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dennis westra
Nov17-04, 06:38 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I have a basic question on the computation of the de Rahm cohomology\non a group manifold. What I need to know is how to choose the basis of\nclosed forms which are not exact.\n\n[Moderator\'s note: There are many different ways how can you choose\na basis of a linear space. Some bases are more convenient, but others\nare bases, too. ;-) LM]\n\nCan I take just the left-invariant forms and multiply them with constant\nnumbers, or should these constant also depend on the coordinates on the\ngroup manifold.\n\n[Moderator\'s note: I thought that if something depends on coordinates of\nyour manifold, then it\'s exactly what\'s called a "non-constant". ;-)\nAlso, if you multiply a closed form by a scalar function i.e. a 0-form\n(which is what you call a constant), the product will\nnot be closed, but will be proportional to the wedge product\nof the original closed form and the gradient of the 0-form. LM]\n\nIn other words; does it suffice to calculate the betti numbers for forms\n\\Phi_{a_1 ... a_p}\\sigma^a_1 \\wedge ...\\wedge \\sigma^a_p with \\Phi\nconstants and the \\sigma\'s left-invariant Maurer-Cartan forms? Since forms\nare allowed to depend on the coordinates I would think it differs from\ndoing the analysis with \\Phi non-constant... if I take the group manifold\nto be compact, can a general statement be made?\n\n[Moderator\'s note: Please, who has more time, answer this! Tx, LM]\n\nthanks for yoyr time,\n\\dw\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>I have a basic question on the computation of the de Rahm cohomology
on a group manifold. What I need to know is how to choose the basis of
closed forms which are not exact.

[Moderator's note: There are many different ways how can you choose
a basis of a linear space. Some bases are more convenient, but others
are bases, too. ;-) LM]

Can I take just the left-invariant forms and multiply them with constant
numbers, or should these constant also depend on the coordinates on the
group manifold.

[Moderator's note: I thought that if something depends on coordinates of
your manifold, then it's exactly what's called a "non-constant". ;-)
Also, if you multiply a closed form by a scalar function i.e. a 0-form
(which is what you call a constant), the product will
not be closed, but will be proportional to the wedge product
of the original closed form and the gradient of the 0-form. LM]

In other words; does it suffice to calculate the betti numbers for forms
\Phi_{a_1 ... a_p}\sigma^a_1 \wedge ...\wedge \sigma^a_p with \Phi
constants and the \sigma's left-invariant Maurer-Cartan forms? Since forms
are allowed to depend on the coordinates I would think it differs from
doing the analysis with \Phi non-constant... if I take the group manifold
to be compact, can a general statement be made?

[Moderator's note: Please, who has more time, answer this! Tx, LM]

