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Hi,
This is about Witten's paper on "Mirror Manifolds and Topological String Theory" , 9112056.
What is the best way to visualise powers, both positive and negative, of the canonical line bundle? What are the meanings of "K^1/2" and "K^-1"?
What is the relation between twisting the fermions and rotational symmetry on the world sheet?
Why are the twisting the only ones we could do? If we took the fermons to be sections of other powers of the canonical line bundle on sigma ( of course tensored with the pullback of the tangent bundle on X) what would be the effect on the world sheet? Would the sigma-model lagrangian remain invariant?
What is the relation between twisting with different powers of the line bundle and the associated U(1) gauge symmetry?
Does twisting the fermions change the spins of the G+- in the superconformal algebra? What exactly is the relation between changing the relevant bundles of the fermions and the effect on the superconformal algebra- for example the shift in the energy-momentum tensor?
A.J. Tolland
Dec9-04, 02:43 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 Wed, 8 Dec 2004, boltu wrote:\n\n> This is about Witten\'s paper on "Mirror Manifolds and Topological String\n> Theory" , 9112056.\n\n> What is the best way to visualise powers, both positive and negative, of\n> the canonical line bundle? What are the meanings of "K^1/2" and "K^-1"?\n\nRemember that for line bundles, dual and inverse are the same\nthing.\nFor one dimensional complex curves, K = T^*X. So K^{-1} = TX.\nAnd K^{-1/2} is a square root of the tangent bundle, one of the\nconventional spinor bundles, so K^{1/2} is the dual of\n\n> What is the relation between twisting the fermions and rotational symmetry\n> on the world sheet?\n\nTwisting the fermions corresponds to declaring a subgroup of SO(2)\nx U(1)_L x U(1)_R other than SO(2) to be the rotational symmetry group.\nThe twistings correspond to using, instead of J,\n\nJ - 1/2 (J_L - J_R) (A-model)\nJ - 1/2 (J_L + J_R) (B-model)\n\n> Why are the twisting the only ones we could do? If we took the fermons to\n> be sections of other powers of the canonical line bundle on sigma (of\n> course tensored with the pullback of the tangent bundle on X) what would\n> be the effect on the world sheet? Would the sigma-model lagrangian remain\n> invariant?\n\nThe point of doing this is that we want to treat Q as a BRST\noperator, on any curve Sigma. Q has to act globally, so we need the\nfermionic parameters of the Lie algebra to be well-defined sections of\nsome line bundles. This is easy locally, in some small neighborhood.\nBut it\'s difficult globally: many line bundles don\'t have _any_ global\nsections. So we twist and put our fermionic parameters in the functions,\nwhere there is always at least the constant sections.\n\nIt\'s a lovely coincidence that this forces us to choose the A or B\ntwistings. Any twisting would be a symmetry fo the classical Lagrangian.\nBut only the A and B twistings don\'t become anomalous when we move to the\nquantum theory.\n\n> Does twisting the fermions change the spins of the G+- in the\n> superconformal algebra? What exactly is the relation between changing\n> the relevant bundles of the fermions and the effect on the\n> superconformal algebra- for example the shift in the energy-momentum\n> tensor?\n\nYes, twisting changes the spins. The spins are encoded in the\neigenvalues h and \\bar{h} of the grading operators L_0 and \\bar{L}_0.\nspin = h - \\bar{h}. So twisting alters the line bundles by changing how\ntheir fibers transform under rotations; this means that the spins of\nfields which arise as sections of these line bundles are altered.\nChanging the spins really means changing how we measure them, so we must\nalter the components L and \\Lbar of the energy momentum tensor.\n\nHope that helped.\n\n--A.J.\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 Wed, 8 Dec 2004, boltu wrote:
> This is about Witten's paper on "Mirror Manifolds and Topological String
> Theory" , 9112056.
> What is the best way to visualise powers, both positive and negative, of
> the canonical line bundle? What are the meanings of "K^1/2" and "K^-1"?
Remember that for line bundles, dual and inverse are the same
thing.
For one dimensional complex curves, K = T^*X. So K^{-1} = TX.
And K^{-1/2} is a square root of the tangent bundle, one of the
conventional spinor bundles, so K^{1/2} is the dual of
> What is the relation between twisting the fermions and rotational symmetry
> on the world sheet?
Twisting the fermions corresponds to declaring a subgroup of SO(2)
x U(1)_L x U(1)_R other than SO(2) to be the rotational symmetry group.
The twistings correspond to using, instead of J,
J - 1/2 (J_L - J_R) (A-model)
J - 1/2 (J_L + J_R) (B-model)
> Why are the twisting the only ones we could do? If we took the fermons to
> be sections of other powers of the canonical line bundle on \sigma (of
> course tensored with the pullback of the tangent bundle on X) what would
> be the effect on the world sheet? Would the \sigma-model lagrangian remain
> invariant?
