Re: Nonabelian 2-forms [Archive]

Peter Woit

Jun24-04, 10:18 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>Urs Schreiber wrote:\n\n>>You probably mean that these groups/algebras can be embedded in a\n>>suitable version of SU(infinity) / U(infinity). That would be the same\n>>as saying that they have unitary representations.\n>>\n>>\n>\n>Zunger claims to show that they are not just embedded, but in fact equal.\n>His point is that there is only one "simple peudocompact algebra", in his\n>terminology, and that both SU(\\infty) as well as Poisson/U(1) are simple\n>pseudocompact.\n\n\nI took a quick look at Zunger\'s paper, and as far as I can tell he is\nclaiming to prove that the symplectomorphism groups of all compact\nsymplectic manifolds are isomorphic. This can\'t be right since they are\ntopologically different.\n\nOne reason it\'s hard to figure out exactly what he is saying in that\npaper is that he doesn\'t bother to distinguish between a Lie group and its\nLie algebra, saying things like "the Poisson algebra is U(1) times a\nsimple pseudo-compact algebra" where I guess you have to interpret "U(1)"\nas its Lie algebra, i.e. the real numbers.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Urs Schreiber wrote:

>>You probably mean that these groups/algebras can be embedded in a
>>suitable version of SU(infinity) / U(infinity). That would be the same
>>as saying that they have unitary representations.
>>
>>
>
>Zunger claims to show that they are not just embedded, but in fact equal.
>His point is that there is only one "simple peudocompact algebra", in his
>terminology, and that both SU(\infty) as well as Poisson/U(1) are simple
>pseudocompact.

I took a quick look at Zunger's paper, and as far as I can tell he is
claiming to prove that the symplectomorphism groups of all compact
symplectic manifolds are isomorphic. This can't be right since they are
topologically different.

One reason it's hard to figure out exactly what he is saying in that
paper is that he doesn't bother to distinguish between a Lie group and its
Lie algebra, saying things like "the Poisson algebra is U(1) times a
simple pseudo-compact algebra" where I guess you have to interpret "U(1)"
as its Lie algebra, i.e. the real numbers.

Urs Schreiber

Jun24-04, 10:28 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>On Thu, 24 Jun 2004, Peter Woit wrote:\n\n> I took a quick look at Zunger\'s paper, and as far as I can tell he is\n> claiming to prove that the symplectomorphism groups of all compact\n> symplectic manifolds are isomorphic. This can\'t be right since they are\n> topologically different.\n\nOn p.3 the theorem is stated and it explicitly refers to the algebra, not\nthe group.\n\n> One reason it\'s hard to figure out exactly what he is saying in that\n> paper is that he doesn\'t bother to distinguish between a Lie group and its\n> Lie algebra, saying things like "the Poisson algebra is U(1) times a\n> simple pseudo-compact algebra" where I guess you have to interpret "U(1)"\n> as its Lie algebra, i.e. the real numbers.\n\nI don\'t know if the proof is correct, but from the context it seems\nalways to be clear where the author refers to the algebra and where to the\ngroup.\n\n\n\n\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Thu, 24 Jun 2004, Peter Woit wrote:

> I took a quick look at Zunger's paper, and as far as I can tell he is
> claiming to prove that the symplectomorphism groups of all compact
> symplectic manifolds are isomorphic. This can't be right since they are
> topologically different.

On p.3 the theorem is stated and it explicitly refers to the algebra, not
the group.

> One reason it's hard to figure out exactly what he is saying in that
> paper is that he doesn't bother to distinguish between a Lie group and its
> Lie algebra, saying things like "the Poisson algebra is U(1) times a
> simple pseudo-compact algebra" where I guess you have to interpret "U(1)"
> as its Lie algebra, i.e. the real numbers.

I don't know if the proof is correct, but from the context it seems
always to be clear where the author refers to the algebra and where to the
group.

Urs Schreiber

Jun24-04, 11:23 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>On Thu, 24 Jun 2004, Urs Schreiber wrote:\n\n> I don\'t know if the proof is correct, but from the context it seems\n> always to be clear where the author refers to the algebra and where to the\n> group.\n\nEr, I don\'t seem to understand the apparently most trivial point: On p.7\nit says: "Projections on the Poisson algebra are clearly also projections\non the algebra of smooth functions under multiplication". Why?\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Thu, 24 Jun 2004, Urs Schreiber wrote:

> I don't know if the proof is correct, but from the context it seems
> always to be clear where the author refers to the algebra and where to the
> group.

Er, I don't seem to understand the apparently most trivial point: On p.7
it says: "Projections on the Poisson algebra are clearly also projections
on the algebra of smooth functions under multiplication". Why?

Peter Woit

Jun24-04, 08:51 PM

<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>Urs Schreiber wrote:\n\n>> PW: I took a quick look at Zunger\'s paper, and as far as I can tell he is\n>> claiming to prove that the symplectomorphism groups of all compact\n>> symplectic manifolds are isomorphic. This can\'t be right since they are\n>> topologically different.\n>\n> On p.3 the theorem is stated and it explicitly refers to the algebra, not\n> the group.\n\nWell, he\'s explicitly claiming a result about the group: "we will show\nthat the group of area-preserving diffeomorphisms of any Riemann surface\nis SU(infinity)". If his results are only about the Lie algebra, he should\nsay so. In any case I\'m still having trouble believing that all such Lie\nalgebras are isomorphic.\n\nI tried to follow his proof, but have trouble from the very beginning. He\nwants to use standard techniques about roots and weights, which rely on\nstudying how a maximal abelian subgroup acts on the adjoint\nrepresentation. He seems to be claiming that in his case such a group\nexists and is compact. I don\'t see why that should be true.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Urs Schreiber wrote:

>> PW: I took a quick look at Zunger's paper, and as far as I can tell he is
>> claiming to prove that the symplectomorphism groups of all compact
>> symplectic manifolds are isomorphic. This can't be right since they are
>> topologically different.
>
> On p.3 the theorem is stated and it explicitly refers to the algebra, not
> the group.

Well, he's explicitly claiming a result about the group: "we will show
that the group of area-preserving diffeomorphisms of any Riemann surface
is SU(infinity)". If his results are only about the Lie algebra, he should
say so. In any case I'm still having trouble believing that all such Lie
algebras are isomorphic.

I tried to follow his proof, but have trouble from the very beginning. He
wants to use standard techniques about roots and weights, which rely on
studying how a maximal abelian subgroup acts on the adjoint
representation. He seems to be claiming that in his case such a group
exists and is compact. I don't see why that should be true.

Charlie Stromeyer Jr.

Jun25-04, 07:13 PM

<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>Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote in message news:\n\n> Er, I don\'t seem to understand the apparently most trivial point: On p.7\n> it says: "Projections on the Poisson algebra are clearly also projections\n> on the algebra of smooth functions under multiplication". Why?\n\nCan you perhaps clarify some what it is that you suspect you don\'t\nunderstand? For example, is it why only the product rule is different,\nor why the projections on the algebra of smooth functions are\npiecewise constant etc.?\n\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote in message news:

> Er, I don't seem to understand the apparently most trivial point: On p.7
> it says: "Projections on the Poisson algebra are clearly also projections
> on the algebra of smooth functions under multiplication". Why?

Can you perhaps clarify some what it is that you suspect you don't
understand? For example, is it why only the product rule is different,
or why the projections on the algebra of smooth functions are
piecewise constant etc.?

