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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, $\omega$.

Mtheorist
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  • 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.
    • – ACuriousMind Jul 27 '16 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. – Mtheorist Jul 27 '16 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 $K$. – ACuriousMind Jul 27 '16 at 17:12
  • As explained on page 542 of Nakahara's Geometry, Topology and Physics, the Conformal Killing Vectors which generate the conformal group can be identified with the OVERLAP of Diff_0 and Weyl, hence, it is highly unlikely that Weyl plays a trivial role here. – Mtheorist Jul 27 '16 at 17:22
  • 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 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 and as a consequence to get an invariance of the action just in term of spacetime symmetries. – Harold Jul 27 '16 at 21:01
  • I would suggest these two papers that bring interesting insights on these questions: hep-th/9607110, 1510.08042. – Harold Jul 27 '16 at 21:01
  • While the Weyl transformations are conceptually different from conformal transformations, one can identity a subset of Weyl transformations with conformal transformations- notice that in order for $w$ to be able to be written as $\log |\partial_z f|$, $w$ needs to be harmonic. On the other hand, conformal transformations can only give rise to $e^{2w}$ with harmonic $w$. With this identification one thinks conformal transformations as a subgroup as Weyl transformations. – user110373 Jul 27 '16 at 21:46
  • @ACuriousMind In this case, the picture should be that $H$ is defined as a subgroup of $G$ but is isomorphic to a subgroup of $K$. – user110373 Jul 27 '16 at 21:51
  • @user110373 I agree with this, which brings me back to my original question, why the discrepancy with Lubos Motl's statement? – Mtheorist Jul 27 '16 at 23:21
  • @Harold Then there is an inconsistency in Polchinski's definitions, on one hand the conformal group includes holomorphic reparametrizations together with Weyl transformations which preserve the unit metric, on the other hand he also defines it to be the set of all holomorphic reparametrizations. I will take a look at the references, thanks. – Mtheorist Jul 27 '16 at 23:25
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    @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. – user110373 Jul 28 '16 at 01:47
  • @ACuriousMind Please ignore my previous comment. – user110373 Jul 28 '16 at 01:48
  • @Harold From your reference, http://arxiv.org/abs/1510.08042, it is mentioned on page 2 that 'It is clear that Weyl invariance implies conformal invariance, but not the other way around...'. So it seems that conformal transformations are indeed a subset of Weyl transformations, as user110373 has also asserted. – Mtheorist Jul 28 '16 at 01:49
  • @Mtheorist The distinction is similar to the one between the change of coordinate matrix and the automorphism of the vector space represented by the same matrix in some coordinate. – user110373 Jul 28 '16 at 01:53
  • @user110373 Okay, so I think you have answered the question. Weyl transformations can be identified as a SUBSET but not a SUBGROUP of conformal transformations, since the composition law is different. I hope you can answer the question, hopefully with a little more mathematical detail for clarity, once this question is no longer on 'hold'. – Mtheorist Jul 28 '16 at 02:00
  • @Mtheorist: I don't understand this sentence in this way: the 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. – Harold Jul 28 '16 at 15:30
  • @Harold, I have edited the question, I think it can now be seen that one can at least map certain Weyl transformations to conformal transformations, even though Weyl transformations cannot be a subset of conformal transformations due to their different origin, as you say. – Mtheorist Jul 31 '16 at 04:18
  • Possible duplicates: http://physics.stackexchange.com/q/38138/2451 and links therein. – Qmechanic Aug 03 '16 at 07:58
  • Meta. – rob Aug 11 '16 at 05:22
  • @Mtheorist This has been opened now, please ask the comment-answers to answer as answers. – wizzwizz4 Aug 16 '16 at 10:13
  • @user110373 please answer the question, the question has been opened – Mtheorist Aug 16 '16 at 16:36
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    @Harold please answer the question, the question has been opened – Mtheorist Aug 16 '16 at 16:36