John Baez
Apr7-04, 08:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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 <24a23f36.0403240033.1502e1da@posting.google.com>,\nThomas Larsson <thomas_larsson_01@hotmail.com> wrote:\n\n>Josh Willis <jwillis@gravity.psu.edu> skrev i\n>diskussionsgruppsmeddelandet:c3goq8$18le$1@f04n12. cac.psu.edu...\n\n>> (3) When the symmetry group is infinite dimensional, as for instance\n>> Diff(S^1), any neighborhood of the identity contains group elements that\n>> cannot be obtained by exponentiating the Lie algebra. Thus, the algebra\n>> doesn\'t carry the full information about the group.\n\nI wouldn\'t jump to that final conclusion unless I knew some\nexamples where drastically different infinite-dimensional\nLie groups had the same Lie algebra.\n\nOf course, even in the finite-dimensional case the Lie algebra\ndoesn\'t carry the full information about the Lie group!\nYou can take the universal cover of a Lie group, or take its\nidentity component, without changing its Lie algebra.\n\nBut presumably here you\'re not talking about *that*. You\'re\nsuggesting that infinite-dimensional Lie groups are "worse" in\nsome way. Do you know two infinite-dimensional Lie groups whose\nLie algebras are isomorphic, but where their identity components\ndon\'t have isomorphic universal covers? I don\'t.\n\nAnd of course when we get serious about this business, we\nneed to distinguish between various flavors of "infinite-dimensional\nLie groups". Some are better than others.\n\n>There are also exponentiated vector fields that are not\n>diffeomorphisms.\n\nIf the vector field is smooth and the manifold is S^1, or more\ngenerally compact, its exponential will exist and be a diffeomorphism.\n\nIf the manifold is noncompact there will be smooth vector fields\nthat don\'t exponentiate to give well-defined maps, like the vector\nfield x^3 d/dx on the real line. But, this just means you should\nput some growth conditions on your vector fields if you want to\ngo around exponentiating them.\n\n>My impression is that this distinction is a\n>mathematical subtlety without real physical significance.\n\nI tend to agree, though it\'s *always* dangerous to make general\nvague claims that some bit of math has no "physical significance".\n\nHere\'s one reason I tend to agree:\n\nThe problem Josh Willis mentions about Diff(S^1) is really just a\nspinoff of the fact that it\'s a Frechet Lie group - i.e. it\'s covered\nwith charts that are Frechet spaces. Frechet Lie groups are tricky\nin certain ways.\n\nSo, instead of working with diffeomorphisms, work with something\na little less smooth. Use maps\n\nf: S^1 -> S^1\n\nhaving inverses\n\ng: S^1 -> S^1\n\nsuch that f and g have 1st, 2nd,..., nth derivatives in L^2.\nThe maps form a Hilbert Lie group - i.e., a group covered with charts\nthat are Hilbert spaces. For Lie groups like this, elements near\nthe identity are ALWAYS exponentials of guys in the Lie algebra!\n\nPeople who want to get serious about the *analysis* aspects of the\nVirasoro group, loop groups, and other infinite-dimensional groups\ntend to use Hilbert Lie groups.\n\nI\'m completely sick of arguing about loops versus strings, so I will\nconfine my remarks to the above relatively uncontroversial topics. :-)\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>In article <24a23f36.0403240033.1502e1da@posting.google.com>,
Thomas Larsson <thomas_larsson_01@hotmail.com> wrote:
>Josh Willis <jwillis@gravity.psu.edu> skrev i
>diskussionsgruppsmeddelandet:c3goq8$18le$1@f04n12. cac.psu.edu...
>> (3) When the symmetry group is infinite dimensional, as for instance
>> Diff(S^1), any neighborhood of the identity contains group elements that
>> cannot be obtained by exponentiating the Lie algebra. Thus, the algebra
>> doesn't carry the full information about the group.
I wouldn't jump to that final conclusion unless I knew some
examples where drastically different infinite-dimensional
Lie groups had the same Lie algebra.
Of course, even in the finite-dimensional case the Lie algebra
doesn't carry the full information about the Lie group!
You can take the universal cover of a Lie group, or take its
identity component, without changing its Lie algebra.
But presumably here you're not talking about *that*. You're
suggesting that infinite-dimensional Lie groups are "worse" in
some way. Do you know two infinite-dimensional Lie groups whose
Lie algebras are isomorphic, but where their identity components
don't have isomorphic universal covers? I don't.
