View Full Version : Self-dual lattices
This is related to the T-duality question; does anyone know if the following proposition is true?
A lattice is self-dual iff it can be generated by the weight vectors of some Lie algebra.
Thanks
d70yxj
Volker Braun
Sep8-04, 11:08 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>Lubos, he did not ask for the lattice to be even, so there are lots of\nself-dual lattices even if you restrict yourself to euclidean signature.\nStill, the only simple Lie group with self-dual weight lattice is E_8.\n\n[Moderator\'s note: I don\'t think that there are lots of them; in fact,\nthe lattice of n-tuples of integers Z^k seems to be the only simple\nenough case that it could appear as a weight lattice. Well, Z^k\nappears as the weight lattice of U(N). But we rarely consider U(N)\nincluding the U(1) because U(N) is not simple. Which "lots" of\nlattices do you have in mind? Maybe I missed a large class... LM]\n\nOn Tue, 07 Sep 2004 15:55:28 -0400, d70yxj wrote:\n> A lattice is self-dual iff it can be generated by the weight vectors of\n> some Lie algebra.\n>\n> [Moderator\'s note: Well, everyone is invited to post a different answer,\n> but the statement is definitely false. Most Lie algebras have weight\n> lattices that are not self-dual. Self-dual lattices are very special,\n> and they can be identified with the weight lattices of E_8 or\n> spin(16.k) / Z_2 - where the Z_2 subtlety is necessary because only\n> "one half" of the weight lattice should be taken - something in\n> between the weight lattice and the smaller root lattice. I did not\n> consider self-dual lattices of mixed signatures - they need a special\n> treatment...\n>\n> This posting was delayed because of communication problems with\n> physicsforums.com. LM]\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, he did not ask for the lattice to be even, so there are lots of
self-dual lattices even if you restrict yourself to euclidean signature.
Still, the only simple Lie group with self-dual weight lattice is E_8.
[Moderator's note: I don't think that there are lots of them; in fact,
the lattice of n-tuples of integers Z^k seems to be the only simple
enough case that it could appear as a weight lattice. Well, Z^k
appears as the weight lattice of U(N). But we rarely consider U(N)
including the U(1) because U(N) is not simple. Which "lots" of
lattices do you have in mind? Maybe I missed a large class... LM]
On Tue, 07 Sep 2004 15:55:28 -0400, d70yxj wrote:
> A lattice is self-dual iff it can be generated by the weight vectors of
> some Lie algebra.
>
> [Moderator's note: Well, everyone is invited to post a different answer,
> but the statement is definitely false. Most Lie algebras have weight
> lattices that are not self-dual. Self-dual lattices are very special,
> and they can be identified with the weight lattices of E_8 or
> spin(16.k) / Z_2 - where the Z_2 subtlety is necessary because only
> "one half" of the weight lattice should be taken - something in
> between the weight lattice and the smaller root lattice. I did not
> consider self-dual lattices of mixed signatures - they need a special
> treatment...
>
> This posting was delayed because of communication problems with
> physicsforums.com. LM]
Volker Braun
Sep8-04, 04:34 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>> [Moderator\'s note: I don\'t think that there are lots of them; in fact,\n> the lattice of n-tuples of integers Z^k seems to be the only simple\n> enough case that it could appear as a weight lattice. Well, Z^k\n> appears as the weight lattice of U(N). But we rarely consider U(N)\n> including the U(1) because U(N) is not simple. Which "lots" of\n> lattices do you have in mind? Maybe I missed a large class... LM]\n\nI thought that giving up the even-ness would make it a lot easier to find\na self-dual lattice but its not so easy to write one down. So I cheated:\n\nhttp://www.research.att.com/~njas/lattices/unimodular.html\n\nIn that table the first unimodular lattice that is not the weight lattice\nof a Lie group is in dimension 12, and the "shorter Leech" (in 23d) is\nthe first one that is not even and not a weight lattice.\n\n[Moderator\'s note: That\'s a cool lattice. LM]\n\nSo for now, I\'ll take back the "lot". Well, depends on how you count :-)\n-Volker\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>> [Moderator's note: I don't think that there are lots of them; in fact,
> the lattice of n-tuples of integers Z^k seems to be the only simple
> enough case that it could appear as a weight lattice. Well, Z^k
> appears as the weight lattice of U(N). But we rarely consider U(N)
> including the U(1) because U(N) is not simple. Which "lots" of
> lattices do you have in mind? Maybe I missed a large class... LM]
I thought that giving up the even-ness would make it a lot easier to find
a self-dual lattice but its not so easy to write one down. So I cheated:
http://www.research.att.com/~njas/lattices/unimodular.html
In that table the first unimodular lattice that is not the weight lattice
of a Lie group is in dimension 12, and the "shorter Leech" (in 23d) is
the first one that is not even and not a weight lattice.