Urs Schreiber

Jun29-04, 06:07 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>"Charlie Stromeyer Jr." <cstromey@hotmail.com> schrieb im Newsbeitrag\nnews:61773ed7.0406250948.19f505e5-100000@posting.google.com...\n> Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote in message news:\n>\n> > Er, I don\'t seem to understand the apparently most trivial point: On p.7\n> > it says: "Projections on the Poisson algebra are clearly also\nprojections\n> > on the algebra of smooth functions under multiplication". Why?\n>\n> Can you perhaps clarify some what it is that you suspect you don\'t\n> understand? For example, is it why only the product rule is different,\n> or why the projections on the algebra of smooth functions are\n> piecewise constant etc.?\n\nI understand that we are dealing with the Poisson algebra on the one side,\nwith the product being f*g = {f,g} and with the usual product of smooth\nfunctions on the other side and I understand that homomorphisms on the\nalgebra of smooth functions under ordinary multiplication which are\nprojections, are projections onto piecewise constant functions taking values\nin {0,1}. What I don\'t see yet how this proves that the only homomorphisms\non the Poisson algebra which are at the same time projections are the same\nprojections as above.\n\nOf course I understand that the constant functions are a proper subalgebra\nof the Poisson algebra. What I don\'t understand yet is Zunger\'s proof that\nit is the only proper subalgebra.\n\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Charlie Stromeyer Jr." <cstromey@hotmail.com> schrieb im Newsbeitrag
news:61773ed7.0406250948.19f505e5-100000@posting.google.com...
> Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote in message news:
>
> > Er, I don't seem to understand the apparently most trivial point: On p.7
> > it says: "Projections on the Poisson algebra are clearly also
projections
> > on the algebra of smooth functions under multiplication". Why?
>
> Can you perhaps clarify some what it is that you suspect you don't
> understand? For example, is it why only the product rule is different,
> or why the projections on the algebra of smooth functions are
> piecewise constant etc.?

I understand that we are dealing with the Poisson algebra on the one side,
with the product being f*g = {f,g} and with the usual product of smooth
functions on the other side and I understand that homomorphisms on the
algebra of smooth functions under ordinary multiplication which are
projections, are projections onto piecewise constant functions taking values
in {0,1}. What I don't see yet how this proves that the only homomorphisms
on the Poisson algebra which are at the same time projections are the same
projections as above.

Of course I understand that the constant functions are a proper subalgebra
of the Poisson algebra. What I don't understand yet is Zunger's proof that
it is the only proper subalgebra.

Charlie Stromeyer Jr.

Jun29-04, 06:08 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>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:\n\n> Well, he\'s explicitly claiming a result about the group: "we will show\n> that the group of area-preserving diffeomorphisms of any Riemann surface\n> is SU(infinity)". If his results are only about the Lie algebra, he should\n> say so. In any case I\'m still having trouble believing that all such Lie\n> algebras are isomorphic.\n>\n> I tried to follow his proof, but have trouble from the very beginning. He\n> wants to use standard techniques about roots and weights, which rely on\n> studying how a maximal abelian subgroup acts on the adjoint\n> representation. He seems to be claiming that in his case such a group\n> exists and is compact. I don\'t see why that should be true.\n\nAh, it turns out that his paper works for the adjoint rep but not for\nthe fundamental rep, as explained in:\n\nhttp://arxiv.org/abs/hep-th/0206157\n\nTo honor the spirit of Peter Woit\'s relentless and powerful lack of\ngullibility, I would now like to say that Woit is also correct in that\nit is still unclear how string theory/M-theory could be made\ncompatible with low energy physics, and there may be at least two\nfundamental reasons why.\n\nThe first reason here is well known, but the second is not.\nPersonally, I continue to hope that eventually considering such\nreasons will not kill string theory, but instead help to make string\ntheory stronger and more well defined.\n\n1) The very latest astrophysical evidence is compatible with the idea\nof there being a small, positive and very stable cosmological\nconstant. Following Occam\'s razor, the simplest vacuum solution of\nEinstein\'s equations with the presence of such a lambda is de Sitter\nspacetime. However, string theorists do not yet understand how to\nproperly define for this dS spacetime notions of Hilbert space, their\noperators etc.\n\n2) By definition, string theorists (and also other quantum gravity\ntheorists and cosmologists) study high energy physics which means that\nthey might not have enough overall great expertise about low energy\nphysics. I myself also lack such expertise but, as I have already\nsuggested, there might be some non-trivial quantum weirdness which is\nrevealed by various analyses of different quantum optics experiments.\n\n--------------------------------------------------------------------------------\n\n"As for everything else, so for a mathematical theory: beauty can be\nperceived but not explained."\n\n– Arthur Cayley\n\n--------------------------------------------------------------------------------\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:

> Well, he's explicitly claiming a result about the group: "we will show
> that the group of area-preserving diffeomorphisms of any Riemann surface
> is SU(infinity)". If his results are only about the Lie algebra, he should
> say so. In any case I'm still having trouble believing that all such Lie
> algebras are isomorphic.
>
> I tried to follow his proof, but have trouble from the very beginning. He
> wants to use standard techniques about roots and weights, which rely on
> studying how a maximal abelian subgroup acts on the adjoint
> representation. He seems to be claiming that in his case such a group
> exists and is compact. I don't see why that should be true.

Ah, it turns out that his paper works for the adjoint rep but not for
the fundamental rep, as explained in:

http://arxiv.org/abs/http://www.arxiv.org/abs/hep-th/0206157

To honor the spirit of Peter Woit's relentless and powerful lack of
gullibility, I would now like to say that Woit is also correct in that
it is still unclear how string theory/M-theory could be made
compatible with low energy physics, and there may be at least two
fundamental reasons why.

The first reason here is well known, but the second is not.
Personally, I continue to hope that eventually considering such
reasons will not kill string theory, but instead help to make string
theory stronger and more well defined.

1) The very latest astrophysical evidence is compatible with the idea
of there being a small, positive and very stable cosmological
constant. Following Occam's razor, the simplest vacuum solution of
Einstein's equations with the presence of such a \lambda is de Sitter
spacetime. However, string theorists do not yet understand how to
properly define for this dS spacetime notions of Hilbert space, their
operators etc.

2) By definition, string theorists (and also other quantum gravity
theorists and cosmologists) study high energy physics which means that
they might not have enough overall great expertise about low energy
physics. I myself also lack such expertise but, as I have already
suggested, there might be some non-trivial quantum weirdness which is
revealed by various analyses of different quantum optics experiments.

--------------------------------------------------------------------------------

"As for everything else, so for a mathematical theory: beauty can be
perceived but not explained."

– Arthur Cayley

--------------------------------------------------------------------------------

Peter Woit

Jun29-04, 10:05 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>Charlie Stromeyer Jr. wrote:\n\n>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:\n>\n>>Well, he\'s explicitly claiming a result about the group: "we will show\n>>that the group of area-preserving diffeomorphisms of any Riemann surface\n>>is SU(infinity)". If his results are only about the Lie algebra, he should\n>>say so. In any case I\'m still having trouble believing that all such Lie\n>>algebras are isomorphic.\n>>\n>>I tried to follow his proof, but have trouble from the very beginning. He\n>>wants to use standard techniques about roots and weights, which rely on\n>>studying how a maximal abelian subgroup acts on the adjoint\n>>representation. He seems to be claiming that in his case such a group\n>>exists and is compact. I don\'t see why that should be true.\n>>\n>>\n>\n>Ah, it turns out that his paper works for the adjoint rep but not for\n>the fundamental rep, as explained in:\n>\n>http://arxiv.org/abs/hep-th/0206157\n>\n>\n>\nThe paper you mention only deals with the case of a torus. In the case of\nan arbitrary Riemann surface I still don\'t see what the Cartan subalgebra\nis supposed to be. You always have an adjoint representation, the algebra\nacting on itself, but I don\'t see how you even define a "fundamental"\nrepresentation in this case.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Charlie Stromeyer Jr. wrote:

>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:
>
>>Well, he's explicitly claiming a result about the group: "we will show
>>that the group of area-preserving diffeomorphisms of any Riemann surface
>>is SU(infinity)". If his results are only about the Lie algebra, he should
>>say so. In any case I'm still having trouble believing that all such Lie
>>algebras are isomorphic.
>>
>>I tried to follow his proof, but have trouble from the very beginning. He
>>wants to use standard techniques about roots and weights, which rely on
>>studying how a maximal abelian subgroup acts on the adjoint
>>representation. He seems to be claiming that in his case such a group
>>exists and is compact. I don't see why that should be true.
>>
>>
>
>Ah, it turns out that his paper works for the adjoint rep but not for
>the fundamental rep, as explained in:
>
>http://arxiv.org/abs/http://www.arxiv.org/abs/hep-th/0206157
>
>
>
The paper you mention only deals with the case of a torus. In the case of
an arbitrary Riemann surface I still don't see what the Cartan subalgebra
is supposed to be. You always have an adjoint representation, the algebra
acting on itself, but I don't see how you even define a "fundamental"
representation in this case.