The point of doing this is that we want to treat Q as a BRST
operator, on any curve \Sigma. Q has to act globally, so we need the
fermionic parameters of the Lie algebra to be well-defined sections of
some line bundles. This is easy locally, in some small neighborhood.
But it's difficult globally: many line bundles don't have _any_ global
sections. So we twist and put our fermionic parameters in the functions,
where there is always at least the constant sections.
It's a lovely coincidence that this forces us to choose the A or B
twistings. Any twisting would be a symmetry fo the classical Lagrangian.
But only the A and B twistings don't become anomalous when we move to the
quantum theory.
> Does twisting the fermions change the spins of the G+- in the
> superconformal algebra? What exactly is the relation between changing
> the relevant bundles of the fermions and the effect on the
> superconformal algebra- for example the shift in the energy-momentum
> tensor?
Yes, twisting changes the spins. The spins are encoded in the
eigenvalues h and \bar{h} of the grading operators L_0 and \bar{L}_0.
spin = h - \bar{h}. So twisting alters the line bundles by changing how
their fibers transform under rotations; this means that the spins of
fields which arise as sections of these line bundles are altered.
Changing the spins really means changing how we measure them, so we must
alter the components L and \Lbar of the energy momentum tensor.
Hope that helped.
--A.J.
calabisix
Jan28-05, 03:10 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>Beginner:\n\nI have been studying this paper to get a better understanding of\nTQFT\'s.\nI proceed fine until I get to the definition of the fermi field.\nwhat is the significance of defining fermi fields as the product of\nK^1/2 and the pullback of the tangent bundle?\n\nIs this a standard Differential Geometry concept?\n\nI have browsed the papers of Aspinwall and Morrison (TFT and Rational\nCurves) as well as the papers of Witten (TQFT and Sigma models)...even\nhis earlier paper on the Susy sigma model. I have yet to find an\nintuitive explanation of "why" one creates the product above to proceed\nwith the theory.\n\nThis may be a standard Differential Geometry concept that I just\nhaven\'t crossed yet (hence the "Beginner" heading of my post).\n\n\nThank you,\n\ncalabisix\n\n\nA.J. Tolland wrote:\n> On Wed, 8 Dec 2004, boltu wrote:\n>\n> > This is about Witten\'s paper on "Mirror Manifolds and Topological\nString\n> > Theory" , 9112056.\n>\n> > What is the best way to visualise powers, both positive and\nnegative, of\n> > the canonical line bundle? What are the meanings of "K^1/2" and\n"K^-1"?\n>\n> Remember that for line bundles, dual and inverse are the same\n> thing.\n> For one dimensional complex curves, K = T^*X. So K^{-1} = TX.\n> And K^{-1/2} is a square root of the tangent bundle, one of the\n> conventional spinor bundles, so K^{1/2} is the dual of\n>\n> > What is the relation between twisting the fermions and rotational\nsymmetry\n> > on the world sheet?\n>\n> Twisting the fermions corresponds to declaring a subgroup of SO(2)\n> x U(1)_L x U(1)_R other than SO(2) to be the rotational symmetry\ngroup.\n> The twistings correspond to using, instead of J,\n>\n> J - 1/2 (J_L - J_R) (A-model)\n> J - 1/2 (J_L + J_R) (B-model)\n>\n> > Why are the twisting the only ones we could do? If we took the\nfermons to\n> > be sections of other powers of the canonical line bundle on sigma\n(of\n> > course tensored with the pullback of the tangent bundle on X) what\nwould\n> > be the effect on the world sheet? Would the sigma-model lagrangian\nremain\n> > invariant?\n>\n> The point of doing this is that we want to treat Q as a BRST\n> operator, on any curve Sigma. Q has to act globally, so we need the\n> fermionic parameters of the Lie algebra to be well-defined sections\nof\n> some line bundles. This is easy locally, in some small neighborhood.\n> But it\'s difficult globally: many line bundles don\'t have _any_\nglobal\n> sections. So we twist and put our fermionic parameters in the\nfunctions,\n> where there is always at least the constant sections.\n>\n> It\'s a lovely coincidence that this forces us to choose the A or B\n> twistings. Any twisting would be a symmetry fo the classical\nLagrangian.\n> But only the A and B twistings don\'t become anomalous when we move to\nthe\n> quantum theory.\n>\n> > Does twisting the fermions change the spins of the G+- in the\n> > superconformal algebra? What exactly is the relation between\nchanging\n> > the relevant bundles of the fermions and the effect on the\n> > superconformal algebra- for example the shift in the\nenergy-momentum\n> > tensor?\n>\n> Yes, twisting changes the spins. The spins are encoded in the\n> eigenvalues h and \\bar{h} of the grading operators L_0 and \\bar{L}_0.\n> spin = h - \\bar{h}. So twisting alters the line bundles by changing\nhow\n> their fibers transform under rotations; this means that the spins of\n> fields which arise as sections of these line bundles are altered.\n> Changing the spins really means changing how we measure them, so we\nmust\n> alter the components L and \\Lbar of the energy momentum tensor.\n>\n> Hope that helped.\n>\n> --A.J.\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>Beginner:
I have been studying this paper to get a better understanding of
TQFT's.
