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Let $T_{ab}$ be the stress tensor of a 2D conformal field theory with metric, $dzd\bar{z}$. If the same conformal field theory is defined on a manifold with metric $\Omega(z,\bar{z})^{-2}dzd\bar{z}$, what is the relationship between the new stress tensor and the old stress tensor? This paper claims that the answer is,

$$T_{zz}^{new}= T_{zz}^{old}-\frac{c}{12\pi}\frac{\partial_{z}^2\Omega}{\Omega},$$

in equation 4.15 ($c$ is the central charge). There is a similar equation for the antiholomorphic component.

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  • $\begingroup$ Just wanted to add that the following question is not equivalent to mine $\endgroup$ – Sounak Sinha Jan 2 at 10:41
  • $\begingroup$ This result seems surprising to me $\endgroup$ – Wakabaloola Jan 2 at 15:22
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    $\begingroup$ No I haven't, the paper is somewhat lengthy so I think it will take me some time. It would be really helpful if you could summarise the whole argument in a short answer, if I need any more details I'll just look at the paper. $\endgroup$ – Sounak Sinha Jan 3 at 12:09
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    $\begingroup$ OK, I will write an answer that summarises the main points of the derivation and leave some details to the long paper. Although I am quite busy I will try to write it asap. In the meanwhile, do you know what the implications might be (if any) of the term $\partial_z^2\ln\Omega$ vs $(\partial_z^2\Omega)/\Omega$? $\endgroup$ – Wakabaloola Jan 3 at 12:58
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    $\begingroup$ I think some results will need to be changed slightly, but the final conclusion should still hold. $\endgroup$ – Sounak Sinha Jan 3 at 13:50
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The question

Given an energy-momentum tensor, $T_{zz}[g]$, in a general CFT with central charge, $c$, defined with respect to a metric, $g$, and conformal frame coordinates, $z$, $$ T_{zz}[g]\quad \textrm{wrt}\quad g=\rho(z,\bar{z})dz d\bar{z},\tag{1}\label{Tg} $$ how does $T_{zz}[g]$ change under a Weyl transformation, $g\rightarrow g'$, while keeping frame coordinates, $z$, fixed; i.e. what is $T_{zz}[g']$, $$ T_{zz}[g']\quad \textrm{wrt}\quad g'= e^{\delta\phi(z,\bar{z})}\rho(z,\bar{z})dz d\bar{z}.\tag{2}\label{Tg'} $$ The OP asks about the case $\rho(z,\bar{z})=1$, but I will discuss the more general case (since it requires only a tiny amount of additional effort) and require (for the derivation) that $\rho(z,\bar{z})=1$ only at a specific point, say $z=z_1$. In particular, it is always possible to choose $\rho$ such that at a point, $z=z_1$: $$ \rho(z_1,\bar{z}_1)=1,\quad {\rm and}\quad \partial_z^n\rho(z_1,\bar{z}_1)=0,\quad n>0 \tag{3}\label{HNC} $$ which includes the case of interest (where $\rho$ is $z$-independent and equal to 1 everywhere), but has an advantage that we have made no assumption about the underlying scalar curvature$^*$, $R_{(2)}$. By covariance of the final answer it will follow that the result holds for arbitrary initial conformal metric $g$ (and by a famous theorem therefore also arbitrary metric, not necessarily conformal).

An observation

To compute $T_{zz}[g']$ in terms of $T_{zz}[g]$ note primarily that classically (since we are considering a CFT) in a given chart the energy-momentum tensor is independent$^{**}$ of $\rho$ and hence also of $\delta\phi$. Therefore, any distinction between $T_{zz}[g]$ and $T_{zz}[g']$ must be quantum in nature. This means that it should be possible to rephrase the distinction between $T_{zz}[g]$ and $T_{zz}[g']$ as a change in normal ordering keeping frame coordinates, $z$, and metric component, $\rho$, fixed. This is because the only distinction between classical and quantum energy-momentum tensors is that the latter is normal ordered$^{***}$ to subtract the infinities that arise from self contractions and make it well-defined.$^{\&}$

We make the normal ordering explicit in the notation as follows: $$ \begin{aligned} T_{zz}[g]\, = \,\, :\!T^{(z)}(z_1)\!:_{z}\quad{\rm and}\quad T_{zz}[g']\, = \,\, :\!T^{(z)}(z_1)\!:_{w} \end{aligned}\tag{4}\label{notation} $$

  • The subscripts, $z$ and $w$ in $:(\dots):_{z}$ and $:(\dots):_{w}$ respectively label the normal ordering prescription which (as discussed in the footnote $^{***}$) we identify with Weyl normal ordering (WNO).
  • The superscript, $z$, in $T^{(z)}(z_1)$ labels the holomorphic frame coordinate (that we are keeping fixed under $g\rightarrow g'$).
  • The argument, $z_1$, in $T^{(z)}(z_1)$ labels the value of the holomorphic frame coordinate, $z$, at which the operator is inserted.
  • The second relation in (\ref{notation}) is only consistent with the first when $w$ is related to $z$ in a very special way (which is derived in (A) below).

