Drorbn - User contributions [en]

0708-1300/Class notes for Tuesday, March 18

2008-05-01T02:30:08Z

Evan.decorte: /* First Hour */

{{0708-1300/Navigation}}

==Typed Notes==

<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>

===First Hour===

Recall we had defined <math>\tilde{H}(X):= ker\epsilon_*</math> where <math>\epsilon_*:X\rightarrow\{pt\}</math>, <math>\tilde{H}(X,A) = H(X,A)</math>

For <math>X\neq\empty</math>, <math>\tilde{H}_p(X) = H_p(X)</math> for <math>p\neq 0</math> and equals <math>G\oplus\tilde{H}_0(X)</math> for <math>p=0</math>

This homology definition satisfies the axioms with the following changes: Exactness only for <math>X\neq\empty</math> and the dimension axiom being <math>\tilde{H}_*(pt) = 0</math>. Furthermore, instead of additivity we have, under mild conditions of <math>b_0\in X</math> and <math>b_1\in Y</math> (ie non empty) define <math>X\vee Y:= X\cup Y / b_0\sim b_1</math> for a disjoint union. Then, <math>\tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)
</math>

We can actually get the above isomorphism in the following way. There are natural projection maps <math>p_x</math> and <math>p_y</math> from <math>X\vee Y</math> to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps <math>i_x</math> and <math>i_y</math> going the other way. Then, <math>(p_{x*},p_{y*})</math> and <math>i_{x*}+i_{y*}</math> are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it.

<math>\tilde{H}</math> is "kinda" natural:

We have a chain complex where <math>C_p = <\sigma:\Delta_p\rightarrow X></math> where <math>\Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}</math>

We thus get that <math>\Delta_{-1} = \empty</math> since <math>\sum x_1 = 0\neq 1</math> vacuously.

Therefore, <math>C_{-1}(X) = <\sigma:\Delta_{-1}\rightarrow X> = \mathbb{Z}</math>

So, <math>\tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)</math>

Note: We have never actually specified that p is positive axiomatically. In fact, <math>\tilde{H}_p(S^n) = G</math> for p=n and 0 for <math>p\neq n</math> works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p.

'''Degrees'''

<math>(G=\mathbb{Z})</math>

<math>f:S^n\rightarrow S^n</math> then get <math>f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}
</math>

We thus define:
<math>deg(f):=d = f_*(1)</math>

<math>
f_{1,2}:S^0\rightarrow S^0</math> has

<math>1) f_1 = I</math>

2) <math>f_2</math> = flip, ie <math>x_0\rightarrow -x_0</math>

<math>deg f_1 = deg I = 1</math> in all dimensions

deg <math>f_2</math> = deg flip = -1

'''Proposition'''

Let <math>f:S^n\rightarrow S^n</math> be <math>x_0\mapsto-x_0</math> and <math>x_i\mapsto x_i</math> for i>0

then def f= -1

''Proof:''

We get two rows of the following sequence, with the induced maps from f going vertically between them:

<math>\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)</math>

The resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism <math>\partial</math>

'''Propositions:'''

1) if <math>f\sim g: S^n\rightarrow S^n</math> then deg f = deg g

2) <math>S^n\rightarrow^f s^n\rightarrow^g S^n</math> then <math>deg (g\circ f) = deg (f) deg (g)
</math>

3) deg a where a is the antipodal map <math>x\mapsto -x</math> has <math>deg a = (-1)^{n+1}</math> on <math>S^n</math>

===Second Hour===

'''Corollary'''

If n is even, a is not homotopic to I

'''Corollary''' Every <math>f:S^2\rightarrow S^2</math> has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believed true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)

''Proof''

Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. ''Q.E.D''

'''Corollary'''

Every vector field on <math>S^{2n}</math> has a zero, i.e., "on earth there must be a windless points" or "you can't comb the hair on a coconut"

''Proof'' A non zero vector field induces a homotopy of I to a which is impossible.

