existence of such retracts implies ''Brouwer's Theorem''

2) '''Fundamental Theorem of Algebra'''

if <math>p\in\mathbb{C}[z]</math> is a polynomial with degree greater than zero then <math>\exists z_0\in\mathbb{C}</math> such that <math>p(z_0) =0</math>

''Proof:''

Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, <math>p = z^n +</math> lower order terms.

Define a homotopy of paths <math>S^1\rightarrow S^1</math> by first setting <math>q(z) = p(z)/||p(z)||]\in S^1</math>

Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial <math>p_{\alpha}(z) = z^n + \alpha</math>(lower order terms) and let <math>\alpha</math> tend from 1 to 0 in the second half of the homotopy.

Hence, at the end of the homotopy we are left with <math>z^n/||z^n||</math> on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction.

3) '''Boruk-Ulam'''

If <math>f:S^2\rightarrow \mathbb{R}^2</math> is continuous then <math>\exists p\in S^2</math> such that <math>f(p) = f(-p)</math>

Corollary:

1) <math>S^2</math> is not a subset of <math>\mathbb{R}^2</math> (cannot be embedded) as f is not 1:1

2) If <math>S^2 =A_1\cup A_2\cup A_3</math>, with <math>A_i</math> closed then at least one <math>A_i</math> contains a pair of antipodes.

'''Example:'''

This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.

===Second Hour===

''Proof of Corollary:''

Consider <math>f:S^2\rightarrow\mathbb{R}</math> by <math>p\mapsto</math> dist to <math>A_1</math> + dist to <math>A_2</math>

If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same <math>A_i</math>.

'''Definition:'''

1) <math>\gamma:S^1\rightarrow S^1</math> is even if <math>\gamma(-x) = \gamma(x)\ \forall x</math>

2) <math>\gamma</math> is odd if <math>\gamma(-x) = -\gamma(x)\ \forall x</math>

'''Lemma:'''

1) If <math>\gamma</math> is even then ind <math>\gamma</math> is even

2) If <math>\gamma</math> is odd then ind <math>\gamma</math> is odd

''Proof of Borsuk-Ulam:''

Assume <math>f:S^2\rightarrow\mathbb{R}</math> has no p with f(p)=f(-p)

Define <math>g:S^2\rightarrow S^1</math> by <math>g(p) = (f(p)-f(-p))/||f(p)-f(-p)||</math>

<math>\gamma = g|_{equator}:S^1\rightarrow S^1</math> is odd. Therefore, <math>[\gamma]\in\pi_1(S^1)</math> is non zero.

But <math>\gamma</math> is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)

''Proof of part 1 of lemma:''

Suppose <math>\gamma:S^1\rightarrow S^1</math> is even

We can make a commuting diagram where <math>\gamma:S^1\rightarrow S^1</math> is the same as going from <math>S^1\rightarrow S^1</math> first via <math>z\mapsto z^2</math> and then from <math>S^1\rightarrow S^1</math> via <math>\lambda(z) = \gamma(\sqrt{z})</math>

We now consider this diagram under the functor <math>\pi_1</math>

We get <math>\mathbb{Z}\rightarrow^{\times 2}\mathbb{Z}\rightarrow^{\lambda_*}\mathbb{Z}</math>

which commutes with <math>\mathbb{Z}\rightarrow^{\gamma_*}\mathbb{Z}</math> along the bottom

<math>[\gamma] = \gamma_*[\gamma_1] = \lambda_*(2[\gamma_1])= 2\lambda_*([\gamma_1])</math> which is even.

(Note, by <math>\gamma_1:S^1\rightarrow S^1</math> we mean the identity)

Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first.

'''Definition'''

A ''topological group'' is a topological space G with a group structure such that all group operations are continuous

'''Examples:'''

1) <math>S^1</math> (with rotation giving the group structure)

2) SO(3) (See a previous lecture for more info)

If G is a topological group there are two "products" we can define on <math>\pi_1(G)</math>:

1) <math>[\gamma][\lambda] = [\gamma\circ\lambda]</math> with <math>\circ</math> being simply concatenation

2) <math>[\gamma]*[\lambda] = [\gamma *\lambda]</math> where <math>(\gamma*\lambda)(t) = \gamma(t)\cdot\lambda(t)</math>

Claim:

These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here.

We return to the proof of the odd case of the lemma:

Assume <math>\gamma</math> is odd. Then <math>\gamma*\gamma_1</math> is even so <math>2\mathbb{Z}\in[\gamma*\gamma_1]=[\gamma\circ\gamma_1] = [\gamma] + [\gamma_1] = [\gamma] + 1</math>. Therefore, <math>[\gamma]</math> is odd.

Note: The addition above is done in <math>\pi_1(S^1)</math> which is just <math>\mathbb{Z}</math>

''Q.E.D.''

'''Claim:'''

If <math>f,g:X\rightarrow Y</math> are homotopic then <math>f_* = g_*:\pi_1(X)\rightarrow\pi_1(Y)</math>

Consider a loop <math>\gamma</math> in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop.

Then, <math>f\circ\gamma</math> is homotopic to <math>g\circ\gamma</math> using <math>H\circ\gamma</math> as the homotopy.

'''Theorem'''

1) <math>\sim</math> is an equivalence relation.

2) It is an ''ideal'' in the category of topological spaces. That is, <math>f\sim g \Rightarrow f\circ h\sim g\circ h</math> and <math>h\circ f \sim h\circ g</math> for h's such that these make sense.

'''Definition'''

X and Y are "homotopy equivalent" if they are isomorphic in {Topological Spaces} / {homotopy of maps}

I.e., <math>\exists f,g</math> so that <math>f\circ g\sim I_Y</math> and <math>g\circ f\sim I_X</math> for <math>f:X\rightarrow Y</math> and <math>g:Y\rightarrow X</math>

'''Example'''

1) <math>\mathbb{R}^n</math> is homotopy equivalent to a point via f which takes <math>\mathbb{R}^n</math> to a point and g is the zero map

2) An annulus is homotopy equivalent to <math>S^1</math>. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity.

3) Likewise for the figure "thick A" which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity.

4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle.

This last example gives us some idea of the limitation, or "lack of subtlety" to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference.

'''Claim'''

If X and Y are homotopically equivalent then <math>\pi_1(X)\cong\pi_1(Y)</math>

''Proof:''

Consider <math>X\rightarrow^f Y</math> and <math>Y\rightarrow^g X</math> forming a commuting diagram and lets consider its image under <math>\pi_1</math>

Then we get <math>\pi_1(X)\rightarrow^{f_*} \pi_1(Y)</math> and <math>\pi_1 (Y)\rightarrow^{g_8} \pi_1(X</math>) as a commuting diagram.

We know that <math>f\circ g</math> is homotopic to the identity, and thus we also get <math>f_*\circ g_*</math> homotopic to the identity

''Q.E.D.''

'''Theorem'''

<math>\pi_1(X+Y)\cong\pi_1(X)+\pi_1(Y)</math>

Example: <math>\pi_1(\mathbb{T}^2) = \pi_1(S^1\times S^1) = \mathbb{Z}\times\mathbb{Z} = \mathbb{Z}^2</math>

0708-1300/Class notes for Tuesday, January 8

2008-01-09T05:16:18Z

Megan:

{{0708-1300/Navigation}}

==Algebraic Topology==

Temporarily, at least, we will no longer assume that all structures are manifolds and that all functions are smooth. However, functions will still be assumed to be continuous.

We will also, temporarily, assume that all spaces are pointed spaces. That is, a space X will be assumed to have a distinguished point x, whether mentioned explicitly or not.

General idea of algebraic topology: to find "functors" from topology to algebra.

'''Informal definition:'''
A category C consists of

1) a collection Obj(C) of objects

2) a class hom(C) of morphisms between these objects, so that for each X1 and X2 in Obj(C) we get a set mor{X1, X2}

3) Composition laws:

a) a binary function on hom(C): mor(X1, X2) x mor (X2, X3 → mor (X1, X3)

b) Identities (for every X, there is an IX in mor(X,X))

4) Compatibility laws:

a) associativity of composition

b) identities are in fact identities.

"Functors" preserve categories.

'''Example:'''
Recall that the Brouwer Fixed Point Theorem is implied by the statement that there is no retract from the n-dimensional disc Dn to the (n-1)-dimensional sphere Sn-1. Later, we will prove this last statement in its categorical reformulation: we will find a functor H such that H(D) = {0} and H(Sn-1) = '''Z'''. So the existence of a retract would imply that there are some functions f: '''Z''' → {0} and g: {0} → '''Z''' such that gf is the identity. Clearly this is false, so no retract exists.

'''Definition:'''
Given (X, x0) (which, by the above comment, we will sometimes write simply as X) we define the "fundamental group" or "Poincaré group" of X:

π1 (X, x0) := { [γ]: γ : [0,1] → X, γ(0) = γ(1) = x0}

Now we have to:

1) Define homotopy and prove that it's an equivalence relation

2) Define a binary operation, show that it's well-defined, and then show that the fundamental group is in fact a group.

3) Demonstrate functoriality

'''Definition:'''
If f, g : (X, x0) → (Y, y0), we say that "f is homotopic to g" (and we write f ~ g) if there exists a function H : [0,1] × X → Y such that H restricted to {0}×X is f, H restricted to {1}×X is g, and H at any time t is in our category (ie H(t, x0) = y0.)

We know now go through four boring claims, labeled BC1 - BC4. Many have simple visual proofs which I will not reproduce here.

'''BC1:'''

~ is an equivalence relation.

'''Proof of BC1:'''

a) f ~ f for all functions f. Indeed, set H(t, x) = f(x) for all x.

b) If f ~ g (by a function H) then g ~ f. Indeed, set H'(t,x) = H(1-t, x).

c) If f ~ g (by a function H1) and g ~ h (by a function H2, then f ~ h. Indeed, set H(t, x) = H1(2t,x) if t < ½ and H(t,x) = H2(2t-1, x) if t > ½.

'''Definition:'''
We define a binary operation on the equivalence classes of paths:

[γ] • [γ'] = [γ • γ']

where the • represents the concatenation of paths. Intuitively, we follow the first path at twice its normal speed, then follow the second, again at twice its normal speed.

'''BC2:'''
This is well-defined,

'''BC3:'''
π1 (X, x0) is a group.

'''Proofette of BC3:'''

Associativity: Let γ0, γ1 and γ2 be paths.
Note that while (γ0 • γ1) • γ2 is not the same path as γ0 • (γ1 • γ2), they are homotopic as paths.

Identity: [e] = [x0], where x0 is the constant path.

Inverse: [γ]-1 = [γ "tilda"], where γ"tilda"(t) = γ(t).

'''BC4:''' π1 is a functor.

<math> f_* </math> [γ] = [<math> f_* </math> γ] = [f • γ] defines <math> f_* </math> : π1(f) : π1(X) → π1(Y).

To check (all are clear):

1) This is well-defined.

2) It respects compositions.

3) It respects the identity.

4) <math> (gf)_* </math> = <math> f_* </math> • <math> g_* </math>.

'''Examples:'''

1) π1(x0, x0) = {e}.

2) π1 ('''R'''n, 0) = {e}

3) π1 (S1, 1) = '''Z'''.

0708-1300/Class notes for Tuesday, January 8

2008-01-09T05:14:33Z

Megan:

==Algebraic Topology==

Temporarily, at least, we will no longer assume that all structures are manifolds and that all functions are smooth. However, functions will still be assumed to be continuous.

We will also, temporarily, assume that all spaces are pointed spaces. That is, a space X will be assumed to have a distinguished point x, whether mentioned explicitly or not.

General idea of algebraic topology: to find "functors" from topology to algebra.

'''Informal definition:'''
A category C consists of

1) a collection Obj(C) of objects

2) a class hom(C) of morphisms between these objects, so that for each X1 and X2 in Obj(C) we get a set mor{X1, X2}

3) Composition laws:

a) a binary function on hom(C): mor(X1, X2) x mor (X2, X3 → mor (X1, X3)

b) Identities (for every X, there is an IX in mor(X,X))

4) Compatibility laws:

a) associativity of composition

b) identities are in fact identities.

"Functors" preserve categories.

'''Example:'''
Recall that the Brouwer Fixed Point Theorem is implied by the statement that there is no retract from the n-dimensional disc Dn to the (n-1)-dimensional sphere Sn-1. Later, we will prove this last statement in its categorical reformulation: we will find a functor H such that H(D) = {0} and H(Sn-1) = '''Z'''. So the existence of a retract would imply that there are some functions f: '''Z''' → {0} and g: {0} → '''Z''' such that gf is the identity. Clearly this is false, so no retract exists.

