Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

December 30, 2003

Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

Posted by Urs Schreiber

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It has been shown in Part I (see also hep-th/0401175) that the modes of the $K$ -deformed exterior derivative ${\mathbf d}_{K,\xi}$ on loop space together with their adjoints ${\mathbf d}^+_{K,\xi}$ generate the classical super Virasoro algebra. In the following deformations of ${\mathbf d}_{K,\xi}$ are studied under which the form of the superconformal algebra is preserved. The new algebra representations obtained this way are identified as corresponding to the massless NS and NS-NS background fields. A further 2-form background is found and T-duality is studied for all these algebras.

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(2.2) Isomorphisms of the superconformal algebra

From section 2.1.2 of [3] it follows that the general continuous isomorphism of the 0-mode sector ( $\xi = 1$ ) of the superconformal algebra is induced by some operator

(1)

\mathbf{W} = \int d\sigma\, W(\sigma) \,,

where $\mathbf{W}$ is an operator on loop space of unit reparametrization weight, and looks like

(2)

\mathbf{d}_{K,1} \mapsto \mathbf{d}^{\mathbf{W}}_{K,1} := e^{-\mathbf{W}} \mathbf{d}_{K,1} e^{\mathbf{W}}

This construction obviously immediately generalizes to the full algebra including all modes

(3)

\mathbf{d}_{K,\xi} \mapsto \mathbf{d}^{\mathbf{W}}_{K,\xi} := e^{-\mathbf{W}} \mathbf{d}_{K,\xi} e^{\mathbf{W}}

if the crucial relation

(4)

\mathbf{\Delta}^{\mathbf{W}}_{K,\xi_1\xi_2} = \{ \mathbf{d}^{\mathbf{W}}_{K,\xi_1}, (\mathbf{d}^{\mathbf{W}})^{&dagger;}_{K,\xi_2} \}

remains well defined (i.e. if the modes combine multiplicatively on the left hand side.)

Every operator $\mathbf{W}$ which satisfies these conditions therefore induces a classical algebra isomorphism of the superconformal algebra. However, two different $\mathbf{W}$ need not induce two different isomorphisms. In particular, anti-hermitean $\mathbf{W}^{&dagger;} = -\mathbf{W}$ induce pure gauge transformations in the sense that all algebra elements are transformed by the same unitary similarity transformation. Examples for such unitary transformations are given below. They are related to background gauge transformations as well as to string dualities. For a detailed discussion of the role of such automorphism in the general framework of string duality symmetries see section 7 of [8].

2.3 Gravitational background by means of algebra isomorphisms

First we reconsider the purely gravitational background from the point of view that its superconformal algebra derives from the superconformal algebra for flat cartesian target space by a deformation of the above form. For the point particle limit this was discussed in equations (38)-(42) of [3] and the generalization to loop space is straightforward:

Denote by

(5)

\mathbf{d}_{K,1}^{\eta} := \mathcal{E}^{&dagger; (\mu,\sigma)} \partial_{(\mu,\sigma)} + i \mathcal{E}_{(\mu,\sigma)} X^{\prime(\mu,\sigma)}

the $K$ -deformed exterior derivative on flat loop space and define the deformation operator by

(6)

\mathbf{W} = \mathcal{E}^{&dagger;} \cdot (\ln E) \cdot \mathcal{E} \,.

where $\ln E$ is the logarithm of a vielbein on loop space, regarded as a matrix. This $\mathbf{W}$ is constructed so as to satisfy

(7)

e^{\mathbf{W}} \mathcal{E}^{&dagger;a} e^{-\mathbf{W}} = \sum_\nu e^a{}_\nu \mathcal{E}^{&dagger; (b=\nu)} \,,

which yields

(8)

e^{\mathbf{W}} \mathcal{E}^{&dagger;\mu} e^{-\mathbf{W}} = \mathcal{E}^{&dagger; (b=\mu)} \,.

It is because of the fact that $e^{\mathbf{W}}$ interchanges between two different vielbein fields which define two different metric tensors that the index structure becomes a little awkward in the above equations. Since we won’t need these transformations for the further developments we don’t bother to introduce special notation to deal with this issue more cleanly. The point here is just to indicate that a $\mathbf{W}$ with the above properties does exist. $e^\mathbf{W}$ transforms all $p$ -forms with respect to $E$ to $p$ -forms with respect to the flat metric and hence

(9)

\mathbf{d}_{K,\xi} = e^{-{\mathbf{W}}} \mathbf{d}_{K,\xi}^{\eta} e^{\mathbf{W}} \,,

so that, indeed, this $\mathbf{W}$ induces a gravitational field on the target space.

As was discussed on p. 10 of [3] we need to require $\mathrm{det}e = 1$ , and hence

(10)

\mathrm{tr}\,\ln e = 0

in order that $\mathbf{d}^{&dagger;\mathbf{W}}_{K,\xi} = (\mathbf{d}^\mathbf{W}_{K,\xi^*})^&dagger;$ . This is just a condition on the nature of the coordinate system with respect to which the metric is constructed by the above deformation. Also, recall that, while as an abstract operator $\mathbf{d}_{K,\xi}$ is of course independent of any metric, its representation in terms of the operators $X^{(\mu,\sigma)},\partial_{(\mu,\sigma)}, {\mathcal{E}^{&dagger; \mu}}, \mathcal{E}^\mu$ is not, which is what the above is all about.

Note furthermore, that

(11)

\mathbf{W}^&dagger; = \pm \mathbf{W} \Leftrightarrow (\ln e)^{\mathrm{T}} = \pm \ln e \,.

According to the discussion in section (2.2) this implies that the antisymmetric part of $\ln e$ generates a pure gauge transformation and only the (traceless) symmetric part of $\ln e$ is responsible for a perturbation of the gravitational background. A little reflection shows that the gauge transformation induced by antisymmetric $\ln e$ is a rotation of the vielbein frame. For further discussion of this point see pp. 58 of [9].

(2.4) B-field background

As in section 2.1.3 of [3] we now consider the Kalb-Ramond $B$ -field 2-form

(12)

B = \frac{1}{2}B_{\mu\nu}dx^\mu \wedge dx^\nu

on target space with field strength $H = dB$ . This induces on loop space the 2-form

(13)

B_{(\mu,\sigma)(\nu,\sigma^\prime)} (X) = B_{\mu\nu}(X(\sigma)) \delta_{\sigma,\sigma^\prime} \,.

We will study the deformation operator

(14)

\mathbf{W}^{(B)}(X) := \frac{1}{2} B_{(\mu,\sigma)(\nu,\sigma^\prime)} (X) \mathcal{E}^{&dagger; (\mu,\sigma)} \mathcal{E}^{&dagger; (\nu,\sigma^\prime)}

on loop space (which is manifestly of reparametrization weight 1) and show that the superconformal algebra that it induces is indeed that found by a canonical treatment of the usual supersymmetric $\sigma$ -model with gravitational and Kalb-Ramond background.

When calculating the deformation induced by this $\mathbf{W}$ one finds

(15)

\mathbf{d}_{K,\xi}^{(B)} = \int d\sigma\, \xi \left( \mathcal{E}^{&dagger;\mu}{\hat \nabla}_\mu + i T\mathcal{E}_\mu X^{\prime\mu} + \frac{1}{6} H_{\alpha\beta\gamma} \mathcal{E}^{&dagger; \alpha} \mathcal{E}^{&dagger; \beta} \mathcal{E}^{&dagger; \gamma} - iT\mathcal{E}^{&dagger;\mu} B_{\mu\nu}X^{\prime\nu} \right) \,.

Supercommuting this with its adjoint shows that the consistency condition is satisfied, i.e. the modes of the deformed Laplace-Beltrami operator are well defined.

With hindsight this is no surprise, because the above are precisely the superconformal generators in functional form as found by canonical analysis of the non-linear supersymmetric $\sigma$ -model

(16)

S = \frac{T}{2} \int d^2 \xi d^2 \theta \, \left( G_{\mu\nu} + B_{\mu\nu} \right) D_+ \mathbf{X}^\mu D_- \mathbf{X}^\nu \,,

where $\mathbf{X}^\mu$ are worldsheet superfields. The calculation can be found in section 2 of [10] . (See eqs. (32), (33).)

(2.5) Dilaton background

The deformation operator which induces the gravitational background was of the form $\mathbf{W} = \mathcal{E}^&dagger; \cdot M \cdot \mathcal{E}$ with $M$ a traceless symmetric matrix. If instead we consider a deformation of the same form but for pure trace we get

(17)

\mathbf{W}^{(D)} = - \frac{1}{2} \int d\sigma\, \Phi(X) \mathcal{E}^{&dagger;\mu}\mathcal{E}_\mu \,,

for some real scalar field $\Phi$ on target space. This should therefore induce a dilaton background.

The associated superconformal generators are (we suppress the $\sigma$ dependence and the mode functions $\xi$ from now on)

(18)

\mathbf{d}_{K}^{(\Phi)} = e^{\Phi/2}\mathcal{E}^{&dagger; \mu} \left( {\hat \nabla}_\mu - \frac{1}{2} (\partial_\mu \Phi) \mathcal{E}^{&dagger; \nu} \mathcal{E}_\nu \right) +iT e^{-\Phi/2} X^{\prime \mu} \mathcal{E}_\mu

and their adjoints. It is readily seen that for this deformation the consistency condition is satisfied, so that these operators indeed generate a superconformal algebra.

Comparison of the superpartners of $\Gamma_{\pm,\mu}$

(19)

\mp\frac{1}{2} \{ \mathbf{d}_K^{(\Phi)} \pm \mathbf{d}_K^{&dagger;(\Phi)} , \Gamma_{\mp,\mu} \} = e^{\Phi/2} \partial_\mu \mp iT e^{-\Phi/2} G_{\mu\nu} X^{\prime\mu} + {fermionic terms}

with the bosonic currents obtained from the Born-Infeld action

(20)

S = -T \int e^{-\Phi}\sqrt{\mathrm{det}\,G}

shows that this has the form expected for the dilaton coupling of a D-string.