Urs Schreiber

Jun29-04, 10:12 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>On Tue, 29 Jun 2004, Peter Woit wrote:\n\n> an arbitrary Riemann surface I still don\'t see what the Cartan subalgebra\n> is supposed to be.\n\nWhat is it that you think is problematic? That there are infinitely\nmany mutually comuting elements?\n\n\nP.S.\nI am currently in Paris at Strings_{04} and have little time (and notebook\nbattery power).\nNext week I\'ll be in Sweden on a NCG conference. But after all that I hope\nto have more time for this discussion. I\'ll contact Yonatan Zunger to see\nif he can help us understanding his proof.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 29 Jun 2004, Peter Woit wrote:

> an arbitrary Riemann surface I still don't see what the Cartan subalgebra
> is supposed to be.

What is it that you think is problematic? That there are infinitely
many mutually comuting elements?

P.S.
I am currently in Paris at Strings_{04} and have little time (and notebook
battery power).
Next week I'll be in Sweden on a NCG conference. But after all that I hope
to have more time for this discussion. I'll contact Yonatan Zunger to see
if he can help us understanding his proof.

Peter Woit

Jun29-04, 02:44 PM

<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>Urs Schreiber wrote:\n\n>> P.Woit wrote:\n>> an arbitrary Riemann surface I still don\'t see what the Cartan subalgebra\n>> is supposed to be.\n>\n> What is it that you think is problematic? That there are infinitely\n> many mutually comuting elements?\n\nIn the case of a compact Lie group it\'s a (not completely trivial) theorem\nthat all choices of a Cartan subalgebra are conjugate, of dimension the\nrank of the group. Here it seems the rank is supposed to be infinite (he\nhas a countably infinite set of simple roots). Is there really an infinite\nset of linearly independent mutually commuting vector fields? Even if\nthere is, how do I know that everything doesn\'t depend on how I choose\nthis set?\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Urs Schreiber wrote:

>> P.Woit wrote:
>> an arbitrary Riemann surface I still don't see what the Cartan subalgebra
>> is supposed to be.
>
> What is it that you think is problematic? That there are infinitely
> many mutually comuting elements?

In the case of a compact Lie group it's a (not completely trivial) theorem
that all choices of a Cartan subalgebra are conjugate, of dimension the
rank of the group. Here it seems the rank is supposed to be infinite (he
has a countably infinite set of simple roots). Is there really an infinite
set of linearly independent mutually commuting vector fields? Even if
there is, how do I know that everything doesn't depend on how I choose
this set?

Lubos Motl

Jun29-04, 03:21 PM

<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>On Tue, 29 Jun 2004, Peter Woit wrote:\n\n> In the case of a compact Lie group it\'s a (not completely trivial) theorem\n> that all choices of a Cartan subalgebra are conjugate, of dimension the\n> rank of the group.\n\nYes.\n\n> Here it seems the rank is supposed to be infinite ...\n\nOf course.\n\n> (he has a countably infinite set of simple roots). Is there really an\n> infinite set of linearly independent mutually commuting vector fields?\n\nYou bet. For example, on a genus g Riemann surface, let\'s locally choose\ncoordinates theta,phi that resemble those on a sphere near (in a vicitity\nof) its Northern Pole, and I reparameterize phi,theta to make the volume\nform constant (it\'s easy to see that it\'s always possible). Any vector\nfield of the form\n\nf(\\theta) \\times \\partial / \\partial \\phi\n\nis obviously a generator of a volume-preserving-diffeomorphism, and all\ndiffeomorphisms of this kind commute with one another because the\n\\phi-derivative of a function of \\tbeta vanishes. If there are any\nproblems with locality, I can require f(\\theta) to vanish outside the disk\ni.e. for \\theta > \\theta_0. Even the diffeomorphisms that only act on a\nsmall disk within the Riemann surface contain an infinite-dimensional\ncommuting subgroup.\n\nThis example of commuting generators of the diffeomorphisms is an\ninfinite-dimensional commuting Cartan subalgebra. Its dimension is the\nsame as the dimension of the space of functions f(\\theta) of an\nappropriate type. One can choose a countable basis, but also a continuous\nbasis of this space - via technology known as the Fourier transform. ;-)\n\nConcerning Peter Woit\'s question whether the groups of volume-preserving\ndiffeomorphisms are isomorphic to SU(\\infty) globally - well, the answer\nis probably NO. If all of them were isomorphic to some group, they would\nalso have to be isomorphic to each other. I think that one can prove that\nthe (area-preserving) diffeomorphisms of a sphere (g=0) are different\nglobally from the diffeomorphisms of a torus (g=1). The former contains an\nSO(3) subgroup (isometries of a sphere), for example. Such an SO(3)\nsubgroup does not exist for g>0 although I don\'t know how to quickly\nprove this statement.\n\nWell, let me prove something else: the area-preserving (continuous)\ndiffeomorphisms of a sphere contain no compact U(1)^2 group - which is a\ngroup easily found for a torus. Every Killing vector on a sphere must\nvanish at least at one point (you can\'t comb a sphere). If the other\nKilling vector - candidate for U(1)^2 - does not vanish at this point,\nthey will fail to commute. If it vanishes at the same point, you will be\nable to see that they\'re either not independent, or they don\'t have the\nrequired periodicity.\n\nBack to the non-existing SO(3) subgroup for the torus.\n\nIf you tried to apply the dictionary resulting from the SU(\\infty) Lie\nalgebra, you would probably obtain a version of an infinite cover of\nSO(3): no rotations by "2.pi" (not even "4.pi") would be mapped to the\nidentity. Such a version of an SO(3) group (well, a group with the same\nalgebra) can have non-integer and non-half-integer spins (such irreps are\ninevitably infinite-dimensional).\n\n> Even if there is, how do I know that everything doesn\'t depend on how\n> I choose this set?\n\nIt depends what you mean by "everything". The finiteness and continuity\nconditions on the elements of SU(\\infty), necessary to reproduce the\narea-preserving diffeomorphism Lie algebras, depend on the genus. But if\nwe say "a group of area-preserving diffeomorphisms", we know very well\nwhat it means.\n\nIn Matrix theory logic, it is sort of straightforward to see that\nSU(\\infty), the gauge group of Matrix theory in its glory, must reduce to\nthe area-preserving diffeomorphisms in various "corners". One can\nexplicitly construct a fuzzy sphere (g=0) and a fuzzy torus (g=1). The\ntopology of a membrane can change through the "fuzzy" configuration space,\nand because it can happen between g=0 and g=1 and because this "throat\nnucleation" can be seen to be local on the membrane (and in the\n\\infty-dimensional representation of the unitary group), one can generate\nthe higher genus membranes, too, whose local physics must be identical\nlike for the g=0, g=1 membranes.\n\nThe paper you discussed probably offers a very explicit construction of\nthe map for g>1, which may be fun (something related to higher genus phase\nspaces and geometric quantization?), but it is probably very\nnon-canonical, unlike the sphere and the torus.\n\nBest wishes\nLubos\n___________________________________ ___________________________________________\nE-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/\neFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 29 Jun 2004, Peter Woit wrote:

> In the case of a compact Lie group it's a (not completely trivial) theorem
> that all choices of a Cartan subalgebra are conjugate, of dimension the
> rank of the group.

Yes.

> Here it seems the rank is supposed to be infinite ...

Of course.