'''Definition:'''
Given (X, x0) (which, by the above comment, we will sometimes write simply as X) we define the "fundamental group" or "Poincaré group" of X:

π1 (X, x0) := { [γ]: γ : [0,1] → X, γ(0) = γ(1) = x0}

Now we have to:

1) Define homotopy and prove that it's an equivalence relation

2) Define a binary operation, show that it's well-defined, and then show that the fundamental group is in fact a group.

3) Demonstrate functoriality

'''Definition:'''
If f, g : (X, x0) → (Y, y0), we say that "f is homotopic to g" (and we write f ~ g) if there exists a function H : [0,1] × X → Y such that H restricted to {0}×X is f, H restricted to {1}×X is g, and H at any time t is in our category (ie H(t, x0) = y0.)

We know now go through four boring claims, labeled BC1 - BC4. Many have simple visual proofs which I will not reproduce here.

'''BC1:'''

~ is an equivalence relation.

'''Proof of BC1:'''

a) f ~ f for all functions f. Indeed, set H(t, x) = f(x) for all x.

b) If f ~ g (by a function H) then g ~ f. Indeed, set H'(t,x) = H(1-t, x).

c) If f ~ g (by a function H1) and g ~ h (by a function H2, then f ~ h. Indeed, set H(t, x) = H1(2t,x) if t < ½ and H(t,x) = H2(2t-1, x) if t > ½.

'''Definition:'''
We define a binary operation on the equivalence classes of paths:

[γ] • [γ'] = [γ • γ']

where the • represents the concatenation of paths. Intuitively, we follow the first path at twice its normal speed, then follow the second, again at twice its normal speed.

'''BC2:'''
This is well-defined,

'''BC3:'''
π1 (X, x0) is a group.

'''Proofette of BC3:'''

Associativity: Let γ0, γ1 and γ2 be paths.
Note that while (γ0 • γ1) • γ2 is not the same path as γ0 • (γ1 • γ2), they are homotopic as paths.

Identity: [e] = [x0], where x0 is the constant path.

Inverse: [γ]-1 = [γ "tilda"], where γ"tilda"(t) = γ(t).

'''BC4:''' π1 is a functor.

<math> f_* </math> [γ] = [<math> f_* </math> γ] = [f • γ] defines <math> f_* </math> : π1(f) : π1(X) → π1(Y).

To check (all are clear):

1) This is well-defined.

2) It respects compositions.

3) It respects the identity.

4) <math> (gf)_* </math> = <math> f_* </math> • <math> g_* </math>.

'''Examples:'''

1) π1(x0, x0) = {e}.

2) π1 ('''R'''n, 0) = {e}

3) π1 (S1, 1) = '''Z'''.

0708-1300/Class notes for Tuesday, January 8

2008-01-09T03:59:23Z

Megan:

Algebraic Topology

Temporarily, at least, we will no longer assume that all structures are manifolds and that all functions are smooth. However, functions will still be assumed to be continuous.

We will also, temporarily, assume that all spaces are pointed spaces. That is, a space X will be assumed to have a distinguished point x, whether mentioned explicitely or not.

General idea of algebraic topology: to find "functors" from topology to algebra.

'''Informal definition:'''
A category C consists of

1) a collection Obj(C) of objects

2) a class hom(C) of morphisms between these objects, so that for each X1 and X2 in Obj(C) we get a set mor{X1, X2}

3) Composition laws:

a) a binary function on hom(C): mor(X1, X2) x mor (X2, X3 --> mor (X1, X3)

b) Identities (for every X, there is an IX in mor(X,X))

4) Compatibility laws:

a) associativity of composition

b) identities are in fact identities.

Template:0708-1300/Navigation

2008-01-09T03:30:25Z

Megan:

0708-1300/Barnie the polar bear

2007-11-19T16:47:17Z

Megan:

Barnie is a polar bear. Every morning, Barnie walks to the south x steps then x steps to the east, x to the north and finally x to the west. He do this to keep its shape. He is a little bit lacy and every day he walks for a smaller time, so x is decreasing every time. Of course, every day Barnie ends his trip at home.

But, we are going to change a little bit Barnie's fairy world. We are going to give Barnie two linearly independent vector fields (<math>X</math> and <math>Y</math>) on the earth and Barnie should follow the local flows generated by these vector fields during his journey.

Poor Barney! It can happen that at the end of his walk he won't end at home and he will have to walk a little more to go home. But not too much. In fact, all the ends of his walks form a curve which is almost constant at home!! (Its derivative is zero). The only problem is the second derivative which is twice <math>[X,Y]</math> at home. Even more if <math>[X,Y]</math> is zero then Barnie will be happy again, ending every day at home at the end of his walk.

'''References'''

Spivak M. ''A Comprehensive Introduction to Differential Geometry'' Vol 1

0708-1300/Class notes for Thursday, October 25

2007-11-05T19:59:25Z

Megan: /* Typed Notes */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General Comments'''

1) Exam is on Nov 8th

2) Specs for the exam will be given next thursday.

'''Aside 2'''

''Classification of Surfaces''

1) Every 2 dimensional manifold that is compact connected and without boundary is smoothly equivalent to one of the following:

a) There is a family of such surfaces consisting of the sphere <math>\mathbb{S}^2</math>, the torus <math>\mathbb{R}^2</math> and various connected sums of many tori.

b) A connected sum of <math>\mathbb{RP}^2</math>'s, the real projective plane. This is formed by taking the sphere <math>\mathbb{S}^2</math> and identifying antipodal points.

''Definition''

A manifold without boundary has every point locally homeomorphic to a (relatively) open subset of <math>H^n=\{(x_1,\ldots,x_n)| x_n\geq 0\}</math>

2) If we now consider such manifolds WITH boundary they will be equivalent to something in case one only with open disks removed.

'''Aside 1'''

''Classification of 1-manifolds''

1) A compact 1-manifold without boundary is smoothly equivalent to a circle

2) A compact 1-manifold with boundary is smoothly equivalent to a closed interval

3) A non compact 1-manifold with boundary is smoothly equivalent to a half open interval

4) A non compact 1-manifold without boundary is smoothly equivalent to an open interval

'''Continuing with the Proof of the Whitney Embedding Theorem'''

We recall we are in part 3a of the proof.

The main steps we did in this last class were repeated again and so I won't do that here; however, there was an analogy to the proof presented that I will comment on.

Recall we have the idea of a ''remoteness function'' (that last class we called s) that we call r. If we were to consider the slice of the manifold with a certain value of r we will have problems with the fact that an embedding on one slice won't smoothly transition to an embedding on an adjacent slice. This problem is not solved from moving from a slice to disjoint intervals of values of r. However we get a nice analogy for this case:

We consider a roll of film on which each section (being the subset (not submanifold) of the manifold with values of r in some interval) occurring on each image in the roll of film. Thus as we move along the film we get to see the section with larger and larger r values. Indeed we could cut each image on the film up separately and lay them on top of each other in a stack. We would like this stack to correspond to an atlas. The problem of course is in the smoothness of the transition from one image to another.

The way we resolve this problem is to think of it as follows. We take an image and then shrink it down to zero and when we blow it back up again it is a new image. This is akin to turning the bulb off, then changing the film and then turning the bulb back on again.

In computer graphics, there is a problem where one tries to draw an image but that the processing time to draw the image is greater then the rate at which images are displayed. And hence part of the image is redrawn while part remains the same resulting in a distorted image. The solution is two have two buffers in memory. The image is displayed from one buffer while the next image is calculated and stored in the second buffer. Once the new image is completely drawn in it displayed on the screen and the original buffer is used to draw/compute the next image.

We now return to last class where we have defined the function

<math>\Phi(p) = (\Phi_{even},\ \Phi_{odd},\ r(p))</math>
that went from the manifold into <math>\mathbb{R}^{4m+3}</math>. (as each <math>\Phi</math> has dimension 2m+1 and an extra dimension for the r)

Here the <math>\Phi_{even}</math> and <math>\Phi_{odd}</math> correspond to the two buffers in the analogy.

Now this will turn out to be an embedding and we can reproduce our point on the manifold through use of the r. Indeed, the function is 1:1. Indeed for two points x ≠ y:

a) if we are near the same value of n then we are in the same <math>\phi</math>. Since each <math>\phi</math> is an embedding (hence one-to-one) <math>\Phi(x)</math> ≠ <math> \Phi(y)</math>.

b) if we are not near the same n, then we will have different r's necessarily, so again <math>\Phi(x)</math> ≠ <math> \Phi(y)</math>.

'''Claim'''

A proper 1:1 continuous function is a homeomorphism onto its image.

i.e., we need to show a proper function <math>\Rightarrow</math> closed.

''Note''

This claim shows that our map <math>\Phi</math> is an embedding (provided such a proper r exists)

''Summary of what is left to do:''

1) Prove this claim

2) Prove the existence of such an r

3) Show how we can reduce from <math>\mathbb{R}^{4m+3}</math> to <math>\mathbb{R}^{2m+1}</math>

''Proof of 3''

We can repeat the arguments used in part 2 of this proof again (the lack of compactness is not problem, as Sard's theorem and the dimension reducing argument doesn't depend on it)

However, we do have to make the following modification. In addition to not choosing a projection direction in im<math>\beta_1</math> <math>\cup</math> im<math>\beta_2</math> we have to also make sure not to project in the vertical direction down into the r axis. But this is still quite possible by Sard's Theorem and hence we can reduce down to an embedding into <math>\mathbb{R}^{2m+1}</math>

''Proof of 2''

Consider <math>\{\lambda_{\alpha}\}</math> a partition of unity subordinant to a countable cover by sets with compact closure. We can make this partition be countable as each chart with compact closure only needs finitely many <math>\lambda_{\alpha}</math>'s.

Hence we get a partition of unity <math>\{\lambda_k\}_{k\in\mathbb{N}}</math> with supp<math>\lambda_k</math> being compact.

Set <math>r:=\sum k\lambda_k</math>

''Claim:''
r is proper

''Proof'

Assume <math>p\notin\bigcup_{k<n} supp \lambda_k</math>

then <math>1=\sum_{k=1}^{\infty}\lambda_k (p) = \sum_{k=n}^{\infty}\lambda_k (p)</math>

<math>\Rightarrow</math>

<math>\sum_{k=1}^{\infty}k\lambda_k(p) =\sum_{k=n}^{\infty}k\lambda_k(p)\geq \sum_{k=n}^{\infty}n\lambda_k(p) = n\sum_{k=n}^{\infty}\lambda_k(p) =n</math>

so, <math>r^{-1}([0,n])\subset\bigcup_{k=1}^{n-1}supp\lambda_k</math>.

So compact

''Q.E.D''

0708-1300/Class notes for Thursday, October 25

2007-11-05T19:42:58Z

Megan: /* Typed Notes */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General Comments'''

1) Exam is on Nov 8th

2) Specs for the exam will be given next thursday.

'''Aside 2'''

''Classification of Surfaces''

1) Every 2 dimensional manifold that is compact connected and without boundary is smoothly equivalent to one of the following:

a) There is a family of such surfaces consisting of the sphere <math>\mathbb{S}^2</math>, the torus <math>\mathbb{R}^2</math> and various connected sums of many tori.

b) A connected sum of <math>\mathbb{RP}^2</math>'s, the real projective plane. This is formed by taking the sphere <math>\mathbb{S}^2</math> and identifying antipodal points.

''Definition''

A manifold without boundary has every point locally homeomorphic to a (relatively) open subset of <math>H^n=\{(x_1,\ldots,x_n)| x_n\geq 0\}</math>

2) If we now consider such manifolds WITH boundary they will be equivalent to something in case one only with open disks removed.

'''Aside 1'''

''Classification of 1-manifolds''

1) A compact 1-manifold without boundary is smoothly equivalent to a circle

2) A compact 1-manifold with boundary is smoothly equivalent to a closed interval

3) A non compact 1-manifold with boundary is smoothly equivalent to a half open interval

4) A non compact 1-manifold without boundary is smoothly equivalent to an open interval

'''Continuing with the Proof of the Whitney Embedding Theorem'''

We recall we are in part 3a of the proof.

The main steps we did in this last class were repeated again and so I won't do that here; however, there was an analogy to the proof presented that I will comment on.

