Fefferman-Graham Expansion

01-28

In this note, we will review some basic stuff regarding Fefferman-Graham expansion which is always used to obtain quantum anomaly, central charges and counterterm. We mainly focus on Einstein theory to give the basic idea.

1. FG Expansion

For d+1 Einstein theory, i.e.

$L=sqrt{-G}(R+dfrac{d(d-1)}{l^2})$

We have ansatz for solution in the form of

$ds^{2}=G_{ij}dx^{i}dx^{j}=dfrac{l^2}{4}rho^{-2}drho^{2}+rho^{-1}g_{ij}dx^{i}dx^{j}$

Its obvious we have

$Gamma^{rho}_{rhorho}=-dfrac{1}{rho}, Gamma^{rho}_{ij}=dfrac{2}{l^2}(g_{ij}-rho g_{ij}), Gamma^{i}_{rho j}=dfrac{1}{2}(-dfrac{1}{rho}delta^{i}_{j}+g_{kj}g^{ik})$

Therefore Einstein equation reads (To draw reference for most of basic work on this subject[1], we adopted conventions $R_{munu}=Gamma^{sigma}_{sigmanu,mu}+cdotcdotcdot$ , which is actually opposite to our custom.)

$g_{ij}+dfrac{1}{2}(g^{lk}g_{lk}-dfrac{d-2}{rho})g_{ij}-dfrac{1}{2rho}g^{lk}g_{lk}g_{ij} -g^{kl}g_{k(i}g_{j)l}+dfrac{l^2}{2rho}hat{R}_{ij}=0, dfrac{1}{2}g^{kl}g^{ij}g_{li}g_{kj}-g^{ik}g_{ik}=0, D_{[i}g_{j]k}=0$ where prime means derivative for $rho, D_{i}$ is covariant derivative for $g_{ij}$ and $hat{R}_{ij}$ is Ricci tensor of $g_{ij}$ .

We are interested in solutions behaviour as $rhorightarrow0$ , therefore we focus on series solution.

According to differential equation theory, for any second order differential equation

$y(rho)+p(rho)y(rho)+q(rho)y(rho)=0$ ,

Index equation calls

$alpha(alpha-1)+a_{0}alpha+b_{0}=0, where~~p=a_{0}rho^{-1}+cdotcdotcdot ,q=b_{0}rho^{-2}+cdotcdotcdot$

Call its two solutions are $alpha_{1}$ and $alpha_{2}$

When $alpha_{1}-alpha_{2}$ is not an integer, general solution is just ordinary Taylor series.
When $alpha_{1}-alpha_{2}$ is an integer, general solution must contain $Log$ term in addition to Taylor series.

Note for Einstein equation

$a_{0}=-dfrac{d-2}{2},b_{0}=0$

then $alpha_{1}-alpha_{2}=dfrac{d}{2}$

When $d$ is odd, solution is $g_{ij}=g^{(0)}_{ij}+rho g^{(2)}_{ij}+cdotcdotcdot$
When $d$ is even, previous solution breaks down at $rho^{d/2}$ , instead we have

$g_{ij}=g^{(0)}_{ij}+rho g^{(2)}_{ij}+cdotcdotcdot+rho^{d/2}g^{(d)}_{ij}+rho^{d/2}bar{g}^{(d)}Log(rho)+cdotcdotcdot$

This is called FG expansion.

2. Coefficients

We are allowed to obtain coefficients $g^{(n)}_{ij}$ order by order.

We denote

$g^{lk}g_{lk}:=A=A^{(0)}+rho A^{(1)}+cdotcdotcdot, g^{kl}g_{k(i}g_{j)l}:=B=B^{(0)}+cdotcdotcdot$

$D_{i}=D^{(0)}_{i}+cdotcdotcdot, hat{R}_{ij}=hat{R}^{(0)}_{ij}+rho hat{R}^{(1)}_{ij}+cdotcdotcdot$

where $D^{(0)}_{ij}$ is covariant derivative for $g^{(0)}_{ij}$ .

Substitute them back, rearrange order by order, we could have access to obtaining coefficients.

2.1. First Coefficient

The first coefficient is pretty easy, we have

$g^{(2)}_{ij}=dfrac{l^2}{d-2}(hat{R}^{(0)}_{ij}-dfrac{1}{2(d-1)}hat{R}^{(0)}g^{(0)}_{ij})$

where $hat{R}_{ij}^{(0)}$ and $hat{R}_{ij}^{(0)}$ is Ricci tensor and scalar curvature of $g^{(0)}_{ij}$ .

2.2. Second Coefficient

The second one is far more cumbersome, and we would like to exphibit details over this subsection.