I proceed fine until I get to the definition of the fermi field.
what is the significance of defining fermi fields as the product of
K^1/2 and the pullback of the tangent bundle?
Is this a standard Differential Geometry concept?
I have browsed the papers of Aspinwall and Morrison (TFT and Rational
Curves) as well as the papers of Witten (TQFT and \Sigma models)...even
his earlier paper on the Susy \sigma model. I have yet to find an
intuitive explanation of "why" one creates the product above to proceed
with the theory.
This may be a standard Differential Geometry concept that I just
haven't crossed yet (hence the "Beginner" heading of my post).
Thank you,
calabisix
A.J. Tolland wrote:
> On Wed, 8 Dec 2004, boltu wrote:
>
> > This is about Witten's paper on "Mirror Manifolds and Topological
String
> > Theory" , 9112056.
>
> > What is the best way to visualise powers, both positive and
negative, of
> > the canonical line bundle? What are the meanings of "K^1/2" and
"K^-1"?
>
> Remember that for line bundles, dual and inverse are the same
> thing.
> For one dimensional complex curves, K = T^*X. So K^{-1} = TX.
> And K^{-1/2} is a square root of the tangent bundle, one of the
> conventional spinor bundles, so K^{1/2} is the dual of
>
> > What is the relation between twisting the fermions and rotational
symmetry
> > on the world sheet?
>
> Twisting the fermions corresponds to declaring a subgroup of SO(2)
> x U(1)_L x U(1)_R other than SO(2) to be the rotational symmetry
group.
> The twistings correspond to using, instead of J,
>
> J - 1/2 (J_L - J_R) (A-model)
> J - 1/2 (J_L + J_R) (B-model)
>
> > Why are the twisting the only ones we could do? If we took the
fermons to
> > be sections of other powers of the canonical line bundle on \sigma
(of
> > course tensored with the pullback of the tangent bundle on X) what
would
> > be the effect on the world sheet? Would the \sigma-model lagrangian
remain
> > invariant?
>
> The point of doing this is that we want to treat Q as a BRST
> operator, on any curve \Sigma. Q has to act globally, so we need the
> fermionic parameters of the Lie algebra to be well-defined sections
of
> some line bundles. This is easy locally, in some small neighborhood.
> But it's difficult globally: many line bundles don't have _any_
global
> sections. So we twist and put our fermionic parameters in the
functions,
> where there is always at least the constant sections.
>
> It's a lovely coincidence that this forces us to choose the A or B
> twistings. Any twisting would be a symmetry fo the classical
Lagrangian.
> But only the A and B twistings don't become anomalous when we move to
the
> quantum theory.
>
> > Does twisting the fermions change the spins of the G+- in the
> > superconformal algebra? What exactly is the relation between
changing
> > the relevant bundles of the fermions and the effect on the
> > superconformal algebra- for example the shift in the
energy-momentum
> > tensor?
>
> Yes, twisting changes the spins. The spins are encoded in the
> eigenvalues h and \bar{h} of the grading operators L_0 and \bar{L}_0.
> spin = h - \bar{h}. So twisting alters the line bundles by changing
how
> their fibers transform under rotations; this means that the spins of
> fields which arise as sections of these line bundles are altered.
> Changing the spins really means changing how we measure them, so we
must
> alter the components L and \Lbar of the energy momentum tensor.
>
> Hope that helped.
>
> --A.J.
A.J. Tolland
Jan31-05, 04:02 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, 28 Jan 2005, calabisix wrote:\n\n> I proceed fine until I get to the definition of the fermi field.\n> what is the significance of defining fermi fields as the product of\n> K^1/2 and the pullback of the tangent bundle?\n\nIt\'s commonly used. The idea is that we want worldsheet fermions\nwhich take values in vector fields on the target space. But you can\'t\nmultiply bundles on different spaces (for roughly the same reason that you\ncan\'t add vectors which live in different fibers) , so you pull the\ntangent bundle back to the worldsheet, using the only map available.\n\nThat help?\n\n--A.J.\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, 28 Jan 2005, calabisix wrote:
> I proceed fine until I get to the definition of the fermi field.
> what is the significance of defining fermi fields as the product of
> K^1/2 and the pullback of the tangent bundle?
It's commonly used. The idea is that we want worldsheet fermions
which take values in vector fields on the target space. But you can't
multiply bundles on different spaces (for roughly the same reason that you
can't add vectors which live in different fibers) , so you pull the
tangent bundle back to the worldsheet, using the only map available.
That help?
--A.J.
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