(2.6) Gauge field background

A gauge field background $A = A_\mu dx^\mu$ should express itself via $B \to B + \frac{1}{T}F$ , where $F = dA$ , if we assume $A$ to be a $U(1)$ connection for the moment. Since the present discussion so far refers only to closed strings and since closed strings have trivial coupling to $A$ it is to be expected that an $A$ -field background manifests itself as a pure gauge transformation in the present context. This motivates to investigate the deformation induced by the anti-hermitean

(21)

\mathbf{W} = -i A_{(\mu,\sigma)}(X) X^{\prime(\mu,\sigma)} \,.

The associated superconformal generators are found to be

(22)

\mathbf{d}_{K}^{(A)(B)} = \mathbf{d}^{(B)}_{K} - i \mathcal{E}^{+ \mu}F_{\mu\nu} X^{\prime\nu}

and the respective adjoints. Comparison with the form of the generators found for a B-field background shows that indeed

(23)

\mathbf{d}_{K}^{(A)(B)} = \mathbf{d}_K^{(B+\frac{1}{T}F)} \,,

so that we can identify the background induced by the above $\mathbf{W}$ with that of the NS $U(1)$ gauge field. Since $e^{\mathbf{W}}(X)$ is nothing but the Wilson loop of $A$ around $X$ , it is natural to conjecture that for a general (non-abelian) gauge field background $A$ the corresponding deformation is the Wilson loop as well:

(24)

\mathbf{d}_K^{(A)} = \left( \mathrm{Tr}\,\mathcal{P}\, e^{i \int A_\mu X^{\prime\mu}} \right) \mathbf{d}_K \left( \mathrm{Tr}\,\mathcal{P}\, e^{-i \int A_\mu X^{\prime\mu}} \right) \,,

where $\mathcal{P}$ indicates path ordering and $\mathrm{Tr}$ the trace in the Lie algebra, as usual.

(2.8) $C$ -field background

So far we have found deformation operators for all massless NS and NS-NS background fields. One notes a close similarity between the form of these deformation operators and the form of the corresponding vertex operators: The deformation operators for $G$ , $B$ and $\Phi$ are bilinear in the form creation/annihilation operators on loop space, with the bilinear form (matrix) seperated into its traceless symmetric, antisymmetric and trace part.

Interestingly, though, there is one more deformation operator obtainable by such a bilinear in the form creation/annihilation operators. It is

(25)

\mathbf{W}^{(C)} := \frac{1}{2} \int d\sigma\, C_{\mu\nu}(X) \mathcal{E}^{\mu} \mathcal{E}^\nu \,.

It induces the generators

(26)

\mathbf{d}^{(C)}_{K,\xi} = \int d\sigma\, \xi \left( \mathcal{E}^{&dagger; \mu} {\hat \nabla}_\mu + i\mathcal{E}_\mu X^{\prime \mu} - \mathcal{E}^\nu C_\nu{}^\mu {\hat \nabla}_\mu + \frac{1}{2} \mathcal{E}^{&dagger; \alpha} \mathcal{E}^\mu \mathcal{E}^\nu (\nabla_\alpha C_{\mu\nu}) - \frac{1}{2} C_\nu{}^\mu \mathcal{E}^\nu \mathcal{E}^\alpha \mathcal{E}^\beta (\nabla_\mu C_{\alpha\beta}) + \frac{1}{2}C^\alpha{}_\beta \mathcal{E}^\beta \mathcal{E}^\mu \mathcal{E}^\nu (\nabla_\alpha C_{\mu\nu}) \right) \,.

It turns out that these generators, too, respect the consistency condition on the modes of the deformed Laplace-Beltrami operator, so this yields yet another superconformal algebra.

What, though, is the physical interpretation of the field $C$ on spacetime? It is apparently not the NS 2-form field, because the generators are different and don’t seem to be unitarily equivalent. A possible guess would therefore be that it is the RR 2-form $C_2$ . The corresponding SCFT should therefore describe a D1-string instead of an F-string. This needs to be further examined. A further hint in this direction is that under a duality transformation which changes the sign of the dilaton the $C$ -field changes roles with the $B$ -field. This, and other relations among the classical SCFTs constructed here, is discussed in Part III.

Posted at December 30, 2003 1:17 PM UTC

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Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs!

Trying to follow along is hurting my poor brain, but I’m definitely trying! :)

I’m presently attempting to interpret these deformations physically. One thought I’m mulling over is what if instead of deforming the exterior derivative and its adjoint, if you deformed the states they are acting on? Kind of like Schrodinger vs Heisenberg representations.

(1)

|\psi(e)\rangle = e^{-e W} |\psi(0)\rangle

Then the deformed operators

(2)

d_e = e^{-e W} d e^{e W}

on these states would seem to be the same as the undeformed operators acting on undeformed states (kind of) :)

(3)

d_e |\psi(e)\rangle = e^{-eW} d |psi(0)\rangle.

It seems like it may be an equivalent way to view the same thing.

Yes, someone is reading your blog entries! :)

Best regards,
Eric

Posted by: Eric on December 31, 2003 1:56 AM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Eric!

You wrote:

I`m presently attempting to interpret these deformations physically.

Yes, I think that is a crucial question. As I have discussed, by using appropritate deformation operators $\mathbf{W}$ one can obtain functional representations of various superconformal algebras. Except for the obvious cases an important question will always be: What target space backgrounds, what target space physics, do these deformations describe?

Of course this is a familiar question: CFTs don’t necessarily come with a Lagrangian, and given some CFT one may always ask if it comes from some sigma-model.

For instance I should try to figure out if there is a supersymmetric Lagrangian that reproduces the superconformal constraints which I obtained by deforming with

(1)

\mathbf{W} = C_{(\mu,\sigma)(\nu,\sigma^\prime)} \mathcal{E}^{(\mu,\sigma)} \mathcal{E}^{(\nu,\sigma^\prime)} \,.

Having this Lagrangian would help clarify the nature of the field that I called $C_{\mu\nu}$ , I’d hope. (But there should be other ways, too, to clarify this.)

But maybe your question is more general: What do these deformations mean physically in general? The answer is that they induce certain potentials on target space.

In its most elementary form this is exhibited by the deformation of the $D=1$ , $N=2$ theory (simple supersymmetric quantum mechanics of a point particle) originally introduced by Witten in his work on Morse theory:

(2)

d \to d^W := e^{-W}\,d\, e^{W} \,.

When you compute the deformed Laplace-Beltrami operator $H = \{d^W,(d^{W})^+\}$ you see that it looks like

(3)

H = \{d,d^+\} + (grad W)^2 + {fermionic terms} \,.

The first term is just the kinetic term. The second term appears due to the deformation and it is just an ordinary scalar potential. Depending on the state that this Hamiltonian is acted on there are further contributions to the potential from the last term.

So a scalar deformation yields a scalar potential. You could now replace the scalar function $W$ with something else, for instance a 2-form $B$ . Calculating the deformation of $d$ , $d^+$ and $\{d,d^+\}$ as above shows that the resulting theory is that describing a point particle in a background with torsion. In fact this is the point particle limit of a string in a Kalb-Ramond field background.

So in general the deformations should create fields in spacetime that interact with the point/string which is described by the operators that are being deformed.

It is well known how such background field interaction is described in terms of vertex operators in the path integral formulation. There should therefore be a close relation between the deformation operators $\mathbf{W}$ that I have listed and the corresponding vertex operators. As I have mentioned, one indeed sees many similarities. In particular the scalar coefficients that enter the $\mathbf{W}$ s for gravitational, Kalb-Ramond, dilaton, and gauge field backgrounds are precisely those that enter the corresponding vertex operators. But the remianing operator structure is a little different, though not totally unrelated. Closed string backgrounds are given by deformation operators that are bilinears in the loop space form creators/annihilators, while the open string gauge field background is given by something poportional to $X^\prime$ (the tangent to the loop).

It should also be possibe to relate this to the technique of marginal deformations of CFTs, somehow.

You furthermore wrote:

One thought I`m mulling over is what if instead of deforming the exterior derivative and its adjoint, if you deformed the states they are acting on? Kind of like Schrodinger vs Heisenberg representations.

(4) $|\psi_e\rangle = \exp(-eW) |\psi_0\rangle$

Then the deformed operators

(5) $d_e = \exp(-eW) \,d\, \exp(eW)$

on these states would seem to be the same as the undeformed operators acting on undeformed states (kind of) :)

(6) $d_e |\psi_e\rangle = exp(-eW) \,d \,|psi_0\rangle.$

It seems like it may be an equivalent way to view the same thing.

Yes, definitely. Note however, that what you write in general only holds for $d_e$ , not for $d_e^+$ , or vice versa. A very crucial point of this whole deformation business is that unless $W$ is anti-hermitean the operators $d$ and $d^+$ are not subject to the same similarity transformation. That is because

(7)

d_e = e^{-e W}\,d\,e^{eW}

but

(8)

d_e^+ = e^{e W^+}\,d^+\,e^{- e W^+} \,,

which again is due to the requirement

(9)

(d_e)^+ = d_e^+ ,

that is necessary to ensure that the Hamiltonian is self-adjoint. (Here $(\cdot)^+$ is of course supposed to be the adjoint of $(\cdot)$ . I do not know how to produce the TeX \dag with itex.)

For non anti-hermitean $W$ you can choose states in a way that cancels the deformation on $d_e$ , as you noted:

(10)

d_e |\psi_e\rangle = exp(-eW) \,d \,|psi_0\rangle\,.

But for these same states there will be no such cancellation for $d_e^+$ :

(11)

d^+_e |\psi_e\rangle = exp(eW^+) \,d^+ e^{-e W^+}e^{-eW} \,|psi_0\rangle

unless it happens that $W$ is anti-hermitean $W^+ = -W$ .

Therefore the deformations where $W$ is anti-hermitean are special. They correspond to unitary transformations which affect all operators of the constraint algebra in the same way. Indeed the fact that a unitary transformation of the operators can be completely canceled by a unitary transformation of the states, the way that you have indicated, shows that these deformation don’t produce new phyics, but lead to pure gauge transformations in a generalized sense.

