⇤Math diary
May 2, 2023: Proof of the index theorem
May 2 2023
Today was the last day of my index theory class. We learned about the Atiyah-Singer index theorem, a landmark result relating topology with analysis on manifolds. Start with a Elliptic differential operator $D$ on a manifold $M$. Elliptic operators are things which act like a laplacian, and they are the nicest operators you ever will see. For one, the solutions to $Df=0$ are automatically smooth, by a little bit of magic called elliptic regularity. In the land of PDEs, where so much time is spent juggling the specific regularity and number of allowed derivatives of a function, elliptic regularity is a cheat code. Also, the space of solutions (the kernel of $D$) is always a finite dimensional vector space. The cokernel, which is the target of $D$ modulo the image of $D$, is also finite dimensional. The difference of dimensions of the kernel and cokernel is called the Index of $D$.
For example, take the laplacian $\Delta$ acting on differential forms. Hodge theory tells you that every solution to the Laplace equation $\Delta f=0$ has to circle a hole in the manifold $M$, and the number of solutions counts the number of holes. More precisely, the the kernel of $\Delta$ is isomorphic to the cohomology $H^*(M)$.
But notice, even though the laplacian depends on the metric of $M$, the dimension of its kernel does not. It is a topological quantity. The Atiyah-Singer index theorem states that a similar fact is true for all Elliptic operators: The index (dimension of kernel minus cokernel) is a topological invariant, and there is an explicit formula using characteristic classes. This generalize a great number of other theorems. Applied to the the exterior derivative $\text{d}$ on a surface, the index (an integer) equals the integral of the curvature. For higher dimensional manifolds, it says index of $\text{d}$ is the euler charecteristic. Applied to the Dolbeaut operator $\overline{\partial}$, we get the Riemann-Roch theorem.
The goal of this class was to prove the Atiyah-Singer index theorem, with no gaps, starting from basic differential geometry. This involves a huge tower of structures: Principle bundles and their curvature, characteristic classes, spin geometry and the Dirac operator, and PDE theory of elliptic operators. And it culminated in todays 2 and a half hour lecture, proving the index theorem for Dirac operators and deriving the formula for their index, the $\hat{A}$ genus.
How to prove the index theorem
First, some background about Dirac operators. They were born from need in physics. Dirac wanted to understand how quantum mechanics worked relativistically. The quantum world is governed by the Schrodinger equation, $i \partial_t \Psi = \Delta \Psi$ where $\Delta$ is the spacial laplacian. This bothered Dirac, because in special relativity, time and space should be in equal footing. How can time be degree 1, and space degree 2? To fix this, Dirac needed a first order differential operator $D$ such that $D^2 = \Delta$. Then, the defining equation $i\partial_t s = Ds$ is purely degree 1, and all kosher. Unfortunately, no such $D$ exists… for scalar functions. But we can make a matrix valued operator $D$ with $D^2 = \Delta$! The algebraic structure underlying these matrices are called Clifford algebras, and this realization is the birth of spin and related ideas.
Dirac operators are somehow the example of degree 1 elliptic operators. They included the exterior derivative $\text{d}$ and dolbeaut operator $\overline{\partial}$ as special cases. In fact, if you understand them very well, you understand most things for higher degree too. So let’s focus on the index theorem for the dirac operator $D$ canonically associated to a spin manifold$M$. The statement is roughly: \(\text{index} D = \hat{A}(M) = \int_M \frac{\sqrt{R}/2}{\sinh \sqrt{R}/2}\) Where $R$ is a collection of terms involving the curvature of the metric on $M$, and the right hand side is evaluated by expanding its talyor series, and using characteristic classes of $TM$ to evaluate the term of degree equal to the dimension of $M$. This looks like such a weird formula: where did the sinh come from? what’s going on?
Let me do my best to sketch a proof. Fundamentally, we will write down an integral and solve it in a couple different ways: One analytic (the index), and the other topological.
