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I am new to AdS/CFT. I know that the dual of the bulk metric is the CFT stress tensor but what about the dual of the bulk stress tensor? I mean in principle one can extrapolate whatever bulk fields to the boundary and then compute the stress tensor's dual on the boundary, but is there a simple form of the dual or some general properties that is independent of the type of matter fields.

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2 Answers 2

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A short answer is that the boundary limit of the bulk stress tensor is simply the boundary stress tensor. But we could say a bit more in addition, coming from the 1999 paper by Balasubramanian and Kraus.

As usual, the stress tensor of the theory is $T^{\mu\nu} = \frac{2}{\sqrt{-\gamma}} \frac{\delta S}{\delta \gamma_{\mu \nu}}$ where $\gamma$ is the boundary metric (which is the boundary limit of the bulk metric $g$). To relate this to the bulk theory, we'd need to use the AdS/CFT dictionary facts that $S = S_{CFT} = S_{bulk}$ and that by the dictionary, $S_{bulk}$ is sourced by fields on the boundary. The sources include bulk fields and the boundary metric $\gamma$. So we can write $S_{bulk}(\gamma_{\mu\nu},fields)$. This lets us write $T^{\mu\nu} = \frac{2}{\sqrt{-\gamma}} \frac{\delta S_{bulk}(\gamma_{\mu\nu},fields)}{\delta \gamma_{\mu \nu}}$ as a functional of the bulk action.

But this isn't the entire story. In the Kraus paper, they argue that bulk action has to be regularized by additional counterterms that also depend on the bulk metric, leading to an action $S_{eff}$ which is $S_{bulk}$ + additional counterterms. These counterterms lead to a lot of consistencies with AdS/CFT, including an accounting of the trace anomaly and the identification of the boundary central charge $c$ with the AdS length scale $l$, as $c = \frac{3 l}{2G}$

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  • $\begingroup$ so what is the boundary dual of the bulk stress tensor? $\endgroup$
    – Shadumu
    Commented Jun 28, 2019 at 8:25
  • $\begingroup$ The boundary limit of the bulk stress tensor is the boundary stress tensor. Is this what you're asking? $\endgroup$
    – Joe
    Commented Jun 28, 2019 at 12:32
  • $\begingroup$ I am asking the boundary dual not the boundary limit. $\endgroup$
    – Shadumu
    Commented Jun 29, 2019 at 4:14
  • $\begingroup$ A bulk field will in general be some smeared combination of boundary operators, and will be rather nonlocal. Are you familiar with the HKLL reconstruction? There would be a similar idea for the stress tensor. To obtain the bulk stress tensor from the boundary, you'd need to solve the equations of gravity given the boundary data, and it'd be a messy combination of boundary fields. $\endgroup$
    – Joe
    Commented Jun 29, 2019 at 4:51
  • $\begingroup$ so there is no simple dual then. $\endgroup$
    – Shadumu
    Commented Jun 29, 2019 at 5:35
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From our understanding of AdS/CFT, Stress Tensor at the boundary acts as the source of gravity deformation in the bulk. It is in the same spirit as Scalar operator, current as a source for Scalar and vector field respectively.

Coming back to the question, Stress Tensor of the bulk fields (Scalar, Vector) will deform the geometry. But to the leading order, It looks like Scalar field in the Pure AdS. This is the usual story with AdS/CFT correspondence.

There is one instance what I can think of where Bulk Stress Tensor gives more information in the context of Fluid/Gravity Correspondance. I am going to write directly from this paper "Nonlinear Fluid Dynamics from Gravity(arXiv:0712.2456)".

I will start from a boosted black brane solution in Eddington- Finkelstein coordinates so that Horizon is a smooth surface. These are $AdS_5$ solution. \begin{equation} \begin{split} ds^{2} = &-2u_{\mu} dx^{\mu} dr +[-r^2 f(br)u_{\mu} u_{\nu} +r^2 P_{\mu\nu}] \end{split} \end{equation} \begin{equation} f(r)=1-\frac{1}{r^4} \end{equation} And $u_{\mu}$ are the usual boost velocity normalised to one, $T=1/\pi b$ and $P^{\mu\nu}=\eta^{\mu\nu}+u^{\mu}u^{\nu}$. So the non-trivial part is to make these velocity and Temperataure to be the local field of $x$ (Which is $v,x,y,z$ as coordinates at fixed $r$ surface). Then our metric becomes \begin{equation} \begin{split} ds^{2} = &-2u_{\mu}(x) dx^{\mu} dr +[-r^2 f(b(x)r)u_{\mu}(x) u_{\nu}(x) +r^2 P_{\mu\nu}(x)] \end{split} \end{equation} So This solution is obviously not going to satisfy Einstein equation. So what these authors have done is to solve the Einstein equation order by order in the derivative expansion of these Fluid velocity varibales. \begin{equation} g=g^0(\beta^i,b)+ \epsilon g^1(\beta^i,b)+\epsilon^2 g^1(\beta^i,b)\\ \beta_i=\beta^0_i+\epsilon \beta^1_i\\ b=b^o+\epsilon b^1 \end{equation} And all the $\beta_i$ and $b$ are the functions of $x$. So Put these ansatz back in to the Einstein equation. The Einstein constraint equations at first order require that the zero order velocity and temperature fields obey the equations of perfect fluid dynamics $$\partial_{\mu}T^{\mu\nu}_0=0$$ By solving Constraint and dynamical equation, We can find the metric in the derivative expansion and these constriant equations are just Fluid dynmaics equation. $\partial_vb^0=1/3\partial_i \beta_i^0$ and $\partial_i b^0= \partial_v \beta_i^0$. Of course these results are of the first order for metric and zeroth order for velocities. But can be generalised to second order.

The relevance of this is to your question is basically the conservation of Stress Tensor in the bulk(ar fixed r hypersurface) is the constraint Einstein equation. These are inturn related to Fluid dynmaics equation written above. The validity of using fluid variables is related to the fact that in large wavelength limit any quantum field theory looks like a fluid. We are taking Bulk Einstein equation and solve it at any fixed $r$ hypersurface. This is universal behaviour and worked out in many of the cases like Charged case, Non relativistic case etc. Look at the that paper for explicit calculation or result.

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  • $\begingroup$ so in short what is the boundary dual to a bulk stress tensor, which can be the stress tensor of any matter field not necessarily scalar matter? $\endgroup$
    – Shadumu
    Commented Jun 29, 2019 at 4:12

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