> (he has a countably infinite set of simple roots). Is there really an
> infinite set of linearly independent mutually commuting vector fields?

You bet. For example, on a genus g Riemann surface, let's locally choose
coordinates \theta,\phi that resemble those on a sphere near (in a vicitity
of) its Northern Pole, and I reparameterize \phi,\theta to make the volume
form constant (it's easy to see that it's always possible). Any vector
field of the form

f(\theta) \times \partial / \partial \phi

is obviously a generator of a volume-preserving-diffeomorphism, and all
diffeomorphisms of this kind commute with one another because the
\phi-derivative of a function of \tbeta vanishes. If there are any
problems with locality, I can require f(\theta) to vanish outside the disk
i.e. for \theta > \theta_0. Even the diffeomorphisms that only act on a
small disk within the Riemann surface contain an infinite-dimensional
commuting subgroup.

This example of commuting generators of the diffeomorphisms is an
infinite-dimensional commuting Cartan subalgebra. Its dimension is the
same as the dimension of the space of functions f(\theta) of an
appropriate type. One can choose a countable basis, but also a continuous
basis of this space - via technology known as the Fourier transform. ;-)

Concerning Peter Woit's question whether the groups of volume-preserving
diffeomorphisms are isomorphic to SU(\infty) globally - well, the answer
is probably NO. If all of them were isomorphic to some group, they would
also have to be isomorphic to each other. I think that one can prove that
the (area-preserving) diffeomorphisms of a sphere (g=0) are different
globally from the diffeomorphisms of a torus (g=1). The former contains an
SO(3) subgroup (isometries of a sphere), for example. Such an SO(3)
subgroup does not exist for g>0 although I don't know how to quickly
prove this statement.

Well, let me prove something else: the area-preserving (continuous)
diffeomorphisms of a sphere contain no compact U(1)^2 group - which is a
group easily found for a torus. Every Killing vector on a sphere must
vanish at least at one point (you can't comb a sphere). If the other
Killing vector - candidate for U(1)^2 - does not vanish at this point,
they will fail to commute. If it vanishes at the same point, you will be
able to see that they're either not independent, or they don't have the
required periodicity.

Back to the non-existing SO(3) subgroup for the torus.

If you tried to apply the dictionary resulting from the SU(\infty) Lie
algebra, you would probably obtain a version of an infinite cover of
SO(3): no rotations by "2.\pi" (not even "4.\pi") would be mapped to the
identity. Such a version of an SO(3) group (well, a group with the same
algebra) can have non-integer and non-half-integer spins (such irreps are
inevitably infinite-dimensional).

> Even if there is, how do I know that everything doesn't depend on how
> I choose this set?

It depends what you mean by "everything". The finiteness and continuity
conditions on the elements of SU(\infty), necessary to reproduce the
area-preserving diffeomorphism Lie algebras, depend on the genus. But if
we say "a group of area-preserving diffeomorphisms", we know very well
what it means.

In Matrix theory logic, it is sort of straightforward to see that
SU(\infty), the gauge group of Matrix theory in its glory, must reduce to
the area-preserving diffeomorphisms in various "corners". One can
explicitly construct a fuzzy sphere (g=0) and a fuzzy torus (g=1). The
topology of a membrane can change through the "fuzzy" configuration space,
and because it can happen between g=0 and g=1 and because this "throat
nucleation" can be seen to be local on the membrane (and in the
\infty-dimensional representation of the unitary group), one can generate
the higher genus membranes, too, whose local physics must be identical
like for the g=0, g=1 membranes.

The paper you discussed probably offers a very explicit construction of
the map for g>1, which may be fun (something related to higher genus phase
spaces and geometric quantization?), but it is probably very
non-canonical, unlike the sphere and the torus.

Best wishes
Lubos
__{_______________________________________________ _____________________________}
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^

zunger

Jun29-04, 04:18 PM

<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>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:<cber50\\$rvu\\$1-100000@newsmaster.cc.columbia.edu>...\n\n> ...\n> One reason it\'s hard to figure out exactly what he is saying in that\n> paper is that he doesn\'t bother to distinguish between a Lie group and its\n> Lie algebra, saying things like "the Poisson algebra is U(1) times a\n> simple pseudo-compact algebra" where I guess you have to interpret "U(1)"\n> as its Lie algebra, i.e. the real numbers.\n\nI\'ve been wishing I had been more clear about that for a while, and I\nhaven\'t had time to write up the detailed explanation of this point.\nYou\'re completely right about the topology, of course; the proof in\nthe paper is only about the Lie algebras, not about the groups.\n\nAs far as the topologies, I\'m not 100% certain, but Jeff Harvey and I\ntalked this over a while ago and came up with the following notion.\nThere\'s an infinite series of "SU(infty)" groups, all of which have\nthe same Lie algebra (or at least, commutation relations) but which\ndiffer in being restricted to things with finite Tr A^n for all A in\nthat group, for each natural n. The case n=0 imposes no restriction,\nand that particular group is the directed limit of SU(N) at large N;\nit has messy topology. The limit n->infty of this sequence is simply\nSU(K), the group of compact special unitary operators on a countable\nHilbert space. This group has very simple topology. The proof in\n0106030 doesn\'t distinguish between them, calling them all\n"SU(infty)."\n\nTo resolve this, one just has to decide which operators one is\ninterested in for the Poisson algebra. Finiteness of the Yang-Mills\naction clearly would require at least n=2, but the full DBI action\nalmost certainly would require n=infty, i.e. requiring compact\noperators. In essence, the requirement of a finite action should\neliminate precisely those modes which lead to the messy topology of\nthe n=0 SU(infty), which is the topology that\'s been worrying a lot of\npeople in recent papers.\n\nYonatan Zunger\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:<cber50$rvu$1-100000@newsmaster.cc.columbia.edu>...

> ...
> One reason it's hard to figure out exactly what he is saying in that
> paper is that he doesn't bother to distinguish between a Lie group and its
> Lie algebra, saying things like "the Poisson algebra is U(1) times a
> simple pseudo-compact algebra" where I guess you have to interpret "U(1)"
> as its Lie algebra, i.e. the real numbers.

I've been wishing I had been more clear about that for a while, and I
haven't had time to write up the detailed explanation of this point.
You're completely right about the topology, of course; the proof in
the paper is only about the Lie algebras, not about the groups.

As far as the topologies, I'm not 100% certain, but Jeff Harvey and I
talked this over a while ago and came up with the following notion.
There's an infinite series of "SU(\infty)" groups, all of which have
the same Lie algebra (or at least, commutation relations) but which
differ in being restricted to things with finite Tr A^n for all A in
that group, for each natural n. The case n=0 imposes no restriction,
and that particular group is the directed limit of SU(N) at large N;
it has messy topology. The limit n->\infty of this sequence is simply
SU(K), the group of compact special unitary operators on a countable
Hilbert space. This group has very simple topology. The proof in
0106030 doesn't distinguish between them, calling them all
"SU(\infty)."

To resolve this, one just has to decide which operators one is
interested in for the Poisson algebra. Finiteness of the Yang-Mills
action clearly would require at least n=2, but the full DBI action
almost certainly would require n=\infty, i.e. requiring compact
operators. In essence, the requirement of a finite action should
eliminate precisely those modes which lead to the messy topology of
the n=0 SU(\infty), which is the topology that's been worrying a lot of
people in recent papers.

Yonatan Zunger

zunger

Jun29-04, 04:19 PM

<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>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:<cbf7im\\$89d\\$1-100000@newsmaster.cc.columbia.edu>...\n\n> Well, he\'s explicitly claiming a result about the group: "we will show\n> that the group of area-preserving diffeomorphisms of any Riemann surface\n> is SU(infinity)". If his results are only about the Lie algebra, he should\n> say so. In any case I\'m still having trouble believing that all such Lie\n> algebras are isomorphic.\n\nGah. If the text said "group," that\'s a typo that got through editing.\nIt should have said "algebra."\n\n> I tried to follow his proof, but have trouble from the very beginning. He\n> wants to use standard techniques about roots and weights, which rely on\n> studying how a maximal abelian subgroup acts on the adjoint\n> representation. He seems to be claiming that in his case such a group\n> exists and is compact. I don\'t see why that should be true.\n\nNot compact, only "pseudocompact." (Basically that the root lattice is\nbounded, although infinite-dimensional) The idea is that showing this\nweaker condition is enough to prove a cousin of the classification\ntheorem; the proof mostly mimics the usual derivation of Dynkin\ndiagrams.\n\nYonatan\n\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:<cbf7im$89d$1-100000@newsmaster.cc.columbia.edu>...

