Why does the trace of equation A.4 give 4α instead of 6α?

[nenyan]

Is there anyone can give me a hand?

http://arxiv.org/abs/astro-ph/0210603

When I read this paper I can not get Eq.(A.5) from (A.4). Why it is $4\alpha$? If we take the trace of Eq.(A.4), why not it give us $6\alpha$?

Eq. (A.3):

[Edited by a mentor to fix a small problem in the Latex formatting]

 

[fzero]

The trace of (A.4) is

$$\delta\hat{h}^r_{ii} + 2 \partial^2 \epsilon = 6 \alpha,$$

while $\partial_i\partial_j$ gives

$$ \partial_i\partial_j \hat{h}^r_{ij} + 2 \partial^4 \epsilon = 2 \partial^2\alpha.$$

Multiplying the 2nd by $\partial^{-2}$ and subtracting gives the first line of (A.5).

 

[nenyan]

The trace of (A.4) is

$$\delta\hat{h}^r_{ii} + 2 \partial^2 \epsilon = 6 \alpha,$$

while $\partial_i\partial_j$ gives

$$ \partial_i\partial_j \hat{h}^r_{ij} + 2 \partial^4 \epsilon = 2 \partial^2\alpha.$$

Multiplying the 2nd by $\partial^{-2}$ and subtracting gives the first line of (A.5).

Thank you fzero! Could you please give me some detail? How to get
$$ \partial_i\partial_j \hat{h}^r_{ij} + 2 \partial^4 \epsilon = 2 \partial^2\alpha.$$
from
$$\delta\hat{h}^r_{ii} + 2 \partial^2 \epsilon = 6 \alpha,$$
And how to get the frist line of (A.5). Actually, I do not understand the "\partial^{-2}". Thank you again.

 

[nenyan]

Oh, I see. Thank you very much, fzero.