0708-1300/Class notes for Tuesday, November 13 - Revision history

Trefor: /* Second Hour */

2008-02-09T03:37:38Z

Second Hour

← Older revision		Revision as of 23:37, 8 February 2008
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	2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point.		2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point.

	3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the x_2, x_1 plane the tenancy for it to rotate clockwise would be given by <math>\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}</math>		3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the <math>x_2</math>, <math>x_1</math> plane the tenancy for it to rotate clockwise would be given by <math>\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}</math>

Trefor: /* Second Hour */

2008-02-09T03:33:55Z

Second Hour

← Older revision		Revision as of 23:33, 8 February 2008
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	Now let us compute <math>d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} ~~dx_3~~\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3</math> + 6 more terms representing the 3 partials of each of the last 2 terms.		Now let us compute <math>d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_2\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3</math> + 6 more terms representing the 3 partials of each of the last 2 terms.

	As each <math>dx_i\wedge dx_i</math> term vanishes we are left with just,		As each <math>dx_i\wedge dx_i</math> term vanishes we are left with just,

Trefor: /* First Hour */

2008-02-09T03:07:52Z

First Hour

← Older revision		Revision as of 23:07, 8 February 2008
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	Given M, <math>\exists</math> ! linear map <math>d:\Omega^(M)\rightarrow\Omega^{k+1}(M)</math> satisfies		Given M, <math>\exists</math> ! linear map <math>d:\Omega^k(M)\rightarrow\Omega^{k+1}(M)</math> satisfies

	1) If <math>f\in\Omega^0(M)</math> then <math>df(X) = X(f)</math> for <math>X\in TM</math>		1) If <math>f\in\Omega^0(M)</math> then <math>df(X) = X(f)</math> for <math>X\in TM</math>

Trefor: /* First Hour */

2008-02-09T02:53:16Z

First Hour

← Older revision		Revision as of 22:53, 8 February 2008
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	Now, <math>~~dx_2~~(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}~~x_2~~ =\delta_{i2}</math> and hence we get a basis.		Now, <math>dx_j(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_j =\delta_{ij}</math> and hence we get a basis.

Aroesler: /* Second Hour */

2008-02-07T22:02:49Z

Second Hour

← Older revision		Revision as of 18:02, 7 February 2008
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	We are left with computing d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>		We are left with computing d from <math>\Omega^1(M)</math> to <math>\Omega^2(M)</math>

	Computing, <math>d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2</math>		Computing, <math>d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2</math>

Aroesler: /* Second Hour */

2008-02-07T22:01:24Z

Second Hour

← Older revision		Revision as of 18:01, 7 February 2008
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	As each <math>dx_i\wedge dx_i</math> term vanishes we are left with just,		As each <math>dx_i\wedge dx_i</math> term vanishes we are left with just,

	<math>= \frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3</math>		<math>= (\frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3</math>

	I.e., d takes <math>(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial ~~dx_i~~}</math>		I.e., d takes <math>(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial x_i}</math>

	this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>.		this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>.

Aroesler: /* Second Hour */

2008-02-07T22:00:15Z

Second Hour

← Older revision		Revision as of 18:00, 7 February 2008
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	Returning, we thus get <math>d(\sum_I f_I dx_I) = \sum_{i,J} \frac{\partial f_I}{\partial x_j} ~~dx_k~~\wedge dx_I</math>		Returning, we thus get <math>d(\sum_I f_I dx_I) = \sum_{I,j} \frac{\partial f_I}{\partial x_j} dx_j\wedge dx_I</math>

	Thus our d takes functions to vector fields by <math>f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})</math>		Thus our d takes functions to vector fields by <math>f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})</math>

Trefor at 00:42, 26 November 2007

2007-11-26T00:42:04Z

← Older revision		Revision as of 20:42, 25 November 2007
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	Now differential forms pull back, i.e., for <math>\phi:M\rightarrow N</math> then <math>\~~phi_~~*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)</math>		Now differential forms pull back, i.e., for <math>\phi:M\rightarrow N</math> then <math>\phi^*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)</math>
	via		via
	<math>\~~phi_~~(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_ v_1,\ldots \phi_* v_k)</math>		<math>\phi^(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_ v_1,\ldots \phi_* v_k)</math>


	The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via <math>\~~phi_~~(\omega\wedge\lambda)=\~~phi_~~(\omega)\wedge\~~phi_~~*(\lambda)</math>		The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via <math>\phi^(\omega\wedge\lambda)=\phi^(\omega)\wedge\phi^*(\lambda)</math>

Aptikuis: /* First Hour */

2007-11-25T22:59:01Z

First Hour

← Older revision		Revision as of 18:59, 25 November 2007
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	The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via <math>\~~phi^~~(\omega\wedge\lambda)=\~~phi^~~(\omega)\wedge\~~phi^~~*(\lambda)</math>		The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via <math>\phi_(\omega\wedge\lambda)=\phi_(\omega)\wedge\phi_*(\lambda)</math>

Aptikuis: /* First Hour */

2007-11-25T22:58:15Z

First Hour

← Older revision		Revision as of 18:58, 25 November 2007
Line 92:		Line 92:


	Now differential forms pull back, i.e., for <math>\phi:M\rightarrow N</math> then <math>\~~phi^~~*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)</math>		Now differential forms pull back, i.e., for <math>\phi:M\rightarrow N</math> then <math>\phi_*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)</math>
	via		via
	<math>\~~phi^~~(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_ v_1,\ldots \phi_* v_k)</math>		<math>\phi_(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_ v_1,\ldots \phi_* v_k)</math>