'''Theorem'''

If <math>f:S^n\rightarrow S^n \ni y_0</math> is smooth (and every map may be approximated by one) and <math>y_0\in S^n</math> is a regular value (which occurs almost everywhere by Sard's Theorem) and <math>f^{-1}(y_0) = \{x_1,\cdots, x_n\}</math> then deg f = <math>\sum_{j=1}^k \pm 1 = \sum</math> sign (det(<math>df_{x_i}</math>))

.ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near <math>x_i</math> with coordinates near <math>y_0</math> using an orientation preserving rotation of <math>S^n</math>

''Examples''

1) <math>S^1\rightarrow S^1</math> via <math>z\mapsto z^k</math>

This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign.

So, <math>deg f = \sum_k +1 = k</math>

2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the <math>deg f = +1 -1 +1-1\cdots = 0</math>

''Proof of Theorem''

1) Let <math>T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}</math> be linear and norm preserving, <math>T\in M_{n+1\times n+1}</math> and <math>T^{T}T = I</math>. Then, deg T = det T

Proof: Every rigid rotation is a product of reflections.

2) Let <math>A:\mathbb{R}^N\rightarrow\mathbb{R}^n</math> be any <math>n\times n</math> non singular matrix so that <math>A(\infty)=\infty</math> so this induces a map <math>\tilde{A}:S^n\rightarrow S^n</math>. Then, <math>deg\tilde{A} = sign(det A)</math>

Proof: Gaussian elimination results in making A a product of "elementary matrices" which come in three types: A matrix with 1's along the diagonal except one diagonal entry being <math>\lambda</math>. A matrix which is the identity only with two rows interchanged. A matrix which is the identity with a <math>\lambda</math> in some non diagonal location.

The latter of these is clearly homotopic to the identity by simply turning the <math>\lambda</math> off.

The middle of these is just a reflection. The former of these, if <math>\lambda>0</math> it is clearly homotopic to the identity. But if <math>\lambda<0</math> then it is homotopic to a reflection.

3) <math>f:\mathbb{R}^n\rightarrow\mathbb{R}^n</math> such that <math>f^{-1}(0) = 0</math>, <math>df|_0 = A</math> non singular, <math>f(\infty) = \infty</math> so f defines <math>\tilde{f}:S^n\rightarrow S^n</math> then deg f = sign(det <math>df|_0)</math>

Proof: Consider for <math>t\geq 1</math>, <math>f_t(x) := tf(x/1)</math>. Then, <math>f_1 = f</math>, <math>f_{\infty}=A</math>. This is a homotopy as it makes good sense for <math>t\in[0,\infty]</math>. So, <math>deg f = deg\tilde{A}</math>

4) All that remains to prove the theorem is the shift from (0,0) to <math>(x_0,y_0)</math> which induces a rotation at <math>df|_{x_0}</math>

==Homologies with non trivial negative p's?==

In the axiomatic definition of homology there is no specification that the homology groups <math>H_p(X)</math> must be necessarily trivial for negative p's. That said, in the singular homology that we are developing using CW complexes we only get non trivial homologies for positive p's. The question was raised in class of when you could possibly have non trivial homology at negative p's.

There are infact homology theories that DO have non trivial homology for negative p. In particular, Dr. Putnam from the University of Victoria has defined a homology theory on Smale space and this theory does in fact have non trivial homology groups at negative p's. While Smale spaces have a complicated axiomized definition, very loosely they are a topological space equipped with a metric and a homeomorphism from the space to itself such that locally you can write the space as the direct sum of a section of the space that increases under the homeomorphism(in terms of the metric) and one that decreases. I.e., it is like you have a local coordinate system that tells you where the space is expanding and where it is contracting under the homeomorphism. The classic example of this are Shifts of Finite Type from Dynamical systems. I computed the homology of a particular Smale space consisting of a torus as the space and a particular homeomorphism of it, and found it to have a non trivial homology at <math>H_{-1}
</math>

0708-1300/Homework Assignment 9

2008-04-30T23:48:41Z

Evan.decorte: /* Just for Fun */

{{0708-1300/Navigation}}

==Reading==
Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!

==Doing==
(Problems 1,2,4,5 below are taken with slight modifications from Hatcher's book, pages 79-80).