[Moderator's note: That's a cool lattice. LM]
So for now, I'll take back the "lot". Well, depends on how you count :-)
-Volker
Hi, thanks, yes I did mean just self-dual, and not necessarily even. I have in mind the lattice that a target space might be compactified on (which need not be either even or self-dual, is that right?), rather than the (related) lattice on which closed string momenta live (which I think has to be both even and self dual).
But I thought that when the compactification lattice _is_ self-dual, it is then that one finds an extension of the abelian gauge symmetry to a non-abelian symmetry. In some simple examples it seems that this self-dual lattice is the maximal torus of a Lie algebra, and then the corresponding (full) Lie algebra generates the appropriate extended gauge symmetry.
From what you're saying this can't be general - so is there not necessarily an extended gauge symmetry even when the compactification lattice itself is self-dual? Or if there always is, what is it? Or am I getting this completely wrong?
>>Hi, thanks, yes I did mean just self-dual, and not necessarily even. I
>>have in mind the lattice that a target space might be compactified on
>>(which need not be either even or self-dual, is that right?), rather
>>than the (related) lattice on which closed string momenta live (which I
>>think has to be both even and self dual).
>[Moderator's note: In toroidal compactification, the momenta always live
> on a lattice that is dual to the winding lattice - up to some 2\pi
> factors - and the lattice of windings is the quotient of R^k that defines
> the target space torus. The two lattices you talk about are always
> related, and stringy consistency requires that they are even self-dual.
Right, sorry, I may not have expressed this very clearly. So, I compactify on a d-dimensional lattice, \Lambda, and then the windings live on this lattice, while momentum modes live on the dual, \Lambda* to this lattice. However, looking at Polchinski Ch8.4, one defines a related 2d-dimensional lattice \Gamma, and it is _this_ lattice that's constrained by modular invariance to be even self-dual, right?
But I didn't get the impression there is a reason for \Lambda to be even self-dual, necessarily. Again, am I wrong in thinking that it is in the special case where \Lambda _itself_ is also self-dual, that you have the enhanced non-abelian gauge symmetry, instead of U(1)^d? (For example, in compactification on a circle, you have U(1) going to SU(2) but only at the self-dual radius...I have a feeling I am seeing this incorrectly, but at least this is my impression).
Oh, and I thought that in the heterotic string this constraint on \Gamma then also becomes a constraint of even self-duality on \Lambda, but that's because you're only compactifying on one side, left or right movers. Hence you are automatically forced to get the enhanced gauge symmetry there, E_8xE_8 rather than the generic U(1)^16.
>[Moderator's note: The lattice sites
> whose length squared is exactly equal to two give new massless
> "W" gauge bosons that make the gauge group non-Abelian.
> LM]
I can see that there will be new massless states, but if I compactify on a general self-dual lattice, how do I know what gauge group these new massless states generate?
PS there is something odd in that your moderator's notes appear on the newsgroup posting but not on the PF web-based version of my posting.