> Well, he's explicitly claiming a result about the group: "we will show
> that the group of area-preserving diffeomorphisms of any Riemann surface
> is SU(infinity)". If his results are only about the Lie algebra, he should
> say so. In any case I'm still having trouble believing that all such Lie
> algebras are isomorphic.

Gah. If the text said "group," that's a typo that got through editing.
It should have said "algebra."

> I tried to follow his proof, but have trouble from the very beginning. He
> wants to use standard techniques about roots and weights, which rely on
> studying how a maximal abelian subgroup acts on the adjoint
> representation. He seems to be claiming that in his case such a group
> exists and is compact. I don't see why that should be true.

Not compact, only "pseudocompact." (Basically that the root lattice is
bounded, although infinite-dimensional) The idea is that showing this
weaker condition is enough to prove a cousin of the classification
theorem; the proof mostly mimics the usual derivation of Dynkin
diagrams.

Yonatan

Peter Woit

Jun29-04, 06:48 PM

<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>Lubos Motl wrote:\n\n>On Tue, 29 Jun 2004, Peter Woit wrote:\n>\n>\n>\n>>Is there really an\n>>infinite set of linearly independent mutually commuting vector fields?\n>>\n>>\n>\n>You bet. For example, on a genus g Riemann surface, let\'s locally choose\n>coordinates theta,phi that resemble those on a sphere near (in a vicitity\n>of) its Northern Pole, and I reparameterize phi,theta to make the volume\n>form constant (it\'s easy to see that it\'s always possible). Any vector\n>field of the form\n>\n> f(\\theta) \\times \\partial / \\partial \\phi\n>\n>is obviously a generator of a volume-preserving-diffeomorphism, and all\n>diffeomorphisms of this kind commute with one another because the\n>\\phi-derivative of a function of \\tbeta vanishes. If there are any\n>problems with locality, I can require f(\\theta) to vanish outside the disk\n>i.e. for \\theta > \\theta_0. Even the diffeomorphisms that only act on a\n>small disk within the Riemann surface contain an infinite-dimensional\n>commuting subgroup.\n>\n>\n>\nHi Lubos,\n\nThanks for the explanation. Let me see if I understand this. These\nvector fields\ngenerate an infinite number of commuting R actions, but in general these\naren\'t periodic, they don\'t exponentiate to U(1) actions. One problem I\nsee with\nZunger\'s argument that the Lie algebras for all g are isomorphic is that\nit relies on constructing SU(2) subgroups of\nthe group of diffeomorphisms, and studying how the Lie algebra decomposes\nunder the actions of these SU(2) subgroups. But (for g>0), as you point out,\nthere are no SU(2) subgroups, no?\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Lubos Motl wrote:

>On Tue, 29 Jun 2004, Peter Woit wrote:
>
>
>
>>Is there really an
>>infinite set of linearly independent mutually commuting vector fields?
>>
>>
>
>You bet. For example, on a genus g Riemann surface, let's locally choose
>coordinates \theta,\phi that resemble those on a sphere near (in a vicitity
>of) its Northern Pole, and I reparameterize \phi,\theta to make the volume
>form constant (it's easy to see that it's always possible). Any vector
>field of the form
>
> f(\theta) \times \partial / \partial \phi
>
>is obviously a generator of a volume-preserving-diffeomorphism, and all
>diffeomorphisms of this kind commute with one another because the
>\phi-derivative of a function of \tbeta vanishes. If there are any
>problems with locality, I can require f(\theta) to vanish outside the disk
>i.e. for \theta > \theta_0. Even the diffeomorphisms that only act on a
>small disk within the Riemann surface contain an infinite-dimensional
>commuting subgroup.
>
>
>
Hi Lubos,

Thanks for the explanation. Let me see if I understand this. These
vector fields
generate an infinite number of commuting R actions, but in general these
aren't periodic, they don't exponentiate to U(1) actions. One problem I
see with
Zunger's argument that the Lie algebras for all g are isomorphic is that
it relies on constructing SU(2) subgroups of
the group of diffeomorphisms, and studying how the Lie algebra decomposes
under the actions of these SU(2) subgroups. But (for g>0), as you point out,
there are no SU(2) subgroups, no?

Thomas Larsson

Jul4-04, 12:18 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>Lubos Motl <motl@feynman.harvard.edu> wrote in message news:<Pine.LNX.4.31.0406291553001.30237-100000@feynman.harvard.edu>...\n> On Tue, 29 Jun 2004, Peter Woit wrote:\n>\n> > In the case of a compact Lie group it\'s a (not completely trivial) theorem\n> > that all choices of a Cartan subalgebra are conjugate, of dimension the\n> > rank of the group.\n>\n> Yes.\n>\n> > Here it seems the rank is supposed to be infinite ...\n>\n> Of course.\n>\n> > (he has a countably infinite set of simple roots). Is there really an\n> > infinite set of linearly independent mutually commuting vector fields?\n>\n> You bet. For example, on a genus g Riemann surface, let\'s locally choose\n> coordinates theta,phi that resemble those on a sphere near (in a vicitity\n> of) its Northern Pole, and I reparameterize phi,theta to make the volume\n> form constant (it\'s easy to see that it\'s always possible). Any vector\n> field of the form\n>\n> f(\\theta) \\times \\partial / \\partial \\phi\n>\n> is obviously a generator of a volume-preserving-diffeomorphism, and all\n> diffeomorphisms of this kind commute with one another because the\n> \\phi-derivative of a function of \\tbeta vanishes.\n\nActually, one needs more than a Cartan subalgebra to construct\nirreps - one needs a triangular decomposition,\n\ng = b_- + h + b_+ ,\n\nwhere h is the CSA and b_- and b_+ are Borel subalgebras. The\ng irreps are constructed by induction from a one-dimensional\nrepresentation of h + b_+. IOW, one introduces a highest-\nweight state with well-defined h eigenvalues and is\nannihilated by all of b_+, and then let b_- act on it.\n\nWhat is important here is that h, b_-, b_+, and h + b_+ are\nall subalgebras, i.e. closed under commutation. This is what\nallows you to induce from h + b_+ to g.\n\nIt is true that vector fields of the form f(x^1) d/dx^2\ncommute with each other and therefore can be considered as a\nCSA, but this is pretty useless since there are no\ninteresting Borel subalgebras. If one induces from the CSA h\nalone, one gets a huge and highly reducible beast; to ensure\nirreducibility, a Borel subalgebra of annihilation operators\nis needed.\n\nInstead, the right way to construct irreps of infinite-\ndimensional algebras is as follows. In the spirit of Lie-\nCartan-Sternberg-Kac and others, we may restrict attention to\nthe Lie algebras of polynomial vector fields in a single\nchart. The algebras which are relevant here are graded by the\ndilatation eigenvalue:\n\ng = g_-1 + g_0 + g_1 + g_2 + g_3 + ...\n\nThis includes:\n\ng = vect(n), g_0 = gl(n). Arbitrary diffeomorphisms.\n\ng = svect(n), g_0 = sl(n). Volume-preserving diffs.\n\ng = H_n, g_0 = sp(n). Symplectomorphisms.\n\nWe thus have a decomposition of the form\n\ng = g_- + g_0 + g_+,\n\nwhere g_- = g_-1 and g_+ = g_1 + g_2 + g_3 + .... We can\nextend each g_0 irrep R to g_0 + g_+ by requiring that g_+\nacts trivially, and we then get a g representation by\ninduction from this g_0 + g_+ rep. It turns out that this\nmodule is equivalent to (more precisely, contragredient to)\nthe module of tensor fields of type R, although why this is\ntrue is not completely obvious to me.\n\nTo make the analogy with the finite-dimensional case\ncomplete, we need to add the Borel subalgebras of g_0 to g_-\nand g_+. We then get a triangular decomposition\ng = b_- + h + b_+, where h is the CSA of g_0. Thus the\nirreps of g are labelled by the weights of g_0.\n\nOur current interest is g = svect(2) = H_2, with g_0 = sl(2)\n= sp(2). In what sense do we have a correspondence between\nsl(2) and sl(infinity) representations?\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Lubos Motl <motl@feynman.harvard.edu> wrote in message news:<Pine.LNX.4.31.0406291553001.30237-100000@feynman.harvard.edu>...
> On Tue, 29 Jun 2004, Peter Woit wrote:
>
> > In the case of a compact Lie group it's a (not completely trivial) theorem
> > that all choices of a Cartan subalgebra are conjugate, of dimension the
> > rank of the group.
>
> Yes.
>
> > Here it seems the rank is supposed to be infinite ...
>
> Of course.
>
> > (he has a countably infinite set of simple roots). Is there really an
> > infinite set of linearly independent mutually commuting vector fields?
>
> You bet. For example, on a genus g Riemann surface, let's locally choose
> coordinates \theta,\phi that resemble those on a sphere near (in a vicitity
> of) its Northern Pole, and I reparameterize \phi,\theta to make the volume
> form constant (it's easy to see that it's always possible). Any vector
> field of the form
>
> f(\theta) \times \partial / \partial \phi
>
> is obviously a generator of a volume-preserving-diffeomorphism, and all
> diffeomorphisms of this kind commute with one another because the
> \phi-derivative of a function of \tbeta vanishes.