# Show that if <math>p_1\colon X_1\to B_1</math> and <math>p_2\colon X_2 \to B_2</math> are covering spaces, then so is their product <math>p_1\times p_2\colon X_1\times X_2\to B_1\times B_2</math>.
# Construct (i.e., describe in explicit terms) a simply-connected covering space of the space <math>X\subset\mathbb{R}^3</math> that is the union of a sphere and a diameter. Do the same when <math>X</math> is the union of a sphere and a circle intersecting it in two points.
# Do the same to the space <math>Y</math> of the term test: <math>Y=\{z\in{\mathbb C}\colon|z|\leq 1\}/(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1)</math>.
# Find all the connected 2-sheeted and 3-sheeted covering spaces of the "figure eight space" <math>S^1\vee S^1</math> (two circles joined at a point), up to isomorphism of covering spaces without base points.
# Let <math>a</math> and <math>b</math> be the generators of <math>\pi_1(S^1\vee S^1)</math> corresponding to the two <math>S^1</math> summands. Draw a picture of the covering space of <math>S^1\vee S^1</math> corresponding to the normal subgroup generated by <math>a^2</math>, <math>b^2</math>, and <math>(ab)^4</math>, and prove that this covering space is indeed the correct one.

==Due Date==
This assignment is due in class on Thursday February 28, 2008.

==Just for Fun==
* What happens if in problem 1 we consider infinitely many covering spaces. That is, is the product of an infinite family of covering spaces a covering space? [[0708-1300/Covering_Product|Here]] is an idea but don't look at it until you have think on the problem for a while.
* This raises another question. A "pathwise totally disconnected space" is a space in which every path is a constant path. How much of the theory of covering spaces can be generalized to "coverings" in which the fibers are pathwise totally disconnected, instead of discrete?
* Here is a short introduction to regular covering spaces and an application of them to the last problem on this homework: [[Media:0708-1300-Regular-Covering-Spaces.pdf|Regular Covering Spaces]] (PDF).

0708-1300/Homework Assignment 9

2008-04-30T23:35:40Z

Evan.decorte: /* Just for Fun */

0708-1300/Homework Assignment 9

2008-04-30T23:23:49Z

Evan.decorte: /* Just for Fun */

File:0708-1300-Regular-Covering-Spaces.pdf

2008-04-30T23:21:53Z

Evan.decorte:

0708-1300/Homework Assignment 9

2008-04-30T23:13:19Z

Evan.decorte: /* Just for Fun */

{{0708-1300/Navigation}}

==Reading==
Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!

==Doing==
(Problems 1,2,4,5 below are taken with slight modifications from Hatcher's book, pages 79-80).

# Show that if <math>p_1\colon X_1\to B_1</math> and <math>p_2\colon X_2 \to B_2</math> are covering spaces, then so is their product <math>p_1\times p_2\colon X_1\times X_2\to B_1\times B_2</math>.
# Construct (i.e., describe in explicit terms) a simply-connected covering space of the space <math>X\subset\mathbb{R}^3</math> that is the union of a sphere and a diameter. Do the same when <math>X</math> is the union of a sphere and a circle intersecting it in two points.
# Do the same to the space <math>Y</math> of the term test: <math>Y=\{z\in{\mathbb C}\colon|z|\leq 1\}/(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1)</math>.
# Find all the connected 2-sheeted and 3-sheeted covering spaces of the "figure eight space" <math>S^1\vee S^1</math> (two circles joined at a point), up to isomorphism of covering spaces without base points.
# Let <math>a</math> and <math>b</math> be the generators of <math>\pi_1(S^1\vee S^1)</math> corresponding to the two <math>S^1</math> summands. Draw a picture of the covering space of <math>S^1\vee S^1</math> corresponding to the normal subgroup generated by <math>a^2</math>, <math>b^2</math>, and <math>(ab)^4</math>, and prove that this covering space is indeed the correct one.

==Due Date==
This assignment is due in class on Thursday February 28, 2008.