thanks for yoyr time,
\dw

Peter Woit
Nov17-04, 08:15 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>dennis westra wrote:\n\n&gt;I have a basic question on the computation of the de Rahm cohomology\n&gt;on a group manifold. What I need to know is how to choose the basis of\n&gt;closed forms which are not exact.\n&gt;\n&gt;[Moderator\'s note: There are many different ways how can you choose\n&gt; a basis of a linear space. Some bases are more convenient, but others\n&gt; are bases, too. ;-) LM]\n&gt;\n&gt;Can I take just the left-invariant forms and multiply them with constant\n&gt;numbers, or should these constant also depend on the coordinates on the\n&gt;group manifold.\n&gt;\n&gt;[Moderator\'s note: I thought that if something depends on coordinates of\n&gt; your manifold, then it\'s exactly what\'s called a "non-constant". ;-)\n&gt; Also, if you multiply a closed form by a scalar function i.e. a 0-form\n&gt; (which is what you call a constant), the product will\n&gt; not be closed, but will be proportional to the wedge product\n&gt; of the original closed form and the gradient of the 0-form. LM]\n&gt;\n&gt;In other words; does it suffice to calculate the betti numbers for forms\n&gt;\\Phi_{a_1 ... a_p}\\sigma^a_1 \\wedge ...\\wedge \\sigma^a_p with \\Phi\n&gt;constants and the \\sigma\'s left-invariant Maurer-Cartan forms? Since forms\n&gt;are allowed to depend on the coordinates I would think it differs from\n&gt;doing the analysis with \\Phi non-constant... if I take the group manifold\n&gt;to be compact, can a general statement be made?\n&gt;\n&gt;\n&gt;\nIf your group is compact, the subcomplex of left-invariant forms has\nthe same cohomology as the full de Rham complex. You show this\nby showing that averaging over left translations is an "exact\nfunctor" (doesn\'t change the cohomology). This only works\nif the group is compact so the integral you do when you average\nconverges.\n\nThe cohomology of the complex of left-invariant forms on the\ngroup is exactly the Lie algebra cohomology. Now, taking\nthe right-invariant subcomplex of this complex again doesn\'t\nchange the cohomology. This gives a sequence with zero\ndifferential, its terms exactly represent the cohomology.\n\nSo, for a compact Lie group, the cohomology is represented\nby bi-invariant (both left and right invariant) forms on the group.\n\nYou can also think about this using Hodge theory, the cohomology\nis represented by harmonic forms, again they\'re the bi-invariant ones.\n\nFor more about this, try reading two things by Raoul Bott\n\n"On Characteristic Classes and Continuous Cohomology" in\nSupersymmetry and its Applications, Cambridge University\nPress, 1986\n\nand\n\n"Some Aspects of Invariant Theory in Differential Geometry"\nin\nDifferential Operators on Manifolds" Edizioni Cremonese (1975)\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>dennis westra wrote:

>I have a basic question on the computation of the de Rahm cohomology
>on a group manifold. What I need to know is how to choose the basis of
>closed forms which are not exact.
>
>[Moderator's note: There are many different ways how can you choose
> a basis of a linear space. Some bases are more convenient, but others
> are bases, too. ;-) LM]
>
>Can I take just the left-invariant forms and multiply them with constant
>numbers, or should these constant also depend on the coordinates on the
>group manifold.
>
>[Moderator's note: I thought that if something depends on coordinates of
> your manifold, then it's exactly what's called a "non-constant". ;-)
> Also, if you multiply a closed form by a scalar function i.e. a 0-form
> (which is what you call a constant), the product will
> not be closed, but will be proportional to the wedge product
> of the original closed form and the gradient of the 0-form. LM]
>
>In other words; does it suffice to calculate the betti numbers for forms
>\Phi_{a_1 ... a_p}\sigma^a_1 \wedge ...\wedge \sigma^a_p with \Phi
>constants and the \sigma's left-invariant Maurer-Cartan forms? Since forms
>are allowed to depend on the coordinates I would think it differs from
>doing the analysis with \Phi non-constant... if I take the group manifold
>to be compact, can a general statement be made?
>
>
>
If your group is compact, the subcomplex of left-invariant forms has
the same cohomology as the full de Rham complex. You show this
by showing that averaging over left translations is an "exact
functor" (doesn't change the cohomology). This only works
if the group is compact so the integral you do when you average
converges.

The cohomology of the complex of left-invariant forms on the
group is exactly the Lie algebra cohomology. Now, taking
the right-invariant subcomplex of this complex again doesn't
change the cohomology. This gives a sequence with zero
differential, its terms exactly represent the cohomology.

So, for a compact Lie group, the cohomology is represented
by bi-invariant (both left and right invariant) forms on the group.

You can also think about this using Hodge theory, the cohomology
is represented by harmonic forms, again they're the bi-invariant ones.

For more about this, try reading two things by Raoul Bott

"On Characteristic Classes and Continuous Cohomology" in
Supersymmetry and its Applications, Cambridge University
Press, 1986

and

"Some Aspects of Invariant Theory in Differential Geometry"
in
Differential Operators on Manifolds" Edizioni Cremonese (1975)