2
$\begingroup$

I couldn't help but notice a formal similarity between the Lefschetz index $$ \mathrm{ind}(f)=\sum_k (-1)^k\operatorname{tr}(f_*|H_k) $$ and the Witten index $$ Z=\operatorname{tr}((-1)^Fe^{-\beta H}) $$

Indeed, if $f$ has no fixed points then $\mathrm{ind}(f)$ vanishes, and if SUSY is spontaneously broken then $Z$ vanishes. And both indices agree with the Euler characteristic for suitable choices of $f$ and the quantum system under consideration. Am I reading too much into these expressions? Or is there any underlying reason for this apparent similarity? If Lefschetz is not the right place to look for a mathematical definition of $Z$, is there any other index (previously defined by mathematicians) that is closer to the Witten index?

$\endgroup$
  • $\begingroup$ In Witten's famous paper "Supersymmetry and Morse theory", he relates it to the index at critical points of a Morse function. $\endgroup$ – Danu May 30 at 22:39
1
$\begingroup$

The Witten index is a generic index. A clever choice of the quantum system and the Hamiltonian should provide a quantum mechanical model, whose Witten index is equal to the analytical index under consideration. In fact, the index can be exactly computed by means of a path integral which provides the topological formula for the index. This procedure is sometimes called a "quantum mechanical proof of an index formula".

There is a lot of literature on this subject; for the Lefschetz fixed-point formula, please see the following article by Si Li and the following work by Niemi and Palo.

Please let me give you some physical intuition on how the quantum mechanical proofs work and how the supersymmetry comes into action.

In physics, we are interested in quantum systems with degenerate ground states, whose gap from the excited states is very large or infinite. Thus effectively, the dynamics is restricted to the ground states. A prototype of such a system is a charged particle on a plane in a very large magnetic field. In this case the ground states are the lowest Landau levels separated by a huge gap from the excited levels due to large magnetic field. There are many operators which have nontrivial limits under this restriction, thus can generate nontrivial dynamics in the restricted system. The Schrodinger operator (which in general can be a pseudo-differential operator) can be used as a projector to this limit as follows: $$\mathcal{O}_{LLL} = \lim_{\beta\rightarrow \infty}e^{-\beta H} \mathcal{O} e^{-\beta H} $$ In the limit all matrix elements with excited states are killed.

May be the most important example are the translations along the axes which are just multiplication operators on the full Hilbert space, become generators of a Heisenberg-Weyl algebra after projection: $$[x_{LLL}, y_{LLL}] = i \frac{\hbar c}{e B}$$ Thus, the projected physics is interesting, and we can be interested, for example in quantities of the form: $$\mathrm{Tr}_{LLL} \left(e^{- i \theta Q} \right)= \lim_{\beta\rightarrow \infty} \mathrm{Tr}_{ L^2(\mathbb{R}^2)} \left( e^{-\beta H} e^{- i \theta Q} \right)$$ The trace in the left-hand side is over the zero-mode spectrum only, while on the right-hand side it is over the whole $L^2(\mathbb{R}^2)$. The left-hand side is a character computation, while the right-hand side has the form of a grand partition function.

As very well known, there are standard rules to convert the right-hand side into a path integral over closed trajectories. The limit $\beta\rightarrow \infty$ indicates that we may compute the integral in the WKB approximation.

Now, for a large class of Hamiltonians, there exist a supersymmetric extension obtained by changing the differentials in the Hamiltonian into Lie derivatives and extending the domain from ordinary square integrable functions into square integrable differential forms (in our case: $L^2(\Lambda\mathbb{R}^2)$, and in our case having the property that its zero-modes are bosonic and coincide with the original bosonic zero modes of the original Hamiltonian.

Thus, we get: $$\mathrm{Tr}_{LLL} \left(e^{- i \theta Q} \right)= \mathrm{Tr}_{ L^2(\Lambda\mathbb{R}^2)} (-)^F \left( e^{-\beta H_S} e^{- i \theta Q} \right)$$

Where $H_S$ is the super-symmetrized Hamiltonian and $(-)^F$ here is $+1$ on even forms and $-1$ on odd forms.

Here, we don't need the limit $\beta\rightarrow \infty$ anymore, because, in supersymmetry, the spectrum of the non-zero modes is identical among odd and even forms thus, thus due to $(-)^F$, the contribution from the excited spectrum vanishes and the sum is independent of $\beta$. (The $(-)^F$ implies periodic boundary conditions for the Fermions).

In fact, actually, we can take the opposite limit $\beta\rightarrow 0$, which implies that only the constant configurations survive in the path integral, and the solution of the path integral becomes quite trivial.

In the case of the Lefschetz formula, the natural choice of the Hamiltonian is the Laplacian $$\Delta = d \delta + \delta d$$ The zero modes of the Laplacian are just the Harmonic forms which additively generate the de-Rham cohomology group. (This time the zero modes are not restricted to the zero forms). Finally, we can replace the charge operator $ e^{- i \theta Q}$ by a diffeomorphism $f$. The effect of adding a diffeomorphism is twisting the boundary conditions of the final point with respect to the initial one by the action of $f$. The closed trajectories which localize the path integral are just those which pass through the fixed points.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.