In my discussion I had given two mildly interesting examples of such gauge tranformations. One is the deformation

(12)

\mathbf{W} = \mathcal{E}^+\cdot M \cdot \mathcal{E}

for $M$ a real antisymmetric matrix $M^{\mathrm{T}} = -M$ . This deformation induces Lorentz rotations on the vielbein field which of course leave the metric unaffacted and hence have no effect on physics. This is the gauge freedom of local Lorentz rotations.

Another example is the deformation which creates an NS gauge field on spacetime

(13)

\mathbf{W} = -i A_{(\mu,\sigma)}X^{\prime(\mu,\sigma)} \,.

Obviously this $\mathbf{W}$ is anti-hermitean. The reason is that the closed string (and all what I said applies to the closed string only so far) does not couple to the $A_\mu$ field. So turning on this field on spacetime must not change the physics of closed strings propagating in this background - and indeed it does not. When moving this deformation from the operators to the states one sees that the deformation induces a pure phase shift on the states, which is not observable.

Posted by: Urs Schreiber on December 31, 2003 3:21 PM | Permalink | Reply to this

dagger

I do not know how to produce the TeX \dag with itex

Hmmm. That could be a good symbol to add to itex.

If you can’t find the symbol you want in the WebTeX manual, you can always use the corresponding XHTML or numeric entity (in this case, &dagger; or †).

Posted by: Jacques Distler on December 31, 2003 3:55 PM | Permalink | Reply to this

MathML entities

Oh yeah. Here’s the complete list of MathML Named Entities. Theoretically, each of these should have an itex equivalent. Alas, only a fraction of them are currently implemented. And some of them are probably implemented incorrectly (i.e., the TeX symbol gets mapped to the wrong, or even to a nonexistent, MathML entity).

There is much debugging to be done in itex, so please let me know when you find missing, or incorrect symbols. I’ll try to correct them, as time permits.

Posted by: Jacques Distler on December 31, 2003 4:07 PM | Permalink | Reply to this

Re: MathML entities

$MathML-enabled post (click for more details).$

Ok, I have replaced the $(\cdot)^+$ with $(\cdot)^&dagger;$ using &dagger; in the blog entry.

The only other thing that I have noted so far is that \; is not accepted, only \, is.

I have to run now, the party is about to start!! :-)

Have a happy new year!

Posted by: Urs Schreiber on December 31, 2003 4:45 PM | Permalink | Reply to this

Re: MathML entities

Ok, I have replaced the (•)⁺ with (•)^† using &dagger; in the blog entry.

I’ve added \dagger → &dagger; and \ddagger → &ddagger; to the itex2MML conversions.

The only other thing that I have noted so far is that \; is not accepted, only \, is.

The WebTeX manual on spaces lists what’s available in that regard. Personally, I feel we’re so far from pixel-perfect rendering of MathML, that I’m not about to spend alot of time tweaking the spacing in my formulas (or the itex2MML support for others to tweak their formulas).

I have to run now, the party is about to start!! :-)

Happy New Year!

Posted by: Jacques Distler on December 31, 2003 8:42 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs!

Thanks for your reply. I had a feeling I would get in trouble with the adjoint :)

I’m still searching for an alternative way to view this because it seems troubling to me to deform the exterior derivative. The reason being that I associate the exterior derivate with Stokes theorem, i.e. $d$ is that thing which makes Stokes’ theorem valid. If you deform $d$ , then it seems like you are also deforming the boundary map $\partial$ .

Seeing how this deformation can take you from Minkowski space to a more general semi-Riemannian manifold makes me wonder if ALL deformations can be cast into the form of a modification of the Hodge star. Something like

(1)

\star_e = \star \exp(eW)

maybe. Then with the undeformed $d$ and the deformed adjoint

(2)

d^\dagger_e = \exp(-eW) d^\dagger \exp(eW)

it would seem to me that you’d still have a completely valid supersymmetry [note: what little I know about supersymmetry is limited to a few pages of Witten’s Morse theory paper :)]. Would we be losing anything else? If you could somehow reproduce your results with a mere deformation of the Hodge star, then that would seem more natural to me and would leave Stokes’ theorem (the most beautiful theorem of all mathematics) untouched :)

Cheers!

Eric

Posted by: Eric on January 2, 2004 2:38 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Ok. I was thinking about this some more.

The Dirac operator is basically the square root of the Laplacian, but you don’t call

(1)

D = d + d^\dagger

a modified $d$ , right? Similarly, you can think of

(2)

d_K = d + i_K

as a square root of the Lie derivative along K. It is not really a modified exterior derivative either, just as the Dirac operator is not a modified exterior derivative. Is there a name for the square root of a Lie derivative? :)

Eric

Posted by: Eric on January 2, 2004 9:16 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

Is there a name for the square root of a Lie derivative? :)

Yes! It is called a supersymmetry generator! :-)

At least that’s true when the Lie derivative is along a Killing vector field.

Posted by: Urs Schreiber on January 2, 2004 9:20 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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I am tired, having spent the whole day with preparing, solving and typing exercise sheets for the course that I am tutoring. There is some backlog from christmas and I have to work ahead because I’ll be in Barcelona at the RTN Winter School on Strings, Supergravity and Gauge Fields next week. (If I recall correctly Robert Helling is involved in the organization, so maybe I’ll meet him there. (?)) But before quitting I need to do at least one expedient thing today, so let me try to say something appropriate in response to your comments:

I’m still searching for an alternative way to view this because it seems troubling to me to deform the exterior derivative. The reason being that I associate the exterior derivate with Stokes theorem, i.e. $d$ is that thing which makes Stokes’ theorem valid. If you deform d , then it seems like you are also deforming the boundary map $\partial$ .

Yes, we have talked about this in private email. If we let $\langle \omega,S\rangle$ be the pairing of the differential form $\omega$ with the chain $S$ and let $(\cdot)^\mathrm{T}$ denote the dual of an operator with respect to this pairing, so that

(1)

d^\mathrm{T} = \partial

then the deformation

(2)

d_e = e^{-eW} \,d\,e^{eW}

of the exterior derivative induces a similar deformation on the boundary operator $\partial$ which sends chains to their boundary:

(3)

\partial_e := (d_e)^\mathrm{T} = e^{eW^\mathrm{T}}\partial e^{-eW^\mathrm{T}} \,.

The deformation of the meaning of boundary and coboundary is of course completely equivalent to that of $d$ and $d^\dagger$ .

Seeing how this deformation can take you from Minkowski space to a more general semi-Riemannian manifold makes me wonder if ALL deformations can be cast into the form of a modification of the Hodge star.

I need to think more about what this does really mean. Right now I can come up only with these two observations, which are maybe related to what you have in mind:

If you have a physical system governed by constraints of the form

(4)

e^{-W}\,d\,e^W |\psi\rangle = 0

(5)

e^{W^\dagger}\,d^\dagger\,e^{-W^\dagger} |\psi\rangle = 0

and you cannot stand the sight of the first equation then you can of course always perform an isomorphism

(6)

A \to e^W \, A \, e^{-W}

on the entire algebra and send states to

(7)

|\psi\rangle \to |\tilde \psi\rangle := e^{W}|\psi\rangle \,.

The above constraints are equivalent to

(8)

d |\tilde\psi\rangle = 0

(9)

d^\dagger \, e^{-W^\dagger}e^{-W} |\tilde \psi\rangle = 0 \,.

This even preserves the supersymmetry algebra, up to the isomorphism.

If you want to preserve the form of the relation

(10)

d^\dagger = \pm \star^{-1}\, d\, \star

while deforming, then you could define

(11)

\star_e := e^{-W}\,\star\,e^{-W^\dagger} \,.

This gives

(12)

d_e^\dagger = \pm \star_e^{-1}\,d_e\,\star_e \,.

But maybe you are looking for something deeper than the above equivalences.

Posted by: Urs Schreiber on January 2, 2004 9:46 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hello!

I hope my elementary questions are not too boring for everyone.

I have no idea what “D = 1, N = 2” means, but in your paper hep-th/0311064, you say that the “D = 1, N = 2 supersymmetry algebra may be represented by operators $d$ and $d^\dagger$ , which satisfy

(1)

\{d,d\} = \{d^\dagger,d^\dagger\} = 0,\quad \{d,d^\dagger\} = \Delta,

as well as $(d)^\dagger = d^\dagger$ .”

When I then think about your example deformation that produces a semi-Riemannian manifold from flat Minkowski space, the thought that comes to my mind is that this seems like it may be achieved simply by a deformation of the Hodge star, which incorporates information about the metric. On the condition that this statement makes any sense at all (which I’m not sure it does), then I am wondering if the other example deformations that you demonstrate can also be expressed as simple deformations of the Hodge star.

The obvious choice for a deformed Hodge star would seem to be

(2)

\star_e = \star exp(eW),

which would give rise to a modified inner product

(3)

[A,B]_e := \int_M A\wedge\star_e B.

At the risk of rambling nonsensically, I’d then speculate that the adjoint exterior derivative of $d$ with respect to this modified inner product (which may not even be an inner product any more), is given by

(4)

d^\dagger_e = exp(-eW) d^\dagger exp(eW).

Unless I am missing something obvious (which is likely), then the operators $d$ and $d^\dagger_e$ satisfy the conditions above for a D = 1, N = 2 supersymmetry algeba (whatever that is) :)

If this makes any sense so far, then I’m wondering if any of the deformations you consider might also be obtained simply from $d$ and $d^\dagger_e$ ?

Phew! :)

Thanks again,

Eric

Posted by: Eric on January 3, 2004 4:38 AM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hello again,

Sorry for responding to myself, but a thought just occured to me soon after I hit the “submit” button.

I was thinking about in what situations my deformed inner product

(1)

[A,B]_e = \int_M A\wedge\star_e B

would actually be an inner product. The answer is obvious to me now. Since

(2)

[A,B]_e = [A,exp(eW) B],

it follows that a necessary condition for $[A,B]_e$ to be an inner product is that $exp(eW)$ must be Hermitian with respect to the undeformed inner product.

