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Consider a system on a Riemannian manifold with the Lagrangian $$L = \frac{1}{2}g_{IJ} \dot{\phi}^I \dot{\phi}^J + \frac{i}{2}g_{IJ}(\overline{\psi}^I D_t \psi^J - D_t \overline{\psi}^I \psi^J) - \frac{1}{2}R_{IJKL} \psi^I \overline{\psi}^J \psi^K \overline{\psi}^L,\tag{10.198}$$

with $g_{IJ}$ the metric, $R_{IJKL} $ the Riemann curvature tensor, $\phi^i$ bosonic variables, $\psi^i$ fermionic variables and $D_t$ the covariant derivative. The Witten Index is defined as $${\rm Tr}(-1)^F = \int_{PBC} \mathcal{D}\phi \mathcal{D}\psi \mathcal{D} \bar{\psi}e^{-S},$$ where $PBC =$ periodic boundary conditions. I now have to derive an expression of the Witten index in terms of an integral involving the Riemann curvature tensor but i do not now how to handle this path integral?

The question is from Exercise 10.4.1, https://www.claymath.org/library/monographs/cmim01c.pdf page 210.

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The trick here is to notice that the Witten index for a finite temperature $\beta$ is given by

$$\text{Tr}\left\{(-1)^Fe^{-\beta H}\right\}=\int_{\text{PBC}}\mathcal{D}\phi\mathcal{D}\overline\psi\mathcal{D}\psi\,e^{-S},$$

where the boundary conditions are on a circle of circumference $\beta$. Next, we know that the Witten index is independent of the temperature (it computes the Euler characteristic of the Riemannian manifold), and so we can take the $\beta\to 0$ limit of this expression. In this case, all non-constant modes of the fields $\phi$ and $\psi$ have energy proportional to $1/\beta$, and thus will be exponentially suppressed in the $\beta\to 0$ limit. Thus, the path integral will localize only to those modes which are constant in time, namely

$$\text{Tr}\,(-1)^F\propto\int_{\mathcal{M}}\mathrm{d}\phi\,\sqrt{g}\int\mathrm{d}\overline{\psi}\,\mathrm{d}\psi\,\exp\left(-\frac{\beta}{2}R_{IJKL}\psi^I\overline{\psi}^J\psi^K\overline{\psi}^L\right),$$

where we have traded out our path integral for a standard integral over constant modes only, and the $\sqrt{g}$ term comes from the integral over non-constant modes in the Guassian limit (a factor of $1/\sqrt{g}$ from the bosonic fields and a factor of $g$ from the fermionic ones). The constant of proportionality can be worked out by being careful with the suppression of non-constant modes and working explicitly with the path integral measure over fourier components. However, this is quite technical.

Now, as a warm-up, if the manifold is $2$-dimensional, we have

$$\text{Tr}\,(-1)^F\propto\int\mathrm{d}^2\phi\,\sqrt{g}\int\mathrm{d}^2\overline{\psi}\mathrm{d}^2\psi\exp\left(-\frac{\beta}{2}R_{IJKL}\psi^I\overline{\psi}^J\psi^K\overline{\psi}^L\right).$$

I will leave it to you to show that, when you bring the Grassmann coordinates down from the exponential and integrate over them, the result is

$$\text{Tr}\,(-1)^F\propto\int\mathrm{d}^2\phi\,\sqrt{g}\,R,$$

where $R$ is the Ricci scalar. Since $\text{Tr}\,(-1)^F=\chi(M)$ is the Euler characteristic, this is exactly the statement of the Gauss-Bonnet theorem, up to a constant of proportionality ($1/4\pi$). The technique for higher dimensions can be worked out in a similar fashion.

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