Lubos Motl
Sep17-04, 02:25 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 Fri, 17 Sep 2004, d70yxj wrote:\n\n> ... this lattice. However, looking at Polchinski Ch8.4, one defines a\n> related 2d-dimensional lattice \\Gamma, and it is _this_ lattice that\'s\n> constrained by modular invariance to be even self-dual, right?\n\nRight, in the non-heterotic case, the lattice that must be self-dual\ncontains both momenta as well as windings, or, equivalently, left-moving\nas well as right-moving bosons. This lattice containing both, if you\nconstruct it from momenta and winding, is always self-dual because it is\n\\Gamma_{d,d} if you do it right. I was focusing on the non-trivial case of\nthe chiral bosons and self-dual lattices of the heterotic string. Of\ncourse, the lattice of *momenta* only in a theory of non-chiral bosons,\nwith independent windings, does not have to be self-dual - one can\ncompactify string theory on any torus, roughly speaking.\n\n> But I didn\'t get the impression there is a reason for \\Lambda to be\n> even self-dual, necessarily. Again, am I wrong in thinking that it is\n> in the special case where \\Lambda _itself_ is also self-dual, that you\n> have the enhanced non-abelian gauge symmetry, instead of U(1)^d?\n\nI\'ve told you the universal criterion how to decide whether you obtain an\n(unbroken) non-Abelian gauge symmetry: you look for (massless) spin-one\nexcitations of the string that are charged under the Cartan subalgebra.\nThey will be the W-bosons. The existence of the right states is more or\nless equivalent to what you say, but it is more correct to look at the\nlattice of allowed left-moving (or right-moving) momenta, which contains\nvectors that are the sum (or difference) of the allowed momenta (and\nwindings). I think it is important to say the result for the most usual\ntheories: the bosonic string can have various extended symmetries,\nespecially SU(2) x SU(2) at the self-dual radius. However, type II\nsuperstring theories do not have such extended non-Abelian gauge\nsymmetries. The heterotic string is a compromise between the bosonic and\nsuperstring case and you can obtain a *single* extended SU(2) gauge group\n(from the bosonic sector, let\'s call it the left-moving one) at the\nself-dual radius, as well as some other options.\n\n> (For example, in compactification on a circle, you have U(1) going to\n> SU(2) but only at the self-dual radius...I have a feeling I am seeing\n> this incorrectly, but at least this is my impression).\n\nIn this case you are correct if properly interpreted. The extended SU(2)\nappears when the radius is self-dual under T-duality. Whether it is the\nsame radius as a "self-dual lattice" depends on your choice of units and\nyour treatment of 2\\pi - the lattice of windings contains "dimensionful"\nsites, and therefore its dual lattice has different units, and such a\nlattice cannot really be self-dual.\n\n> Oh, and I thought that in the heterotic string this constraint on\n> \\Gamma then also becomes a constraint of even self-duality on \\Lambda,\n> but that\'s because you\'re only compactifying on one side, left or right\n> movers. Hence you are automatically forced to get the enhanced gauge\n> symmetry there, E_8xE_8 rather than the generic U(1)^16.\n\nRight.\n\n> I can see that there will be new massless states, but if I compactify\n> on a general self-dual lattice, how do I know what gauge group these\n> new massless states generate?\n\nYou find all stringy excitations that are vector-like (spin one) massless\nexcitations in spacetime, which are charged under the Cartan subalgebra.\nYou must the find a non-Abelian group whose roots are exactly the charges\nof the states from the previous sentence under the Cartan subalgebra.\nString theory is a powerful structure; it always leads to consistent\nspacetime physics if the worldsheet consistency rules are satisfied, which\nmakes it easier to comprehend the small "miracle" that the charged spin\none massless particles that you find always form a root system - it must\nbe so because Yang-Mills theory is the only way to couple spin one\nparticles consistently.\n\n> PS there is something odd in that your moderator\'s notes appear on the\n> newsgroup posting but not on the PF web-based version of my posting.\n\nThat\'s a possible job for the masters of PF.\n_____________________________________________ _________________________________\nE-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/\neFax: +1-801/454-1858 work: +1-617/384-9488 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 Fri, 17 Sep 2004, d70yxj wrote:
> ... this lattice. However, looking at Polchinski Ch8.4, one defines a
> related 2d-dimensional lattice \Gamma, and it is _this_ lattice that's
> constrained by modular invariance to be even self-dual, right?
Right, in the non-heterotic case, the lattice that must be self-dual
contains both momenta as well as windings, or, equivalently, left-moving
as well as right-moving bosons. This lattice containing both, if you
construct it from momenta and winding, is always self-dual because it is
\Gamma_{d,d} if you do it right. I was focusing on the non-trivial case of
the chiral bosons and self-dual lattices of the heterotic string. Of
course, the lattice of *momenta* only in a theory of non-chiral bosons,
with independent windings, does not have to be self-dual - one can
compactify string theory on any torus, roughly speaking.
> But I didn't get the impression there is a reason for \Lambda to be
> even self-dual, necessarily. Again, am I wrong in thinking that it is
> in the special case where \Lambda _itself_ is also self-dual, that you
> have the enhanced non-abelian gauge symmetry, instead of U(1)^d?