This seems slightly interesting because I think you mentioned that if it were anti-Hermitian, it could be gauged away so only the Hermitian part leads to new physics.

Best regards,

Eric

Posted by: Eric on January 3, 2004 5:08 AM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Eric,

you write:

I hope my elementary questions are not too boring for everyone.

Please continue asking elementary questions if you feel like it! :-) Even though I am eager to sort out the more string specific implications of what I have written in that draft (and hoping for comments by the string theorists at the Coffe Table) there are of course also more general questions raised by the insight that the original deformation by Witten which may have seemed like a mere weird mathematical trick without further meaning, turns out to be the general recipe for producing superconformal algebras for all kinds of string background fields. This fact certainly seems to make a further analysis of the ‘moral meaning’ of these deformations worthwhile. And with your insistence on clarifying the precise implication of these deformations on inner products and Hodge duality you put your finger precisely at the heart of the matter, I believe.

I have no idea what “D = 1, N = 2” means,

This designates the supersymmetry algebra of $D=1$ dimensions with $N=2$ supercharges. This is the rather trivial algebra

(1)

\{\mathbf{D}^A,\mathbf{D}^B\} = 2 \delta^{AB}\mathbf{\Delta}

where $A,B\in \{1,\cdots,N\}$ are indices, $\mathbf{D}^A$ are the $N=2$ supersymmetry generators and $\mathbf{\Delta}$ is the single (because of $D=1$ ) bosonic generator.

When I then think about your example deformation that produces a semi-Riemannian manifold from flat Minkowski space, the thought that comes to my mind is that this seems like it may be achieved simply by a deformation of the Hodge star, which incorporates information about the metric.

I see how it may seem to be this way, but I am not sure if this can be given a precise meaning. This deformation which gives curved space from flat space is both simple and a little subtle. The subtle point is that an operator $d$ which is quite independent of the metric still receives a deformation when the metric is modified. This is due to the fact that it is represented in terms of operators which are not independent of the metric, and the deformation has to take account of that. I feel that a clean mathematical formulation of this is missing and desireable, but so far the best I have come up with is what I have written in that paper. In the context of what I have written there the Hodge star operator receives no deformation under the action of $\mathbf{W}^G$ .

To see what I mean by this, use the formulation of Hodge star deformations that I have mentioned in another recent comment in this thread here: For arbitrary deformations the relation

(2)

d^\dagger = - {\bar \star} \,d\,{\bar\star}^{-1}

is preserved if we introduce the deformed Hodge star ${\bar \star}_e$ defined by

(3)

\bar \star_e := e^{\mathbf{W}^\dagger} \, \bar \star \, e^{\mathbf{W}} \,.

It is easy to see that this operator makes the follwoing relation hold true:

(4)

d^\dagger_e = - {\bar \star}_e \, d_e \, {\bar\star}_e^{-1} \,.

As I said, this holds for arbitrary deformations. So let’s see what happens in the case of gravitational deformations with $\mathbf{W} = \mathbf{W}^{(G)}$ : Due to the special form of this deformation operator together with the tracelessness condition (p. 13 of my draft or p. 10 of hep-th/0311064) one sees that $\mathbf{W}^{(B)}$ is anti Hodge self dual:

(5)

{\bar \star} \mathbf{W}^{(G)} {\bar \star} = - \mathbf{W}^{(G)} \,.

But this implies that

(6)

{\bar \star}^{(G)} = e^{\mathbf{W}^\dagger} \, {\bar \star} \, e^{\mathbf{W}} = e^{\mathbf{W}} e^{-\mathbf{W}} {\bar\star} = \bar \star

and hence the Hodge star remains invariant under the gravitational deformation. This should not be too surprising if you recall how the Hodge star operator is constructed from the ONB form creators and annihilators (as in equation (A.14) of hep-th/0311064) and that what was an ONB form creator/annihilator before the deformation is one also after the deformation. The deformation actually only affects how the coordinate basis creators/annihilators are defined in terms of the ONB creators/annihilators.

Posted by: Urs Schreiber on January 3, 2004 4:24 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

I am not involved with the RTN school. At some point I thought about attending but I will stay here in Cambridge although the program looks very interesting.

Robert

Posted by: Robert on January 6, 2004 12:41 PM | Permalink | Reply to this

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Weblog: Musings
Excerpt: A minor update to the itex2MML executable, used in my plugin. I added a few more MathML entities.
Tracked: January 1, 2004 7:52 PM

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs and everyone,

I’m writing a new comment because for some reason I am getting some kind of parsing error when I try to respond to your Jan 3, 4:24 PM UTC post.

Regarding my elementary questions…

Well, if my elementary questions are detracting from some possibly more interesting string discussions then it really would not hurt my feelings at all if you (or anyone else) asked me to take it to private email (like we usually do anyway) :) I just thought that this might be a place to discuss things in the open.

Back to semi technical things…

I know I am horrible at explaining myself clearly and I don’t know if I am getting my point across very well here either. My point is that I see the “geometrical language” as providing a nice way to separate topological concepts from metric dependent concepts. In the case of (sourcefree) Maxwell’s equations we have

(1)

dF = 0,

which is purely topological and

(2)

\delta F = 0,

which is metric dependent (I am switching to $\delta = d^\dagger$ for convenience). I don’t know if I can prove this, but I am pretty sure that any deformations of the geometry (which would seem to give rise to potentials) should be incorporated into the Hodge star via the transformation

(3)

\star_e = \star exp(eW).

In your responses, you have given a different deformation of the Hodge star, which I can’t seem to understand. In my opinion, a modified Hodge star should give rise to a modified inner product and a modified $\delta$ . I guess I am trying to push the idea that perhaps all of the deformations you present can be cast into a mere deformation of the Hodge star (and corresponding deformed inner product). It seems to me that keeping $d$ undeformed while deforming the Hodge star (and consequently the inner product and adjoint exterior derivative) would not affect the supersymmetry algebra.

I guess I should buckle down and see if this is a case for Witten’s simple deformation :)

On a slightly different note…

I was reading your hep-th/0311064 again and another thought came to me. Essentially, it seems like you are trying to use some Killing vector to define a flow of time and then you separate the equations of motion into a Schrodinger-like term plus a spatial constraint. My experience has shown me that artificially separating a “spacetime” phenomena into “space” + “time” typically ends up making things excessively complicated.

I’m sure that the following computation is not original, but I haven’t seen it before and thought it was kind of cute.

If my hunch is correct and we should not be deforming $d$ , then we can try to do some perturbation theory based on the field $F$ and the adjoint $\delta$ , i.e.

(4)

F = \sum_m F^{(m)}

and

(5)

\delta = \sum_m \delta^{(m)},

where the superscript $(m)$ denotes terms of order $m$ in some small perturbation parameter. The topological equation remains unchanged and we have

(6)

dF^{(m)} = 0

for all $m$ . However, the metric dependent equation becomes

(7)

\delta F = \delta^{(0)} F^{(0)} + [\delta^{(0)} F^{(1)} + \delta^{(1)} F^{(0)}] + O(e^2) = 0

so that

(8)

\delta^{(0)} F^{(0)} = 0,

which is not all that surprising, but the inetresting thing is we have

(9)

\delta^{(0)} F^{(1)} = -\delta^{(1)} F^{(0)}.

The physical interpretation of this is kind of neat. The first order perturbation of the electromagnetic field $F^{(1)}$ is the field due to the presence of a source derived from the deformed adjoint $\delta^{(1)}$ applied to the undeformed field $F^{(0)}$ . In other words, the deformation of $\delta$ produces a current

(10)

j^{(1)} = -\delta^{(1)} F^{(0)}

so that at first order in the perturbation we have

(11)

dF^{(1)} = 0

and

(12)

\delta^{(0)} F^{(1)} = j^{(1)}.

Of course the pattern is now obvious. The field $F^{(1)}$ is going to give rise to a new current $j^{(2)}$ and so on so that we have

(13)

dF^{(m)} = 0

and

(14)

\delta^{(0)} F^{(m)} = j^{(m)}.

This is pretty neat.

Sorry if I bored you with my reinventing the wheel, but I can’t help but think this is somehow important for what you are doing :)

Best regards,

Eric

Posted by: Eric on January 3, 2004 6:01 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Eric,

you wrote:

I’m writing a new comment because for some reason I am getting some kind of parsing error when I try to respond to your Jan 3, 4:24 PM UTC post.

Maybe there is a limit to the depth of nested comments?

I just thought that this might be a place to discuss things in the open.

Yes, certainly, I believe this is what this blog is intended for. No, you are not distracting me, for sure!

In your responses, you have given a different deformation of the Hodge star, which I can’t seem to understand. In my opinion, a modified Hodge star should give rise to a modified inner product and a modified $\delta$ . I guess I am trying to push the idea that perhaps all of the deformations you present can be cast into a mere deformation of the Hodge star (and corresponding deformed inner product). It seems to me that keeping $d$ undeformed while deforming the Hodge star (and consequently the inner product and adjoint exterior derivative) would not affect the supersymmetry algebra.

Ok. In a previous comment I had offered two interpretations of your desire to keep $d$ undeformed. It seems like you are complaining about the second one. ;-) What you write above sounds to me like it harmonizes with the first one, however.

So let’s start with the ordinary version of the deformation

(1)

d_e = e^{-\mathbf{W}}\,d\,e^{\mathbf{W}}

(2)

\delta_e = e^{\mathbf{W}^\dagger}\, \delta\,e^{-\mathbf{W}^\dagger}

and then apply the isomorphism

(3)

A \mapsto A_i := e^{\mathbf{W}}\,A\, e^{-\mathbf{W}}

to every operator in sight. This manifestly preserves all the algebraic relations between all operators and yields

(4)

d_e \mapsto d_{e,i} = d

(5)

\delta_e \mapsto \delta_{e,i} = e^{\mathbf{W}} e^{\mathbf{W}^\dagger} \, \delta \, e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} \,.