Recall we have the idea of a ''remoteness function'' (that last class we called s) that we call r. If we were to consider the slice of the manifold with a certain value of r we will have problems with the fact that an embedding on one slice won't smoothly transition to an embedding on an adjacent slice. This problem is not solved from moving from a slice to disjoint intervals of values of r. However we get a nice analogy for this case:

We consider a roll of film on which each section (being the subset (not submanifold) of the manifold with values of r in some interval) occurring on each image in the roll of film. Thus as we move along the film we get to see the section with larger and larger r values. Indeed we could cut each image on the film up separately and lay them on top of each other in a stack. We would like this stack to correspond to an atlas. The problem of course is in the smoothness of the transition from one image to another.

The way we resolve this problem is to think of it as follows. We take an image and then shrink it down to zero and when we blow it back up again it is a new image. This is akin to turning the bulb off, then changing the film and then turning the bulb back on again.

In computer graphics, there is a problem where one tries to draw an image but that the processing time to draw the image is greater then the rate at which images are displayed. And hence part of the image is redrawn while part remains the same resulting in a distorted image. The solution is two have two buffers in memory. The image is displayed from one buffer while the next image is calculated and stored in the second buffer. Once the new image is completely drawn in it displayed on the screen and the original buffer is used to draw/compute the next image.

We now return to last class where we have defined the function

<math>\Phi(p) = (\Phi_{even},\ \Phi_{odd},\ r(p))</math>
that went from the manifold into <math>\mathbb{R}^{4m+3}</math>. (as each <math>\Phi</math> has dimension 2m+1 and an extra dimension for the r)

Here the <math>\Phi_{even}</math> and <math>\Phi_{odd}</math> correspond to the two buffers in the analogy.

Now this will turn out to be an embedding and we can reproduce our point on the manifold through use of the r. Indeed, the function is 1:1 as if we are near the same value of n then we are in the same <math>\phi</math> and so 1:1. If we are not near the same n, then we will have different r's necessarily.

'''Claim'''

A proper 1:1 continuous function is a homeomorphism onto its image.

i.e., we need to show a proper function <math>\Rightarrow</math> closed.

''Note''

This claim shows that our map <math>\Phi</math> is an embedding (provided such a proper r exists)

''Summary of what is left to do:''

1) Prove this claim

2) Prove the existence of such an r

3) Show how we can reduce from <math>\mathbb{R}^{4m+3}</math> to <math>\mathbb{R}^{2m+1}</math>

''Proof of 3''

We can repeat the arguments used in part 2 of this proof again (the lack of compactness is not problem, as Sard's theorem and the dimension reducing argument doesn't depend on it)

However, we do have to make the following modification. In addition to not choosing a projection direction in im<math>\beta_1</math> <math>\cup</math> im<math>\beta_2</math> we have to also make sure not to project in the vertical direction down into the r axis. But this is still quite possible by Sard's Theorem and hence we can reduce down to an embedding into <math>\mathbb{R}^{2m+1}</math>

''Proof of 2''

Consider <math>\{\lambda_{\alpha}\}</math> a partition of unity subordinant to a countable cover by sets with compact closure. We can make this partition be countable as each chart with compact closure only needs finitely many <math>\lambda_{\alpha}</math>'s.

Hence we get a partition of unity <math>\{\lambda_k\}_{k\in\mathbb{N}}</math> with supp<math>\lambda_k</math> being compact.

Set <math>r:=\sum k\lambda_k</math>

''Claim:''
r is proper

''Proof'

Assume <math>p\notin\bigcup_{k<n} supp \lambda_k</math>

then <math>1=\sum_{k=1}^{\infty}\lambda_k (p) = \sum_{k=n}^{\infty}\lambda_k (p)</math>

<math>\Rightarrow</math>

<math>\sum_{k=1}^{\infty}k\lambda_k(p) =\sum_{k=n}^{\infty}k\lambda_k(p)\geq \sum_{k=n}^{\infty}n\lambda_k(p) = n\sum_{k=n}^{\infty}\lambda_k(p) =n</math>

so, <math>r^{-1}([0,n])\subset\bigcup_{k=1}^{n-1}supp\lambda_k</math>.

So compact

''Q.E.D''

0708-1300/Class notes for Tuesday, October 23

2007-11-05T19:38:46Z

Megan: /* Second Hour */

{{0708-1300/Navigation}}
==Dror's Notes==

* You're all invited to my talk today at 12, {{Home Link|Talks/UofT-GS-071023/index.html|"Non-Commutative Gaussian Elimination and Rubik's Cube"}}.
* Today's office hours will go 1-2.
* Our handout today is a printout of a Mathematica notebook that demonstrates a "space-filling" Peano curve. Here's the {{Home Link|classes/0708/GeomAndTop/APeanoCurve.nb|notebook}}, and here's a {{Home Link|classes/0708/GeomAndTop/APeanoCurve.pdf|PDF}} version. Also, here's the main picture on that notebook:

[[Image:0708-1300-A Peano Curve.png|center|540px]]

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

'''Diversion'''

It is noted that there are no smooth curves that cover the plane. However, there ARE continuous curves that cover the plane.

As an example we consider the continuous function from the unit interval to the unit square that is defined iteratively that looks like the function above.

The construction is done as follows:

<math>f_0</math> draws a diagonal line from the bottom left corner to the top right corner of the unit square.
For <math>f_1</math> one breaks the unit interval into 7 sections. The map <math>f_1</math> takes the 1st, 3rd, 5th, 7th sections respectively to a diagonal line in the bottom left, bottom right, top left and top right subboxes respectively where the diagonal line goes from the bottom left to the top right corner of each subbox. The 2nd, 4th and 6th "filler" sections of the unit interval simply draw the lines that map the end of one diagonal to the start of the other. The <math>f_i</math> is defined iteratively. See the diagram above to see what this looks like.

We note that this is obviously continuous and we get uniform convergence to a continuous function into the unit square. As every point in the square get approached arbitrary close to a point in the image of one of the iterates of the function, compactness tells us the entire square is covered.

'''Definition'''

A ''general'' definition of "locally something" is typically that every point has a neighborhood in which this something property holds. Or perhaps a neighborhood basis where this property holds.

Hence, precisely:

1) A cover <math>\{U_{\alpha}\}</math> is ''locally finite'' if every point has a neighborhood V such that V intersects only finitely many <math>U_{\alpha}</math>'s.

2) A space is ''paracompact'' if:

a) Every open cover has a locally finite refinement

b) If a cover <math>\{U_{\alpha}\}</math> is locally finite then <math>\exists\ \{V_{\alpha}\}</math> so that <math>\{V_{\alpha}\}</math> still covers the space but that <math>\bar{V_{\alpha}}\subset U_{\alpha}</math>

Note: Manifolds are paracompact

'''Whitney Embedding Theorem'''

We recall the 3 steps to prove this theorem mentioned last class. It is noted that step three will actually be broken up into two steps:

3a: For an arbitrary <math>M^m</math> we can embed <math>M^m</math> in <math>\mathbb{R}^{4m+3}</math>

3b: We can then embed it in <math>\mathbb{R}^{2m+1}</math>

'''Proof of Theorem'''

We had previously proved most of part 1, but what we still had to show was that such partitions of unity actually exist.

''Reminder of Definition:''

A partition of unity subordinate to <math>\{U_{\alpha}\}</math> is a collection of functions <math>\lambda_{\alpha}:M\rightarrow\mathbb{R}</math> such that:

1) Supp <math>\lambda_{\alpha}\subset U_{\alpha}</math>

2) <math>\sum_{\alpha} \lambda_{\alpha} = 1</math>

'''Claim:''' If <math>\phi:U\rightarrow\mathbb{R}^n</math> is a chart and <math>K\subset U</math> is compact then we can find a function <math>\lambda:M\rightarrow\mathbb{R}</math> that is compact and <math>K\subset</math> supp <math>\lambda \subset U</math>

''Proof of Claim''

Because we are inside a chart, it is enough to just do this in <math>\mathbb{R}^n</math>.

For every <math>p\in K</math> we can find a radius r such that <math>B_{r(p)}(p)\subset U</math>. By compactness we can take only finitely many such p's. Hence, <math>\{B_{r(p_i)}(p_i)\}</math> cover K.

We want to put a bump function of each ball and sum them up to give us our <math>\lambda</math>.

Let <math>f_r(x) = e^{\frac{1}{x^2-r^2}}</math> for x<r and 0 otherwise.

Now let <math>\lambda(p):=\sum_i f_{r(p_i)}(d^2(p,p_i))</math>

''Q.E.D''

'''Theorem'''

On a manifold, given an open cover, you can find a partition of unity subordinate to a locally finite refinement of it.

''Proof''

WLOG, the cover is by charts and each one is bounded and the cover is locally finite <math>\{U_{\alpha}\}</math>

By paracompactness, find <math>V_{\alpha}\subset U_{\alpha}</math> such that <math>\bar{V_{\alpha}}\subset U_{\alpha}</math> and <math>\cup V_{\alpha} = M</math>. By the previous claim can find <math>\bar{V_{\alpha}}\subset</math> supp <math>\lambda^t_{\alpha}\subset U_{\alpha}</math>

Now consider <math>\sum \lambda_{alpha}^t =: \lambda^t</math>. This is a finite sum.

By local finiteness, it is smooth and so we define <math>\lambda_{\alpha} :=\frac{\lambda^{t}_{\alpha}}{\lambda}</math>

''Q.E.D. for part 1 of Whitney''

===Second Hour===

'''Proof of Part 2'''

''Claim'': Suppose <math>\Phi:M^m\rightarrow\mathbb{R}^N</math> is an embedding of a compact manifold <math>M^m</math> for a large <math>N>2m+1</math>. Then <math>\exists</math> an embedding <math>\Phi:M^m\rightarrow\mathbb{R}^{N-1}</math>.

''Note:'' We will then backwardly induct down until it is embedded in dimension 2m+1.

''Proof of Claim''

We begin by noting the similarity of this with a homework problem.

Let <math>V\in S^{N-1}\subset\mathbb{R}^N</math>. Such a v defines an orthogonal N-1 dimensional hyperplane. Let <math>\pi_v:\mathbb{R}^N\rightarrow\mathbb{R}^{N-1}</math> be the projection onto this N-1 dimensional hyperplane parallel to v.

Constructing the map <math>\Phi' := \pi_v\circ\Phi</math> , Sard's Theorem is going to show that for most such v's, this will be the embedding we are looking for.

Let us consider the v's where this does NOT work. There are two ways this will not work, either <math>\Phi</math> is not 1:1 or <math>\Phi_*</math> is not 1:1. Hence we will construct two function <math>\beta_1</math> and <math>\beta_2</math> so that the v's fail if they are in im<math>\beta_1</math> <math>\cup</math> im<math>\beta_2</math>.

Define <math>\beta_1:M\times M -\{diagonal\}\rightarrow S^{N-1}</math> by

<math>\beta_1(p_1,p_2) = \frac{\Phi(p_2) - \Phi(p_1)}{||\Phi(p_2) - \Phi(p_1)||}</math>

The image thus consists of points in <math>S^{N-1}</math> and hence define the projection direction. Intuitively we see that this should not work because if one were to project in this direction then the two separate points <math>p_1</math> and <math>p_2</math> would be mapped to the same spot, thus the map <math>\Phi'</math> would not be 1:1 and would change the topology of the resulting space. Indeed, the reverse is true,if <math>\Phi'</math> is not 1:1 for a given projection direction then that projection direction will be in the image of <math>\beta_1</math>

For <math>\Phi'</math> to be an embedding we also require that <math>\Phi'_* = \pi_v\circ\Phi_*</math> (as <math>\pi_v</math> is linear) to be 1:1 and thus <math>\Phi'</math> will be an immersion.

This is equivalent to saying that <math>\pi_v</math> does not "kill" anything in the image of <math>\Phi_*</math>

<math>\Leftrightarrow</math>

<math>v\notin</math> im<math>\Phi_* \Leftrightarrow v\notin</math> im<math>\beta_2</math> where <math>\beta_2</math> is defined as follows:

<math>\beta_2:TM = \cup_p T_p M - \{0\}\rightarrow S^{N-1}</math>

give by, for <math>p\in M</math> and <math>w\in T_p M</math>, <math>(p,w)\mapsto \frac{\Phi_{*p}(w)}{||\Phi_{*p}(w)||}</math>

However the domains of both <math>\beta</math>'s have dimension 2m and so by Sard's Theorem, Im<math>\beta_1</math> <math>\cup</math> Im<math>\beta_2</math> is of measure zero.

We thus choose any other <math>v\in S^{N-1}</math> and this forms a perfectly fine projection direction so the composition map <math>\Phi' = \pi_v\circ\Phi</math> is an embedding into <math>R^{N-1}</math>.