Actually, one needs more than a Cartan subalgebra to construct
irreps - one needs a triangular decomposition,

g = b_- + h + b_+ ,

where h is the CSA and b_- and b_+ are Borel subalgebras. The
g irreps are constructed by induction from a one-dimensional
representation of h + b_+. IOW, one introduces a highest-
weight state with well-defined h eigenvalues and is
annihilated by all of b_+, and then let b_- act on it.

What is important here is that h, b_-, b_+, and h + b_+ are
all subalgebras, i.e. closed under commutation. This is what
allows you to induce from h + b_+ to g.

It is true that vector fields of the form f(x^1) d/dx^2
commute with each other and therefore can be considered as a
CSA, but this is pretty useless since there are no
interesting Borel subalgebras. If one induces from the CSA h
alone, one gets a huge and highly reducible beast; to ensure
irreducibility, a Borel subalgebra of annihilation operators
is needed.

Instead, the right way to construct irreps of infinite-
dimensional algebras is as follows. In the spirit of Lie-
Cartan-Sternberg-Kac and others, we may restrict attention to
the Lie algebras of polynomial vector fields in a single
chart. The algebras which are relevant here are graded by the
dilatation eigenvalue:

g = g_-1 + g_0 + g_1 + g_2 + g_3 + ...

This includes:

g = vect(n), g_0 = gl(n). Arbitrary diffeomorphisms.

g = svect(n), g_0 = sl(n). Volume-preserving diffs.

g = H_n, g_0 = sp(n)[/itex]. Symplectomorphisms.

We thus have a decomposition of the form

[itex]g = g_- + g_0 + g_+,

where g_- = g_-1 and g_+ = g_1 + g_2 + g_3 + .... We can
extend each g_0 irrep R to g_0 + g_+ by requiring that g_+
acts trivially, and we then get a g representation by
induction from this g_0 + g_+ rep. It turns out that this
module is equivalent to (more precisely, contragredient to)
the module of tensor fields of type R, although why this is
true is not completely obvious to me.

To make the analogy with the finite-dimensional case
complete, we need to add the Borel subalgebras of g_0 to g_-
and g_+. We then get a triangular decomposition
g = b_- + h + b_+, where h is the CSA of g_0. Thus the
irreps of g are labelled by the weights of g_0.

Our current interest is g = svect(2) = H_2, with g_0 = sl(2)= sp(2). In what sense do we have a correspondence between
sl(2) and sl(infinity) representations?

Maarten Bergvelt

Jul4-04, 12:12 PM

<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>In article <24a23f36.0407022205.5306a283-100000@posting.google.com>,\nThomas Larsson wrote:\n> Actually, one needs more than a Cartan subalgebra to construct\n> irreps - one needs a triangular decomposition,\n>\n> g = b_- + h + b_+ ,\n>\n> where h is the CSA and b_- and b_+ are Borel subalgebras. The\n> g irreps are constructed by induction from a one-dimensional\n> representation of h + b_+. IOW, one introduces a highest-\n> weight state with well-defined h eigenvalues and is\n> annihilated by all of b_+, and then let b_- act on it.\n\nWell, b_- need not be a Borel subalgebra, for instance for semi-simple Lie\nalgebras the complement of h+b_+ is n_-, the lower nilpotent\nsubalgebra (the commutator of b_-).\n\n> What is important here is that h, b_-, b_+, and h + b_+ are\n> all subalgebras, i.e. closed under commutation. This is what\n> allows you to induce from h + b_+ to g.\n\nTo do induction you just need a subalgebra, which does not need to\nhave a complementary subalgebra. In examples it does happen that the\ncomplement is a subalgebra, which has the nice effect that the induced\nmodule is the universal envelopping algebra of the complement.\n\n--\nMaarten Bergvelt\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <24a23f36.0407022205.5306a283-100000@posting.google.com>,
Thomas Larsson wrote:
> Actually, one needs more than a Cartan subalgebra to construct
> irreps - one needs a triangular decomposition,
>
> g = b_- + h + b_+ ,
>
> where h is the CSA and b_- and b_+ are Borel subalgebras. The
> g irreps are constructed by induction from a one-dimensional
> representation of h + b_+. IOW, one introduces a highest-
> weight state with well-defined h eigenvalues and is
> annihilated by all of b_+, and then let b_- act on it.

Well, b_- need not be a Borel subalgebra, for instance for semi-simple Lie
algebras the complement of h+b_+ is n_-, the lower nilpotent
subalgebra (the commutator of b_-).

> What is important here is that h, b_-, b_+, and h + b_+ are
> all subalgebras, i.e. closed under commutation. This is what
> allows you to induce from h + b_+ to g.

To do induction you just need a subalgebra, which does not need to
have a complementary subalgebra. In examples it does happen that the
complement is a subalgebra, which has the nice effect that the induced
module is the universal envelopping algebra of the complement.

--
Maarten Bergvelt

Thomas Larsson

Jul5-04, 05:46 PM

<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>Maarten Bergvelt <bergv@math.uiuc.edu> wrote in message news:<slrncegdtr.dij.bergv-100000@u00.math.uiuc.edu>...\n> In article <24a23f36.0407022205.5306a283-100000@posting.google.com>,\n> Thomas Larsson wrote:\n>\n> > What is important here is that h, b_-, b_+, and h + b_+ are\n> > all subalgebras, i.e. closed under commutation. This is what\n> > allows you to induce from h + b_+ to g.\n>\n> To do induction you just need a subalgebra, which does not need to\n> have a complementary subalgebra. In examples it does happen that the\n> complement is a subalgebra, which has the nice effect that the induced\n> module is the universal envelopping algebra of the complement.\n\nYes, sorry. What I meant is that for induction to produce something\nuseful, i.e. something which is irreducible or almost so, you need\nto induce from a large subalgebra. For simple Lie algebras, both\nfinite-dimensional and infinite-dimensional ones of polynomial vector\nfields, it is possible to find a triangular decomposition\ng = n_- + h + n_+, such that each irrep sits inside the module\ninduced from a one-dimensional rep of h + n_+, where h is finite-\ndimensional. In particular, a single number is needed to label the\nirreps of the algebra of area-preserving diffs in 2D.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Maarten Bergvelt <bergv@math.uiuc.edu> wrote in message news:<slrncegdtr.dij.bergv-100000@u00.math.uiuc.edu>...
> In article <24a23f36.0407022205.5306a283-100000@posting.google.com>,
> Thomas Larsson wrote:
>
> > What is important here is that h, b_-, b_+, and h + b_+ are
> > all subalgebras, i.e. closed under commutation. This is what
> > allows you to induce from h + b_+ to g.
>
> To do induction you just need a subalgebra, which does not need to
> have a complementary subalgebra. In examples it does happen that the
> complement is a subalgebra, which has the nice effect that the induced
> module is the universal envelopping algebra of the complement.