This way $d$ remains $d$ , as desired. The resulting relation between $d_{e,i}$ and $\delta_{e,i}$ can now indeed be expressed by means of a modified inner product. As you have already indicated, let

(6)

\langle \cdot|\cdot\rangle_i := \langle \cdot | e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} \cdot\rangle \,,

where on the right we have the ordinary Hodge inner product, and write $A^{\dagger_i}$ for the adjoint of an operator $A$ with respect to $\langle \cdot|\cdot\rangle_i$ .

This is nice, because $\langle\cdot|\cdot\rangle_i$ is just the inner product on the deformed states

(7)

\psi \mapsto \psi_i = e^{\mathbf{W}}\psi

that reproduces the ordinary inner product on the ordinary states.

Now, indeed, we get

(8)

\delta_{e,i} = d^{\dagger_i} \,.

As you have already pointed out, since

(9)

\langle \alpha |\beta \rangle = \int \alpha \wedge \star \beta

we have

(10)

\langle \alpha |\beta \rangle_i = \int \alpha \wedge \star e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} \beta

and hence the entire deformation may be regarded as a deformation of the Hodge star alone:

(11)

\star_i = \star e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} \,,

(12)

d_{e,i} = d \,,

(13)

\delta_{e,i} = \pm \star_i^{-1} d \star_i \,.

Yes, I agree that this looks interesting. I have to think about it.

My experience has shown me that artificially separating a “spacetime” phenomena into “space” + “time” typically ends up making things excessively complicated.

True, but sometimes it happens that one is interested in the energy of a physical system and this requires measuring it’s momentum with respect to a timelike Killing vector. This is what all the gymnastics are supposed to accomplish. And indeed things may become complicated. But I believe that the method that I am trying to propose there may improve in some situations on the only alternative method so far, which requires analyzing the GS string in light-cone gauge. For one, the method that I am proposing is applicable to arbitrary backgrounds, which need not even have a lighlike Killing vector (something that is required to even go to light cone gauge).

I still have to write up the paper where I do the prove-of-principle of this method for the case of superstrings on $AdS_3\times S^3$ . It turns out that the method does have a certain elegance. But it also has some drawbacks. It is a tradeoff.

I’m sure that the following computation is not original, but I haven’t seen it before and thought it was kind of cute.

I agree, your computation demonstrates an interesting way to look at deformations of Dirac-Kaehler equations. I would need to further think about how much of this general concept carries over to superstring dynamics.

This requires in particular a better understanding of how to obtain the quantum (normal ordered) superalgebra given the classical algebra in functional form. This seems to be a non-trivial step in general, since it is one in the cases where I fully understand it: Namely for the case of string dynamics in WZW backgrounds I know the functional form of the algebra as well as the quantized form. There it turns out that the primary bosonic fields (the bosonic currents) of the quantum SCFT are rather complicated combinations of bosonic and fermionic functional operators. Already in the case of general $G$ and $B$ backgrounds which are not group manifolds it is not clear to me how the classical functional generators translate into quantum SCFT fields.

Posted by: Urs Schreiber on January 3, 2004 7:12 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs :)

Excellent! Unless I’m confused, I think you just proved that my intuition (for once) was correct :)

It does seem that the entire deformation can be stuffed into a simple deformation of the Hodge star

(1)

\star_e = \star exp(eW).

for some Hermitian W. Note that the adjoint may also be expressed as

(2)

\delta_e = exp(-eW) \delta exp(eW).

The thing I really like about this route is that it maintains the clear distinction between topology and geo”metry”. Not to mention, it doesn’t touch Stokes’ theorem :)

Another feature is that it makes it clear that an anti-Hermitian W has no effect on the physics because if W is anti-Hermitian, then (shifting to your notation for a second)

(3)

\star_i = \star exp(-W^\dagger) exp(-W) = \star exp(W) exp(-W) = \star.

Assuming this carries over to infinite dimensional loop space, then could this mean that the dualities in string theory may be somehow related simply via deformations of a Hodge star? Ok. I am way out of my league now and I’m sure this question doesn’t even make sense :)

Best regards,

Eric

Posted by: Eric on January 4, 2004 3:27 AM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Eric wrote:

Excellent! Unless I’m confused, I think you just proved that my intuition (for once) was correct :)

Yes, you were and are perfectly right.

The good thing about discussing things is that other people, like you, will have other perspectives on the same issue. I focused on the deformed $d$ and $d^\dagger$ because they give the desired functional form of the superconformal generators the way one derives them from the sigma-model action. You are more concerned with the fundamental nature of $d$ and $d^\dagger$ and point out that it is somewhat unnatural to deform $d$ .

I believe that both points of view are viable. Indeed, as I have shown, they are, (at least ‘classically’) related by a global similarity transformation. I.e. your point of view is the correct one when one acts on all states with $e^{\mathbf{W}}$ . It is a sort of picture changing (to borrow a notion from the CFT of superconformal ghosts, where one can also write the same expressions in various equivalent forms, called ‘pictures’).

It does seem that the entire deformation can be stuffed into a simple deformation of the Hodge star

(1) $\star_e = \star exp(eW).$

for some Hermitian W. Note that the adjoint may also be expressed as

(2) $\delta_e = exp(-eW) \delta exp(eW).$

Yes. Here you are using a slightly different notation than I have been, so far. I would suggest that we stick to the $\mathbf{W}$ s the way I have used them so far (because even in the deformed-Hodge-reformulation there are the states which are related to the original ‘picture’ by applying $e^{\mathbf{W}}$ ) and maybe introduce a new letter for the deformation of the Hodge star. Maybe $V$ ? I.e.

(3)

e^{-\mathbf{V}} = e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} \,.

The thing I really like about this route is that it maintains the clear distinction between topology and geo”metry”. Not to mention, it doesn’t touch Stokes’ theorem

I agree, this has a certain appeal.

Another feature is that it makes it clear that an anti-Hermitian W has no effect on the physics because if W is anti-Hermitian, then (shifting to your notation for a second)

(4) $\star_i = \star exp(-W^\dagger) exp(-W) = \star exp(W) exp(-W) = \star.$

Yes!

Assuming this carries over to infinite dimensional loop space, then could this mean that the dualities in string theory may be somehow related simply via deformations of a Hodge star?

Yes and no. It is an old observation that for instance T-duality manifests itself as an algeba isomorphism on the vertex operator algebra, or, classically, on the algebra of operators on loop space. But this is of the form with anti-Hermitian $\mathbf{W}$ ! This was first shown in

M. Evans & I. Giannakis, T-Duality in Arbitrary String Backgrounds

(see equation (3.8))

and is reviewed in the section Duality Transformations as Inner Automorphisms and Gauge Symmetries on pp. 46 of

F. Lizzi & R. Szabo, Duality Symmetries and Noncommutative Geometry of String Spacetime.

So you are right in your guess that string dualities (at least T-duality, maybe also S-duality) can be expressed this way, but the point is, being a duality means that it leaves (in our language) the Hodge star invariant. A duality in this sense is a generalized form of a gauge symmetry.

As I show in the draft, nontrivial deformations of the Hodge star (in ‘your picture’) correspond to modifications of the background fields that are not obtainable by duality transformations. For such transformations string spectra, for instance, will change, while under a duality they remain unmodified.

Finally I want to note a relation of algebra deformations in the ‘Forgy-picture’ to what we have done in metric discrete geometry:

Recall that the deformation operator $\mathbf{W}^{(G)}$ that induces a gravitational background is of the form

(5)

\mathbf{W}^{(G)} = \mathcal{E}^\dagger \cdot M \cdot \mathcal{E}

with $M$ coming from a symmetric, traceless matrix. This guy is obviously self-adjoint, so that

(6)

e^{-\mathbf{V}} = e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} = e^{-2 \mathbf{W}^{(G)}} \,.

Now, if you look at the modified inner product produced by this deformation

(7)

\langle \cdot|\cdot \rangle_i = \langle \cdot|e^{-2 \mathbf{W}^{(G)}}\cdot \rangle

you’ll note that this is precisely the deformed inner product which we used to induce a metric on discrete spaces in section 4.5 of our draft

E. Forgy & U. Schreiber, Discrete Differential Geometry on $n$ -Diamond Complexes .

See equation (253)!

This neatly shows the big picture that is emerging here. In particular we now know the physical interpretation of using self-adjoint, invertible operators $g^{-1}$ in equation (193) of section 4.2 of that paper which are not of the form $\exp(e^\dagger \cdot \ln g \cdot e)$ . These other deformations induce other background fields!

We should really finish up this paper soon…

Posted by: Urs Schreiber on January 4, 2004 4:12 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs :)

I believe that both points of view are viable. Indeed, as I have shown, they are, (at least ‘classically’) related by a global similarity transformation. I.e. your point of view is the correct one when one acts on all states with $e^W$ .

“Give an inch, and they’ll take a mile.” :)

It seems you are saying that there is some existing model based on some existing states and that my model would be correct if you act on the existing states via $e^W$ . My question is then, “How do you know that MY states are not correct already?” Perhaps for the existing model to be correct, you need to act on MY states with $e^{-W}$ :) Perhaps the existing model is flawed *gasp!* :)

[snip of stuff about notation]

Sure, I have no problem with that. I will follow your suggestion, i.e.

(1)

exp(-V) = exp(-W^\dagger) exp(-W).

[snip of stuff about string dualities]

Ok. I think I had the wrong idea of what string dualities were. I thought that they constituted different physics, but if they are related via anti-Hermitian deformations, then it seems pretty obvious that they are merely different guises for the same physics. Did I understand this correctly? Deforming the algebra via an anti-Hermitian deformation is essentially a (generalized) gauge transformation. New physics enter only via Hermitian deformations.

As I was about to hit the “submit” button, a whacky thought occurred to me (surprise!). In the above paragraph, I used the word “physics” to describe some preconceived notion of what “physics” is and that the dualities are just (generalized) gauge transformations of some underlying “physics”. Is it possible that what I am calling “physics” above is actually what people usually refer to as “M-theory” and that the string dualities are just (generalized) gauge transformations of some underlying M-theory = “physics”?

