# susy QM and Atiyah-Singer index theorem

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Consider maps $t\mapsto x^i(t)$ from circle to some Riemannian (spin) manifold and lagrangian
$$\mathcal L = \frac12 g_{ij}(x) \partial_t x^i \partial_t x^j + \frac12 g_{ij} \psi^j \left(\delta^i_k \partial_t + \Gamma^i_{mk} \partial_t x^m\right)\psi^k,$$
where $\psi^k$ are real Grassmann variables. This is supersymmetric under
$$\delta x^i =\epsilon \psi^i, \qquad \delta\psi^i=\epsilon \partial_t x^i.$$
We want to compute
$$\operatorname{Tr}(-)^Fe^{-\beta H}=\int_\text{periodic}[dx][d\psi] \exp \left(-\int_0^\beta dt \mathcal L\right),$$
in the limit $\beta \to 0$.

My question is: to see that the lagrangian for quadratic fluctuations around constant configurations $\xi^i=x^i -x^i_0$, $\eta^i=\psi^i-\psi^i_0$ (namely the one surviving in $\beta \to 0$ limit) is
$$\mathcal L^{(2)}=\frac12 g_{ij}(x_0) \partial_t \xi^i \partial_t \xi^j - \frac14 R_{ijkl}\xi^i\partial_t\xi^j \psi_0^k\psi_0^l +\frac{i}2\eta^a\partial_t\eta^a,$$
what are the right substitutions to make, besides using Riemann normal coordinates and vielbein $e_i^a e_j^b \eta_{ab}=g_{ij}$?

asked Nov 4, 2014
edited Nov 5, 2014

I'm a bit confused. There are infinite number of constant configurations, so what exactly are $x_0$ and $\psi_0$?

Right, just fix one of them, and expand around it; the path integral measure splits as $[dx][d\psi]=[dx_0][d\psi_0][d\xi][d\eta]$.

To be honest I can't see how $[dx][d\psi]=[dx_0][d\psi_0][d\xi][d\eta]$. Let's take lattice regularization then the integral is just a finite dimensional multiple integral, then clearly for each fixed $x_0$ and $\psi_0$, we have $[dx][d\psi]=[d\xi][d\eta]$, and then shouldn't $[dx_0][d\psi_0][d\xi][d\eta]=\text{some volume}\cdot[dx][d\psi]$?

What I meant is: expand $x=x_0 + \xi$ around constant configuration $x_0(t)=x_0$, and the same for fermions. Then the measure splits as $[dx]=\mathcal N [dx_0][d\xi]$, where $\mathcal N$ is unknown normalization; my point is that, whatever $\beta$-dependent substitution you make, it should leave that measure invariant. Do you agree?

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