An answer

We will take advantage of these observations to answer the question. The answer will consist of two steps, (A) and (B).


(A) A holomorphic change of frame induced by Weyl transformations

The first step towards computing $T_{zz}[g']$ in terms of $T_{zz}[g]$ is to search for a new conformal frame coordinate, $w$, such that $g'=\rho(w,\bar{w})dw d\bar{w}$, which according to (\ref{Tg'}) implies that at a generic point, $z$, (as opposed to $z=z_1$) on the Riemann surface: $$ e^{\delta\phi(z,\bar{z})}\rho(z,\bar{z})dz d\bar{z} = \rho(w,\bar{w})dw d\bar{w}\tag{5}\label{wz} $$ Since we want to keep the component $\rho$ fixed under $g\rightarrow g'$ it is useful to let $\rho$ have the same functional forms on the left- and right-hand sides of (\ref{wz}). Given we want to identify both $z$ and $w$ with holomorphic frame coordinates it should be possible to express $w$ as a holomorphic function of $z$. We will construct this $w(z)$ by Taylor series. Working infinitesimally is easiest, so we define a small variation $\delta z(z)$, $$ w(z) = z+\delta z(z),\tag{6}\label{w(z)} $$ at a generic point $z$, subject to $\delta z(z_1)=0$ so that $w(z_1)=z_1$ is fixed (the results in the more general case where $w(z_1)$ is not specified is given in a footnote$^{@}$). Then substitute (\ref{w(z)}) into (\ref{wz}), and expand all terms to leading order in the variation. This results in the following relation at a generic point, $z$, (see Sec. 2.4.4 of the long paper): $$ \delta\phi(z,\bar{z}) = \nabla_z\delta z(z)+\nabla_{\bar{z}}\delta\bar{z}(\bar{z})\tag{7}\label{dphi} $$ The quantity $\nabla_z$ is associated to a covariant derivative with respect to $g$, $\nabla_z\delta z = \partial_z\delta z+\delta z\,\partial_z\ln\rho$ (see Appendix D.2 of the long paper). Notice that at $z=z_1$ according to (\ref{HNC}) $\nabla_z\delta z(z_1)= \partial_z\delta z(z_1)$, with similar relations for all higher purely holomorphic (or purely anti-holomorphic) derivatives. Taking this into account, compute the $n^{\rm th}$ (for $n\geq1$) covariant derivative of $\delta \phi$ using (\ref{dphi}), and evaluate it at $z=z_1$, where covariant and partial derivatives are interchangeable. This leads to:$^{\#}$ $$ \partial_z^{n+1}\delta z(z_1)=\nabla_z^{n}\delta\phi(z_1),\qquad n\geq1\tag{8}\label{dzdeltazz1} $$ where we took into account that $\delta z(z_1)=0$. Since $w(z)$ is holomorphic in $z$ so is $\delta z(z)$, which in turn means there exists a convergent Taylor expansion. Choosing to expand around $z=z_1$, according to (\ref{w(z)}), (\ref{dphi}) and (\ref{dzdeltazz1}), to leading order in the variation,$^{****}$ $$ \boxed{ \begin{aligned} w(z)&=w(z_1)+e^{\frac{1}{2}\delta\phi(z_1)}\Big(z-z_1+\sum_{n=1}^\infty\frac{1}{(n+1)!}\nabla_z^{n}\delta\phi(z_1)(z-z_1)^{n+1}\Big) \end{aligned} }\tag{9}\label{w(z)Taylor} $$ Notice from (\ref{w(z)}) that $w(z_1)=z_1$, which ensures the point $z=z_1$ is fixed under $z\rightarrow w$. (As mentioned, the more general case where $w(z_1)$ is not specified is given in a footnote$^{@}$.) There are some points worth emphasising before proceeding:

  1. The quantity $w(z)$ is holomorphic in $z$, but it is not holomorphic in $z_1$
  2. We are neglecting an immaterial potential phase, $e^{i{\rm Im}\nabla_z\delta z(z_1)}$, (see the footnote)
  3. The relation (\ref{w(z)Taylor}) generates the holomorphic change of frame, $z\rightarrow w(z)$, induced by a Weyl transformation, $\rho\rightarrow e^{\delta\phi}\rho$

(B) Change in energy-momentum tensor under a holomorphic change of normal ordering keeping frame coordinates fixed

The change in an energy-momentum tensor induced by a holomorphic change in normal ordering keeping frame coordinates fixed was computed in a previous post (see the starred boxed equation there). More general and detailed derivations are given in Sec. 4.2, Sec. 4.3 and Sec. 4.4 of the long paper. The derivations of the following equation are given there$^{*****}$. The result for a general CFT with central charge $c$ and for an arbitrary holomorphic change in normal ordering, $z\rightarrow w(z)$ at fixed frame coordinates, z, is: $$ :\!T^{(z)}(z_1)\!:_w\,\,=\,\,:\!T^{(z)}(z_1)\!:_z-\frac{c}{12}\Big[\frac{\partial_{z}^3w}{\partial_{z}w}-\frac{3}{2}\Big(\frac{\partial_{z}^2w}{\partial_{z}w}\Big)^2\Big](z_1)\tag{10}\label{TwTz} $$ This is true for any holomorphic $w(z)$; to ensure we are computing the change induced by a Weyl transformation, $g\rightarrow g'=e^{\delta\phi}g$, we need the specific choice (\ref{w(z)Taylor}) according to which, $$ \begin{aligned} \partial_zw(z_1)&=e^{\frac{1}{2}\delta\phi(z_1)}\\ \partial_z^2w(z_1)&=e^{\frac{1}{2}\delta\phi(z_1)}\nabla_z\delta\phi(z_1)\\ \partial_z^3w(z_1)&=e^{\frac{1}{2}\delta\phi(z_1)}\nabla_z^{2}\delta\phi(z_1)\\ \end{aligned} $$ Substituting the latter into (\ref{TwTz}) and recalling the identifications (\ref{notation}) we learn that to leading order in the variation: $$ T_{zz}[g']\,=\,T_{zz}[g]-\frac{c}{12}\nabla_z^2\delta\phi(z_1)+\dots\tag{11}\label{TwTz2} $$ where the dots denote terms of higher order in $\delta\phi$. The OP phrased the question in terms of $\Omega$, related to $\delta\phi$ by: $e^{\delta\phi}=\Omega^{-2}$. In terms of $\Omega$ the relation we have been aiming at according to (\ref{TwTz2}) at $z=z_1$ (where $\rho=1$) reads: $$ \boxed{\,\,T_{zz}[g']\,=\,T_{zz}[g]+\frac{c}{3!}\nabla_z^2\ln\Omega+c\,\mathcal{O}((\ln\Omega)^2),\quad {\rm when}\quad g'=\Omega^{-2}g\,\,}\tag{12}\label{TwTz3} $$ It is important to emphasise that the derivation here has assumed $\Omega$ is close to the identity, I do not know how/if the result is modified if higher order terms in $\delta\phi$ are included. The expected corrections are of order $(\ln\Omega)^2$ as indicated. Because of (\ref{HNC}) we could also replace the covariant derivative, $\nabla_z^2$, with a partial derivative, $\partial_z^2$, when $z$ corresponds to a holomorphic normal coordinate where (\ref{HNC}) holds. But since the resulting relation (\ref{TwTz3}) is covariant it holds in any conformal coordinate system, and so also holds for arbitrary metric component $\rho(z,\bar{z})$ (generalising beyond (\ref{HNC})); and by a famous theorem it also therefore holds for arbitrary metric (not necessarily conformal, see, e.g., Sec. 2.1.2 of the long paper). Of course it also remains true for arbitrary local curvature, $R_{(2)}$. (For further details on holomorphic normal coordinates see Sec. 2.4 of the long paper.)

In the comments I mentioned I was slightly surprised by the result quoted by the OP: it clearly differs from the result I have derived here because $(\partial_z^2\Omega)/\Omega$ appears rather than $\partial_z^2\ln\Omega$. Recall however that I have assumed $\delta\phi$ is small, I don't know what higher order terms would look like. Also, the factor of $-1/2\pi$ that is absent in my derivation is presumably due to a different normalisation convention for the energy-momentum tensor (mine being consistent with Polchinski's textbook).



