So, i've been reading volume 1 of Polchinski's String Theory text book and have a doubt. His first derivation of the Weyl anomaly goes as follows:

From dimensional analysis, we know that:

$$\begin{align}T_{z\bar{z}} = \frac{a_1}{2}g_{z\bar{z}}R.\end{align}\tag{3.4.9}$$

Where $T_{ab}$ is the energy-momentum tensor and $R$ is the scalar curvature of the worldsheet. Taking covariant derivatives and using the conservation of $T_{ab}$, it is direct to show that $$ \nabla^zT_{zz}=-\frac{a_1}{2}\partial_zR. \tag{3.4.11}$$

We can now fix the constant $a_1$ by comparing the Weyl transformations in both sides. Expanding around a flat world-sheet, the RHS becomes $$ a_1\partial_z\nabla^2\delta\omega \approx 4a_1\partial^2_z\partial_\bar{z}\delta\omega. \tag{3.4.12}$$

Moreover, from the $TT$ OPE one shows that, under conformal transformations, $T_{zz}$ transforms as

$$ \epsilon^{-1}\delta T_{zz} = -\frac{c}{12}\partial^3_z v^{z}(z) - 2\partial_z v^z(z)T_{zz}(z)-v^z(z)\partial_zT_{zz}(z). \tag{3.4.13}$$

Now, from previous considerations in the chapter (section 3.3) we know that this conformal transformation is equivalent to a coordinate transformation $\delta z = \epsilon v(z)$ plus a Weyl transformation $$2\delta\omega = \epsilon\partial_z v(z)+(\epsilon\partial_z v(z))^*.$$ Note that the last two terms in the equation above are just the tensor transformation under diffeomorphisms, which means that, under Weyl transformations, $T_{zz}$ transforms as $$ \delta_W T_{zz} = -\frac{c}{12}\partial^3_zv(z) = -\frac{c}{6}\partial_z^2\delta\omega,\tag{3.4.14}$$

where the fact that $v$ is holomorphic was used in the second equality. Now, since we are expanding around a flat metric, equating the Weyl tranformation of both sides of equation (3.4.11) gives

$$\begin{align} &-\frac{c}{6}\partial^z\partial^2_z\delta\omega = 4a_1\partial^2_z\partial_{\bar{z}}\delta\omega \\ &-\frac{c}{3}\partial_\bar{z}\partial^2_z\delta\omega = 4a_1\partial^2_z\partial_{\bar{z}}\delta\omega, \end{align}\tag{2}$$

and this would imply that $$ a_1 = -\frac{c}{12}, $$ which concludes his derivation.

But, in the derivation above, we used a $\delta\omega$ which is the sum of a holomorphic function with an anti-holomorphic one (remember that $2\delta\omega = \epsilon\partial_z v(z)+(\epsilon\partial_z v(z))^*$). Doesn't this means that $$ \partial_z\partial_\bar{z}\delta\omega = 0 $$ and, thus, equation (2) is satisfied regardless of the relation between $a_1$ and $c$ (since the LHS and RHS are zero)? It seems to me that this fact spoils the demonstration. Does it? If not, why?

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    $\begingroup$ a Weyl transformation, $\delta \omega$ is an arbitrary smooth function of $z,\bar{z}$. $\endgroup$ Commented Sep 16, 2021 at 4:54
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    $\begingroup$ i can see why this might be confusing; i will try to find some time to write up an answer, unless someone beats me to it. i hope it isn’t too urgent. $\endgroup$ Commented Sep 16, 2021 at 18:45
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    $\begingroup$ Oh, this would be great. Since Polchinski does two derivations (and there are others in other references), I don't think it is urgent. But I'd really like to undestand this point. Thank you! $\endgroup$
    – RodPhys
    Commented Sep 16, 2021 at 19:16
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    $\begingroup$ More on Weyl anomaly: physics.stackexchange.com/q/442846/2451 $\endgroup$
    – Qmechanic
    Commented Sep 17, 2021 at 14:20
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    $\begingroup$ I just remembered that an answer to this question was given here: physics.stackexchange.com/a/605267/83405. The OP there was also confused about whether $\delta v(z)$ (or $\delta z(z)$) was holomorphic (see also the comments below that answer). Briefly, it is holomorphic in $z$ but smooth in the base point $z_1$ where $\delta_WT_{zz}$ is evaluated. Does that answer your question? $\endgroup$ Commented Sep 18, 2021 at 7:17