I know the chance of this making any sense is close to nil, but if there is a shred of truth to this, then I’d speculate that M-theory should really be called EM-theory and perhaps all the different string theories are merely different (generalized) gauges of Maxwell’s equations (on loop space) :) Could it be that the differences between different string theories are analogous to different gauges in EM, e.g. Lorentz vs Coulomb? Of course this is vague and crude (as usual), but I hope you can see what I mean :)

Now, if you look at the modified inner product produced by this deformation

(2) $\langle \cdot | \cdot \rangle_i = \langle \cdot | e^{-2W^{(G)}} \cdot \rangle,$

you’ll note that this is precisely the deformed inner product which we used to induce a metric on discrete spaces in section 4.5 of our draft

Right. I knew that you could get from flat space to curved space via a deformation of the inner product in the discrete theory, so I was fairly confident it would work in the continuum. The leap of faith was in conjecturing that ALL the deformations you discuss might be expressible in this form. The cool thing is now we can probably incorporate all of your deformation work to the discrete theory directly :)

Has anyone looked into T-duality and S-duality on a lattice? :)

Best regards,

Eric

PS: In case anyone is reading this besides Urs and me, you may or may not get a kick out of the fact that I am an electrical engineer :)

Posted by: Eric on January 4, 2004 6:30 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Eric wrote:

It seems you are saying that there is some existing model based on some existing states and that my model would be correct if you act on the existing states via $e^W$ . My question is then, “How do you know that MY states are not correct already?” Perhaps for the existing model to be correct, you need to act on MY states with $e^{-W}$ :)

Yes, you could view it that way, too. It makes no difference! :-) Let me try to recapitulate what we have been discussing:

The set of constraints

(1)

e^{-\mathbf{W}}\,d\,e^{\mathbf{W}} |\psi\rangle = 0

(2)

e^{\mathbf{W}^\dagger}\,\delta\, e^{-\mathbf{W}^\dagger} |\psi\rangle = 0

is equivalent to the set of constraints

(3)

d |\phi\rangle = 0

(4)

e^{\mathbf{W}} e^{\mathbf{W}^\dagger}\,\delta\, e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} |\phi\rangle = 0

because the solutions are mapped to each other 1-1 by the relation

(5)

|\psi\rangle = e^{-\mathbf{W}}|\phi\rangle \,.

Furthermore the inner products agree in the sense that

(6)

\langle \psi_1|\psi_2\rangle = \langle \phi_1|\phi_2\rangle_i \,,

where, as before,

(7)

\langle \cdot|\cdot \rangle_i = \langle \cdot| e^{-\mathbf{W}^\dagger} e^{-\mathbf{W}} \cdot \rangle \,.

Hence the physics described within both frameworks is equivalent.

I think that for gravitational deformations the $\phi$ -picture is more natural, while for other deformations the $\psi$ -picture may be more natural. It does not really matter, however, because both are equivalent.

Ok. I think I had the wrong idea of what string dualities were. I thought that they constituted different physics, but if they are related via anti-Hermitian deformations, then it seems pretty obvious that they are merely different guises for the same physics. Did I understand this correctly?

Yes, in terms of these algebra-automorphism the dualities look pretty trivial.

Deforming the algebra via an anti-Hermitian deformation is essentially a (generalized) gauge transformation. New physics enter only via Hermitian deformations.

Yes.

Is it possible that what I am calling “physics” above is actually what people usually refer to as “M-theory” and that the string dualities are just (generalized) gauge transformations of some underlying M-theory = “physics”?

Well, it is at least conjectured that the dualities found are symmetries of the full nonperturbative theory. There are many hints, but as far as I understand this is still a conjecture. Hopefully someone else chimes in and gives you a more detailed answer than I am currently able to.

The cool thing is now we can probably incorporate all of your deformation work to the discrete theory directly :)

Yes! This is the general idea: Formulate as much of perturbative string theory as possible in terms of noncommutative geometry and then see how much of the resulting system can be kept while changing the algebra in the first slot of the spectral triple.

Has anyone looked into T-duality and S-duality on a lattice? :)

Following Evans, Giannakis, Froehlich, Lizzi, and Szabo we know that T-duality is algebraically fully captured by noting that the canonical algebra

(8)

\left[ \partial_{(\mu,\sigma)}, X^{\prime(\nu,\sigma^\prime)} \right] = \delta_\mu^\nu \delta^\prime(\sigma,\sigma^\prime)

is unaffacted by the map

(9)

\partial_{(\mu,\sigma)} \mapsto X^{\prime(\mu,\sigma)}

(10)

X^{\prime((\mu,\sigma)} \mapsto \partial_{(\mu,\sigma)} \,.

Now does this generalize to a discretized loop space? Yes, I think it does if we use the symmetric q-calculus, i.e. if we set

(11)

X^{\prime(\mu,\sigma)} = \frac{1}{2\epsilon} \left( X^{(\mu,\sigma+\epsilon)} - X^{(\mu,\sigma-\epsilon)} \right) \,.

That’s because in this case

(12)

\delta^\prime(\sigma,\sigma^\prime) = -\delta^\prime(\sigma^\prime,\sigma) \,,

as in the continuum. Therefore the above isomorphism of the canonical algebra goes through for a discretized loop space (polygon space) just as well, and hece we have T-duality in this case. (Unless I am confused and have made an error somewhere, that is.)

Posted by: Urs Schreiber on January 4, 2004 7:53 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Great! I think I am beginning to understand :) Thank you for your patience.

Another question…

If you are someone like me who is intimidated by infinite dimensional loop spaces, is it possible that by just studying deformations of the Hodge star on target space that we can still learn quite alot about these dualities?

For example, there is nothing holding us back from just deforming the regular inner product with W being a 2-form B on target space and forgetting about loop space altogether. Similarly, we can deform via Hodge duality. Is there anything interesting to be learned here?

Would this be like studying a “ground state” of string theory or something? There was something you said at some point that made me think this might be the case.

One more thing…

You commented about how my approach seems more natural for gravitational deformations but the orginal approach may be better for other deformations. Is it possible that this is due to the fact that, with the other (anti-Hermitian) deformations, my approach doesn’t even show a difference at all among the various dualities? If so, is this really a bad thing? Maybe the various dualities SHOULD be hidden :)

Have fun in Barcelona!

Eric

PS: Anyone else is certainly more than welcome to join Urs’ and my conversation.

Posted by: Eric on January 4, 2004 9:03 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Eric wrote:

If you are someone like me who is intimidated by infinite dimensional loop spaces, is it possible that by just studying deformations of the Hodge star on target space that we can still learn quite alot about these dualities?

First of all you are right that all the loop space formalism restricts consistently to target space where we are left with the point-particle limit of superstring dynamics, the superparticle. But recall that the dualities are inherently stringy in nature and are lost in the point particle limit. For instance T-duality exchanges momentum ( $\partial_{(\mu,\sigma)}$ ) with string winding ( $X^{\prime(\mu,\sigma)}$ ). This requires that we are not dealing with points, but with loops.

You commented about how my approach seems more natural for gravitational deformations but the orginal approach may be better for other deformations.

The ‘original approach’ has the advantage that at least for the cases where I understand the transition, the functional constraints in this approach translate directly to the constraints of the quantum SCFT. I don’t know if the similarity transformation involved in going from the ‘original approach’ to ‘your approach’ survives the translation to the quantized algebra. See, we have in this discussion been dealing with the ‘classical’ algebra, where we naively write down ‘Poisson brackets’ like $[\partial_{(\mu,\sigma)},X^{(\nu,\sigma^\prime)}] = \delta_{(\mu,\sigma)}^{(\nu,\sigma^\prime)}$ . But since this is really field theory there are furthermore normal-ordering effects, which may modifiy these classical relations.

Have fun in Barcelona!

Thanks! But let me clarify: I am leaving for Barcelona this week - but not before next Sunday! :-) In other words: I’ll be here this week and be gone (and probably off-line) next week, Jan. 12th-18th.

Posted by: Urs Schreiber on January 5, 2004 10:05 AM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Good morning!

But recall that the dualities are inherently stringy in nature and are lost in the point particle limit.

Well, it is hard for me to “recall” this because I never knew this. I don’t know the first thing about string theory, remember? :)

All I know is that you seem to indicate that some of the dualities in string theory may be obtained by anti-Hermitian deformations. You went through some illustrations: deforming by scalar, gravitational deformation, deforming by a 2-form, etc.

I know (or I am pretty sure) that the gravitational (or metric) deformation goes through directly if you are dealing solely with the target space (just like we did for the discrete theory). I also convinced myself that if you deform by a scalar that is proportional to time, then it is equivalent to adding an “Ohm’s Law” type loss term so that the fields decay exponentially (or grow if you get the sign wrong :)). That is kind of neat. I can explain if it is interesting enough to anyone :) I also understand how deforming by Hodge duality works. However, I haven’t (yet) been able to interpret the deformation via a 2-form B.

In other words, all of the deformations above (scalar, metric, Hodge, 2-form) can be carried out on the target space while being completely oblivious to loop space. If you did do this, you’d get a kind of non-stringy duality (I think). I was wondering if the dualities that would appear under these deformations would have anything to do with the dualities that apear in string theory.

I am leaving for Barcelona this week - but not before next Sunday! :-)

Ah hah! Great! :) For some reason I though you’d be in Barcelona this week :)

Cheers,

Eric

Posted by: Eric on January 5, 2004 2:41 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Eric wrote:

All I know is that you seem to indicate that some of the dualities in string theory may be obtained by anti-Hermitian deformations.

Yes.

You went through some illustrations: deforming by scalar, gravitational deformation, deforming by a 2-form, etc.

Wait, these are not illustrations of dualities. These are illustrations of new backgrounds unrelated by dualities (in general). The deformation operator $\mathbf{W}$ for gravity, the 2-form field, etc. is not anti-Hermitian.

The only anti-Hermitian deformation operator which induces a duality that I have explicitly mentioned is that inducing T-duality. See this comment for references.