Yes, sorry. What I meant is that for induction to produce something
useful, i.e. something which is irreducible or almost so, you need
to induce from a large subalgebra. For simple Lie algebras, both
finite-dimensional and infinite-dimensional ones of polynomial vector
fields, it is possible to find a triangular decomposition
g = n_- + h + n_+, such that each irrep sits inside the module
induced from a one-dimensional rep of h + n_+, where h is finite-
dimensional. In particular, a single number is needed to label the
irreps of the algebra of area-preserving diffs in 2D.

Robert C. Helling

Jul9-04, 05:46 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>On Tue, 29 Jun 2004 11:05:17 -0400, Peter Woit <woit@cpw.math.columbia.edu> wrote:\n> Charlie Stromeyer Jr. wrote:\n>\n>>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:\n>>\n>>>Well, he\'s explicitly claiming a result about the group: "we will show\n>>>that the group of area-preserving diffeomorphisms of any Riemann surface\n>>>is SU(infinity)". If his results are only about the Lie algebra, he should\n>>>say so. In any case I\'m still having trouble believing that all such Lie\n>>>algebras are isomorphic.\n>>>\n\nI didn\'t follow this thread but coming back from Paris and CERN, let\nme just throw in some mathematical facts: There are several things\nthat can be called SU(oo) but that are not the same. IIRC in his book\non Kac-Moody-Algebras, Kac mentions that there are two obvious ways to\nextend the Dynkin diagram of SU(N) (aka A_{N-1)) to infintite N:\n\nO-O-O-O-O-...\n\nand\n\n...-O-O-O-O-O-...\n\nand that those algebras are not isomorphic. More genernal, there is a\nmathematical way of taking (direct) limits of groups and that amounts\nto specify a groups G_N for each N and a nice embedding\n\nf_N : G_N -> G_{N+1}\n\nand what you get can also depend on your choice of f\'s. This is the\nmaths story. Then there is Jens Hoppe\'s membrane story which takes the\nalgebras g_N (which are finite algebras of divergence free vector\nfields that generate APDs) and picks some basis for it in which the structure\nconstants are\n\n[a_i,a_j] = c(N)_ijk a_k\n\nand then he shows that for fixed i,j and k, these converge to some\nstructure constants for some version of SU(oo). This obviously depends\non the choice of basis and at least to me it is not clear that this\nhas anything to do with the math procedure described above.\n\nI do not doubt that for reasonable 3 manifolds there are the groups of\nAPDs but I don\'t see how this limiting procedure shows that they are\nall isomorphic.\n\nRobert\n\n\n\n\n--\n..oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO o.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO\nRobert C. Helling Department of Applied Mathematics and Theoretical Physics\nUniversity of Cambridge\nprint "Just another Phone: +44/1223/766870\nstupid .sig\\n"; http://www.aei-potsdam.mpg.de/~helling\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 29 Jun 2004 11:05:17 -0400, Peter Woit <woit@cpw.math.columbia.edu> wrote:
> Charlie Stromeyer Jr. wrote:
>
>>Peter Woit <woit@cpw.math.columbia.edu> wrote in message news:
>>
>>>Well, he's explicitly claiming a result about the group: "we will show
>>>that the group of area-preserving diffeomorphisms of any Riemann surface
>>>is SU(infinity)". If his results are only about the Lie algebra, he should
>>>say so. In any case I'm still having trouble believing that all such Lie
>>>algebras are isomorphic.
>>>

I didn't follow this thread but coming back from Paris and CERN, let
me just throw in some mathematical facts: There are several things
that can be called SU(oo) but that are not the same. IIRC in his book
on Kac-Moody-Algebras, Kac mentions that there are two obvious ways to
extend the Dynkin diagram of SU(N) (aka A_{N-1)) to infintite N:

O-O-O-O-O-...

and

...-O-O-O-O-O-...

and that those algebras are not isomorphic. More genernal, there is a
mathematical way of taking (direct) limits of groups and that amounts
to specify a groups G_N for each N and a nice embedding

f_N : G_N -> G_{N+1}

and what you get can also depend on your choice of f's. This is the
maths story. Then there is Jens Hoppe's membrane story which takes the
algebras g_N (which are finite algebras of divergence free vector
fields that generate APDs) and picks some basis for it in which the structure
constants are

[a_i,a_j] = c(N)_ijk a_k

and then he shows that for fixed i,j and k, these converge to some
structure constants for some version of SU(oo). This obviously depends
on the choice of basis and at least to me it is not clear that this
has anything to do with the math procedure described above.

I do not doubt that for reasonable 3 manifolds there are the groups of
APDs but I don't see how this limiting procedure shows that they are
all isomorphic.

Robert

--
..oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo. oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO
Robert C. Helling Department of Applied Mathematics and Theoretical Physics
University of Cambridge
print "Just another Phone: +44/1223/766870
stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling

Urs Schreiber

Jul9-04, 09:50 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>On Fri, 9 Jul 2004, Robert C. Helling wrote:\n\n> There are several things that can be called SU(oo) but that are not the\n> same.\n\nNote that Zunger does address this issue in his paper.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Fri, 9 Jul 2004, Robert C. Helling wrote:

> There are several things that can be called SU(oo) but that are not the
> same.

Note that Zunger does address this issue in his paper.

Thomas Larsson

Jul11-04, 12:45 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>"Robert C. Helling" <helling@ariel.physik.hu-berlin.de> wrote in message news:<2l7b7hF9rng5U1-100000@uni-berlin.de>...\n\n> I do not doubt that for reasonable 3 manifolds there are the groups of\n> APDs but I don\'t see how this limiting procedure shows that they are\n> all isomorphic.\n\nActually, groups of APDs are interesting for 2-manifolds - in\n3D, there are groups of volume-preserving diffs instead.\nReasonable here means that a volume form exists, which I think\nmeans that the manifold is orientable.\n\nMy point was that a good limiting procedure must produce the\nright irreps. The group of APDs in 2D has two generalizations\nto higher dimensions: groups of volume-preserving diffs\n(algebras of divergence-free vector fields), and\nsymplectomorphism groups (algebras of Hamiltonian vector\nfields). The algebra irreps act on spaces of tensor fields.\nThe corresponding groups act on spaces of sections of the\nappropriate tensor bundles over the manifold in question.\n\nOn the torus we don\'t need this fancy language, because we can\ndo everything very explicitly in a Fourier basis. In 2D, the\ngenerators are labelled by momenta m = (m_1, m_2) \\in Z^2. Let\nw^ij be the symplectic metric (w^ij = epsilon^ij in 2D), and\ndenote\n\nm^i = w^ij m_j, mxn = w^ij m_i n_j = m_i n^i.\n\nThe symplectomorphism algebra takes the form\n\n[T_m, T_n] = mxn T_m+n.\n\nFor simplicity, let\'s write down the module action on a 2-tensor\nfield, with Fourier components u^ij_m:\n\nT_m u^ij_m = mxn u^ij_m+n - m^i m_k u^kj_m+n - m^j m_k u^ik_m+n.\n\nIt is straightforward to verify that this formula defines a\nrepresentation (to ensure irreducibility, we also need the\ncondition w_ij u^ij_m = 0). Analogously we have a represenation\non totally symmetric p-tensor fields u^ijk..p_m. Such a field\nhas p+1 independent components (in 2D), and corresponds to the\nsu(2) irrep with spin j = p/2.\n\nNow, I can follow the construction of the Moyal algebra from\nsu(N), with shift and clock matrices, and I agree that the limit\nN -> infinity formally gives the symplectomorphism algebra above.\nBut I cannot see which su(infinity) irreps correspond to the\np-tensor fields above, and I don\'t understand what happens to\nthe other su(N) irreps in the limit.\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Robert C. Helling" <helling@ariel.physik.hu-berlin.de> wrote in message news:<2l7b7hF9rng5U1-100000@uni-berlin.de>...

