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What does it mean to put a CFT on different surfaces? I am initially motivated by the question of comparing same CFT on plane and cylinder but I believe this question can be generalized. Can a CFT put on every two dimensional surface? What about higher dimensions?

Qmechanic
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  • Hints: What does it mean to put a QFT on a generic manifold? What does it mean for a QFT to be a CFT? – ɪdɪət strəʊlə Aug 17 '22 at 14:11
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    As far as I know, this only makes sense if there is a path integral definition. – Bronsteinx Aug 17 '22 at 14:14
  • No. For example Marcos Mariño discusses the canonical quantisation of scalar fields in curved spacetime here https://www.marcosmarino.net/uploads/1/3/3/5/133535336/polycopie-qfcs.pdf – ɪdɪət strəʊlə Aug 17 '22 at 14:22
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    As far as I could see, that is done for free theory. Moreover, a CFT is not defined through an action. How could you use same prescriptions for a CFT ? – Bronsteinx Aug 17 '22 at 14:34
  • There are some interesting applications for CFTs on nontrivial surfaces in $d>2$. Radial quantization in higher dimensions relates the scaling dimensions of operators to the energy spectrum on a spatial $(d-1)-$sphere (Cardy, J. Phys. Math. Gen. 17, L385, 1984). In $d=3$, the F-theorem establishes the the free-energy on the Euclidean three-sphere (also related to entanglement entropy) as an RG monotone (https://arxiv.org/abs/1202.5650). It is also useful to consider $d>2$ CFTs on a spatial torus, e.g. to compare with numerics on lattice models (https://arxiv.org/abs/1701.03111). – Seth Whitsitt Aug 20 '22 at 07:02

1 Answers1

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Easiest case

Conformal transformations are locally composed of a rotation and rescaling. So if the space you're interested in is related to flat space by a Weyl rescaling factor $\Omega(x)^2$, the theory will have its $n$-point functions change according to \begin{equation} \left < \mathcal{O}_1(x_1) \dots \mathcal{O}_n(x_n) \right >_g = \prod_{i = 1}^n \Omega(x_i)^{\Delta_i} \left < \mathcal{O}_1(x_1) \dots \mathcal{O}_n(x_n) \right >_{\Omega^2 g} \quad (91) \end{equation} where I've used the numbering from TASI lectures by David Simmons-Duffin.

Harder case

Perturbations to the metric are governed by insertions of the stress tensor. So if you can go from flat space to the new space by a diffeomorphism, you can use \begin{equation} \left < \mathcal{O}_1(x_1) \dots \mathcal{O}_n(x_n) \right >_g = \left < \exp\left ( \int (g_{\mu\nu} - \eta_{\mu\nu}) T^{\mu\nu}(x) dx \right ) \mathcal{O}_1(x_1) \dots \mathcal{O}_n(x_n) \right >_\eta. \quad (8) \end{equation} The Taylor expansion of this involves infinitely many correlation functions. So in $d > 2$, all we can say is that solving a theory in flat space allows you to solve it in a deformed space. In $d = 2$, things are nicer because stress tensor insertion obey the Virasoro Ward identity. Therefore, solving for a correlator in flat space allows you to solve it in a deformed space.

Hardest case

If the curved manifold you're looking at has a non-trivial topology, the CFT might fail to be consistent there. One necessary condition called modular invariance reads \begin{equation} Z(\tau) = Z(\tau + 1) = Z(-1 / \tau) \end{equation} and is discussed in most standard references. Here, $Z$ is the torus partition function defined by \begin{equation} Z = \mathrm{Tr} \left [ e^{2\pi i \tau (L_0 - c/24)} e^{2\pi i \bar{\tau} (\bar{L}_0 - \bar{c} / 24)} \right ] \end{equation} while $\tau$ is the complex structure of the torus. To make this sufficient, one also needs a modular covariance condition to be obeyed by the one-point functions. Moreover, Moore and Seiberg showed that 2d CFTs consistent on the sphere and the torus are automatically consistent on all Riemann surfaces. In higher dimensions, the analogous problem is wide open.

Update

For a small diffeomorphism $\delta g_{\mu\nu}$, you can take the derivative of (8) at $\eta_{\mu\nu}$ to get \begin{equation} \delta \left < \mathcal{O}_1(x_1) \dots \mathcal{O}_n(x_n) \right >_\eta = \int \delta g_{\mu\nu} \left < T^{\mu\nu}(x) \mathcal{O}_1(x_1) \dots \mathcal{O}_n(x_n) \right >_\eta dx. \end{equation} This also makes the connection with (91) easy to see because in the case of a Weyl transformation $\delta g_{\mu\nu} = \omega(x) \eta_{\mu\nu}$ and the whole thing becomes an integral involving the trace of the stress tensor. Using \begin{equation} T^\mu_\mu(x) \mathcal{O}_i(x_i) = \Delta_i \delta(x - x_i) \mathcal{O}_i(x_i), \quad (92) \end{equation} we end up with an overall factor of $\sum_i \omega_i(x_i) \Delta_i$ which is exactly the result of taking $\Omega(x) = e^{\omega(x)}$ to first order.