And of course when we get serious about this business, we
need to distinguish between various flavors of "infinite-dimensional
Lie groups". Some are better than others.
>There are also exponentiated vector fields that are not
>diffeomorphisms.
If the vector field is smooth and the manifold is S^1, or more
generally compact, its exponential will exist and be a diffeomorphism.
If the manifold is noncompact there will be smooth vector fields
that don't exponentiate to give well-defined maps, like the vector
field x^3 d/dx on the real line. But, this just means you should
put some growth conditions on your vector fields if you want to
go around exponentiating them.
>My impression is that this distinction is a
>mathematical subtlety without real physical significance.
I tend to agree, though it's *always* dangerous to make general
vague claims that some bit of math has no "physical significance".
Here's one reason I tend to agree:
The problem Josh Willis mentions about Diff(S^1) is really just a
spinoff of the fact that it's a Frechet Lie group - i.e. it's covered
with charts that are Frechet spaces. Frechet Lie groups are tricky
in certain ways.
So, instead of working with diffeomorphisms, work with something
a little less smooth. Use maps
f: S^1 -> S^1
having inverses
g: S^1 -> S^1
such that f and g have 1st, 2nd,..., nth derivatives in L^2.
The maps form a Hilbert Lie group - i.e., a group covered with charts
that are Hilbert spaces. For Lie groups like this, elements near
the identity are ALWAYS exponentials of guys in the Lie algebra!
People who want to get serious about the *analysis* aspects of the
Virasoro group, loop groups, and other infinite-dimensional groups
tend to use Hilbert Lie groups.
I'm completely sick of arguing about loops versus strings, so I will
confine my remarks to the above relatively uncontroversial topics. :-)
Thomas Larsson <thomas_larsson_01@hotmail.com> wrote:
>Josh Willis <jwillis@gravity.psu.edu> skrev i
>diskussionsgruppsmeddelandet:c3goq8$18le$1@f04n12. cac.psu.edu...
>> (3) When the symmetry group is infinite dimensional, as for instance
>> Diff(S^1), any neighborhood of the identity contains group elements that
>> cannot be obtained by exponentiating the Lie algebra. Thus, the algebra
>> doesn't carry the full information about the group.
I wouldn't jump to that final conclusion unless I knew some
examples where drastically different infinite-dimensional
Lie groups had the same Lie algebra.
Of course, even in the finite-dimensional case the Lie algebra
doesn't carry the full information about the Lie group!
You can take the universal cover of a Lie group, or take its
identity component, without changing its Lie algebra.
But presumably here you're not talking about *that*. You're
suggesting that infinite-dimensional Lie groups are "worse" in
some way. Do you know two infinite-dimensional Lie groups whose
Lie algebras are isomorphic, but where their identity components
don't have isomorphic universal covers? I don't.
And of course when we get serious about this business, we
need to distinguish between various flavors of "infinite-dimensional
Lie groups". Some are better than others.
>There are also exponentiated vector fields that are not
>diffeomorphisms.
If the vector field is smooth and the manifold is S^1, or more
generally compact, its exponential will exist and be a diffeomorphism.
If the manifold is noncompact there will be smooth vector fields
that don't exponentiate to give well-defined maps, like the vector
field x^3 d/dx on the real line. But, this just means you should
put some growth conditions on your vector fields if you want to
go around exponentiating them.
>My impression is that this distinction is a
>mathematical subtlety without real physical significance.
I tend to agree, though it's *always* dangerous to make general
vague claims that some bit of math has no "physical significance".
Here's one reason I tend to agree:
The problem Josh Willis mentions about Diff(S^1) is really just a
spinoff of the fact that it's a Frechet Lie group - i.e. it's covered
with charts that are Frechet spaces. Frechet Lie groups are tricky
in certain ways.
So, instead of working with diffeomorphisms, work with something
a little less smooth. Use maps
f: S^1 -> S^1
having inverses
g: S^1 -> S^1
such that f and g have 1st, 2nd,..., nth derivatives in L^2.
The maps form a Hilbert Lie group - i.e., a group covered with charts
that are Hilbert spaces. For Lie groups like this, elements near
the identity are ALWAYS exponentials of guys in the Lie algebra!
People who want to get serious about the *analysis* aspects of the
Virasoro group, loop groups, and other infinite-dimensional groups
tend to use Hilbert Lie groups.
I'm completely sick of arguing about loops versus strings, so I will
confine my remarks to the above relatively uncontroversial topics. :-)