13
$\begingroup$

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?

$\endgroup$
4
  • 4
    $\begingroup$ 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 $\endgroup$
    – user10001
    Commented Aug 16, 2013 at 17:22
  • $\begingroup$ 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? $\endgroup$
    – user26143
    Commented Aug 16, 2013 at 20:07
  • 2
    $\begingroup$ 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 $\endgroup$
    – user10001
    Commented Aug 16, 2013 at 22:01
  • $\begingroup$ Sorry, how to see the cancellation of unphysical degree of freedom using the invariances. Would you suggest any reference? Thank you very much! $\endgroup$
    – user26143
    Commented Aug 16, 2013 at 23:21

1 Answer 1

12
$\begingroup$

In the answer below, I will only try to motivate why $\mathrm{Weyl}\times\mathrm{Diff}$ invariance is (thought to be) necessary in (bosonic) string theory.

Consider a (classical) string in a $D$-dimensional spacetime with coordinates $X^\mu$ and metric $G_{\mu\nu}$. As the string moves it defines a two dimensional worldsheet surface $S$. Let $g$ denote the 2D metric induced on the surface from the spacetime metric $G$. The area of the surface measured wrt the metric $g$ serves as the (Nambu-Goto) action of a classical string. Parametrizing the surface with some coordinates $\sigma_1,\sigma_2$ we can write the induced metric as

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

and the Nambu-Goto action can be written as

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

To define this action we needed to choose coordinates (more precisely local coordinate charts) on the surface $S$. It is clear that the action ($\propto\int d\mathcal{A} =$ area of the surface) is independent of the choice of the coordinates $\sigma_1,\sigma_2$ on $S$. The 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 could quantize the above action, but to avoid the strange square root we introduce a different version of the action (they are equivalent at the quantum level). This is done by introducing on $S$ an independent worldsheet metric $h$. Do not confuse this with the induced metric. We may choose any metric we like, as long as it has the correct Euclidean/Lorentzian signature. It is known that the Polyakov 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. The 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 the same quantum theory of the 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. Diffeomorphism 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 the 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 2D surfaces). So in the classical theory defined by $S_P$, we can gauge away the metric $h$ completely, by gauging the continuous symmetries of Weyl and diffeomorphism invariance*. Thus the (Euclidean) partition function of the bosonic string is defined

$$Z\equiv \int \frac{[dX dh]}{V_{\mathrm{Weyl}\times \mathrm{Diff}}} \exp(-S_P[X,h])$$

If we make sure that quantization process preserves these gauge symmetries $\mathrm{Weyl}\times\mathrm{Diff}$, then in the quantum theory too they can be used to gauge away the worldsheet metric degrees of freedom.


* If you do not gauge $\mathrm{Weyl}$, the theory is still consistent: you obtain the linear dilaton CFT. Though exotic, it is still useful: see Polchinski I, $\S$3.4, or this question.

$\endgroup$
6
  • 1
    $\begingroup$ @user26143 : (and user10001) : See the difference between strings and membranes $\endgroup$
    – Trimok
    Commented Aug 17, 2013 at 7:51
  • $\begingroup$ 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? $\endgroup$
    – user26143
    Commented Aug 17, 2013 at 16:13
  • 1
    $\begingroup$ @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 $\endgroup$
    – user10001
    Commented Aug 17, 2013 at 16:49
  • $\begingroup$ "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? $\endgroup$
    – user26143
    Commented Aug 17, 2013 at 17:01
  • $\begingroup$ @user26143 I meant its independent of choice of coodinates $\sigma_1,\sigma_2$ on the surface $S$ not of $X^\mu$ $\endgroup$
    – user10001
    Commented Aug 17, 2013 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.