In other words, all of the deformations above (scalar, metric, Hodge, 2-form) can be carried out on the target space while being completely oblivious to loop space.

That’s true, but since these are not dualities this does not give you any notion of string duality when ignoring that strings are loops. In fact, when you look at the above mentioned references, you’ll see that the deformation operator which induces T-duality cannot even be defined on target space, but needs loop space for its definition.

Posted by: Urs Schreiber on January 5, 2004 3:13 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs :)

You went through some illustrations: deforming by scalar, gravitational deformation, deforming by a 2-form, etc.

Wait, these are not illustrations of dualities. These are illustrations of new backgrounds unrelated by dualities (in general).

I’m obviously confused :) I was thinking that if you had an undeformed solution, you could map this to a deformed solution, but (after embarassing myself) I guess I see that this is really only possible for anti-Hermitian deformations. If the deformation is Hermitian, then the undeformed solutions do not map to deformed solutions in any obvious way (because the “physics” is different). Am I correct in understanding that what you call a “duality” is something that is related via an anti-Hermitian deformation?

Is the Hodge star anti-Hermitian? :) I know I would consider $F \to \star F$ to be a duality.

Best regards,

Eric

Posted by: Eric on January 5, 2004 5:33 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Eric wrote:

If the deformation is Hermitian, then the undeformed solutions do not map to deformed solutions in any obvious way (because the “physics” is different).

Exactly!

Am I correct in understanding that what you call a “duality” is something that is related via an anti-Hermitian deformation?

Right!

Note however, that not every anti-Hermitian deformation is called a duality. Some anti-Hermitian deformations can be interpreted instead as target space gauge transformation, like local Lorentz rotations, gauge transformations of the 2-form field and target space diffeomorphisms. It is sort of neat that these can be treated on equal footing with stringy dualities this way.

Is the Hodge star anti-Hermitian? :)

Yes, indeed, when you miltiply the Hodge star by a phase factor such as to make it idempotent then the result is anti-Hermitian. But for the $D=2$ algebra we cannot directly use the Hodge star as a deformation operator $\mathbf{W}$ , because the $\mathbf{W}$ must be of unit reparametrization weight in order to preserves the superconformal algebra. It is still possible to consider the effect of loop space Hodge duality on the background fields in a certain sense. This I have tried to work out in section 4.3 of my draft.

I know I would consider $F \to \star F$ to be a duality.

Yes, certainly this is the appropriate nomenclature if F is the field strength of a YM theory. It is not a priori clear that this must have anything to do with Hodge duality on loop space. As it turns out, though, Hodge duality on loop space has some features which make it tempting to speculate that there is a relation. This requires further thinking! :-)

Posted by: Urs Schreiber on January 5, 2004 5:50 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hey, wait! I wrote:

But for the $D=2$ algebra we cannot directly use the Hodge star as a deformation operator $\mathbf{W}$ , because the $\mathbf{W}$ must be of unit reparametrization weight in order to preserves the superconformal algebra.

That’s imprecise. What I should have said is that $\mathbf{W}$ must commute with the reparametrization Killing Lie derivative $\mathcal{L}_K$ on loop space. But every Killing Lie derivative commutes with the Hodge star! That’s the content of equation (A.78) in my own paper! ;-) It is easy to see that the same holds true for all modes of $\mathcal{L}_K$ .

This means that we can set

(1)

\mathbf{W} = \bar \star \,.

As soon as I finish correcting these exercise sheets I may spend some more CPU time on this…

Posted by: Urs Schreiber on January 5, 2004 6:12 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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You went through some illustrations: deforming by scalar, gravitational deformation, deforming by a 2-form, etc.

Wait, these are not illustrations of dualities. These are illustrations of new backgrounds unrelated by dualities (in general).

Hold on a second. These WOULD be dualities of some kind or another if we just made sure that they were anti-Hermitian, right? For example, deforming by a scalar is a Hermitian deformation, but multiplying by $i = \sqrt{-1}$ , we get an anti-Hermitian deformation. In fact, given ANY Hermitian $W$ , we could get an anti-Hermitian $\tilde{W}$ just by multiplying by $i$ , i.e.

(1)

\tilde{W} = i W.

This is one way to get an anti-Hermitian deformation from a Hermitian one.

On the other hand, a 2-form $B$ is neither Hermitian nor anti-Hermitian. In this case, the adjoint of $B$ would be interior multiplication by the dual 2-vector, i.e.

(2)

B^\dagger = i_{#B},

where $#B$ is to 2-vector dual to the 2-form $B$ . We could then construct an anti-Hermitian deformation

(3)

W = 1/2 (B - B^\dagger).

This must CERTAINLY give rise to a non-stringy (and seemingly non-trivial) duality on target space. Is there any corresponding duality in string theory?

Best regards,

Eric

Posted by: Eric on January 5, 2004 8:27 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Eric wrote:

Hold on a second. These WOULD be dualities of some kind or another if we just made sure that they were anti-Hermitian, right?

Yes, right. For instance in that draft I mention that the antisymmetric part of the gravitational deformation operator $\mathbf{W}$ gives rise to an anti-Hermitian $\mathbf{W}$ and hence does not affect the gravitational field but merely Lorentz-rotates the vielbein.

For example, deforming by a scalar is a Hermitian deformation, but multiplying by $i = \sqrt{-1}$ , we get an anti-Hermitian deformation. In fact, given ANY Hermitian $W$ , we could get an anti-Hermitian $\tilde{W}$ just by multiplying by $i$ , i.e.

(1) $\tilde{W} = i W.$

This is one way to get an anti-Hermitian deformation from a Hermitian one.

True. I don’t know, however, if when $\mathbf{W}$ is a sensible nontrivial Hermitian deformation operator that then $i\mathbf{W}$ will be physically meaningful. This addresses a general question that needs to be better understood (at least I need to better understand it): There are billions and billions of easily-constructable admissable (i.e. rep-invariant) anti-Hermitian $\mathbf{W}$ . What do they all mean, physically?

I think if at all then this question must be addressed taking normal ordering into account, because that will change which operators are really admissable and which are not.

On the other hand, a 2-form $B$ is neither Hermitian nor anti-Hermitian. In this case, the adjoint of $B$ would be interior multiplication by the dual 2-vector, i.e.

(2) $B^\dagger = i_{#B},$

where $#B$ is to 2-vector dual to the 2-form $B$ . We could then construct an anti-Hermitian deformation

(3) $W = 1/2 (B - B^\dagger).$

This must CERTAINLY give rise to a non-stringy (and seemingly non-trivial) duality on target space. Is there any corresponding duality in string theory?

Dunno. But let me emphasize again that the transformation induced by this deformation need not necessarily be a duality in the ordinary sense. It could be just some plain old gauge transformation. I don’t know.

On the other hand, if my interpretation of $B^\dagger$ is correct, then this describes an RR $C_2$ field and the deformation

(4)

\mathbf{W} = B - B^\dagger

induces a background with both NS-NS and RR 2-form fields turned on in just such a way that the string which couples to both of them feels no effect at all.

Now, the above sentence opens a can of worms that definetly needs to be addressed:

If my interpretation of $B^\dagger$ -deformations are correct, then, with kind help by Arvind, we know that the associated SCFT must describe a D-string, not an F string.

This should imply that deforming with $B$ and $B^\dagger$ simultaneously should describe a $(p,q)$ -string, i.e. a ‘bound state’ of $p$ fundamental (ordinary) strings with $q$ D-strings. As soon as I find the time I’ll try to further clarify this. (Help is very much appreciated!) Anyway - it is natural to conjecture that by deforming with $B - B^\dagger$ one obtains a $(1,1)$ string in a background where the forces form the F-strin’s coupling to $B$ precisely cancel the forces from the coupling of the D-string to $C = -B^\dagger$ .

Posted by: Urs Schreiber on January 5, 2004 8:55 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs,

Here is a quick calculations pertaining to my last comment about non-stringy anti-Hermitian deformations on target space.

For a moment, let me assume that spacetime is a (3+1)d Lorentzian manifold (*gasp*) and let’s see what happens if we deform a solution $|F\rangle$ of Maxwell’s equations

(1)

d|F\rangle = 0

and

(2)

\delta|F\rangle = 0

by an anti-Hermitian deformation

(3)

W = \frac{1}{2} (B - B^\dagger),

where $B$ is a 2-form creation operator. Under these conditions, we have

(4)

exp(W) = 1 + \frac{1}{2}(B - B^\dagger) + \frac{1}{8}[B^2 + (B^\dagger)^2 - (B B^\dagger + B^\dagger B)]

Since this deformation is anti-Hermitian, we should be able to directly write down a solution of the deformed equations via

(5)

|F'\rangle = exp(W) |F\rangle

so that

(6)

F' = (1 - \frac{1}{8} ||B||^2) F + \frac{1}{2} B\wedge F + \frac{1}{2} i_{#B} F.

It seems like the deformation effectively saps some energy away from the 2-form $F$ and stuffs it into a 0-form and a 4-form. I wonder what happens when you throw sources into the mix? :) Could there be some interactions among the 0-form, 2-form, and 4-form?

Hmmm… :)

Eric

Posted by: Eric on January 5, 2004 9:00 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Wait, maybe I am too tired, but why do you have

(1)

exp(W) = 1 + \frac{1}{2}(B - B^\dagger) + \frac{1}{8}[B^2 + (B^\dagger)^2 - (B B^\dagger + B^\dagger B)] \,?

I understand that the pure terms stop a quadratic order, but could you give me a hint why there are no higher-order mixed terms?

Posted by: Urs Schreiber on January 5, 2004 9:09 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Yes. You are absolutely correct :)

We’ll definitely get more terms contributing (I think!), but I’m pretty sure the conclusion will be the same. We will have 3 terms: a 0-form, a 2-form, and a 4-form. Unless some miracle happens of course and we get some magical cancellations :)

I’ll think some more about this.

