"Simple" Variation of the action
I'm concerned with the derivation of the quasi-local stress tensor (getting from eqn 2.4 to eqn 2.6 in this paper: http://arxiv.org/pdf/hep-th/0508218v2.pdf). As is the case with all the references I have found, the details of this calculation are glossed over as being easy, simple etc, but unfortunately, I cannot get it to work!
Essentially we need to vary the action with respect to the BOUNDARY metric which they call $g_{ij}$.
Now obviously the last term in 2.4 cancels with the <PERSON>-Hawking term in 2.5 and so really the action to be varied is just $S=\int_M d^dx d \eta \sqrt{g} ({}^{(d)}R + (\text{Tr}K)^2 -\text{Tr}K^2 -2 \Lambda)$.
Varying the boundary <PERSON> scalar and cosmological constant terms with respect to the boundary metric gives the boundary <PERSON> equations which will vanish assuming everything is on-shell (like they say it should be) and so really we are just left to vary $S=\int_M d^dx d \eta \sqrt{g} ((\text{Tr}K)^2 -\text{Tr}K^2)$.
In eqn 2.2, they say $K_{ij} = \frac{1}{2} \partial_\eta g_{ij}$ and so writing $\text{Tr}K = K^{ij} g_{ij}$, we have $\delta \text{Tr} K = K^{ij} \delta g_{ij} + \delta K^{ij} g_{ij} = K^{ij} \delta g_{ij} + \frac{1}{2} g^{ij} \delta \partial_\eta g_{ij}$. This allows us to establish that $\delta ( \text{Tr} K)^2 = 2 \text{Tr}K \delta \text{Tr} K= 2 \text{Tr} K \left(K^{ij} \delta g_{ij} + \frac{1}{2} g^{ij} \delta \partial_\eta g_{ij} \right)$.
Meanwhile the other term is $\text{Tr} K^2 = K^{mn} g_{nk} K^{kl} g_{ml}$ and so $\delta \text{Tr} K^2 = 2 K^{mi} K^{jl} g_{ml} \delta g_{ij} + 2 g^{jk} K_{kl} g^{il} \delta K_{ij} = 2 K^{mi} K^{jl} g_{ml} \delta g_{ij} + K^{ij} \delta \partial_\eta g_{ij}$
We will also need the result $\delta \sqrt{g} = \frac{1}{2} \sqrt{g} g^{ij} \delta g_{ij}$.
We then have $T^{ij}=\frac{\delta S}{\delta g_{ij}} = \frac{1}{2} \sqrt{g} ((\text{Tr} K)^2 - \text{Tr} K^2) g^{ij} \delta g_{ij} + \sqrt{g} \left[ 2 \text{Tr} K \left(K^{ij} \delta g_{ij} + \frac{1}{2} g^{ij} \delta \partial_\eta g_{ij} \right) - 2 K^{mi} K^{jl} g_{ml} \delta g_{ij} + K^{ij} \delta \partial_\eta g_{ij}\right]$
If I isolate the $\delta \partial_\eta g_{ij}$ bits, I can use Gauss Theorem to integrate over $d\eta$ and get $\int_{\partial M} d^dx \sqrt{g} (K^{ij}-\text{Tr}K g^{ij})$ which is the answer I want (eqn 2.6 in the paper), the problem is that the rest of the terms look like they should cancel but I cannot get them to. 818