Connor Behan
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  • Thanks for the great answer. Is (8) correct for finite weyl transformations ? Because i think the expression is not invariant under diffeomorphisms $\delta g_{\mu \nu} = \nabla_{(\mu}\xi_{\nu)}$. – Bronsteinx Aug 19 '22 at 13:40
  • Here we are dealing with changes to the background metric on which the CFT is defined. We would only expect diffeomorphisms to be symmetries of a different type of theory which includes the dynamics of the metric. Assuming this, (8) holds for general deformations including Weyl ones as the update tries to argue. – Connor Behan Aug 19 '22 at 15:18
  • I don't think this is the case. Certainly, a classical covariantized action with a non-dynamical metric would have diffeomorphism symmetry which implies that $\nabla^\mu T_{\mu \nu} = 0$. This relation also I believe holds for a CFT in a curved background. Why it holds if diffeomorphism invariance is not true ? – Bronsteinx Aug 19 '22 at 16:39
  • The relation holds for QFTs in a fixed curved background because "cousin theories" related to them are diffeomorphism invariant. In these latter theories, the metric has been promoted to a field on which transformations can act (I should not have called this "dynamical"). No one should call these objects CFTs as the conformal group is much smaller than the diffeomorphism group. CFTs are precisely what you get when you make one of these theories Weyl invariant as well and then declare the metric to be fixed. A bunch of other Phys.SE questions deal with this. – Connor Behan Aug 19 '22 at 16:59
  • Can you provide a reference? I don't think this way of perturbing metric is same as building it infinitesimally. – Bronsteinx Aug 19 '22 at 17:40
  • For example, $\nabla^\mu T_{\mu \nu} = 0$ holds in Polchinski's treatment, Vol.1 (3.4.11). Partition function is then invariant under $\delta g_{\mu \nu} = \nabla_{(\mu}\xi_{\nu)}$ while $(8)$ is not. – Bronsteinx Aug 19 '22 at 18:01
  • It holds here too. References besides DSD are https://physics.stackexchange.com/questions/7889/conformal-weyl-transformations-apparent-discrepancies-and-confusions and https://physics.stackexchange.com/questions/226464/simple-conceptual-question-conformal-field-theory – Connor Behan Aug 19 '22 at 18:06
  • Maybe it will help to call $S[\phi]$ a theory and $S[\phi, g]$ a set of theories. You can only talk about diff invariance for the latter because you cannot "apply a transformation" to $\mathrm{diag}(-1,1,1,1)$... the metric must be a variable. If this is the case and your "CFT" really is a set of theories, the answer of how to put it on a different surface is trivial... just set $g$ to be the metric of that surface. Therefore, the object you have (which I would agree is a CFT) is probably given only on a constant background and you are asking how it can be promoted to a set of theories. – Connor Behan Aug 19 '22 at 18:14
  • Without an action, I only know one way to find such a formulation. That is to regard the CFT as a list of correlators $L$ and then build a set consisting of the images of $L$ under (8) for all possible metric deformations. Diffeomorphisms will now permute elements of that set meaning the set as a whole is diff invariant as desired. – Connor Behan Aug 19 '22 at 18:19
  • Thank you for the replies. By diffeomorphism, I mean precisely $Z[g] = Z[g^\xi]$ where $\xi$ is a generator of coordinate transformation. I don't see how the conservation holds in $(8)$. If I vary the partition faction, the variation is $\int dx T_{\mu \nu} \nabla_{(\mu}\xi_{\nu)}$ which is non-zero as $\partial^\mu T_{\mu \nu} = 0$ in the flat theory. I am objecting to the finite Weyl scaling and not to the infinitesimal Weyl scaling. I believe you can calculate $T_{\mu \nu}$ for a given metric and use that metric for further scaling the metric. It won't be as simple as $(8)$. – Bronsteinx Aug 19 '22 at 18:41
  • If $g = \delta + h$ where $h$ is infinitesimal, I can calculate $<T_{\mu \nu}>{\delta+h}$: If I do a further infinitesimal deformation of the metric $h'$, partition function becomes $<1+T^{\mu \nu}h'{\mu \nu}>{\delta+h}$. You could also calculate this in the flat space, which gives $<exp(T^{\mu \nu}(h+h'){\mu \nu})>{\delta}$. Equating coeffs of $h\times h'$ gives $<T{\mu \nu}(x)>{\delta+h} = <T{\mu \nu}(x)>{\delta}+\int dy <T{\mu \nu}(x)T^{\rho \sigma}(y)h_{\rho \sigma}(y)>{\delta}$. So, I believe, for each $g$ you should calculate ${g}$ and use that to further vary metric. – Bronsteinx Aug 19 '22 at 18:51
  • Isn't the first order shift of $Z$ away from the flat theory just $\left < \int dx T_{\mu\nu} \partial_{(\mu} \xi_{\nu)} \right >$ because the change of co-ordinates is already infinitesimal? In any case, the procedure you outline sounds right whether or not it ends up being equivalent to (8). – Connor Behan Aug 19 '22 at 23:02
  • Yes, however, I had written it for an arbitrary metric $g$. – Bronsteinx Aug 20 '22 at 19:41