By backwards induction we can repeat this procedure to get an embedding of M into <math>\mathbb{R}^{2m+1}</math>

Note we can not go lower than this with these arguments since then Sard's theorem doesn't apply.

Now, to apply Sard's Theorem it was implicitly assumed that TM was itself a 2m dimensional manifold, a fact we haven't yet seen.

We can equip TM with coordinate charts in the following way.

Given a curve <math>\gamma:\mathbb{R}\rightarrow U\subset M</math> and a coordinate chart on the manifold <math>\phi:U\rightarrow\mathbb{R}^m</math> we construct the function from the equivalence class of curves taking <math>[\gamma]\mapsto (\phi\circ\gamma(0), d(\phi\circ\gamma(0))\in\mathbb{R}^{2m}</math>

These are of course only defined for equivalence classes in some neighborhood and it needs to be checked (easily) that this defined coordinate chart has smooth overlap functions.

''Q.E.D.''

'''Part 3 of the Proof'''

We start by introducing the idea of a "remoteness function"

''Definition:'' A continuous function between topological spaces is called "proper" if the inverse image of every compact set is compact.

Consider <math>s:M\rightarrow \mathbb{R}</math> that is smooth and proper.

So, <math>s^{-1}</math> of a compact set is compact.

Consider,

<math>\Phi = (\cup\Phi_t,s)\rightarrow\mathbb{R}^{2m+1+1} = \mathbb{R}^{2m+2}</math> that embeds <math>s^{-1}(t)</math>. Now this isn't quite right because <math>s^{-1}(t)</math> is only actually a manifold for regular values t. This will be adjusted for later.

Now, for each <math>n\in\mathbb{N}</math>,

let <math>U_n = s^{-1}((n-2/3,n+2/3))</math> and <math>V_n = s^{-1}((n-3/4,n+3/4))</math>

The idea with these definitions is just that so that the <math>U_n\subset V_n</math> and that <math>U_n\cap U_{n+1}\neq\empty</math>, <math>V_n\cap V_{n+1}\neq\empty</math>

We now let <math>\lambda_n</math> be a smooth function such that <math>\lambda_n |_{((n-2/3,n+2/3)} =1</math> and supp<math>\lambda_n\subset</math> (n-3/4,n+3/4)

We now let <math>\Phi_n:U_n\rightarrow \mathbb{R}^{2n+1}</math> be the embedding from part 2.

''Note:'' I think that in the above line it much the closure of <math>U_n</math> that was meant in class because we can only apply part 2 on compact manifolds. Either that or the intervals used in define <math>U_n</math> would be the closed interval instead of the open one which would thus make <math>U_n</math> compact.

Now of course we don't always choose the numbers 2/3 and 3/4 as in the above construction, we merely choose the endpoints to be regular (possible because of Sard's Theorem) and to satisfy the appropriate properties mentioned above. Hence such <math>U_n</math> exist and is a compact manifold.

Now, define
<math>\Phi_{even} = \sum_{even\ k}\lambda_k(s(p)) \Phi_k(p)</math>
and <math>\Phi_{odd}</math> defined analogously.

We then let

<math>\Phi(p) = (\Phi_{even},\ \Phi_{odd},\ s(p))</math>

This will give us the embedding we are interested in. It still remains to be shown why such a map s exists...

0708-1300/Class notes for Thursday, October 18

2007-11-05T03:08:43Z

Megan: /* Definition */ I corrected the definition of locally finite.

{{0708-1300/Navigation}}

==Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===Outline===

Today we stated the Whitney Embedding Theorem and began to discuss its proof. Along the way, we also encountered some related notions that will serve us well in the future. We begin by stating the theorem:

===Theorem (Whitney Embedding)===
Let <math>M\!</math> be a smooth <math>m\!</math>-manifold. Then there exists an embedding <math>\Phi : M \to \mathbb{R}^{2m+1}</math>.

====Proof====
=====Outline=====

We will break the proof of the theorem into three parts:

<ol>
<li> Find an embedding of a compact <math>M\!</math> into <math>\mathbb{R}^N\!</math> for some <math>N\!</math>.
<li> Use Sard's Theorem to reduce <math>N\!</math> to <math>2m+1\!</math>.
<li> Use the "Zebra Trick" to prove the theorem for non-compact <math>M\!</math>.
</ol>

Parts two and three shall be left to the next lecture.

=====Part 1=====

 Suppose that <math>M\!</math> is compact. Let <math>\{\phi_\alpha : U_\alpha \to \mathbb{R}^m \}_{\alpha \in A}</math> be an atlas for <math>M\!</math>, and note that <math>\{U_\alpha\}_{\alpha \in A}\!</math> is an open cover of <math>M\!</math>. Hence it possesses a finite subcover <math>\{U_j\}_{1 \le j \le J}\!</math>, and the corresponding collection <math>\{\phi_j\}_{1 \le j \le J}</math> of charts is an atlas. 

Choose smooth functions <math>\{\lambda_j : M \to \mathbb{R}_{\ge 0} \}_{1 \le j \le J}</math> with the following properties:

<ol>
<li> <math>\lambda_j |_{M \setminus U_j} = 0 </math>
<li> <math>\bigcup_{j=1}^{J} \lambda_j^{-1}\left((0,\infty)\right) = M</math>
<li> <math>\mathrm{supp}\left(\lambda_j\right) \subset U_j</math> for <math>1 \le j \le J </math>
</ol>

where <math>\mathrm{supp}(f)\!</math> for <math>f : M \to \mathbb{R}</math> is the support of <math>f\!</math>, ie. the closure of <math>f^{-1}(\mathbb{R} \setminus \{0\}).</math> The existence of such functions follows from the existence of smooth partitions of unity for manifolds---a concept that will be discussed later on.

Now define <math>\Phi : M \to \mathbb{R}^{J + mJ}</math> by <math>\Phi(p) = \left(\lambda_1(p), \ldots, \lambda_J(p), \lambda_1 \tilde \phi_1 (p), \ldots, \lambda_J \tilde \phi_J(p)\right)</math>, where <math>\tilde \phi_j : M \to \mathbb{R}^m</math> is defined by <math>\tilde \phi_j |_{U_j} = \phi_j</math> and <math>\tilde \phi_j |_{M \setminus U_j} = 0</math>.

We claim that <math>\Phi\!</math> is an embedding. That <math>\Phi\!</math> is smooth follows immediately from its construction (the <math>\lambda_j\!</math>s have been used to smear out the <math>\phi_j\!</math> to smooth functions on all of <math>M\!</math>). That <math>\Phi\!</math> is injective is also clear. It takes a bit of work to show that <math>\Phi\!</math> is an immersion, but this is left as an exercise. It remains to see that <math>\Phi\!</math> is a homeomorphism, but this fact follows from the following topological lemma. <math>\Box\!</math>

===Lemma===

Let <math>(X,\tau)\!</math> and <math>(Y,\sigma)\!</math> be topological spaces. Suppose that <math>X\!</math> is compact, <math>Y\!</math> is Hausdorff, and that <math> f : X \to Y</math> is continuous and injective. Then <math>f : X \to f(X)</math> is a homeomorphism.

====Proof====

 Since <math>f\!</math> is an injection onto its image, it is a bijection. Since <math>f\!</math> is continuous, it remains to show that <math>f^{-1}\!</math> is continuous. Thus, it suffices to see that <math>f\!</math> takes closed sets to closed sets. Let <math>A \subset X</math> be closed. Since <math>X\!</math> is compact, so is <math>A\!</math>. Hence <math>f(A)\!</math> is compact since <math>f\!</math> is continuous. But <math>Y\!</math> is Hausdorff, and every compact subset of a Hausdorff space is closed. Hence <math>f(A)\!</math> is closed. <math>\Box</math>

===Remarks===

 The smearing functions we used in Part 1 of the proof of the Whitney Embedding Theorem are very similar to partitions of unity---collections of functions that break the constant function <math>p \mapsto 1</math> into a bunch of bump functions. We will now formalize this notion and show that such collections of functions exist for smooth manifolds.

===Definition===

Let <math>U = \{U_\alpha\}_{\alpha \in A}</math> be an open cover of a topological space (manifold) <math>M\!</math>. A partition of unity subordinate to <math>U\!</math> is a collection <math>\{\lambda_\beta : M \to \mathbb{R}_{\ge 0} \}_{\beta \in B}</math> of continuous (smooth) functions such that

<ol>
<li> For every <math>\beta \in B</math> there is an <math>\alpha \in A</math> such that <math>\mathrm{supp}(\lambda_\beta) \subset U_\alpha</math>
<li> <math>\{\mathrm{supp}(\lambda_\beta)\}_{\beta \in B}</math> is locally finite, ie. for every <math>p \in M</math>, there exists a neighbourhood <math>U</math> of <math>p</math> such that <math>U </math> intersects <math>\mathrm{supp}(\lambda_\beta)</math> for only finitely many <math>\beta\!</math>.
<li> <math>\sum_{\beta \in B} \lambda_\beta = 1</math>
</ol>

 A local refinement of <math>U\!</math> is an open cover <math>\{V_\gamma\}_{\gamma \in C}</math> of <math>M\!</math> such that for every <math>\gamma \in C</math>, <math>V_\gamma \subset U_\alpha</math> for some <math>\alpha \in A</math>. <math>M\!</math> is called paracompact if every open cover of <math>M\!</math> has a locally finite refinement.

===Remarks===

For further information on paracompactness, we refer the reader to the corresponding [http://en.wikipedia.org/wiki/Paracompact Wikipedia entry]. Note, in particular, that locally compact, second-countable topological spaces---such as manifolds---are paracompact, and that paracompact spaces are shrinking spaces. The following result (which follows immediately from these facts) will be useful for constructing partitions of unity on manifolds:

===Theorem===

Manifolds are paracompact. In particular, if <math>\{U_\alpha\}_{\alpha \in A}</math> is locally finite then there is an open cover <math>\{V_\alpha\}_{\alpha \in A}</math> such that <math>\bar{V}_\alpha \subset U_\alpha</math> for every <math>\alpha \in A</math>.

0708-1300/Class notes for Thursday, October 18

2007-11-05T03:01:36Z

Megan: /* Definition */

{{0708-1300/Navigation}}

==Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===Outline===

Today we stated the Whitney Embedding Theorem and began to discuss its proof. Along the way, we also encountered some related notions that will serve us well in the future. We begin by stating the theorem:

===Theorem (Whitney Embedding)===
Let <math>M\!</math> be a smooth <math>m\!</math>-manifold. Then there exists an embedding <math>\Phi : M \to \mathbb{R}^{2m+1}</math>.

====Proof====
=====Outline=====

We will break the proof of the theorem into three parts:

<ol>
<li> Find an embedding of a compact <math>M\!</math> into <math>\mathbb{R}^N\!</math> for some <math>N\!</math>.
<li> Use Sard's Theorem to reduce <math>N\!</math> to <math>2m+1\!</math>.
<li> Use the "Zebra Trick" to prove the theorem for non-compact <math>M\!</math>.
</ol>

Parts two and three shall be left to the next lecture.

=====Part 1=====

 Suppose that <math>M\!</math> is compact. Let <math>\{\phi_\alpha : U_\alpha \to \mathbb{R}^m \}_{\alpha \in A}</math> be an atlas for <math>M\!</math>, and note that <math>\{U_\alpha\}_{\alpha \in A}\!</math> is an open cover of <math>M\!</math>. Hence it possesses a finite subcover <math>\{U_j\}_{1 \le j \le J}\!</math>, and the corresponding collection <math>\{\phi_j\}_{1 \le j \le J}</math> of charts is an atlas. 

Choose smooth functions <math>\{\lambda_j : M \to \mathbb{R}_{\ge 0} \}_{1 \le j \le J}</math> with the following properties:

<ol>
<li> <math>\lambda_j |_{M \setminus U_j} = 0 </math>
<li> <math>\bigcup_{j=1}^{J} \lambda_j^{-1}\left((0,\infty)\right) = M</math>
<li> <math>\mathrm{supp}\left(\lambda_j\right) \subset U_j</math> for <math>1 \le j \le J </math>
</ol>

where <math>\mathrm{supp}(f)\!</math> for <math>f : M \to \mathbb{R}</math> is the support of <math>f\!</math>, ie. the closure of <math>f^{-1}(\mathbb{R} \setminus \{0\}).</math> The existence of such functions follows from the existence of smooth partitions of unity for manifolds---a concept that will be discussed later on.