According to Einstein equation, $O(1)$ term equation gives rise

$g^{(4)}_{ij}=dfrac{1}{4-d}(dfrac{1}{2}A^{(1)}g^{(0)}_{ij}+B^{(0)}_{ij}+dfrac{l^2}{2}hat{R}^{(1)}_{ij})$

Its naturally to find out

$A^{(1)}=-dfrac{1}{2}B^{(0)}$

where $B^{(0)}$ is trace of $B^{(0)}_{ij}$ .

Furthermore, its simple to get $B^{(0)}_{ij}$

$B^{(0)}_{ij}=dfrac{l^{4}}{(d-2)^{2}}(hat{R}^{(0)}_{ki}hat{R}^{(0)k}_{j}+dfrac{1}{4(d-1)^{2}} hat{R}^{(0)2}g^{(0)}_{ij}-dfrac{1}{d-1}hat{R}^{(0)}hat{R}^{(0)}_{ij}), B^{(0)}=dfrac{l^{4}}{(d-2)^{2}}(hat{R}^{(0)ki}hat{R}^{(0)}_{ki}-dfrac{3d-4}{4(d-1)^{2}}hat{R}^{(0)2})$

$hat{R}^{(1)}_{ij}$ is more tricky.

$hat{R}^{(1)}_{ij}=-dfrac{1}{2}D^{(0)}_{i}D^{(0)}_{j}g^{(0)kl}g^{(2)}_{kl}-dfrac{1}{2}D^{(0)}_{k}D^{(0)k}g^{(2)}_{ij} +g^{(0)kl}D^{(0)}_{k}D^{(0)}_{(i}g^{(2)}_{j)l}$

Substitute $g^{(2)}_{ij}$ , and note the last term requires further simplification: Exchange $D~~D$ to $D^{(0)k}hat{R}^{(0)}_{ki}=D^{(0)}_{i}hat{R}^{0}$

The outcome is

$hat{R}^{(1)}_{ij}=l^{2}(dfrac{1}{4(d-1)}D^{(0)}_{i}D^{(0)}_{j}hat{R}^{(0)}-dfrac{1}{d-2}( D^{(0)}_{k}D^{(0)k}hat{R}^{(0)}_{ij}-dfrac{1}{2(d-1)}D^{(0)}_{k}D^{(0)k}hat{R}^{(0)}g^{(0)}_{ij}) +dfrac{1}{d-2}hat{R}^{(0)}_{ikjl}hat{R}^{(0)kl}-dfrac{1}{d-2}hat{R}^{(0)}_{ik}hat{R}^{(0)k}_{j})$

Substitute back, and straighten it up, we end up with

$g^{(4)}_{ij}=dfrac{l^4}{d-4}(-dfrac{1}{8(d-1)}D^{(0)}_{i}D^{(0)}_{j}hat{R}^{(0)}+dfrac{1}{4(d-2)}( D^{(0)}_{k}D^{(0)k}hat{R}^{(0)}_{ij}-$

$-dfrac{1}{2(d-1)}D^{(0)}_{k}D^{(0)k}hat{R}^{(0)}g^{(0)}_{ij}) -dfrac{1}{2(d-2)}hat{R}^{(0)}_{ikjl}hat{R}^{(0)kl}+$

$dfrac{d-4}{2(d-2)^{2}}hat{R}^{(0)}_{ik}hat{R}^{(0)k}_{j} +dfrac{1}{(d-1)(d-2)^{2}}hat{R}^{(0)}hat{R}^{(0)}_{ij}+dfrac{1}{4(d-2)^{2}}hat{R}^{(0)kl}hat{R}^{(0)}_{kl} g^{(0)}_{ij}-dfrac{3d}{16(d-1)^{2}(d-2)^{2}}hat{R}^{(0)2}g^{(0)}_{ij})$ Moreover, we have

$Tr(g^{(4)})=dfrac{1}{4}Tr(g^{(2)2})$

3. Counterterm

To make variation principle well-defined, we have to add Hawking-Gibbons term

$S=-dfrac{1}{16pi}int_{M}d^{d+1}xsqrt{-G}(R+dfrac{d(d-1)}{l^2})+dfrac{1}{8pi}int_{partial M}d^{d}xsqrt{-h}K$

where $h_{ij}$ is induced metric on $partial M$ and $K$ is trace of extrinsic curvature.

Substitute metric ansatz

$S=dfrac{d}{16pi l}int d^{d}xint_{epsilon}^{infty}drhorho^{-d/2-1}sqrt{g}+dfrac{1}{8pi}int d^{d}x rho^{-d/2}(-dfrac{d}{l}sqrt{g}+dfrac{2rho}{l}partial_{rho}sqrt{g})_{rho=epsilon}$

where $epsilon$ is UV cutoff to regularize the whole theory.