Idea 1: use the trace Alright, first we need to express the index (which basically counts solutions to $Ds=0$) using an integral. For the first clever point of the proof, we use the trace! The space which dirac operators acts on $S$ naturally splits into $S^+ \oplus S^-$, and the dirac operator exchanges the two subspaces. That is, $D = D^+ + D^-$, where $D^\pm : S^\pm \to S^\mp$. This means all nonzero eigenvalues come in pairs. Consider a new operator $e^{tD^2}$, which will similarly split according to $S^\pm$ and have pairs of nonzero eigenvalues. We take the Supertrace: the trace of $e^{-tD^2}$ on $S^+$, minus the trace of $e^{-tD^2}$ on $S^-$. Since eigenvalues come in pairs, all nonzero eigenvalues cancel out! We are left with the trace of $e^{tD^2}$ on the zero eigenspace of $D$. But, $e^{-tD^2}$ equals the identity on the zero eigenspace, and its trace measures the dimension! In conclusion, we have the Mckean-singer formula: For any value of $t$, \(\text{index} D = str \, e^{-tD^2}\) The trace is a very local quantity: we can compute it with respect to the basis of $L^2(M)$ provided by dirac delta functions at the point $x$, denoted by $|x\rangle$. the trace is the integral \(str \, e^{tD^2} = \int_M \langle x | e^{-tD^2}|x\rangle \mathrm{d} x\)
Idea 2: use the heat kernel Now let’s examine $e^{tD^2}$. Like all good exponentials, this operator is the solution of the differential equation $\partial_t \Psi = -D^2 \Psi$. But, $D^2 = \Delta$ the laplacian! (Strictly speaking, on a manifold, $D^2 = \Delta + R$ is a 2nd degree operator equaling the laplacian, plus a zeroth order curvature term. ) So, the equation above is a heat equation. If $s_0$ is the initial configuration of the heat equation, then $e^{-t D^2}s_0$ is the solution after time $t$.
We can express the information of $e^{-t D^2}$ in a heat kernel: Drop a dirac delta at a point $p$, simulate for time $t$ under the heat equation, then define $k_t(p,q)$ to be the resulting function at a point $q$. The supertrace above asks for $\int_M k_t(p,p)$: If you drop a dirac delta at a point $p$, how much remains at $p$ after time $t$?
We can pick any positive value of $t$ we’d like, so let’s look at the short time limit. The heat kernel $k_t(p,\cdot)$ is approximately a gaussian centered about $p$, with some number of additional polynomial terms. Turns out, the supertrace selects out the term with degree equal to half the dimension of $M$, and ignores everything else. Evaluating the supertrace (and thus index) becomes evaluating the special term of the asymptotic of the heat kernel. The term in the asymptotic expansion is a differential form, which when integrated across the manifold, gives a Chern class and a topological invariant.
At this point, we are in principle done. We can choose coordinates, plug in the asymptotic expansion to the defining property of heat kernels, and collate differential equations relating the terms. While annoying, we can hypothetically solve every diffeq and calculate the $\hat{A}$ genus. But this is really ugly - how can we simplify the problem more, and see the origin of the formula for the $\hat{A}$-genus?
Idea 3: Renormalize The last step is one philosophically familiar to physicists. We start with an operator $D^2 = \Delta + R$, where the curvature $R$ varies in a complicated way with position. We rescale our manifold, zooming way in to a point $p$. Then the higher order terms of the curvature all die off, and we are left with a quadratic function. In other words, the schrodinger equation with a harmonic oscillator potential. Here, the exact solution is known. For the 1D harmonic oscsilator with potential $\frac{-a}{16}x^2$, It’s called the Mehler kernel: \(k_t = \frac{1}{t^2} \sqrt{\frac{ta}{\sinh(ta)}} \exp(-a \cosh(at)x^2)\) We can see finally where the $\sinh$ term in the $\hat{A}$ genus above! Using this, we get the index of the deformed oeprator equals the $\hat{A}$ genus.
Now here’s the magic: We know already that the index is a topologically invariant quantity. So, the answer should stay the same as you limit the size to be very large (discussed above) or very small (measuring topological quantities). In particular, the approximate harmonic oscillator potential doesn’t yield an approximate solution: It yields an Exact one.
And that’s the proof! It’s a really beautiful idea, I’ve never seen the full proof before. Though, the details do get quite hairy in implementation. Still, fun and helpful class :).