Now define <math>\Phi : M \to \mathbb{R}^{J + mJ}</math> by <math>\Phi(p) = \left(\lambda_1(p), \ldots, \lambda_J(p), \lambda_1 \tilde \phi_1 (p), \ldots, \lambda_J \tilde \phi_J(p)\right)</math>, where <math>\tilde \phi_j : M \to \mathbb{R}^m</math> is defined by <math>\tilde \phi_j |_{U_j} = \phi_j</math> and <math>\tilde \phi_j |_{M \setminus U_j} = 0</math>.

We claim that <math>\Phi\!</math> is an embedding. That <math>\Phi\!</math> is smooth follows immediately from its construction (the <math>\lambda_j\!</math>s have been used to smear out the <math>\phi_j\!</math> to smooth functions on all of <math>M\!</math>). That <math>\Phi\!</math> is injective is also clear. It takes a bit of work to show that <math>\Phi\!</math> is an immersion, but this is left as an exercise. It remains to see that <math>\Phi\!</math> is a homeomorphism, but this fact follows from the following topological lemma. <math>\Box\!</math>

===Lemma===

Let <math>(X,\tau)\!</math> and <math>(Y,\sigma)\!</math> be topological spaces. Suppose that <math>X\!</math> is compact, <math>Y\!</math> is Hausdorff, and that <math> f : X \to Y</math> is continuous and injective. Then <math>f : X \to f(X)</math> is a homeomorphism.

====Proof====

 Since <math>f\!</math> is an injection onto its image, it is a bijection. Since <math>f\!</math> is continuous, it remains to show that <math>f^{-1}\!</math> is continuous. Thus, it suffices to see that <math>f\!</math> takes closed sets to closed sets. Let <math>A \subset X</math> be closed. Since <math>X\!</math> is compact, so is <math>A\!</math>. Hence <math>f(A)\!</math> is compact since <math>f\!</math> is continuous. But <math>Y\!</math> is Hausdorff, and every compact subset of a Hausdorff space is closed. Hence <math>f(A)\!</math> is closed. <math>\Box</math>

===Remarks===

 The smearing functions we used in Part 1 of the proof of the Whitney Embedding Theorem are very similar to partitions of unity---collections of functions that break the constant function <math>p \mapsto 1</math> into a bunch of bump functions. We will now formalize this notion and show that such collections of functions exist for smooth manifolds.

===Definition===

Let <math>U = \{U_\alpha\}_{\alpha \in A}</math> be an open cover of a topological space (manifold) <math>M\!</math>. A partition of unity subordinate to <math>U\!</math> is a collection <math>\{\lambda_\beta : M \to \mathbb{R}_{\ge 0} \}_{\beta \in B}</math> of continuous (smooth) functions such that

<ol>
<li> For every <math>\beta \in B</math> there is an <math>\alpha \in A</math> such that <math>\mathrm{supp}(\lambda_\beta) \subset U_\alpha</math>
<li> <math>\{\mathrm{supp}(\lambda_\beta)\}_{\beta \in B}</math> is locally finite, ie. for every <math>p \in M</math>, <math>p \in \mathrm{supp}(\lambda_\beta)</math> for only finitely many <math>\beta\!</math>.
<li> <math>\sum_{\beta \in B} \lambda_\beta = 1</math>
</ol>

 A local refinement of <math>U\!</math> is an open cover <math>\{V_\gamma\}_{\gamma \in C}</math> of <math>M\!</math> such that for every <math>\gamma \in C</math>, <math>V_\gamma \subset U_\alpha</math> for some <math>\alpha \in A</math>. <math>M\!</math> is called paracompact if every open cover of <math>M\!</math> has a locally finite refinement.

===Remarks===

For further information on paracompactness, we refer the reader to the corresponding [http://en.wikipedia.org/wiki/Paracompact Wikipedia entry]. Note, in particular, that locally compact, second-countable topological spaces---such as manifolds---are paracompact, and that paracompact spaces are shrinking spaces. The following result (which follows immediately from these facts) will be useful for constructing partitions of unity on manifolds:

===Theorem===

Manifolds are paracompact. In particular, if <math>\{U_\alpha\}_{\alpha \in A}</math> is locally finite then there is an open cover <math>\{V_\alpha\}_{\alpha \in A}</math> such that <math>\bar{V}_\alpha \subset U_\alpha</math> for every <math>\alpha \in A</math>.

0708-1300/Class notes for Tuesday, October 16

2007-11-05T02:18:57Z

Megan: /* Second Hour */

{{0708-1300/Navigation}}
==Dror's Computer Program for C+<math>\alpha</math> C==

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of <math>C\times C</math> in a direction <math>t\in[0,\pi/2]</math> (where <math>C</math> is the standard Cantor set). Here's the {{Home Link|classes/0708/GeomAndTop/CCShadow.nb|notebook}}, and here's a {{Home Link|classes/0708/GeomAndTop/CCShadow.pdf|PDF}} version. Also, here's the main picture on that notebook:

[[Image:0708-1300-CCShadow.png|center|540px]]

==Typed Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== First Hour===

'''Today's Agenda:'''

Proof of Sard's Theorem. That is, for <math>f:M^m\rightarrow N^n</math> being smooth, the measure <math>\mu</math>(critical values of f) = 0.

'''Comments regarding last class'''

1) In our counterexample to Sard's Theorem for the case of <math>C^1</math> functions it was emphasized that there are functions f from ''a'' Cantor set C' to ''the'' Cantor set C. We then let <math>g(x,y) = f(x) + f(y)</math> and hence the critical values are <math>C+C = [0,2]</math> as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a <math>C^n</math> such function where we make the "bumps" in f smoother as needed and so <math>f(C') = C''</math> where <math>C''</math> is a "very thin" Cantor set. But now let <math>g(x,y,z,\ldots) = f(x) + f(y) + f(z) +\ldots</math> which will have an image of <math>C'' + C'' + C'' + \ldots =</math> an interval.

2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of <math>C+\alpha C</math> for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.

'''Proof of Sard's Theorem'''

Firstly, it is enough to argue locally, since a manifold is second countable (that is, a manifold has a countable basis) and a countable union of sets of measure zero has measure zero.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.

We can restrict our neighborhoods so that we can assume <math>M=\mathbb{R}^m</math> and <math>N=\mathbb{R}^n</math>

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.

Now, set <math>D_k = \{p |</math> all partial derivities of f of order <math>\leq k </math> vanish} for <math>k\geq 1</math>

Set <math>D_0 = \{p\ |\ df_p</math> is not onto }. This is just the critical points.

Clearly <math>D_0\supset D_1\supset\ldots\supset D_i\supset\ldots\supset D_m</math>

We will show by backwards induction that <math>\mu(F(D_k)) = 0</math>

''Comment:''

We have not actually defined the measure <math>\mu</math>. We use it merely to denote that <math>F(D_k))</math> has measure zero, a concept that we DID define.

===Second Hour===

'''Claim 1'''

<math>F(D_m)</math> has measure 0.

''Proof''

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size <math>\epsilon</math> in 1D becomes a smaller set of size <math>\epsilon^2</math> in 2D etc. More precisely, for <math>f = (f_1,\ldots,f_n)</math>, <math>f(D_n)\subset f_1(D_m)\times{R}^{n-1}</math>. Applying the proposition that if A is of measure zero in <math>\mathbb{R}</math> then <math>A\times\mathbb{R}^{n-1}</math> is measure zero in <math>\mathbb{R}^n</math> we now see that assuming n=1 is justified.

''Reminder''

Taylor's Theorem: for smooth enough <math>g:\mathbb{R}\rightarrow\mathbb{R}</math> and some <math>x_0</math> then <math>g(x) = \sum_{j=0}^{m} \frac{g^{(j)}(x_0)}{j!}(x-x_0)^j + \frac{g^{(m+1)}(t)}{(m+1)!}(x-x_0)^{m+1}</math>
for some t between x and <math>x_0</math>.

For <math>x_0\in D_m</math> all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times <math>(x-x_0)^{m+1}</math>.

Now let us consider a box B in <math>\mathbb{R}^m</math> containing a section of <math>D_m</math>. We divide B into <math>C_1\lambda^m</math> boxes of side <math>1/\lambda</math>.

By Taylor's Theorem, <math>f(B_i)\subset</math> of an interval of length <math>C_2\frac{1}{\lambda^{m+1}}</math> where the constant <math>C_2</math> is determined by Taylor's Theorem. Call this interval <math>I_i</math>

Hence,

<math>f(D_m)\subset\bigcup_{i:B_i\cap D_m \neq 0} f(B_i)\subset \bigcup_{i:B_i\cap D_m \neq 0}I_i</math>

But <math>\sum_{i} length(I_i)\leq C_1\lambda^m C_2\frac{1}{\lambda^{m+1}}</math> which tends to zero as <math>\lambda</math> tends to infinity.

''Q.E.D for Claim 1''

'''Claim 2'''

<math>f(D_k)</math> has measure zero for <math>k\geq 1</math>. We just proved this for k=m.

Now, W.L.O.G. <math>D_{k+1}</math> is the empty set. If not, just consider <math>M^m - D_{k+1}</math> which is still a manifold as <math>D_{k+1}</math> is closed (as it is determined by the "closed" condition that a determinant equals zero)

So, there is some kth derivative g of f such that <math>dg\neq 0</math>.

So <math>D_k\subset g^{-1}(0)</math> but <math>g^{-1}(0)</math> is at least locally a manifold of dimension 1 less. So, <math>f(D_k)\subset f(D_k\cap g^{-1}(0))</math> which has measure zero due to our induction hypothesis.

''Q.E.D for Claim 2''

'''Claim 3'''

<math>f(D_0)</math> is of measure zero.

Recall <math>D_0</math> is defined differently from the <math>D_k</math> and so requires a different technique to prove.

W.L.O.G. assume that <math>D_1</math> is the empty set. So, some derivative of f is not zero. W.L.O.G. <math>\frac{\partial f_1}{\partial x_1}</math> is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point <math>p\in M</math>. Consider <math>f_1:U\rightarrow\mathbb{R}</math> and let <math>df_1</math> be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume <math>f_1:\mathbb{R}^m\rightarrow\mathbb{R}</math> via <math>(x_1,\ldots,x_m)\mapsto x_1</math>. That is, <math>f_1 = x_1</math>.

Our differential df then is just the matrix whose first row consists of <math>(1,0,\ldots,0)</math>. Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say <math>f =(f_1,f_{rest})</math>.

Now define <math>U_t = \{t\} \times\mathbb{R}^{m-1}</math>.
Also, let us denote "critical points of f" by CP(f) and "critical values of f" by CV(f).

'''Claim 4'''

The <math>CP(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CP(f_{rest}|_{U_t})</math>

<math>\Rightarrow</math>

<math>CV(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CV(f_{rest}|_{U_t})</math>.

But <math>CV(f_{rest}|_{U_t})</math> has measure zero by our induction.

'''Lemma 1'''

If <math>A\subset I^2</math> is closed and has <math>\mu(A\cap(\{t\}\times I)) = 0\ \forall t</math>
then <math>\mu(A)=0</math>.

''Proof''

Note: We prove this significantly differently then in Bredon

''Sublemma''

If <math>\{t\}\times U</math> for an open U cover <math>\{t\}\times I\cap A</math> for a closed A then <math>\exists\epsilon>0</math> such that <math>(t-\epsilon,t+\epsilon)\times U\supset (t-\epsilon,t+\epsilon)\times I\cap A</math>

Indeed, let <math>d:A-(I\times U)\rightarrow\mathbb{R}</math> be <math>d(x) = |x_1-t|</math> then d is a continuous function of a compact set and so obtains a minimum and since d>0 then <math>min(d)>0 \rightarrow d>\epsilon>0</math>. But this <math>\epsilon</math> works for the claim. ''Q.E.D''

''The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class''

0708-1300/Class notes for Tuesday, October 16

2007-11-05T02:07:10Z

Megan: /* Second Hour */

{{0708-1300/Navigation}}
==Dror's Computer Program for C+<math>\alpha</math> C==

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of <math>C\times C</math> in a direction <math>t\in[0,\pi/2]</math> (where <math>C</math> is the standard Cantor set). Here's the {{Home Link|classes/0708/GeomAndTop/CCShadow.nb|notebook}}, and here's a {{Home Link|classes/0708/GeomAndTop/CCShadow.pdf|PDF}} version. Also, here's the main picture on that notebook:

[[Image:0708-1300-CCShadow.png|center|540px]]

==Typed Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== First Hour===

'''Today's Agenda:'''

Proof of Sard's Theorem. That is, for <math>f:M^m\rightarrow N^n</math> being smooth, the measure <math>\mu</math>(critical values of f) = 0.

