11
$\begingroup$

I am trying to understand how the conformal group in two dimensions is a subgroup of the direct product of the diffeomorphism group and the group of Weyl transformations, as explained by Polchinski in 'String Theory', Volume 1.

Now, in the appendix on page 364 of the book, Polchinski defines the conformal group (Conf) in two dimensions to be the set of all holomorphic maps. On page 85, the aforementioned explanation is given as to how Conf is a subgroup of the direct product of the diffeomorphism group (diff) and the group of Weyl transformations (Weyl), denoted as (diff $\times$ Weyl) (here, diffeomorphisms refer to general coordinate transformations, and Weyl transformations are point-dependent rescalings of the metric).

This is what I understand from his explanation so far. Conf is a subgroup of diff, since we can choose the transformation function, $f$ to be holomorphic ($f(z)$). This transforms the flat metric ($ds^2=dz d\overline{z}$) as in equation (2.4.10) of Polchinski, i.e., \begin{equation} ds'^2=dz'd\overline{z}'=\frac{\partial z'}{\partial z}\frac{\partial \overline{z}'}{\partial \overline{z}}dz d\overline{z}, \end{equation} where $z'=f(z)$.

On the other hand, there are Weyl transformations which can be used to effect the same transformation given above, i.e., we rescale the flat metric $ds^2=dz d\overline{z}$ by $e^{2\omega}$, where

\begin{equation} \omega=\textrm{ln}|\partial_zf|, \end{equation} which gives \begin{equation} \begin{aligned} ds'^2&=e^{2\omega}dz d\overline{z}\\ &=e^{2 \textrm{ln}|\partial_zf|}dz d\overline{z}\\ &=|\partial_zf|^2dz d\overline{z}\\ &=\frac{\partial z'}{\partial z}\frac{\partial \overline{z}'}{\partial \overline{z}}dz d\overline{z}, \end{aligned} \end{equation} which is the same as the first equation above.

Now, naively understanding this to imply that conf is a subgroup of Weyl, we arrive at the claim, which is conf is a subgroup of diff $\times$ Weyl, which is what we wanted to show. However, (as noted in the comments below) the conformal group cannot be a subgroup of Weyl, since the transformations compose differently. So how can we say that conf is a subgroup of diff $\times$ Weyl?

Addendum: It cannot be true that conf in this case is a subgroup of diff $\times$ Weyl as a trivial extension of conf being a subgroup of diff, since nontrivial Weyl transformations were used by Polchinski in his proof.

Moreover, on page 542 of Nakahara's 'Geometry, Topology and Physics', it is explained that the the conformal Killing vectors which generate conformal transformations are identified with the overlap between diff and Weyl. They also show this by choosing a specific form for the Weyl function, $\omega$.

$\endgroup$
22
  • 1
    $\begingroup$ 1. Weyl group does not mean what you think it means. 2. I don't understand how you think that the fact that the effect of conformal transformations on the metric can be undone by Weyl transformations implies they are a subset of Weyl transformations, which explicitly only act on the metric, while a conformation transformation is always a coordinate transformation/diffeomorphism by definition. $\endgroup$
    – ACuriousMind
    Commented Jul 27, 2016 at 13:48
  • $\begingroup$ I understand that the conformal transformations are coordinate transformations, while Weyl transformations are local rescalings of the metric. My point is, why say that conformal transformations are a subgroup of diff x Weyl, when they are a subgroup of diff alone, as you say. $\endgroup$
    – Mtheorist
    Commented Jul 27, 2016 at 17:11
  • $\begingroup$ If $H\subset G$ is a subgroup, then $H\times\{1\} \subset G\times K$ is a subgroup of $G\times K$ regardless of $K$. $\endgroup$
    – ACuriousMind
    Commented Jul 27, 2016 at 17:12
  • 1
    $\begingroup$ @Mtheorist I realised I said a wrong statement: while it is certainly true that conformal transformations can be identified as a subset of Weyl transformations, it is not a subgroup. They compose differently. Also there is no discrepancy with Lubos Motl's statement as he did not claim there is not a map that identifies conformal transformations as Weyl transformations. By its definition, conformal transformations are not Weyl transformations. $\endgroup$
    – user110373
    Commented Jul 28, 2016 at 1:47
  • 1
    $\begingroup$ @Harold please answer the question, the question has been opened $\endgroup$
    – Mtheorist
    Commented Aug 16, 2016 at 16:36

1 Answer 1

4
$\begingroup$

(This is basically what has been said in the comments with more details.)

I think the confusion comes from the fact that you define a conformal transformation to be a transformation of the metric up to a scale factor. In fact the complete definition includes that this scale factor is brought up by changing the coordinates (after which the metric changes as a tensor). On the other hand a Weyl transformation transforms only the metric and not the coordinates, so it cannot be a conformal transformation.

Now a word on the relation between the conformal and Weyl symmetries. Conformal transformations are coordinate transformations and as such are a symmetry of theory which are invariant under general coordinate transformations. On the other hand if you consider a theory on a fixed background metric (for example flat) then these conformal transformations will not be a symmetry. For this to be the case you need more: you need that the original diffeomorphism invariant theory is invariant under the Weyl symmetry. This allows you to compensate the transformation of the background metric (since it is not a dynamical field anymore) and as a consequence to get an invariance of the action just in term of spacetime symmetries (which are basically isometries up to Weyl rescalings). (Note that the converse is not true: conformal invariance does not necessarily implies Weyl invariance.) Weyl transformations are useful to obtain a conformal theory from a diffeomorphism invariant theory, but there conformal theory can be built in more general ways. In particular Weyl transformations cannot be a subset of conformal transformations since the latter are coordinate transformations while the first ones are not. In some way you can understand conformal transformations as the isometries of a diffeo invariant theory up to Weyl transformations.

Concerning the discussion of Polchinski I think that the problem is the following. The discussion takes place in the context of gauge fixing. The gauge is fixed using diffeomorphisms and Weyl transformations. This introduces a new metric, flat here. Then he asks the question of what tranformations for this new metric keeps the same form. Since it has been obtained using diffeo and Weyl the answer is: a mixture of both. This is what I explained in the previous paragraph. The fact that he addresses a very special case makes the problem more confused. For a general 2d theory one only has diffeo and the conformal gauge is fixed up to the conformal factor. Then it is simpler to understand what is happening.

I would suggest these two papers that bring interesting insights on these questions: hep-th/9607110, 1510.08042.

$\endgroup$

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.