$^*$ Recall that $R_{(2)} = -4\rho^{-1}\partial_z\partial_{\bar{z}}\log \rho$ and notice that mixed derivatives are not affected by the above specified properties (\ref{HNC}) of $\rho$ at $z_1$. (Clearly, if instead $\rho=1$ in a finite patch rather than at a point then it must be that $R_{(2)}=0$ in that patch.) The proof of existence, that a coordinate $z$ always exists such that (\ref{HNC}) is satisfied is given in Sec. 2.4.2 in a long paper by Luest and Skliros.

$^{**}$ For a concrete example of this classical statement I remind the reader that for the Polyakov action, $T_{ab}[g] = -\frac{1}{\alpha'}(\partial_aX\cdot \partial_b X-\frac{1}{2}g_{ab}g^{cd}\partial_cX\cdot \partial_dX)$, (eqn (1.2.22) in Polchinski's textbook). From this it is immediate that classically $T_{ab}[g]_{\rm classical}=T_{ab}[g']_{\rm classical}$, which remains true for any CFT that has a classical energy-momentum tensor (by definition if you like). Quantum-mechanically there are divergences from self contractions that must be subtracted (which corresponds to a normal ordering prescription) in order to have a well-defined local energy-momentum tensor. Any Weyl dependence is then hidden in these subtraction terms and this is what we wish to expose.

$^{***}$ There are many normal ordering (or regularisation) prescriptions, such as 'conformal normal ordering' (CNO), 'Weyl normal ordering' (WNO), 'geodesic normal ordering' (GNO), 'dimensional regularisation' (DR), etc. The simplest prescription that I am aware of that does not require a regulator and hence preserves the useful analyticity properties of CFTs is perhaps CNO (which was invented by Polchinski), and in particular WNO (also invented by Polchinski) which is equivalent to CNO but subject to a specific choice of conformal coordinates, $z$, for which (\ref{HNC}) is satisfied. CNO and WNO are reviewed and developed in a pedagogical manner in the long paper.

$^{\&}$ A comment: there may be additional infinities that need subtracting in interacting CFTs in order to make the energy-momentum operator well-defined as a local composite operator, but the derivation of (\ref{TwTz}) assumes the CFTs are free - the result (\ref{TwTz}) should I expect be found to be general however given the universality of the Schwarzian derivative in association to holomorphic transformations of the energy-momentum tensor.

$^{\#}$ Notation: I am simplifying the notation slightly for brevity; it is implied that $\delta\phi(z)\equiv\delta\phi(z,\bar{z})$, and $\delta z(z)\equiv\delta z(z;z_1,\bar{z}_1)$, and therefore $\delta z(z_1)\equiv\delta z(z_1;z_1,\bar{z}_1)$. The same applies to $w(z)$, in particular $w(z)\equiv w(z;z_1,\bar{z}_1)$ and $w(z_1)\equiv w(z_1;z_1,\bar{z}_1)$. The important point is that $w(z)$ is holomorphic in $z$ but smooth in $z_1,\bar{z}_1$.

$^{****}$ The intermediate steps in the derivation of (\ref{w(z)Taylor}) are the following. (I suggest the reader derives (\ref{w(z)Taylor}), referring to the following only if necessary.) To leading order in the variation, according to (\ref{w(z)}), (\ref{dphi}) and (\ref{dzdeltazz1}), $$ \begin{aligned} w(z) &=z+\delta z(z)\\ &=z+ \sum_{n=0}^\infty \frac{1}{n!}\partial_z^n\delta z(z_1)\,(z-z_1)^n\\ &=z+\delta z(z_1)+\partial_z\delta z(z_1)(z-z_1)+\sum_{n=1}^\infty\frac{1}{(n+1)!}\partial_z^{n+1}\delta z(z_1)(z-z_1)^{n+1}\\ &=z_1+\delta z(z_1)+(1+\partial_z\delta z(z_1))(z-z_1)+\sum_{n=1}^\infty\frac{1}{(n+1)!}\nabla_z^{n}\delta\phi(z_1)(z-z_1)^{n+1}\\ &=w(z_1)+\Big(1+\frac{1}{2}\big(\nabla_z\delta z(z_1)+\nabla_{\bar{z}}\delta {\bar{z}}({\bar{z}}_1)\big)+\frac{1}{2}\big(\nabla_z\delta z(z_1)-\nabla_{\bar{z}}\delta {\bar{z}}({\bar{z}}_1)\big)\Big)(z-z_1)+\sum_{n=1}^\infty\frac{1}{(n+1)!}\nabla_z^{n}\delta\phi(z_1)(z-z_1)^{n+1}\\ &=w(z_1)+\Big(1+\frac{1}{2}\delta\phi(z_1)+i{\rm Im}\,\nabla_z\delta z(z_1)\Big)(z-z_1)+\sum_{n=1}^\infty\frac{1}{(n+1)!}\nabla_z^{n}\delta\phi(z_1)(z-z_1)^{n+1}\\ &=w(z_1)+e^{\frac{1}{2}\delta\phi(z_1)+i{\rm Im}\nabla_z\delta z(z_1)}\Big(z-z_1+\sum_{n=1}^\infty\frac{1}{(n+1)!}\nabla_z^{n}\delta\phi(z_1)(z-z_1)^{n+1}\Big) \end{aligned} $$ The phase is not globally well-defined (the obstruction being the potentially non-zero Euler number), but we can neglect it when the only combinations of operators that appear in the path integral are such that this phase cancels out. I am assuming this is the case.