'''Comments regarding last class'''

1) In our counterexample to Sard's Theorem for the case of <math>C^1</math> functions it was emphasized that there are functions f from ''a'' Cantor set C' to ''the'' Cantor set C. We then let <math>g(x,y) = f(x) + f(y)</math> and hence the critical values are <math>C+C = [0,2]</math> as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a <math>C^n</math> such function where we make the "bumps" in f smoother as needed and so <math>f(C') = C''</math> where <math>C''</math> is a "very thin" Cantor set. But now let <math>g(x,y,z,\ldots) = f(x) + f(y) + f(z) +\ldots</math> which will have an image of <math>C'' + C'' + C'' + \ldots =</math> an interval.

2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of <math>C+\alpha C</math> for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.

'''Proof of Sard's Theorem'''

Firstly, it is enough to argue locally, since a manifold is second countable (that is, a manifold has a countable basis) and a countable union of sets of measure zero has measure zero.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.

We can restrict our neighborhoods so that we can assume <math>M=\mathbb{R}^m</math> and <math>N=\mathbb{R}^n</math>

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.

Now, set <math>D_k = \{p |</math> all partial derivities of f of order <math>\leq k </math> vanish} for <math>k\geq 1</math>

Set <math>D_0 = \{p\ |\ df_p</math> is not onto }. This is just the critical points.

Clearly <math>D_0\supset D_1\supset\ldots\supset D_i\supset\ldots\supset D_m</math>

We will show by backwards induction that <math>\mu(F(D_k)) = 0</math>

''Comment:''

We have not actually defined the measure <math>\mu</math>. We use it merely to denote that <math>F(D_k))</math> has measure zero, a concept that we DID define.

===Second Hour===

'''Claim 1'''

<math>F(D_m)</math> has measure 0.

''Proof''

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size <math>\epsilon</math> in 1D becomes a smaller set of size <math>\epsilon^2</math> in 2D etc. More precisely, for <math>f = (f_1,\ldots,f_n)</math>, <math>f(D_n)\subset f_1(D_m)\times{R}^{n-1}</math>. Applying the proposition that if A is of measure zero in <math>\mathbb{R}</math> then <math>A\times\mathbb{R}^{n-1}</math> is measure zero in <math>\mathbb{R}^n</math> we now see that assuming n=1 is justified.

''Reminder''

Taylor's Theorem: for smooth enough <math>g:\mathbb{R}\rightarrow\mathbb{R}</math> and some <math>x_0</math> then <math>g(x) = \sum_{j=0}^{m} \frac{g^{(j)}(x_0)}{j!}(x-x_0)^j + \frac{g^{(m+1)}(t)}{(m+1)!}(x-x_0)^{m+1}</math>
for some t between x and <math>x_0</math>.

For <math>x_0\in D_m</math> all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times <math>(x-x_0)^{m+1}</math>.

Now let us consider a box B in <math>\mathbb{R}^m</math> containing a section of <math>D_m</math>. We divide B into <math>C_1\lambda^m</math> boxes of side <math>1/\lambda</math>.

By Taylor's Theorem, <math>f(B_i)\subset</math> of an interval of length <math>C_2\frac{1}{\lambda^{m+1}}</math> where the constant <math>C_2</math> is determined by Taylor's Theorem. Call this interval <math>I_i</math>

Hence,

<math>f(D_m)\subset\bigcup_{i:B_i\cap D_m \neq 0} f(B_i)\subset \bigcup_{i:B_i\cap D_m \neq 0}I_i</math>

But <math>\sum_{i} length(I_i)\leq C_1\lambda^m C_2\frac{1}{\lambda^{m+1}}</math> which tends to zero as <math>\lambda</math> tends to infinity.

''Q.E.D for Claim 1''

'''Claim 2'''

<math>f(D_k)</math> has measure zero for <math>k\geq 1</math>. We just proved this for k=m.

Now, W.L.O.G. <math>D_{k+1}</math> is the empty set. If not, just consider <math>M^m - D_{k+1}</math> which is still a manifold as <math>D_{k+1}</math> is closed (as it is determined by the "closed" condition that a determinant equals zero)

So, there is some kth derivative g of f such that <math>dg\neq 0</math>.

So <math>D_k\subset g^{-1}(0)</math> but <math>g^{-1}(0)</math> is at least locally a manifold of dimension 1 less. So, <math>f(D_k)\subset f(D_k\cap g^{-1}(0))</math> which has measure zero due to our induction hypothesis.

''Q.E.D for Claim 2''

'''Claim 3'''

<math>f(D_0)</math> is of measure zero.

Recall <math>D_0</math> is defined differently from the <math>D_k</math> and so requires a different technique to prove.

W.L.O.G. assume that <math>D_1</math> is the empty set. So, some derivative of f is not zero. W.L.O.G. <math>\frac{\partial f_1}{\partial x_1}</math> is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point <math>p\in M</math>. Consider <math>f_1:U\rightarrow\mathbb{R}</math> and let <math>df_1</math> be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume <math>f_1:\mathbb{R}^m\rightarrow\mathbb{R}</math> via <math>(x_1,\ldots,x_m)\mapsto x_1</math>. That is, <math>f_1 = x_1</math>.

Our differential df then is just the matrix whose first row consists of <math>(1,0,\ldots,0)</math>. Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say <math>f =(f_1,f_{rest})</math>.

Now define <math>U_t = \{t\} \times\mathbb{R}^{m-1}</math>
Also lets denote "critical points of f'' by CP(f) and "critical values of f" by CV(f)

'''Claim 4'''

The <math>CP(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CP(f_{rest}|_{U_t})</math>

<math>\Rightarrow</math>

<math>CV(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CV(f_{rest}|_{U_t})</math>.

But <math>CV(f_{rest}|_{U_t})</math> has measure zero by our induction.

'''Lemma 1'''

If <math>A\subset I^2</math> is closed and has <math>\mu(A\cap(\{t\}\times I)) = 0\ \forall t</math>
then <math>\mu(A)=0</math>.

''Proof''

Note: We prove this significantly differently then in Bredon

''Sublemma''

If <math>\{t\}\times U</math> for an open U cover <math>\{t\}\times I\cap A</math> for a closed A then <math>\exists\epsilon>0</math> such that <math>(t-\epsilon,t+\epsilon)\times U\supset (t-\epsilon,t+\epsilon)\times I\cap A</math>

Indeed, let <math>d:A-(I\times U)\rightarrow\mathbb{R}</math> be <math>d(x) = |x_1-t|</math> then d is a continuous function of a compact set and so obtains a minimum and since d>0 then <math>min(d)>0 \rightarrow d>\epsilon>0</math>. But this <math>\epsilon</math> works for the claim. ''Q.E.D''

''The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class''

0708-1300/Class notes for Tuesday, October 16

2007-11-05T00:59:12Z

Megan: /* First Hour */

{{0708-1300/Navigation}}
==Dror's Computer Program for C+<math>\alpha</math> C==

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of <math>C\times C</math> in a direction <math>t\in[0,\pi/2]</math> (where <math>C</math> is the standard Cantor set). Here's the {{Home Link|classes/0708/GeomAndTop/CCShadow.nb|notebook}}, and here's a {{Home Link|classes/0708/GeomAndTop/CCShadow.pdf|PDF}} version. Also, here's the main picture on that notebook:

[[Image:0708-1300-CCShadow.png|center|540px]]

==Typed Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== First Hour===

'''Today's Agenda:'''

Proof of Sard's Theorem. That is, for <math>f:M^m\rightarrow N^n</math> being smooth, the measure <math>\mu</math>(critical values of f) = 0.

'''Comments regarding last class'''

1) In our counterexample to Sard's Theorem for the case of <math>C^1</math> functions it was emphasized that there are functions f from ''a'' Cantor set C' to ''the'' Cantor set C. We then let <math>g(x,y) = f(x) + f(y)</math> and hence the critical values are <math>C+C = [0,2]</math> as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a <math>C^n</math> such function where we make the "bumps" in f smoother as needed and so <math>f(C') = C''</math> where <math>C''</math> is a "very thin" Cantor set. But now let <math>g(x,y,z,\ldots) = f(x) + f(y) + f(z) +\ldots</math> which will have an image of <math>C'' + C'' + C'' + \ldots =</math> an interval.

2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of <math>C+\alpha C</math> for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.

'''Proof of Sard's Theorem'''

Firstly, it is enough to argue locally, since a manifold is second countable (that is, a manifold has a countable basis) and a countable union of sets of measure zero has measure zero.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.

We can restrict our neighborhoods so that we can assume <math>M=\mathbb{R}^m</math> and <math>N=\mathbb{R}^n</math>

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.

Now, set <math>D_k = \{p |</math> all partial derivities of f of order <math>\leq k </math> vanish} for <math>k\geq 1</math>

Set <math>D_0 = \{p\ |\ df_p</math> is not onto }. This is just the critical points.

Clearly <math>D_0\supset D_1\supset\ldots\supset D_i\supset\ldots\supset D_m</math>

We will show by backwards induction that <math>\mu(F(D_k)) = 0</math>

''Comment:''

We have not actually defined the measure <math>\mu</math>. We use it merely to denote that <math>F(D_k))</math> has measure zero, a concept that we DID define.

===Second Hour===

'''Claim 1'''

<math>F(D_m)</math> has measure 0.

''Proof''

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size <math>\epsilon</math> in 1D becomes a smaller set of size <math>\epsilon^2</math> in 2D etc. More precisely, for <math>f = (f_1,\ldots,f_n)</math>, <math>f(D_n)\subset f_1(D_m)\times{R}^{n-1}</math>. Applying the proposition that if A is of measure zero in <math>\mathbb{R}</math> then <math>A\times\mathbb{R}^{n-1}</math> is measure zero in <math>\mathbb{R}^n</math> we now see that assuming n=1 is justified.

''Reminder''

Taylor's Theorem: for smooth enough <math>g:\mathbb{R}\rightarrow\mathbb{R}</math> and some <math>x_0</math> then <math>g(x) = \sum_{j=0}^{m} \frac{g^{(j)}(x_0)}{j!}(x-x_0)^j + \frac{g^{(m+1)}(t)}{(m+1)!}(x-x_0)^{m+1}</math>
for some t between x and <math>x_0</math>.

For <math>x_0\in D_m</math> all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times <math>(x-x_0)^{m+1}</math>.

Now let us consider a box B in <math>\mathbb{R}^m</math> containing a section of <math>D_m</math>. We divide B into <math>C_1\lambda^m</math> boxes of side <math>1/\lambda</math>.

By Taylor's Theorem, <math>f(B_i)\subset</math> of an interval of length <math>C_2\frac{1}{\lambda^{m+1}}</math> where the constant <math>C_2</math> is determined by Taylor's Theorem. Call this interval <math>I_i</math>

Hence,

<math>f(D_m)\subset\bigcup_{i:B_i\cap D_m \neq 0} f(B_i)\subset \bigcup_{i:B_i\cap D_m \neq 0}I_i</math>

But <math>\sum_{i} length(I_i)\leq C_1\lambda^m C_2\frac{1}{\lambda^{m+1}}</math> which tends to zero as <math>\lambda</math> tends to infinity.

''Q.E.D for Claim 1''

'''Claim 2'''

<math>f(D_k)</math> has measure zero for <math>k\geq 1</math>. We just proved this for k=m.

Now, W.L.O.G. <math>D_{k+1}</math> is the empty set. If not, just consider <math>M^m - D_{k+1}</math> which is still a manifold as <math>D_{k+1}</math> is closed (as it is determined by the "closed" condition that a determinant equals zero)

So, there is some kth derivative g of f such that <math>dg\neq 0</math>.

So <math>D_k\subset g^{-1}(0)</math> but <math>g^{-1}(0)</math> is at least locally a manifold of dimension 1 less. So, <math>f(D_k)\subset f(D_k\cap g^{-1}(0))</math> which has measure zero due to our induction hypothesis.

''Q.E.D for Claim 2''

'''Claim 3'''

<math>f(D_0)</math> is of measure zero.

Recall <math>D_0</math> is defined differently from the <math>D_k</math> and so requires a different technique to prove.

W.L.O.G. lets assume that <math>D_1</math> is the empty set. So, some derivative of f is not zero. W.L.O.G. <math>\frac{\partial f_1}{\partial x_1}</math> is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point <math>p\in M</math>. Consider <math>f_1:U\rightarrow\mathbb{R}</math> and let <math>df_1</math> be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume <math>f_1:\mathbb{R}^m\rightarrow\mathbb{R}</math> via <math>(x_1,\ldots,x_m)\mapsto x_1</math>. That is, <math>f_1 = x_1</math>.

