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I actually have an integral which is used to prove Atiyah-Singer index theorem for spin complex in a path integral fashion. The integral I need to evaluate is following (in simplified form)

$\int \prod_{n=1}^\infty da_n^\mu exp{\frac{1}{4}}\sum_{n=1}^\infty[(2n \pi)^2 (a_n^\mu)^2 + a_n^{\mu} a_n^{\nu} R_{\mu\alpha} R_{\nu}^{\alpha}/M^4]$ where $a_n^\mu$ are fourier coefficients and $R_{\mu\nu}$ is the famous curvature tensor with it's two indices contracted (Ricci tensor),

the result to which should be

$\int \prod_{n=1}^\infty det |1-\frac{R_{\mu}^{\alpha} R_{\nu}^{\alpha}}{M^4 (2n\pi)^2}|^{-1/2}$.

Now I am having difficulty to verify this result. I have tried to diagonalize the Ricci tensor in terms of the eigenvalues $x_i$ and $-x_i$ and hence the product of two tensor in $R_{\mu\alpha} R_{\nu}^{\alpha}$ will give only diagonal terms which can be combined with the first term to give something similar to $[1-\frac{R_{\nu}^{\alpha} R_{\nu}^{\alpha}}{M^4 (2n\pi)^2}](2n \pi)^2 (a_n^\mu)^2$ and then I can perform Gaussian integration (did not work) to get similar but not same result.

Can anyone help me out?

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  • $\begingroup$ The summands of sum in the argument of the exponential are quadratic in $a^\mu$, but not diagonal. The correct expression is $(2\pi n)^2 a^\mu(\delta_{\mu\nu} + \frac{R^\alpha_\mu R^\alpha_\nu}{(2\pi n)^2M^4} )a^\nu$. $\endgroup$
    – user1504
    Sep 17, 2014 at 14:49
  • $\begingroup$ Well, I should have written something like $x_i$ in place of $R_{\nu}^{\alpha}$ in the second last line. I chose to diagonalise so I have a minus sign and your expression is also not making feel me any better. $\endgroup$
    – user44895
    Sep 18, 2014 at 9:12

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