$^{@}$ If we consider the more general case where $\delta z(z_1)$ (and its complex conjugate) is not required to vanish then (\ref{w(z)Taylor}) gets replaced by: $$ \begin{aligned} w(z)&=w(z_1)+e^{\frac{1}{2}\delta\phi(z_1)}\Big(z-z_1+\sum_{n=1}^\infty\frac{1}{(n+1)!}\big(\nabla_z^n\delta\phi+\frac{1}{4}\delta \bar{z}\,\nabla_z^{n-1}\!R_{(2)}\big)(z_1)(z-z_1)^{n+1}\Big) \end{aligned} \tag{13}\label{w(z)Taylorgen} $$ where (as indicated) all quantities appearing in the parenthesis with the covariant derivatives are evaluated at $z=z_1$, and there is a similar replacement in (\ref{TwTz2}), namely: $$ T_{zz}[g']\,=\,T_{zz}[g]-\frac{c}{12}\big(\nabla_z^2\delta\phi+\frac{1}{4}\delta \bar{z}\,\nabla_zR_{(2)}\big)(z_1)\tag{14}\label{TwTz2gen} $$ where $w(z_1)$ is now not required to leave $z_1$ invariant, i.e. the condition $w(z_1)=z_1$ that was assumed in (\ref{TwTz2}) is not assumed in (\ref{TwTz2gen}).

$^{*****}$ Apologies for the slightly different notation I'm using in the current post relative to that used there


5th Jan 2020 update: I just realised that the flat space version of the result (\ref{TwTz2}) derived here was derived in Polchinski vol.1, see eqn (3.4.14) there. The derivation is different (it is rather based on the observation that the conformal transformation of the energy-momentum tensor (3.4.13) consists of a coordinate transformation plus a Weyl transformation, and so identifying the coordinate transformation bit enables one to read off the Weyl piece). It is satisfying to see that the result (3.4.14) there is indeed equivalent to (\ref{TwTz2}) in the flat space limit since $\delta\omega = \frac{1}{2}\delta\phi$. Perhaps it is worth mentioning that there is another curved worldsheet derivation of (\ref{TwTz2}) which can be thought of as the curved-space analogue of Polchinski's flat-space derivation that led to (3.4.14). (Perhaps I will include that also when I find some time or if there is interest).

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  • $\begingroup$ In equation 8 did you choose $\delta\phi$ to be holomorphic in $z$? $\endgroup$ – Sounak Sinha Jan 5 at 1:46
  • $\begingroup$ @SounakSinha no, $delta \phi$ is a smooth function of $z,\bar{z}$. $\endgroup$ – Wakabaloola Jan 5 at 7:32
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    $\begingroup$ @SounakSinha eqn (8) is consistent with delta phi being smooth because of point 1 below eqn (9). notice (8) is evaluated at z1 $\endgroup$ – Wakabaloola Jan 5 at 7:37
  • $\begingroup$ @Wakabaloola What about a more direct formula (3.6.5) given in Polchinski Vol. 1? Did you try to put $\mathscr{F}=T_{zz}=-\partial X\partial X$ to obtain (3.4.14)? I failed to get the prefactor ''1/6'' in the case of scalar. $\endgroup$ – Smart Yao Feb 21 at 11:31
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    $\begingroup$ @Wakabaloola Yes, the geodesic type is quite inconvenient and I guess my failure is due to inadequate numbers of Taylor expansion of geodesic distance (complicated of course). But your approach to make the Weyl transformation to be induced by a coordinate change is quite a clever way. I apologize that I also made a comment on the earlier post you answered before, which makes some inconvenience. $\endgroup$ – Smart Yao Feb 21 at 12:29

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