Our differential df then is just the matrix whose first row consists of <math>(1,0,\ldots,0)</math>. Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say <math>f =(f_1,f_{rest})</math>.

Now define <math>U_t = \{t\} \times\mathbb{R}^{m-1}</math>
Also lets denote "critical points of f'' by CP(f) and "critical values of f" by CV(f)

'''Claim 4'''

The <math>CP(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CP(f_{rest}|_{U_t})</math>

<math>\Rightarrow</math>

<math>CV(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CV(f_{rest}|_{U_t})</math>.

But <math>CV(f_{rest}|_{U_t})</math> has measure zero by our induction.

'''Lemma 1'''

If <math>A\subset I^2</math> is closed and has <math>\mu(A\cap(\{t\}\times I)) = 0\ \forall t</math>
then <math>\mu(A)=0</math>.

''Proof''

Note: We prove this significantly differently then in Bredon

''Sublemma''

If <math>\{t\}\times U</math> for an open U cover <math>\{t\}\times I\cap A</math> for a closed A then <math>\exists\epsilon>0</math> such that <math>(t-\epsilon,t+\epsilon)\times U\supset (t-\epsilon,t+\epsilon)\times I\cap A</math>

Indeed, let <math>d:A-(I\times U)\rightarrow\mathbb{R}</math> be <math>d(x) = |x_1-t|</math> then d is a continuous function of a compact set and so obtains a minimum and since d>0 then <math>min(d)>0 \rightarrow d>\epsilon>0</math>. But this <math>\epsilon</math> works for the claim. ''Q.E.D''

''The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class''

0708-1300/Class notes for Tuesday, October 2

2007-11-03T20:37:25Z

Megan: /* Class Notes */

{{0708-1300/Navigation}}
==English Spelling==
Many interesting rules about [[0708-1300/English Spelling]]
==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General class comments'''

1) The class photo is up, please add yourself

2) A questionnaire was passed out in class

3) Homework one is due on thursday

===First Hour===

Today's Theme: Locally a function looks like its differential

'''Pushforward/Pullback'''

Let <math>\theta:X\rightarrow Y</math> be a smooth map.

We consider various objects, defined with respect to X or Y, and see in which direction it makes sense to consider corresponding objects on the other space. In general <math>\theta_*</math> will denote the push forward, and <math>\theta^*</math> will denote the pullback.

1) points ''pushforward'' <math>x\mapsto\theta_*(x) := \theta(x)</math>

2) Paths <math>\gamma:\mathbb{R}\rightarrow X</math>, ie a bunch of points, ''pushforward'', <math>\gamma\rightarrow \theta_*(\gamma):=\theta\circ\gamma</math>

3) Sets <math>B\subset Y</math> ''pullback'' via <math>B\rightarrow \theta^*(B):=\theta^{-1}(B)</math>
Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map <math>\theta</math>

4) A measures <math>\mu</math> ''pushforward'' via <math>\mu\rightarrow (\theta_*\mu)(B) :=\mu(\theta^*B)</math>

5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely <math>\theta^*f = f\circ\theta</math>

6) Tangent vectors, defined in the sense of equivalence classes of paths, [<math>\gamma</math>] ''pushforward'' as we would expect since each path pushes forward. <math>[\gamma]\rightarrow \theta_*[\gamma]:=[\theta_*\gamma] = [\theta\circ\gamma]</math>

CHECK: This definition is well defined, that is, independent of the representative choice of <math>\gamma</math>

7) We can consider operators on functions to be in a sense dual to the functions and hence should go in the opposite direction. Hence, tangent vectors, defined in the sense of derivations, ''pushforward'' via <math>D\rightarrow (\theta_*D)(f):= D(\theta^*f) = D(f\circ\theta)</math>

CHECK: This definition satisfies linearity and Liebnitz property.

'''Theorem 1'''

The two definitions for the pushforward of a tangent vector coincide.

''Proof:''

Given a <math>\gamma</math> we can construct <math>\theta_{*}\gamma</math> as above. However from both <math>\gamma</math> and <math>\theta_*\gamma</math> we can also construct <math>D_{\gamma}f</math> and <math>D_{\theta_*\gamma}f</math> because we have previously shown our two definitions for the tangent vector are equivalent. We can then ''pushforward'' <math>D_{\gamma}f</math> to get <math>\theta_*D_{\gamma}f</math>. The theorem is reduced to the claim that:

<math>\theta_*D_{\gamma}f = D_{\theta_*\gamma}f</math>

for functions <math>f:Y\rightarrow \mathbb{R}</math>

Now, <math>D_{\theta_*\gamma}f = \frac{d}{dt}f\circ(\theta_*\gamma)|_{t=0} = \frac{d}{dt}f\circ(\theta\circ\gamma)|_{t=0} = \frac{d}{dt}(f\circ\theta)\circ\gamma |_{t=0} = D_{\gamma}(f\circ\theta) =\theta_*D_{\gamma}f</math>

''Q.E.D''

'''Functorality'''

let <math>\theta:X\rightarrow Y, \lambda:Y\rightarrow Z</math>

Consider some "object" s defined with respect to X and some "object u" defined with respect to Z. Something has the property of functorality if

<math>\lambda_*(\theta_*s) = (\lambda\circ\theta)_*s</math>

and

<math>\theta^*(\lambda^*u) = (\lambda\circ\theta)^*u</math>

Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.

Let us consider <math>\theta_*</math> on <math>T_pM</math> given a <math>\theta:M\rightarrow N</math>

We can arrange for charts <math>\varphi</math> on a subset of M into <math>\mathbb{R}^m</math> (with coordinates denoted <math>(x_1,\dots,x_m)</math>)and <math>\psi</math> on a subset of N into <math>\mathbb{R}^n</math> (with coordinates denoted <math>(y_1,\dots,y_n)</math>)such that <math>\varphi(p) = 0</math> and <math>\psi(\theta(p))=0</math>

Define <math>\theta^o = \psi\circ\theta\circ\varphi^{-1} = (\theta_1(x_1,\dots,x_m),\dots,\theta_n(x_1,\dots,x_m))</math>

Now, for a <math>D\in T_pM</math> we can write <math>D=\sum_{i=1}^m a_i\frac{\partial}{\partial x_i}</math>

So,
{|
|-
|<math>(\theta_*D)(f) \!</math>
|<math> = D(\theta^* f)\!</math>
|-
|
|<math>=\sum_{i=1}^m a_i\frac{\partial}{\partial x_i}(f\circ\theta) </math>
|-
|
|<math>=\sum_{i=1}^m a_i \sum_{j=1}^n\frac{\partial f}{\partial y_j}\frac{\partial\theta_j}{\partial x_i} </math>
|-
|
|<math>=\begin{bmatrix}
\frac{\partial f}{\partial y_1} & \cdots & \frac{\partial f}{\partial y_n}\\
\end{bmatrix}
\begin{bmatrix}
\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}
</math>
|}

Now, we want to write <math>\theta_*D = \sum b_j\frac{\partial}{\partial y_j}</math>

and so, <math>b_k = (\theta_*D)y_k =\begin{bmatrix}
0&\cdots & 1 & \cdots &0\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

where the 1 is at the kth location. In other words, <math>\theta_*D = \sum_{j=1}^{n} \sum_{i=1}^{m}a_i \frac{\partial \theta_j}{\partial x_i} \frac{\partial }{\partial y_j} </math>

So, <math>\theta_* = d\theta_p</math>, i.e., <math>\theta_*</math> is the differential of <math>\theta</math> at p

We can check the functorality, <math>(\lambda\circ\theta)_* = \lambda_*\circ\theta_*</math>, then <math>d(\lambda\circ\theta) = d\lambda\circ d\theta</math>
This is just the chain rule.

===Second Hour===

'''Defintion 1'''

An ''immersion'' is a (smooth) map <math>\theta:M^m\rightarrow N^n</math> such that <math>\theta_*</math> of tangent vectors is 1:1. More precisely, <math>d\theta_p: T_pM\rightarrow T_{\theta(p)}N</math> is 1:1 <math>\forall p\in M</math>

'''Example 1'''

Consider the canonical immersion, for m<n given by <math>\iota:(x_1,...,x_m)\mapsto (x_1,...,x_m,0,...,0)</math> with n-m zeros.

'''Example 2'''

This is the map from <math>\mathbb{R}</math> to <math>\mathbb{R}^2</math> that looks like a loop-de-loop on a roller coaster (but squashed into the plane of course!) The map <math>\theta</math> itself is NOT 1:1 (consider the crossover point) however <math>\theta_*</math> IS 1:1, hence an immersion.

'''Example 3'''

Consider the map from <math>\mathbb{R}</math> to <math>\mathbb{R}^2</math> that looks like a check mark. While this map itself is 1:1, <math>\theta_*</math> is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.

'''Example 4'''

Can there be objects, such as the graph of |x| that are NOT an immersion, but are constructed from a smooth function?

Consider the function <math>\lambda(x) = e^{-1/x^2}</math> for x>0 and 0 otherwise.

Then the map <math>x\mapsto \begin{bmatrix}
(\lambda(x),\lambda(x))& x>0\\
(0,0)& x=0\\
(-\lambda(-x),\lambda(-x)) & x<0\\
\end{bmatrix}</math>

is a smooth mapping with the graph of |x| as its image, but is NOT an immersion.

'''Example 5'''

The torus, as a subset of <math>\mathbb{R}^3</math> is an immersion

Now, consider the 1:1 linear map <math>T:V\rightarrow W</math> where V,W are vector spaces that takes <math>(v_1,...,v_m)\mapsto (Tv_1,...,Tv_m) = (w_1,..,w_m,w_{m+1},...,w_n)</math>

From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.

'''Theorem 2'''

Locally, every immersion looks like the inclusion <math>\iota:\mathbb{R}^m\rightarrow \mathbb{R}^n</math>.

More precisely, if <math>\theta:M^m\rightarrow N^n</math> and <math>d\theta_p</math> is 1:1 then there exist charts <math>\varphi</math> acting on <math>U\subset M</math> and <math>\psi</math> acting on <math>V\subset N</math> such that for <math>p\in U, \phi(p) = \psi(\theta(p)) = 0</math> such that the following diagram commutes:

<math>\begin{matrix}
U&\rightarrow^{\phi}&U'\subset \mathbb{R}^m\\
\downarrow_{\theta} &&\downarrow_{\iota} \\
V& \rightarrow^{\psi}& V'\subset \mathbb{R}^n\\
\end{matrix}</math>

that is, <math>\iota\circ\varphi = \psi\circ\theta</math>

'''Definition 2'''

M is a ''submanifold'' of N if there exists a mapping <math>\theta:M\rightarrow N</math> such that <math>\theta</math> is a 1:1 immersion.

'''Example 6'''

Our previous example of the graph of a "loop-de-loop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.

'''Example 6'''

The torus is a submanifold as the natural immersion into <math>\mathbb{R}^3</math> is 1:1

'''Definition 3'''

The map <math>\theta:M\rightarrow N</math> is an embedding if the subset topology on <math>\theta(M)</math> coincides with the topology induced from the original topology of M.

'''Example 7'''

Consider the map from <math>\mathbb{R}\rightarrow \mathbb{R}^2</math> whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively.
The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.

'''Corollary 1 to Theorem 2'''

The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.

Indeed, for all smooth <math>f:M\rightarrow \mathbb{R}</math> and <math>\forall p\in M</math> there exists a neighborhood V of <math>\theta(p)</math> and a smooth <math>g:N\rightarrow \mathbb{R}</math> such that <math>g|_{\theta(M)\bigcap U} = f|_U</math>

''Proof of Corollary 1''

Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of <math>\mathbb{R}^m\subset \mathbb{R}^n</math>. We then consider some function defined on M, and hence on the subset in <math>\mathbb{R}^n</math> which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.

''Proof of Theorem 2''

We start with the normal situation of <math>\theta:M\rightarrow N</math> with M,N manifolds with atlases containing <math>(\varphi_0,U_0)</math> and <math>(\psi_0, V_0)</math> respectively. We also expect that for <math>p\in U_0, \varphi_0(p) = \psi_0(\theta(p)) = 0</math>. I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.

