# supersymmetry and the de Rham complex

In Alvarez-Gaume's paper "Supersymmetry and the index theorem" there is given a certain supersymmetric Lagrangian whose quantization, apparently, leads to the de Rham Laplacian on the exterior algebra of a manifold. For what it's worth, the Lagrangian in question is

$$L = \frac{1}{2} g_{ij}(\phi) \dot{\phi}_i \dot{\phi}_j + \frac{\sqrt{-1}}{2} g_{ij}(\phi)\overline{\psi}^i \gamma^0 \frac{D}{dt} \psi^j + \frac{1}{12} R_{ijkl} \overline{\psi}^i \psi^j \overline{\psi}^k \psi^l$$ where $\phi$ is a map from $\mathbb{R}$ into a target manifold $M$, $g_{ij}$ is a metric on $M$, and $\psi^i$ is a spinor field on $M$ (I guess).

The same Lagrangian appears in several places, e.g. in Witten's paper on "Supersymmetry and Morse theory" (although seemingly with $1/12$ replaced by $1/8$). Witten says How canonical quantization of [this Lagrangian] leads to the exterior algebra was discussed in .'' But I cannot find anything of the sort in ="Dynamical breaking of supersymmetry", nor in any other source that I know about.

Can anyone give an explanation, or a reference, for $L$ above is related to the de Rham Laplacian? I am happy with a physics level of rigor but I am hoping to see some details spelled out.

Many thanks for any help.

• As a reference, see the big yellow Clay mathematics "Mirror Symmetry" text: available at claymath.org/library/monographs/cmim01c.pdf . Particularly in the vicinity of the sections on sigma models (10.4 and beyond). – Tom Mainiero Dec 31 '17 at 21:54
• Thanks this reference which I didn't know about. Curiously enough, it uses yet another value for "1/12", namely -1/2. Indeed the point I find most confusing is exactly what role the Riemann curvature tensor term plays - any pointers on that would be much appreciated. This must be intuitively obvious for physicists but I don't understand it. – SUSY student Dec 31 '17 at 23:38
• It might be helpful for you to work through the supersymmetry transformations. Since the metric is a function of phi (ie, is pulled back from the target manifold via the map phi), when you vary phi, you will pick up some derivatives of the metric. Given that, it’s pretty much inevitable that the Riemann tensor will show up contracted with the fermions. The coefficients can be pushed around in various places due to field redefinition, so that’s probably what’s going on there. – Aaron Bergman Jan 1 '18 at 4:16
• Thanks for the suggestion. I will try the supersymmetry transformations. – SUSY student Jan 1 '18 at 14:51
• What is $D/dt$? – Qfwfq Jan 1 '18 at 14:51

Here is a brief sketch of the relation with the de Rham complex. Fields of the theory are $\phi^i$, $\psi^i$, $\overline{\psi}^i$. Upon canonical quantization, the Hilbert space of the quantum theory is the $L^2$-space of differential forms on $M$, quantum supercharges are $d$ and $d^{*}$ operators and so the quantum Hamiltonian is the Laplacian $\Delta=\{d,d^{*}\})$.
The relation between the explicit form of the Lagrangian and the Laplacian=quantum Hamiltonian is quite subtle because of the general phenomenon of ordering ambiguities in quantization. It is a good exercise to write explicitely the classical Hamiltonian from $L$ but it will not tell you immediately what the quantum Hamiltonian is the Laplacian because there are ordering ambiguities. The point of the supersymmetric story is that in this case the Hamiltonian is constructed from the simplest objects which are the supercharges, and that there is no ordering ambiguity to construct the quantum supercharges.