0708-1300/Class notes for Tuesday, January 22 - Revision history

Drorbn: /* Typed Notes */

2008-04-15T10:37:22Z

Typed Notes

← Older revision		Revision as of 06:37, 15 April 2008
Line 41:		Line 41:

	==Typed Notes==		==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===		===First Hour===

Franklin: /* First Hour */

2008-02-18T20:59:01Z

First Hour

← Older revision		Revision as of 16:59, 18 February 2008
Line 86:		Line 86:
	Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>		Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>

	Of course, we need to know that in [[0708-1300/fact fact]] <math>\mathbb{Z}^m\neq\mathbb{Z}^n</math> if <math>m\neq n</math>		Of course, we need to know that in [[0708-1300/fact\|fact]] <math>\mathbb{Z}^m\neq\mathbb{Z}^n</math> if <math>m\neq n</math>

	As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.		As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.

Franklin: /* First Hour */

2008-02-18T20:58:48Z

First Hour

← Older revision		Revision as of 16:58, 18 February 2008
Line 86:		Line 86:
	Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>		Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>

	Of course, we need to know that in [[~~fact\|~~0708-1300/fact]] <math>\mathbb{Z}^m\neq\mathbb{Z}^n</math> if <math>m\neq n</math>		Of course, we need to know that in [[0708-1300/fact fact]] <math>\mathbb{Z}^m\neq\mathbb{Z}^n</math> if <math>m\neq n</math>

	As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.		As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.

Franklin: /* First Hour */

2008-02-18T20:58:21Z

First Hour

← Older revision		Revision as of 16:58, 18 February 2008
Line 86:		Line 86:
	Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>		Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>

	Of course, we need to know that in ~~fact~~ [[<math>\mathbb{Z}^m\neq\mathbb{Z}^n</math>]] if <math>m\neq n</math>		Of course, we need to know that in [[fact\|0708-1300/fact]] <math>\mathbb{Z}^m\neq\mathbb{Z}^n</math> if <math>m\neq n</math>

	As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.		As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.

Franklin: /* First Hour */

2008-02-18T20:57:25Z

First Hour

← Older revision		Revision as of 16:57, 18 February 2008
Line 86:		Line 86:
	Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>		Hence we know that <math>\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})</math>

	Of course, we need to know that in fact <math>\mathbb{Z}^m\neq\mathbb{Z}^n</math> if <math>m\neq n</math>		Of course, we need to know that in fact [[<math>\mathbb{Z}^m\neq\mathbb{Z}^n</math>]] if <math>m\neq n</math>

	As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.		As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic.

Trefor: /* Second Hour */

2008-02-10T02:10:03Z

Second Hour

← Older revision		Revision as of 22:10, 9 February 2008
Line 197:		Line 197:
	1) <math>T_{p,q}</math> is the knot that wraps around the torus p times one way and q times the other way.		1) <math>T_{p,q}</math> is the knot that wraps around the torus p times one way and q times the other way.

	2) Formally, let <math>\sigma:S^1\times S^1\mathbb{R}^3</math> be standard embedding of a torus. Let <math>\gamma:[0,1]\rightarrow S^1\times S^1</math> be <math>t\rightarrow (e^{2\pi i pt}, e^{i2\pi qt})</math>		2) Formally, let <math>\sigma:S^1\times S^1\rightarrow\mathbb{R}^3</math> be the standard embedding of a torus. Let <math>\gamma:[0,1]\rightarrow S^1\times S^1</math> be <math>t\rightarrow (e^{2\pi i pt}, e^{i2\pi qt})</math>

	Then <math>T_{p,q}</math> is <math>\sigma\circ\gamma</math>		Then <math>T_{p,q}</math> is <math>\sigma\circ\gamma</math>

Trefor: /* Second Hour */

2008-02-06T23:44:54Z

Second Hour

← Older revision		Revision as of 19:44, 6 February 2008
Line 157:		Line 157:

	<math>U_2</math> = the other solid torus, also thickened a bit, under <math>\pi_1</math> yields <math><\beta></math>		<math>U_2</math> = the other solid torus, also thickened a bit, under <math>\pi_1</math> yields <math><\beta></math>
	<math>


	U_1\cap U_2</math> is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)		<math>U_1\cap U_2</math> is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)

	So, <math>\pi_1(U_1\cap U_2)\cong\mathbb{Z}^2 \cong <a,b>/ab=ba</math>		So, <math>\pi_1(U_1\cap U_2)\cong\mathbb{Z}^2 \cong <a,b>/ab=ba</math>

Trefor: /* Second Hour */

2008-02-06T23:43:36Z

Second Hour

← Older revision		Revision as of 19:43, 6 February 2008
Line 158:		Line 158:
	<math>U_2</math> = the other solid torus, also thickened a bit, under <math>\pi_1</math> yields <math><\beta></math>		<math>U_2</math> = the other solid torus, also thickened a bit, under <math>\pi_1</math> yields <math><\beta></math>
	<math>		<math>


	U_1\cap U_2</math> is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)		U_1\cap U_2</math> is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)

Canghel: /* Second Hour */

2008-02-06T05:04:54Z

Second Hour

← Older revision		Revision as of 01:04, 6 February 2008
Line 148:		Line 148:
	Lets now add in the second torus into this picture. We first draw a horizontal line between the two disks. We then "blow" up from beneath so the horizontal line is slightly curved. We imagine continuing to blow yielding larger and larger loops between the two disks until it "pops" forming the pure horizontal line consisting of the loop at infinity. Do the same for the bottom. Hence, the boundaries of the two tori drawn this way clearly are the same, and between the two cover the entire zx plane (and "point at infinity). Rotating this picture about the z axis yields all of S^1 as the union of these two sets.		Lets now add in the second torus into this picture. We first draw a horizontal line between the two disks. We then "blow" up from beneath so the horizontal line is slightly curved. We imagine continuing to blow yielding larger and larger loops between the two disks until it "pops" forming the pure horizontal line consisting of the loop at infinity. Do the same for the bottom. Hence, the boundaries of the two tori drawn this way clearly are the same, and between the two cover the entire zx plane (and "point at infinity). Rotating this picture about the z axis yields all of S^1 as the union of these two sets.

			[[Image:0708-1300_notes_22-01-08b.jpg\|200px]]

	'''Claim:'''		'''Claim:'''

Canghel: /* First Hour */

2008-02-06T05:04:00Z

First Hour

← Older revision		Revision as of 01:04, 6 February 2008
Line 69:		Line 69:
	Consider two such pentagons, gluing along the edge c forms precisely the octagon we had for the two holed torus last class.		Consider two such pentagons, gluing along the edge c forms precisely the octagon we had for the two holed torus last class.

			[[Image:0708-1300_notes_22-01-08a.jpg\|200px]]

	'''Proposition'''		'''Proposition'''
Line 134:		Line 135:

	Considering the composition we get <math>\eta\bar{\eta}\gamma\eta\bar{\eta}\sim\gamma</math>.		Considering the composition we get <math>\eta\bar{\eta}\gamma\eta\bar{\eta}\sim\gamma</math>.


	===Second Hour===		===Second Hour===