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Why Weyl invariance is important for consistent string theory

+ 5 like - 0 dislike
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This post is related to this link. I know there is a Weyl invariance for the Polyakov action at least in classical level. My question arises from obtaining effective action in string theory, such as section 7.3 in this lecture note

A consistent background of string theory must preserve Weyl invariance, which now requires $\beta_{\mu\nu}(G)=\beta_{\mu\nu}(B)=\beta(\Phi)=0 $

later the lecture note tried to find effective action by requiring vanishing beta function.

Also in p110 in Polchinski's string theory vol I,

We have emphasized that Weyl invariance is essential to the consistency of string theory.

Why Weyl invariance is so important for consistency of string theory?

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user26143
asked Aug 16, 2013 in Theoretical Physics by user26143 (360 points) [ no revision ]
In case of string theory the three components of worldsheet metric tensor are unphysical degrees of freedom. Reparametrization invariance along the two directions on a worldsheet can cancel 2 of them. To cancel the third one you need Weyl invarince. If you don't care about preserving Weyl symmetry then one of three degrees of freedom of the metric will not be canceled. This left out degree of freedom is i think called Liouville mode and corresponding string theories are known as noncritical string theories

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
Sorry I don't get it. Maybe I can start from a different question, in p22 of Polchinski's book, there is a divergence term in the zero point energy of open-string (1.3.34), $$ \frac{D-2}{2} \frac{ 2l p^+ \alpha'}{ \epsilon^2 \pi } $$. It is said "In fact, Weyl invariance requires that it be cancelled" Why? How it works?

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user26143
i too don't understand Polchinski's argument;) May be he wants to say that since the counterterm for the divergence is non Weyl invariant so the divergence itself has to do with non-Weyl invariance and so it should not be there. But whether or not the divergence has anything to do with non-Weyl invariance it should anyway be regularized ! However in any case the idea is that Weyl invariance is necessary to get rid of unphysical metric degrees of freedom and so one demands that the Weyl invariance remain preserved under quantization. May be I will write an answer after understanding it properly

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
Sorry, how to see the cancellation of unphysical degree of freedom using the invariances. Would you suggest any reference? Thank you very much!

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user26143

1 Answer

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In the answer below I will only try to motivate why Weyl+diff invariance is (thought to be) necessary in (bosonic) string theory.

Consider a (classical) string in a spacetime with coordinates $X^\mu$ and metric $G_{\mu\nu}$. As the string moves it defines a two dimensional surface $S$. Let $g$ denote the metric induced on the surface from the spacetime metric $G$. Area of the surface measured wrt the metric $g$ serves as the (Nambu Goto) action of classical string. Parametrizing the surface with some coordinates $\sigma_1,\sigma_2$ we can write $g$ as

$g_{\alpha\beta}=\partial_{\alpha}X^\mu \partial_{\beta}X^{\nu}G_{\mu\nu}$

and the action can be written as

$S_{NG}=-T\int d\sigma_1 d\sigma_2 \sqrt {-det(g)}$

For defining this action we needed to choose coordinates (more precisely local coordinate charts) on the surface $S$. It is clear that the action (~ area of the surface) is independent of the choice of the coordinates $\sigma_1,\sigma_2$ on $S$. Choice of coordinates on $S$ only serves as an auxiliary tool for describing the action conveniently rather than being a physical property of the surface. So if we quantize our string we would not want any physical observable in our quantum theory to depend upon the choice of coordinates.

We can quantize above action but for convenience we introduce a different version of the action. This is done by introducing on $S$ a metric $h$. We may choose any metric we like except that it be of signature (-1,1) {where we are assuming that spacetime metric has signature (-1,1,...,1)}. It is known that the action

$S_P=-\frac{T}{2}\int d\sigma_1d\sigma_2\sqrt{-det(h)} h^{\alpha\beta}\partial_{\alpha}X^\mu \partial_{\beta}X^{\nu}G_{\mu\nu}$

defines the same classical theory as $S_{NG}$ except for one main difference. Classical theory defined by $S_{P}$ has additional variables corresponding to the three independent components of the (symmetric) metric $h$. However we know that the physical string itself has no such degrees of freedom because we can describe its classical motion using the action $S_{NG}$ which depends only on $X^\mu$, and $G_{\mu\nu}$. Therefore if we want to get a quantum theory of string by quantizing the action $S_P$ then besides requiring that the physical observables in our quantum theory don't have any dependence on choice of coordinates we must also require that they don't depend on choice of metric $h$. In particular there should not be any physical observables corresponding to the metric degrees of freedom $h_{11},h_{12}=h_{21},h_{22}$ and so we should somehow be able to get rid of them.

To get rid of three continuous degrees of freedom we need three continuous symmetries of the action. Diffemorphism invariance allows us to change the two worldsheet coordinates arbitrarily and hence effectively gives us with two continuous symmetries. We need one more continuous symmetry of action which is given by the Weyl invariance. In two dimensions it is known that using diffeomorphism and Weyl transformations any metric can be (locally) turned into a flat metric (this follows from the fact that one can always find local isothermal coordinates on two dimensional surfaces). So in the classical theory defined by action $S_P$ we can gauge away metric $h$ using the continuous symmetries of diffeomorphism and Weyl invariance. If we make sure that quantization process preserve these gauge symmetries then in quantum theory too they can be used to gauge away the metric degrees of freedom. Also since the action doesn't have any other continuous symmetry which can help us to get rid of $h$ so preserving Weyl+diff invariance is necesarry.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
answered Aug 17, 2013 by user10001 (635 points) [ no revision ]
@user26143 : (and user10001) : See the difference between strings and membranes

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Trimok
Thanks for your answer. I have a naive question, still. I guess in the sentence "It is clear the action (~ area of the surface) is independent of the choice of coordinates" you mean the Nambu-Goto action is reparametrization invariant. But, in a naive sense, an area has dimension L^2. The area will depend on the unit we chosen (1cm^2, 0.0001m^2,etc). The invariant is not that obviously unless one shows the action is indeed reparametrization invariant. How to reconcile this difference?

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user26143
@user26143 Coordinates $\sigma_1,\sigma_2$ on the surface are chosen independently of the spacetime coordinates $X^\mu$. We could rather use coordinates $X^\mu$ themselves to denote each point of the surface and also to express its area. For example think of a 2d spherical surface in 3d space and consider a very fine triangulation of this surface. Area of sphere would be sum of the areas of the triangles and area of each triangle can be known by the spacetime coordinates of its vertices. Thus we don't need to introduce any coordinates on the surface itself to measure its area

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
"It is clear that the action (~ area of the surface) is independent of the choice of coordinates." is independent with respect to $X^{\mu}$? or what?

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user26143
@user26143 I meant its independent of choice of coodinates $\sigma_1,\sigma_2$ on the surface $S$ not of $X^\mu$

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
I see. 1cm$^2$=0.0001m$^2$ would be the answer for my question "...an area has dimension L^2. The area will depend on the unit we chosen (1cm^2, 0.0001m^2,etc)."...

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user26143

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