Eric

Posted by: Eric on January 5, 2004 9:21 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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If you are in need for some exercises ;-), how about this:

Figure out the BCH decomposition (of course the expansion won’t stop in general after the first term)

(1)

e^{B - B^\dagger} = e^{B} e^{-B^\dagger} e^{??}

The interesting thing is that a gravitational background and in general also a dilaton background will be generated this way from a $B$ and a $C = -B^\dagger$ background, because of

(2)

[ B,B^\dagger ] = \mathbf{e}^\dagger \cdot M \cdot \mathbf{e} \,.

Ok, enough for today. On my hemisphere it’s time to go to bed!

Posted by: Urs Schreiber on January 5, 2004 9:35 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Nifty :)

Unless I made a mistake (which is highly likely), I get

(1)

exp[\frac{1}{2}(B - B^\dagger)] = exp(\frac{B}{2})exp(-\frac{B^\dagger}{2})exp(\frac{1}{8}[B,B^\dagger]).

Therefore, I think this deformation gives

(2)

F' = exp(-\frac{1}{8}||B||^2)(F + \frac{1}{2}B\wedge F - \frac{1}{2}i_{#B} F)

No guarantees though :)

Regardless, we still see these inhomogeneous forms popping up.

Eric

Posted by: Eric on January 5, 2004 10:45 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Eric,

your first equation is true if $[B,[B,B^\dagger]] = 0$ (see bottom of this page). I am not sure how you get your second equation, though. But of course you are right that the result is a superposition of forms of even degree.

Posted by: Urs Schreiber on January 7, 2004 1:14 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Urs :)

Well, of course I could be mistaken, but here is the way I’m thinking:

(1)

i_{#B} (B\wedge A) = (i_{#B} B)\wedge A + B\wedge(i_{#B} A)

so that

(2)

[B^\dagger,B] = ||B||^2

is a 0-form, which commutes with everything (in the continuum). I hope this is correct.

The second equation comes from applying $exp(-\frac{B^\dagger}{2})$ and $exp(\frac{B}{2})$ to a solution $|F\rangle$ of the undeformed equations, i.e.

(3)

exp(-\frac{B^\dagger}{2}) F = F - \frac{1}{2} B^\dagger F,

since $(B^\dagger)^2 F = 0$ . Of course, I can now see another mistake

(4)

exp(\frac{B}{2}) (F - \frac{1}{2} B^\dagger F) = F + \frac{1}{2}B\wedge F - \frac{1}{2} i_{#B} F - \frac{1}{8} B\wedge B\wedge(i_{#B} F)

so that

(5)

F' = exp(-\frac{1}{8}||B||^2) [F + \frac{1}{2}B\wedge F - \frac{1}{2} i_{#B} F - \frac{1}{8} B\wedge B\wedge(i_{#B} F)].

I was missing the last term before.

Anyway, I don’t really care what the exact form this deformation manifests itself as. The point is that this anti-Hermitian deformation takes a homogeneous form $|F\rangle$ and produces an inhomogeneous form $|F'\rangle$ . This is unlike any “plain old gauge transformation” I have ever seen. So I am wondering if it is ok to call this a duality? If it is a duality (rather than a plain old gauge transformation), then I am furthermore wondering if this duality has any counterpart in string theory. That is really the point. I am still trying to see how much string theory I can learn just by studying deformation of Maxwell’s equations :)

Best regards,

Eric

Posted by: Eric on January 7, 2004 3:29 PM | Permalink | Reply to this

Re: Classical deformations of 2d SCFTs - Part II: Algebra Isomorphisms

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Hi Eric,

sorry for my slow response! You wrote:

Well, of course I could be mistaken, but here is the way I’m thinking:

(1) $i_{#B} (B\wedge A) = (i_{#B} B)\wedge A + B\wedge(i_{#B} A)$

There is furthermore a mixed term on the right hand side

(2)

\cdots + B_{\mu\nu} (i_{#dx^\mu} B) \wedge (i_{#dx^\nu} A) \,.

You conclude:

so that

(3) $[B^\dagger,B] = ||B||^2$

This is the correct scalar term. When one takes the mixed term into account then one finds one more term in the result. I believe the best way to calculate this is to write

(4)

[B^\dagger,B] = [ \frac{1}{2}B_{\mu\nu} \mathbf{e}^\mu \mathbf{e}^\nu, \frac{1}{2}B_{\mu\nu} \mathbf{e}^{\mu\dagger} \mathbf{e}^{\nu\dagger} ] \,,

and to use the canonical anticommutation relations

(5)

\{ \mathbf{e}^{\mu \dagger}, \mathbf{e}^{\nu} \} = g^{\mu\nu}

satisfied by the form creators $\mathbf{e}^{\mu \dagger}$ and the annihilators $\mathbf{e}^\mu$ .

This gives

(6)

[B^\dagger,B] = \frac{1}{2}B_{\mu\nu}B^{\nu\mu} + B_{\mu\lambda}B^{\nu\lambda} \mathbf{e}^{\mu\dagger} \mathbf{e}^\nu \,.

Eric furthermore wrote:

The point is that this anti-Hermitian deformation takes a homogeneous form $|F\rangle$ and produces an inhomogeneous form $|F'\rangle$ . This is unlike any “plain old gauge transformation” I have ever seen.

Right, but this is due to the fact that we must now distinguish between gauge transformations on target space and those on parameter space. You are surely thinking about the gauge transformation $F \to F + dA$ and are observing that $F_2 \to F_0 + F_2 + F_4$ is clearly not of this kind. This is true. Still, some transformations induced by deformation operators $\mathbf{W}$ give rise to gauge transformations of the form that you are expecting to see, but for forms on target space. This is the content of section 4.1.1 of my draft.

I can imagine that the distinction between parameter space and target space could be a source of confusion at this point. Let me try to re-emphasize the concepts which are involved:

In the loop space formulation a state of the single string is an inhomogeneous differential form on loop space. The string may interact with background fields which are given (for instance), by 2-forms on target space. These induce 2-forms on loop space, which then appear roughly on the same footing as the string states themselves. (This is not too surprising, since the background fields can ultimately be understood as coherent string states).

When I say that a certain deformation induces a gauge transformation on the string theory this means (for instance), that the background field $B$ is gauge transformed $B \to B + dA$ . (For the precise relation see the draft.) Of course for general anti-Hermitian $\mathbf{W}$ the string states themselves will also transform in a way that leaves their spectrum, etc, intact, but this way will not be the usual gauge transformation of a $p$ -form, as you noted above. The point is: Even if the string states (think of the point particle limit, if you whish) do not transform as $F\to F + dA$ , the background field $B$ may transform this way, or, for general deformations, any other background field may receive a gauge transformation acoording to its nature. For instance in section 4.1.2 I discuss how certain deformations induce a diffeomorphism transformation on the metric background.

So I am wondering if it is ok to call this a duality?

Certainly not! String dualities are induced by deformation operators which involve the operator $X^\prime$ on loop space, hence none of the transformations that you can discuss in the point-particle limit can be string dualities.

Probably the reason why you are worried about this point is that I may have said something like that an anti-Hermitian deformation induces either a gauge transformation on target space fields or a string duality transformation. I think this is true if you take the concept of gauge transformation in a meaning which is general enough. For instance I would address the transformation of target space fields induced by $\mathbf{W} = \frac{1}{2}(B - B^\dagger)$ as a gauge transformation. But this is a little subtle, since, as I pointed out previously, this transformation may be unlike familiar ordinary gauge transformations, since it apparently relates fields that the fundamental string couples to ( $B$ ) with fields that the D-string couples to ( $B^\dagger$ ).

I am still trying to see how much string theory I can learn just by studying deformation of Maxwell’s equations :)

In case anyone else is reading this let me hasten to point out that the reason why this statement is not insane is that, indeed, Maxwell’s equations as well as the constraints of the NSR string both have the form of generalized Dirac-Kahler equations $(d\pm d^\dagger)|\psi\rangle = 0$ , which is discussed in this paper and this paper-to-be. This does indeed, maybe surprisingly, allow to draw a few parallels. For instance, One can indeed study the deformation induced by a Kalb-Ramond field $\mathbf{W} = B$ in the point particle limit, which is described by $N=2$ supersymmetric quantum mechanics with torsion and formally nothing but a deformed form of p-form electromagnetism. This is described for instance on p. 25 of

J. Froehlich, O. Grandjean, A. Recknagel, Supersymmetric quantum theory, non-commutative geometry, and gravitation.

Hence, if one is interested in the point particle limit of string dynamics in certain backgrounds it is sometimes possible to look at equations of the form $(e^{-W}d e^{W}\pm e^{W^\dagger}d^\dagger e^{-W^\dagger})|\psi\rangle = 0$ on target space.

Having said that, let me return to the question, how much string theory one can learn by looking at the point particle limit alone. I think the safe answer is, despite of the above mentioned facts: Not much. In particular, all the dualities become invisible when you reduce loop space to target space.

Posted by: Urs Schreiber on January 8, 2004 11:19 AM | Permalink | Reply to this

Buscher rules

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It turns out that the Buscher rules for T-duality along a single direction can very conveniently be derived by using the algebra isomorphism $\partial_{(\mu,\sigma)}\leftrightarrow X^{\prime (\mu,\sigma)}$ . This way one gets the inverse T-dual metric with virtually no computational effort.

Interestingly, in the literature one sees the Buscher rules usually given in terms of the metric with lowered indices, such as in equations (9)-(12) of

I. Bandos & B. Julia, Superfield T-duality rules in ten dimensions with one isometry.

I have always wondered why these expressions are as awkward as they are. It turns out that the inverse of the T-dual metric looks much prettier. It is simply

(1)

\tilde G^{ij} = G^{ij}

(2)

\tilde G^{iy} = T G^{-1} (G \pm B)^i{}_y = \pm T (G^{-1}B)^i{}_y

(3)

\tilde G^{yy} = T^2 (G\mp B)G^{-1}(G\pm B)_{yy} \,.

Here $G$ and $B$ are the original fields, $\tilde G$ and $\tilde B$ are the T-dual fields, and $y$ is the direction which is T-dualized.

Posted by: Urs Schreiber on January 6, 2004 6:20 PM | Permalink | Reply to this

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December 30, 2003