# Conformal group in 2D being a subgroup of Diff x Weyl - Polchinksi's 'String Theory'

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., $$ds'^2=dz'd\overline{z}'=\frac{\partial z'}{\partial z}\frac{\partial \overline{z}'}{\partial \overline{z}}dz d\overline{z},$$ 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

$$\omega=\textrm{ln}|\partial_zf|,$$ which gives \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} 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, $\omega$.

• 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. Commented Jul 27, 2016 at 13:48
• 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. Commented Jul 27, 2016 at 17:11
• If $H\subset G$ is a subgroup, then $H\times\{1\} \subset G\times K$ is a subgroup of $G\times K$ regardless of $K$. Commented Jul 27, 2016 at 17:12
• @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. Commented Jul 28, 2016 at 1:47
• @Harold please answer the question, the question has been opened Commented Aug 16, 2016 at 16:36