<math>
\begin{matrix}
M\supset U_0 & \rightarrow^{\varphi_0} & U_1\subset \mathbb{R}^m & \rightarrow^{Id} & U_2 = U_1 \\
\downarrow_{\theta} & &\downarrow_{\theta_1} & &\downarrow_{\iota}\\
N\supset V_0 & \rightarrow^{\psi_0} & V_1\subset \mathbb{R}^n & \leftarrow^{\xi} & V_2\\
\end{matrix}
</math>

It is very important to note that the <math>\varphi_0</math> and <math>\psi_0</math> are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.

In the diagram above, <math>\theta_1 = \psi_0\circ\theta\circ\varphi^{-1}</math>. So, <math>\theta_1(0) = 0</math> and <math>(d\theta_1)_0 = i</math>. Note the <math>\theta_1</math>, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function <math>\xi</math>

We consider the map <math>\xi:V_2\rightarrow V_1</math> given <math>(x,y)\mapsto \theta_1(x) + (0,y)</math>. We note that <math>\xi</math> corresponds with the idea of "lifting" a flattened image back to its original height.

Claims:

1) <math>\xi</math> is invertible near zero. Indeed, computing <math>d\xi_0 = I</math> which is invertible as a matrix and hence <math>\xi</math> is invertible as a function near zero.

2) Take an <math>x\in U_2</math>. There are two routes to get to <math>V_1</math> and upon computing both ways yields the same result. Hence, the diagram commutes.

Hence, our immersion looks (locally) like the standard immersion between real spaces given by <math>\iota</math> and the charts are the compositions going between <math>U_0</math> to <math>U_2</math> and <math>V_0</math> to <math>V_2</math>

''Q.E.D''

0708-1300/Class notes for Thursday, November 1

2007-11-01T17:12:48Z

Megan:

{{0708-1300/Navigation}}
==Today's Agenda==
* [[0708-1300/Homework Assignment 4|HW4]] and [[0708-1300/Term Exam 1|TE1]].
* Continue with [[0708-1300/Class notes for Tuesday, October 30|Tuesday's]] agenda:
** Debt on proper functions.
** Prove that "the sphere is not contractible".
** Complete the proof of the "tubular neighborhood theorem".
===Proper Implies Closed===
'''Theorem.''' A proper function <math>f:X\to Y</math> from a topological space <math>X</math> to a locally compact (Hausdorff) topological space <math>Y</math> is closed.

'''Proof.''' Let <math>B</math> be closed in <math>X</math>, we need to show that <math>f(B)</math> is closed in <math>Y</math>. Since closedness is a local property, it is enough to show that every point <math>y\in Y</math> has a neighbourhood <math>U</math> such that <math>f(B)\cap U</math> is closed in <math>U</math>. Fix <math>y\in Y</math>, and by local compactness, choose a neighbourhood <math>U</math> of <math>y</math> whose close <math>\bar U</math> is compact. Then
{{Equation*|<math>f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U</math>,}}
so that <math>f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U</math>. But <math>\bar U</math> is compact by choice, so <math>f^{-1}(\bar U)</math> is compact as <math>f</math> is proper, so <math>B\cap f^{-1}(\bar U)</math> is compact as <math>B</math> is closed, so <math>f(B\cap f^{-1}(\bar U))</math> is compact (and hence closed) as a continuous image of a compact set, so <math>f(B)\cap U</math> is the intersection <math>f(B\cap f^{-1}(\bar U))\cap U</math> of a closed set with <math>U</math>, hence it is closed in <math>U</math>.

===Note===
The example of a non-contractible "comb" seen today is, in fact, "Cantor's comb". See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps

0708-1300/Class Photo

2007-10-01T16:02:22Z

Megan:

{{0708-1300/Navigation}}

Our class on September 27, 2007:

[[Image:0708-1300-ClassPhoto.jpg|thumb|centre|600px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn @ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Bazett|first=Trefor|userid=Trefor|email=trefor.bazett @ toronto.ca|location=tallest person a little right of center in a beige shirt|comments=}}
{{Photo Entry|last=Bjorndahl|first=Adam|userid=ABjorndahl|email=adam.bjorndahl @ utoronto.ca|location=back row, fifth from the left, under the "f(tp)dt"|comments=Looking forward to a great year!}}
{{Photo Entry|last=Chow|first=Aaron|userid=aaron.chow|email=aaron @ utoronto.ca|location=Third from right, in a black shirt.|comments=Hope we have a good year together!}}
{{Photo Entry|last=Isgur|first=Abraham|userid=Abisgu|email=abraham.isgur@ math.toronto.edu|location=2nd person in the back row, from the right, the one with the beard and long hair|comments=}}
{{Photo Entry|last=Mann|first=Katie|userid=katiemann|email=katie.mann@ utoronto.ca|location=middle, wearing "Eulers" shirt|comments=}}
{{Photo Entry|last=Mourtada|first=Mariam|userid=Mourtada|email=mariam.mourtada@ utoronto.ca|location=I am the girl in the front middle, wearing a blue shirt and catching my hands|comments=I am not wearing glasses! }}
{{Photo Entry|last=Pym|first=Brent|userid=Bpym|email=bpym @ math.toronto.edu|location=10th from the right (cumulatively), under the <math>T_p(M)\!</math>|comments=Adding this entry was my first-ever edit of a Wiki!}}
{{Photo Entry|last=Snow|first=Megan|userid=megan|email=megansnow @ gmail.com|location=back row, slightly right of centre, wearing a blue shirt over a black one|comments=}}
{{Photo Entry|last=Vera Pacheco|first=Franklin|userid=Franklin|email=franklin.vp @ gmail.com|location=Xth from left to right|comments=To find me you must first go to [[http://www.deathball.net/notpron/]] solve the first 4 pages. Once this done you will know how to find me. Once this done go back to NOTPRON an solve the rest of the puzzle}}
{{Photo Entry|last=Watts|first=Jordan|userid=Jwatts|email=jwatts @ math.toronto.edu|location=in the back, 2nd or 3rd from the left, depending on your convention|comments=My glasses become invisible in pictures.}}
{{Photo Entry|last=Wong|first=Silian|userid=kuramay|email=kurama_y @ hotmail.com|location=One of the Asian-looking girls...with sparkling teeth(??)|comments=I'll write up some comments after their existences}}

|}

0708-1300/Class notes for Tuesday, September 11

2007-09-12T02:10:42Z

Megan: /* Differentiability */

{{0708-1300/Navigation}}
==In Small Scales, Everything's Linear==
{| align=center
|- align=center
|[[Image:06-240-QuiltBeforeMap.png|200px]]
|<math>\longrightarrow</math>
|[[Image:06-240-QuiltAfterMap.png|200px]]
|- align=center
|<math>z</math>
|<math>\mapsto</math>
|<math>z^2</math>
|}

Code in [http://www.wolfram.com Mathematica]:

<pre>
QuiltPlot[{f_,g_}, {x_, xmin_, xmax_, nx_}, {y_, ymin_, ymax_, ny_}] :=
Module[
{dx, dy, grid, ix, iy},
SeedRandom[1];
dx=(xmax-xmin)/nx;
dy=(ymax-ymin)/ny;
grid = Table[
{x -> xmin+ix*dx, y -> ymin+iy*dy},
{ix, 0, nx}, {iy, 0, ny}
];
grid = Map[({f, g} /. #)&, grid, {2}];
Show[
Graphics[Table[
{
RGBColor[Random[], Random[], Random[]],
Polygon[{
grid[[ix, iy]],
grid[[ix+1, iy]],
grid[[ix+1, iy+1]],
grid[[ix, iy+1]]
}]
},
{ix, nx}, {iy, ny}
]],
Frame -> True
]
]

QuiltPlot[{x, y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
QuiltPlot[{x^2-y^2, 2*x*y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
</pre>

See also [[06-240/Linear Algebra - Why We Care]].

==Class Notes==
===Differentiability===
Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>.

'''Definition:'''

We say that <math>f</math> is differentiable (''diffable'') at <math>x</math> if there is a linear map <math>L</math> so that

<math>\lim_{h\rightarrow0}\frac{|f(x+h)-f(x)-L(h)|}{|h|}=0.</math>

In this case we will say that <math>L</math> is a differential of <math>f</math> at <math>x</math> and will denote it by <math>df_{x}</math>.

'''Theorem'''

If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following asertions holds:

# <math>df_{x}</math> is unique.
# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>
# If <math>f</math> is linear then <math>df_{x}=f</math>
# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>
# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

===Implicit Function Theorem===

'''Example'''
Although <math>x^2+y^2=1</math> does not defines <math>y</math> as a function of <math>x</math>, in a neighborhood of <math>(0;-1)</math> we can define <math>g(x)=-\sqrt{1-x^2}</math> so that <math>x^2+g(x)^2=1</math>. Furthermore, <math>g</math> is differentiable with differential <math>dg_{x}=\frac{x}{\sqrt{1-x^2}}</math>. This is a motivation for the following theorem.

'''Notation'''

If f:X\times Y\rightarrow Z then given x\in X we will define f_{[x]}:Y\rightarrow Z by f_{[x]}(y)=f(x;y)

'''Definition'''

<math>C^{p}(V)</math> will be the class of all functions defined on <math>V</math> with continuous partial derivatives up to order <math>p.</math>

'''Theorem'''(''Implicit function theorem'')

Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:

There is an open neighborhood of <math>x</math>, <math>V\subset U</math>, and a <math>diffable</math> function <math>g:V\rightarrow\mathbb{R}^m</math> such that for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.

0708-1300/Class notes for Tuesday, September 11

2007-09-12T02:05:22Z

Megan: /* Differentiability */

{{0708-1300/Navigation}}
==In Small Scales, Everything's Linear==
{| align=center
|- align=center
|[[Image:06-240-QuiltBeforeMap.png|200px]]
|<math>\longrightarrow</math>
|[[Image:06-240-QuiltAfterMap.png|200px]]
|- align=center
|<math>z</math>
|<math>\mapsto</math>
|<math>z^2</math>
|}

Code in [http://www.wolfram.com Mathematica]:

<pre>
QuiltPlot[{f_,g_}, {x_, xmin_, xmax_, nx_}, {y_, ymin_, ymax_, ny_}] :=
Module[
{dx, dy, grid, ix, iy},
SeedRandom[1];
dx=(xmax-xmin)/nx;
dy=(ymax-ymin)/ny;
grid = Table[
{x -> xmin+ix*dx, y -> ymin+iy*dy},
{ix, 0, nx}, {iy, 0, ny}
];
grid = Map[({f, g} /. #)&, grid, {2}];
Show[
Graphics[Table[
{
RGBColor[Random[], Random[], Random[]],
Polygon[{
grid[[ix, iy]],
grid[[ix+1, iy]],
grid[[ix+1, iy+1]],
grid[[ix, iy+1]]
}]
},
{ix, nx}, {iy, ny}
]],
Frame -> True
]
]

QuiltPlot[{x, y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
QuiltPlot[{x^2-y^2, 2*x*y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
</pre>

See also [[06-240/Linear Algebra - Why We Care]].

==Class Notes==
===Differentiability===
Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>

'''Definition:'''

We say that <math>f</math> is differentiable (''diffable'') if there is a linear map <math>L</math> so that

<math>\lim_{h\rightarrow0}\frac{|f(x+h)-f(x)-L(h)|}{|h|}=0.</math>

In this case we will say that <math>L</math> is a differential of <math>f</math> and will denote it by <math>df_{x}</math>.

'''Theorem'''

If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following asertions holds:

# <math>df_{x}</math> is unique.
# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>
# If <math>f</math> is linear then <math>df_{x}=f</math>
# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>
# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

===Implicit Function Theorem===

'''Example'''
Although <math>x^2+y^2=1</math> does not defines <math>y</math> as a function of <math>x</math>, in a neighborhood of <math>(0;-1)</math> we can define <math>g(x)=-\sqrt{1-x^2}</math> so that <math>x^2+g(x)^2=1</math>. Furthermore, <math>g</math> is differentiable with differential <math>dg_{x}=\frac{x}{\sqrt{1-x^2}}</math>. This is a motivation for the following theorem.

'''Notation'''

If f:X\times Y\rightarrow Z then given x\in X we will define f_{[x]}:Y\rightarrow Z by f_{[x]}(y)=f(x;y)

'''Definition'''

<math>C^{p}(V)</math> will be the class of all functions defined on <math>V</math> with continuous partial derivatives up to order <math>p.</math>

'''Theorem'''(''Implicit function theorem'')

Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:

There is an open neighborhood of <math>x</math>, <math>V\subset U</math>, and a <math>diffable</math> function <math>g:V\rightarrow\mathbb{R}^m</math> such that for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.