> I do not doubt that for reasonable 3 manifolds there are the groups of
> APDs but I don't see how this limiting procedure shows that they are
> all isomorphic.

Actually, groups of APDs are interesting for 2-manifolds - in
3D, there are groups of volume-preserving diffs instead.
Reasonable here means that a volume form exists, which I think
means that the manifold is orientable.

My point was that a good limiting procedure must produce the
right irreps. The group of APDs in 2D has two generalizations
to higher dimensions: groups of volume-preserving diffs
(algebras of divergence-free vector fields), and
symplectomorphism groups (algebras of Hamiltonian vector
fields). The algebra irreps act on spaces of tensor fields.
The corresponding groups act on spaces of sections of the
appropriate tensor bundles over the manifold in question.

On the torus we don't need this fancy language, because we can
do everything very explicitly in a Fourier basis. In 2D, the
generators are labelled by momenta m = (m_1, m_2) \in Z^2. Let
w^{ij} be the symplectic metric (w^{ij} = \epsilon^ij in 2D), and
denote

m^i = w^{ij} m_j,[/itex] mxn = w^{ij} m_i n_j = m_i n^i.

The symplectomorphism algebra takes the form

[T_m, T_n] = mxn T_m+n.

For simplicity, let's write down the module action on a 2-tensor
field, with Fourier components u^{ij_m}:T_m u^{ij_m} = mxn u^{ij_m}+n - m^i m_k u^{kj_m}+n - m^j m_k u^{ik_m}+n.

It is straightforward to verify that this formula defines a
representation (to ensure irreducibility, we also need the
condition w_{ij} u^{ij_m} = 0). Analogously we have a represenation
on totally symmetric p-tensor fields u^{ijk}..p_m. Such a field
has p+1 independent components (in 2D), and corresponds to the
su(2) irrep with spin [itex]j = p/2.

Now, I can follow the construction of the Moyal algebra from
su(N), with shift and clock matrices, and I agree that the limit
N -> infinity formally gives the symplectomorphism algebra above.
But I cannot see which su(infinity) irreps correspond to the
p-tensor fields above, and I don't understand what happens to
the other su(N) irreps in the limit.

Robert C. Helling

Jul13-04, 02:18 PM

<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>On Sun, 11 Jul 2004 01:45:42 -0400, Thomas Larsson <thomas_larsson_01@hotmail.com> wrote:\n\n> The symplectomorphism algebra takes the form\n>\n> [T_m, T_n] = mxn T_m+n.\n>\n> For simplicity, let\'s write down the module action on a 2-tensor\n> field, with Fourier components u^ij_m:\n>\n> T_m u^ij_m = mxn u^ij_m+n - m^i m_k u^kj_m+n - m^j m_k u^ik_m+n.\n>\n> It is straightforward to verify that this formula defines a\n> representation (to ensure irreducibility, we also need the\n> condition w_ij u^ij_m = 0). Analogously we have a represenation\n> on totally symmetric p-tensor fields u^ijk..p_m. Such a field\n> has p+1 independent components (in 2D), and corresponds to the\n> su(2) irrep with spin j = p/2.\n>\n> Now, I can follow the construction of the Moyal algebra from\n> su(N), with shift and clock matrices, and I agree that the limit\n> N -> infinity formally gives the symplectomorphism algebra above.\n> But I cannot see which su(infinity) irreps correspond to the\n> p-tensor fields above, and I don\'t understand what happens to\n> the other su(N) irreps in the limit.\n\nHmm, isn\'t the point of a recent paper (the one we are discussing?)\nthat for SU(N) the structure constants are something along the lines\nof\n\nN sin(nxm/N)\n\nwhich converges to nxm for n and m fixed and N->oo but if n and m grow\nas fast as N the "sub-leading" terms do not go 0? So, why do we keep n\nand m fixed and what happens to the elements with large n and m that\ndo not converge to the torus algebra?\n\nRobert\n\n\n--\n..oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO o.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO\nRobert C. Helling Department of Applied Mathematics and Theoretical Physics\nUniversity of Cambridge\nprint "Just another Phone: +44/1223/766870\nstupid .sig\\n"; http://www.aei-potsdam.mpg.de/~helling\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Sun, 11 Jul 2004 01:45:42 -0400, Thomas Larsson <thomas_larsson_01@hotmail.com> wrote:

> The symplectomorphism algebra takes the form
>
> [T_m, T_n] = mxn T_m+n.
>
> For simplicity, let's write down the module action on a 2-tensor
> field, with Fourier components u^{ij_m}:
>
> T_m u^{ij_m} = mxn u^{ij_m}+n - m^i m_k u^{kj_m}+n - m^j m_k u^{ik_m}+n.
>
> It is straightforward to verify that this formula defines a
> representation (to ensure irreducibility, we also need the
> condition w_{ij} u^{ij_m} = 0). Analogously we have a represenation
> on totally symmetric p-tensor fields u^{ijk}..p_m. Such a field
> has p+1 independent components (in 2D), and corresponds to the
> su(2) irrep with spin j = p/2.
>
> Now, I can follow the construction of the Moyal algebra from
> su(N), with shift and clock matrices, and I agree that the limit
> N -> infinity formally gives the symplectomorphism algebra above.
> But I cannot see which su(infinity) irreps correspond to the
> p-tensor fields above, and I don't understand what happens to
> the other su(N) irreps in the limit.

Hmm, isn't the point of a recent paper (the one we are discussing?)
that for SU(N) the structure constants are something along the lines
of

N sin(nxm/N)

which converges to nxm for n and m fixed and N->oo but if n and m grow
as fast as N the "sub-leading" terms do not go ? So, why do we keep n
and m fixed and what happens to the elements with large n and m that
do not converge to the torus algebra?

Robert

--
..oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo. oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO
Robert C. Helling Department of Applied Mathematics and Theoretical Physics
University of Cambridge
print "Just another Phone: +44/1223/766870
stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling

Lubos Motl

Jul18-04, 12:52 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>On Tue, 29 Jun 2004, Peter Woit wrote:\n\n> Thanks for the explanation. Let me see if I understand this. These\n> vector fields generate an infinite number of commuting R actions, but\n> in general these aren\'t periodic, they don\'t exponentiate to U(1)\n> actions. One problem I see with Zunger\'s argument that the Lie\n> algebras for all g are isomorphic is that it relies on constructing\n> SU(2) subgroups of the group of diffeomorphisms, and studying how the\n> Lie algebra decomposes under the actions of these SU(2) subgroups. But\n> (for g>0), as you point out, there are no SU(2) subgroups, no?\n\nPeter, I think that I would fully subscribe to this!\n___________________________________________ ___________________________________\nE-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/\neFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\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">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 29 Jun 2004, Peter Woit wrote:

> Thanks for the explanation. Let me see if I understand this. These
> vector fields generate an infinite number of commuting R actions, but
> in general these aren't periodic, they don't exponentiate to U(1)
> actions. One problem I see with Zunger's argument that the Lie
> algebras for all g are isomorphic is that it relies on constructing
> SU(2) subgroups of the group of diffeomorphisms, and studying how the
> Lie algebra decomposes under the actions of these SU(2) subgroups. But
> (for g>0), as you point out, there are no SU(2) subgroups, no?

Peter, I think that I would fully subscribe to this!
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E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
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