After substituting series solution, divergent part appears clearly

$S_{div}=dfrac{1}{16pi}int d^{d}xsqrt{g^{(0)}}(a_{0}epsilon^{-d/2}+a_{2}epsilon^{-d/2+1}+a_{4}epsilon^{-d/2+2} cdotcdotcdot+a_{d-1}epsilon^{-1/2})$

for $d$ odd

$S_{div}=dfrac{1}{16pi}int d^{d}xsqrt{g^{(0)}}(a_{0}epsilon^{-d/2}+a_{2}epsilon^{-d/2+1}+a_{4}epsilon^{-d/2+2} cdotcdotcdot+a_{d-2}epsilon^{-1}+a_{d}Log(epsilon))$

for $d$ even, with

$a_{0}=dfrac{2(2-d)}{l},a_{2}=-dfrac{(d-1)(d-4)}{(d-2)l}Tr(g^{(2)}), a_{4}=dfrac{d^{2}-9d+6}{4(d-4)l}(Tr(g^{(2)2})-(Tr(g^{(2)}))^{2})$

for $dneq 4~~and~~2.$

$a_{2}=-dfrac{1}{l}Tr(g^{(2)})$

for $d=2$

$a_{4}=dfrac{1}{2l}(Tr(g^{(2)2})-(Tr(g^{(2)}))^{2})$

for $d=4$

The counterterm of this system to cancel the divergence is therefore

$S_{ct}=-dfrac{1}{16pi}int d^{d}xsqrt{g^{(0)}}(a_{0}epsilon^{-d/2}+a_{2}epsilon^{-d/2+1}+a_{4}epsilon^{-d/2+2} cdotcdotcdot+a_{d-1}epsilon^{-1/2})$

for odd $d$

$S_{ct}=-dfrac{1}{16pi}int d^{d}xsqrt{g^{(0)}}(a_{0}epsilon^{-d/2}+a_{2}epsilon^{-d/2+1}+a_{4}epsilon^{-d/2+2} cdotcdotcdot+a_{d-2}epsilon^{-1}+a_{d}Log(epsilon))$

for even $d$ .

Note all terms are covariant except for $Log$ , if it exists. To see this, we would like to express $g^{(0)}$ and $a_{n}$ in terms of covariant Ricci tensor and scalar curvature of $h$ .

Firstly, we have to note

$R_{ij}=rho hat{R}_{ij}$

where $R_{ij}$ is for $h_{ij}$ .

Up to order we are interested, at cutoff $epsilon$ , we have

$sqrt{g^{(0)}}=epsilon^{d/2}sqrt{-h}(1-dfrac{1}{2}epsilon Tr(g^{(2)})+dfrac{1}{8} epsilon^{2}((Tr(g^{(2)}))^{2}+Tr(g^{(2)2}))), hat{R}^{(0)}=hat{R}+rhohat{R}^{ij}g^{(2)}_{ij}$

Therefore

$Tr(g^{(2)})=dfrac{l^{2}}{epsilon}dfrac{1}{2(d-1)}(R+dfrac{l^{2}}{d-2}(R_{ij}R^{ij}-dfrac{1}{2(d-1)} R^{2})$

$Tr(g^{(2)2})=dfrac{l^{4}}{epsilon^{2}}dfrac{1}{(d-1)^{2}}(R_{ij}R^{ij}+dfrac{4-3d}{4(d-1)^{2}}R^{2})$

We end up with

$S_{ct}=-dfrac{1}{16pi l}int_{partial M}d^{d}xsqrt{-h}(2(1-d)+dfrac{l^2 R}{d-2} +dfrac{l^{4}(dR^{2}-4(d-1)R_{ij}R^{ij})}{4(d-4)(d-2)^{2}(d-1)}+cdotcdotcdot)$

for odd $d$

$S_{ct}=-dfrac{1}{16pi l}int_{partial M}d^{d}xsqrt{-h}(2(1-d)+dfrac{l^2 R}{d-2} +dfrac{l^{4}(dR^{2}-4(d-1)R_{ij}R^{ij})}{4(d-4)(d-2)^{2}(d-1)}+cdotcdotcdot+a_{d}Log(epsilon)epsilon^{d/2}l)$

for even $d$ . Note only for $d>2$ , $l^{2}$ term appears; only for $d>4$ , $l^{4}$ term appears.

Note for odd $d$ , all terms are covariant and thus independent of conformal coordinates. However, for even $d$ , Log term does depend on conformal coordinates, which will give rise to Weyl anomaly as one transforms conformal coordinates. We will discuss this in next section.

The total action is thus

$S_{renor}=S+S_{ct}$