Residual conformal symmetry in GSW

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In summary, the conversation discusses the combination of reparametrization and Weyl scaling and how they affect the chosen gauge in the GSW equation (2.1.44). The speaker attempts to prove that the chosen gauge h^ab=n^ab remains unchanged under this combination, but their reasoning is flawed. The responder points out that the correct formula for the transformation of a tensor under small diffeomorphisms is d^aA^b + d^bA^a, and that choosing the vector field A^a and Weyl scaling appropriately can result in a vanishing total variation of the metric.
  • #1
Rene Meyer
Hello,

I hope you don't mind this elementary question, but I got stuck with
it this evening:

GSW state in their equ. (2.1.44) that a combined reparametrisation and
a Weyl scaling obeying

d^aA^b + d^bA^a = Gamma n^ab

don't change the choosen gauge h^ab=n^ab. Now I tried to proof this.
The metric varies according to

delta n^ab = Gamma n^ab

and this should be zero. As n^ab is definately not zero, Gamma = 0.
Then d^aA^b = - d^bA^a, which automatically implies eq. (2.1.19) to
vanish too. But then, in the coordinates sigma^+ and sigma^-,

d/d^sigma^+ A^+ = 0 = d/d^sigma^- A^-

Thus the statement that A^+=A^+(sigma^+) and similarily for A^- is
wrong, though I know is it right from other sources. So what is wrong
with my line of reasoning?

Hope someone can find some time and help me.

René.

--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
 
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  • #2
On Mon, 29 Mar 2004, Rene Meyer wrote:

> GSW state in their equ. (2.1.44) that a combined reparametrisation and
> a Weyl scaling obeying
>
> d^aA^b + d^bA^a = Gamma n^ab
>
> don't change the choosen gauge h^ab=n^ab. Now I tried to proof this...


Dear Rene,

I am afraid that there are too many errors in your reasoning, and I won't
comment on your argument because already your starting point is
incorrect, and therefore all other steps are incorrect, too.

The first fact that you don't seem to appreciate (or know) is that under a
small diffeomorphism given by an infinitesimal vector field A^a, the
metric changes by

delta h^{ab} = d^a A^b + d^b A^a (###)

where "d" is the covariant derivative (one can also lower the indices if
one changes the sign). You seem to say that it (or everything) changes
either by zero or by a multiple of itself, which suggests that you are not
aware of (###). It is not hard to prove (###), for example by
Taylor-expanding the general formula for the transformation of a tensor.

h^{ab} (x) = h^{a'b'} (x(x')) . M^a_a' . M^b_b.

wbere M^a_a' is the matrix of derivatives of x^a with respect to x'^a' -
in this case you want to choose x'^a = x^a + A^a (where A^a is
infinitesimal).

Let me just motivate (###). It has the right structure of indices; it is
dimensionless as the metric should be (A^a cancels the derivative) and it
uses covariant derivatives. And it is ab-symmetric. And if "A^2" depends
linearly on "x^2", for example, you know that you are "stretching" the
coordinate "x^2", and therefore "h^{22}" will change by a constant, which
is exactly what (###) says.

The tensor field d^a A^b + d^b A^a is a general tensor field, but if you
choose the vector field A^a appropriately, it may equal a multiple of the
metric that you started with (at each point of your manifold). In this
case you can "undo" the change of the metric by making a Weyl
transformation. Under a Weyl transformation, the metric changes by a
multiple of itself, i.e.

delta h^{ab} = epsilon . h^{ab} ($$$)

The total variation of the metric is the sum of the (###) piece and the
($$$) piece, and for a special choice of A^a and epsilon, this total
variation can vanish.

Best wishes
Lubos
______________________________________________________________________________
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  • #3


Hello René,

Thank you for your question. The residual conformal symmetry in GSW (Green, Schwarz, and Witten) refers to the fact that certain transformations of the fields in the theory do not change the gauge conditions that have been chosen. In your question, you are trying to prove that a combined reparametrization and Weyl scaling transformation, given by

δA^a = d^aξ + A^b d^bξ - A^b d^aη d^bη + A^a d^bη d^bη,
δn^ab = η n^ab,

do not change the chosen gauge condition h^ab = n^ab. However, your proof is incorrect. The correct variation of the gauge condition is given by

δh^ab = η n^ab + n^ab d^aξ - n^ab d^aη d^bη + n^ab d^bη d^aη.

This variation is not necessarily zero, but it can be shown that it is invariant under the above transformation, i.e. δh^ab = 0. Therefore, the gauge condition is not changed by this transformation.

In the coordinates σ^+ and σ^-, the fields A^+ and A^- are not necessarily equal to A^+(σ^+) and A^-(σ^-), but they can be expressed as functions of σ^+ and σ^-. This is because the gauge condition is not changed by the transformation, so the fields are still constrained by the same equations of motion. This does not contradict the statement that A^+ and A^- are functions of σ^+ and σ^-, it just means that they are not necessarily equal to A^+(σ^+) and A^-(σ^-).

I hope this helps clarify your question. Let me know if you have any further questions.
 

1. What is residual conformal symmetry in GSW?

Residual conformal symmetry in GSW refers to the remaining symmetry of a conformal field theory (CFT) after fixing the gauge symmetry. It is a residual symmetry because it is not a full conformal symmetry, but rather a subset that is preserved after fixing the gauge.

2. Why is residual conformal symmetry important?

Residual conformal symmetry is important because it allows us to study CFTs in a gauge-invariant way. This means that we can make calculations and predictions that are not dependent on the specific gauge choice, making them more general and applicable to a wider range of systems.

3. How is residual conformal symmetry related to gauge symmetry?

Residual conformal symmetry is related to gauge symmetry because it is the subset of conformal symmetry that remains after fixing the gauge symmetry. The gauge symmetry is a local symmetry that can be used to remove unphysical degrees of freedom in a theory, and residual conformal symmetry is what remains after this gauge fixing.

4. How does residual conformal symmetry affect the properties of a CFT?

Residual conformal symmetry affects the properties of a CFT by constraining the form of the correlation functions and operators in the theory. This symmetry can be used to derive Ward identities, which relate the correlation functions of different operators, and can give insight into the physical properties of the theory.

5. Can residual conformal symmetry be broken?

Yes, residual conformal symmetry can be broken in certain cases. This can happen when there are anomalies, which are quantum effects that violate the classical symmetries of a theory. In this case, the Ward identities derived from residual conformal symmetry may not hold, and the properties of the CFT may be different from what is predicted by the residual symmetry.

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