==Just for Fun==
* What happens if in problem 1 we consider infinitely many covering spaces. That is, is the product of an infinite family of covering spaces a covering space? [[0708-1300/Covering_Product|Here]] is an idea but don't look at it until you have think on the problem for a while.
* This raises another question. A "pathwise totally disconnected space" is a space in which every path is a constant path. How much of the theory of covering spaces can be generalized to "coverings" in which the fibers are pathwise totally disconnected, instead of discrete?
* Here is a short introduction to regular covering spaces, along with an application of them to the last problem on this homework. [[0708-1300/Regular_Covering_Spaces.pdf]]

0708-1300/Homework Assignment 9

2008-04-30T23:12:25Z

Evan.decorte: /* Just for Fun */

{{0708-1300/Navigation}}

==Reading==
Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!

==Doing==
(Problems 1,2,4,5 below are taken with slight modifications from Hatcher's book, pages 79-80).

# Show that if <math>p_1\colon X_1\to B_1</math> and <math>p_2\colon X_2 \to B_2</math> are covering spaces, then so is their product <math>p_1\times p_2\colon X_1\times X_2\to B_1\times B_2</math>.
# Construct (i.e., describe in explicit terms) a simply-connected covering space of the space <math>X\subset\mathbb{R}^3</math> that is the union of a sphere and a diameter. Do the same when <math>X</math> is the union of a sphere and a circle intersecting it in two points.
# Do the same to the space <math>Y</math> of the term test: <math>Y=\{z\in{\mathbb C}\colon|z|\leq 1\}/(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1)</math>.
# Find all the connected 2-sheeted and 3-sheeted covering spaces of the "figure eight space" <math>S^1\vee S^1</math> (two circles joined at a point), up to isomorphism of covering spaces without base points.
# Let <math>a</math> and <math>b</math> be the generators of <math>\pi_1(S^1\vee S^1)</math> corresponding to the two <math>S^1</math> summands. Draw a picture of the covering space of <math>S^1\vee S^1</math> corresponding to the normal subgroup generated by <math>a^2</math>, <math>b^2</math>, and <math>(ab)^4</math>, and prove that this covering space is indeed the correct one.

==Due Date==
This assignment is due in class on Thursday February 28, 2008.

==Just for Fun==
* What happens if in problem 1 we consider infinitely many covering spaces. That is, is the product of an infinite family of covering spaces a covering space? [[0708-1300/Covering_Product|Here]] is an idea but don't look at it until you have think on the problem for a while.
* This raises another question. A "pathwise totally disconnected space" is a space in which every path is a constant path. How much of the theory of covering spaces can be generalized to "coverings" in which the fibers are pathwise totally disconnected, instead of discrete?
* Here is a short introduction to regular covering spaces, along with an application of them to the last problem on this homework. [[0708-1300/Regular Covering Spaces.pdf]]

0708-1300/Class Photo

2007-10-19T19:39:30Z

Evan.decorte: /* Who We Are... */

{{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=Choi|first=Brian|userid=Brianchoi|email=brianymc.choi@utoronto.ca|location=In the middle of the front row, the weird looking (!) guy with brown shirt over blue and white|comments=}}
{{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=DeCorte|first=Evan|userid=Evan.decorte|email= |location=third from the left, tall guy|comments=}}
{{Photo Entry|last=Fisher|first=Jonathan|userid=jonathan.fisher|email=jonathan.fisher @ utoronto.ca|location=6th from the right, brown shirt, eyes closed|comments=}}
{{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}}
{{Photo Entry|last=Kinzebulatov|first=Damir|userid=Dkinz|email=dkinz @ math.toronto.edu|location=In the middle in red shirt|comments=}}
{{Photo Entry|last=Liu|first=Xiao|userid=Ninetiger|email=ninetiger.liu@utoronto.ca|location=In the first row. A boy in orange T-shirt|comments=}}
{{Photo Entry|last=Li|first=Zhiqiang|userid=li-zhiqiang|email=lizhiqiangfly @ gmail.com|location=2nd from the left, 1st boy in the front row.|comments=}}
{{Photo Entry|last=Antolin Camarena|first=Omar|userid=oantolin|email=oantolin @ math.utoronto.ca|location=furthest person to the right|comments=I'm the TA}}
|}