I've told you the universal criterion how to decide whether you obtain an
(unbroken) non-Abelian gauge symmetry: you look for (massless) spin-one
excitations of the string that are charged under the Cartan subalgebra.
They will be the W-bosons. The existence of the right states is more or
less equivalent to what you say, but it is more correct to look at the
lattice of allowed left-moving (or right-moving) momenta, which contains
vectors that are the sum (or difference) of the allowed momenta (and
windings). I think it is important to say the result for the most usual
theories: the bosonic string can have various extended symmetries,
especially SU(2) x SU(2) at the self-dual radius. However, type II
superstring theories do not have such extended non-Abelian gauge
symmetries. The heterotic string is a compromise between the bosonic and
superstring case and you can obtain a *single* extended SU(2) gauge group
(from the bosonic sector, let's call it the left-moving one) at the
self-dual radius, as well as some other options.
> (For example, in compactification on a circle, you have U(1) going to
> SU(2) but only at the self-dual radius...I have a feeling I am seeing
> this incorrectly, but at least this is my impression).
In this case you are correct if properly interpreted. The extended SU(2)
appears when the radius is self-dual under T-duality. Whether it is the
same radius as a "self-dual lattice" depends on your choice of units and
your treatment of 2\pi - the lattice of windings contains "dimensionful"
sites, and therefore its dual lattice has different units, and such a
lattice cannot really be self-dual.
> Oh, and I thought that in the heterotic string this constraint on
> \Gamma then also becomes a constraint of even self-duality on \Lambda,
> but that's because you're only compactifying on one side, left or right
> movers. Hence you are automatically forced to get the enhanced gauge
> symmetry there, E_{8xE_8} rather than the generic U(1)^16.
Right.
> I can see that there will be new massless states, but if I compactify
> on a general self-dual lattice, how do I know what gauge group these
> new massless states generate?
You find all stringy excitations that are vector-like (spin one) massless
excitations in spacetime, which are charged under the Cartan subalgebra.
You must the find a non-Abelian group whose roots are exactly the charges
of the states from the previous sentence under the Cartan subalgebra.
String theory is a powerful structure; it always leads to consistent
spacetime physics if the worldsheet consistency rules are satisfied, which
makes it easier to comprehend the small "miracle" that the charged spin
one massless particles that you find always form a root system - it must
be so because Yang-Mills theory is the only way to couple spin one
particles consistently.
> PS there is something odd in that your moderator's notes appear on the
> newsgroup posting but not on the PF web-based version of my posting.
That's a possible job for the masters of PF.
__{_______________________________________________ _____________________________}
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Sorry, I've also just realised that you imply that the enhanced gauge symmetry can occur when \Lambda is not necessarily self-dual...and perhaps also that there may be no enhanced symmetry even when \Lambda is self-dual. So probably there is no reason to expect a connection between the gauge groups and the self-dual lattices....
[QUOTE=Lubos Motl>
> I can see that there will be new massless states, but if I compactify
> on a general self-dual lattice, how do I know what gauge group these
> new massless states generate?[/color]
You find all stringy excitations that are vector-like (spin one) massless
excitations in spacetime, which are charged under the Cartan subalgebra.
You must the find a non-Abelian group whose roots are exactly the charges
of the states from the previous sentence under the Cartan subalgebra.
String theory is a powerful structure; it always leads to consistent
spacetime physics if the worldsheet consistency rules are satisfied, which
makes it easier to comprehend the small "miracle" that the charged spin
one massless particles that you find always form a root system - it must
be so because Yang-Mills theory is the only way to couple spin one
particles consistently.
QUOTE]
Thanks for the help and clarifications. I think I'm happy with everything above. I hadn't realised that the enhanced gauge symmetry can occur when \Lambda is not necessarily self-dual, and presumably even when \Lambda is self-dual there may be no enhanced gauge symmetry. (I guess the examples I'd seen incorrectly gave me the impression that self-duality of \Lambda was associated with a non-abelian symmetry).
While I expected the proposition at the top of this thread to be too strong to be true, somehow there must still be a relation between every lattice giving rise to extra massless states, and the appropriate Lie group? This connection still seems a bit mysterious to me.
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