Index Theory Seminars. February 13, 2011

Transcription

1 Index Theory Seminars February 13,

2 1 Episode 1 - Michael Singer This talk was given to motivate and contextualise the series. More specifically, to explain why we are interested in Dirac operators and their analytic properties. 1.1 Partial Differential Operators Definition 1.1. A differential operator order m on R,x 1,...,x n is a function P of the form: Px, = a α x α α m where α is a multi-index with α = i α i and α1 αn α =.... x 1 x n We also insist the a α are smooth. The symbol σ of P at x R n is defined by isolating the highest order derivatives: σpx,ξ = a α xiξ α. P is elliptic at x if σpx,ξ 0, ξ 0. α =m Example 1.2. The Laplacian = i 2 has symbol σ ξ = ξ ξ 2 n, and is therefore elliptic. The Dirac operator is elliptic PROVE THIS. Now suppose M is a compact manifold and E, F M are complex vector bundles. The definitions above are local and hence extend to manifolds in the obvious way with P : ΓM,E ΓM,F. From now we will consider P to be adifferential operator on manifolds. Note that on manifolds we will require that σ is an invertible matrix not just non-zero, so if P is elliptic we will certainly have ranke = rankf. Definition 1.3. Equip M with a Riemannian metric. P is Fredholm if are both finite dimensional. kerp = {u ΓM,E Pu = 0}, cokerp = PΓM,E Given a fibrewise inner-product, on E we get a global L 2 -inner-product on ΓM,E given by u,u E = ux,u xdµ M. Hence there is a formal adjoint with respect to, such that v,pu F = P v,u E. Claim 1.4. P is elliptic if P is elliptic. Claim 1.5. cokerp = kerp. M P : ΓM,F ΓM,E 2

3 1.2 The Index Problem Definition 1.6. The index of P is IndP = dimkerp dimcokerp = dimkerp dimkerp The index of P is a rather stable quantity and depends only on the signature of P. Example 1.7. Finite dimensional Consider a linear function A : C m C n. Then there is a decomposition C m C n = kera V = ima W Then A V gives an isomorphism V = ima. dimkera = m dimv = m dimima = m n dimw = IndA = m n Remark 1.8. If IndP > 0 then P has at least 1 non-trivial solution. The Index Problem is to calculate IndP in topological terms. Why would this be a sensible thing to do? Examples of quantities that can be expressed as indices include: The Euler characteristic χm of M. The signature sgnm of M, when M is even dimensional. Before the solution of the index problem, the index of the classical Dirac operator D was known to be expressible in topological terms: IndD = ÂM,[M] for M a spin manifold. Â is the topological quantity the Â-genus. So there was certainly some evidence that the index itself was a topological quantity. 1.3 The Heat-Kernel Proof Themethodfollowed byjohnroe sbookisnottheoriginalproofbyatiyahandsinger. Hereisanillustrative example of the method we will follow in this seminar series to prove the Index Theorem. Example 1.9. Finite dimensional again We will reprove the Index Theorem for the finite dimensional case in an unnatural way. Let L : C m+n C m+n be given by 0 A L = A 0 3

4 so that L 2 = As u,a Au = 0, we have Au 2 = 0. Hence AA 0 0 A A Let ǫ : C m+n C m+n be given by kera A = kera, keraa = kera. Im 0 ǫ = 0 I n and define a new function ft = trǫe tl2. Claim f is constant for 0 t <. 2. f0 = m n. 3. f = IndA. Proof. 1. f t = trǫ L 2 e tl2 = tr ǫle tl2 L We may calculate that ǫl = Lǫ. But trǫl = trlǫ. Hence f t = f0 = trǫ = m n. 3. Let e j be a basis of C m by eigenvectors of A A ordered by their corresponding eigenvalues λ j so that λ 1 λ 2... λ m. Then O k k A λ k+1 A =... λm and e taa = 1 e tλ k e tλm This gives us So tre taa = dimkera+oe tλ k+1 as t, tre taa = dimkera +Oe tλ k+1 as t. f = lim t dimkera dimkera +Oe tλ k+1 = IndA. 4

5 We want to use this argument for a Dirac operator D. In this case 0 D L = D 0 and L 2 = D D 0 0 DD = +K K 2. In the argument above, we used that d dt e tl2 = L 2 e tl2. So we will use an analagous idea: t kx,y;t = L2 xkx,y;t for k the heat kernel of the Dirac operator. There will also be some analysis involved to show: ft is well defined. f t = 0. lim t ft = IndD. What happens when t 0. For this we will need to understand: Remark k has a series expansion lim kx,y;t. t M kx,y;t = t n 2 e x y 2 2t 1+a 1 x,yt We will be interested in the coefficient a n/2 in the expansion. Somewhere inside this term is the Â-genus. 5