1 Answer 1


What goes wrong in the above reasoning

This is a good question, the answer (as far as I can see) is a little subtle. The OP reasons that because $\delta z(z)$ is holomorphic in $z$, and also,$^{\dagger}$ $$ \delta \phi(z,\bar{z}) = \partial_z \delta z(z)+(\partial_z \delta z(z))^*, \tag{1}\label{1} $$ that therefore holomorphic derivatives of $\delta\phi$ must also be holomorphic in the coordinate value, say $z=z_1$, at which the energy-momentum variation is inserted. This is generically incorrect (i.e. $\partial_z^n\delta\phi|_{z=z_1}$ is not holomorphic for $n\geq1$) because in order for this holomorphic change of coordinates, $$ w(z) = z + \delta z(z), $$ to specifically generate a Weyl transformation of a local operator inserted at $z=z_1$, it is necessary that there is implicit $z_1,\bar{z}_1$ dependence in $\delta z(z)$ in addition to the holomorphic $z$ dependence; more fully we could therefore write $\delta z(z) = \delta z(z;z_1,\bar{z}_1)$ if the local operator whose Weyl variation we wish to compute is inserted at $z=z_1$. Let us derive this conclusion (after making a couple of comments).

Comment 1: In terms of an auxiliary real coordinate, $\sigma^a$, with $a=1,2$, another way of saying the same thing is that a holomorphic frame, $z$, depends implicitly on the point, say $\sigma_1^a$, at which the frame might be centred, as well as the point, $\sigma^a$, at which the coordinate frame $z$ is evaluated; that is, we could more fully write $z=z_{\sigma_1}(\sigma)$ and $z_1=z_{\sigma_1}(\sigma_1)$ (or $z_1=0$ if the frame is centred at $\sigma_1$).

Comment 2: Unless $\delta z(z_1;z_1,\bar{z}_1) =0 $ there will be an additional obstruction which forbids $\delta z(z;z_1,\bar{z}_1)$ and its holomorphic derivatives from being holomorphic in $z_1$, the obstruction being local Ricci curvature (if present). This obstruction is irrelevant here, because for a Weyl transformation we are to solve (\ref{1}) subject to $\delta z(z_1;z_1,\bar{z}_1)=0$, which in turn ensures that the Weyl variation does not shift the location of the inserted operator (namely the energy momentum tensor, its variation, or any other local operator whose Weyl variation we are computing).

$^\dagger$ I hope you will forgive me for writing the equations in terms of $\delta\phi\equiv 2\delta \omega$ and $\delta z(z) \equiv \epsilon v^z(z)$ rather than $\delta\omega$ and $v^z(z)$.


(I am preparing a detailed derivation that I will post here asap.)

  • $\begingroup$ This answer, in addition to the one you quoted in the comments helped a lot in clarifying the subtleties involved. Thank you! So, if I understood it well, equation (1) is valid only at the point in which $T$ is inserted, where we have set $R=0$. But, one cannot use (1) to compute mixed derivatives, and thus $\partial_{\bar{z}}\partial_z\delta\phi(z)\neq\partial^2_z\partial_{\bar{z}}v(z)+\partial^2_{\bar{z}}\partial_z(v(z))^*$ . However, one can use (1) to compute $\partial^2_z\delta\phi$ in the insertion point, as Polchinski does in eq. (3.4.14), right? $\endgroup$
    – RodPhys
    Commented Sep 18, 2021 at 19:00
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    $\begingroup$ I will clarify things further and update the above answer asap $\endgroup$ Commented Sep 18, 2021 at 19:25
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    $\begingroup$ I might have confused you a bit (I confused myself also): you can in fact use (1) for the entire calculation (as Polchinski does) provided $v(z_1)=0$ (because the curvature terms are only visible if $v(z_1)\neq0$). Setting $v(z_1)=0$ (note that $T$ and $R$ are both ultimately inserted at $z=z_1$) is in turn ok because you don't want the holomorphic change of frame, $w(z)=z+v(z)$, (which generates the Weyl variation) to shift the location of the insertions, $T$ and $R$; see also (13) and (14) in physics.stackexchange.com/a/605267/83405.) I will elaborate on the details asap here. $\endgroup$ Commented Sep 18, 2021 at 19:26
  • $\begingroup$ (If the above clarification confused you more just ignore it, I will explain it properly in the main body) $\endgroup$ Commented Sep 18, 2021 at 19:29

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