Witten s Proof of Morse Inequalities

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1 Witten s Proof of Morse Inequalities by Igor Prokhorenkov Let M be a smooth, compact, oriente manifol with imension n. A Morse function is a smooth function f : M R such that all of its critical points are nonegenerate. A point x M is a critical point of f if f x = 0. A critical point is nonegenerate if [ ] 2 f et Hess f x = et x 0. x i x j Both properties o not epen on the choice of coorinates. The inex in x is the number of negative eigenvalues of Hess f x. Let m p = m p f be the number of critical points of inex p. Let b p = b p M = im H p M be the imension of the p th e Rham cohomology group. 0 Ω 0 M Ω 1 M Ω n M 0 This is calle the e Rham complex. Note that 2 = 0. If ω = α, then ω = 0. So Im : Ω p 1 Ω p ker : Ω p Ω p+1, an H p M = ker Im. Theorem 1. The Morse inequalities are as follows. The Morse polynomial M t an Poincare polynomial P t are efine by M t = P t = m k t k k=0 b k t k. There exists a polynomial R t = R 0 + R 1 t + R 2 t with all R k 0 such that k=0 M t P t = 1 + t R t. For example, consier M = the torus with a sale at the top. Then Thus H 0 = R, so b 0 = 1 H 1 = R, so b 1 = 2 H 2 = R, so b 2 = 1. P t = 1 + 2t + t 2. 1

2 2 Let the Morse function be the height function. Then the critical points are two maxima, three minima, an on minimum. Thus, Thus, Observe that we have m 0 = 1 m 1 = 3 m 2 = 2 M t = 1 + 3t + 2t 2. M t P t = 1 + 3t + 2t 2 1 2t t 2 = t + t 2 = 1 + t t. So R t = t. Some Corollaries: Corollary 2. If t = 1, M 1 = P 1, so 1 k m k = 1 k b k = χ M = 2 2g. In our example, we have g = 1, so 0 = 0 = 0 = 0. Corollary 3. We have m k b k for all k. These are calle the weak Morse inequalities. Mt P t Since R t = = M t P t 1 t + t 2... Equating 1+t the t k coefficient gives the following. Corollary 4. Strong Morse Inequalities For each k, m k m k ± m 0 b k b k ± b 0. We have the following fact. Suppose that we have 0 V 0 V 1 V n 0, 2 = 0, im V k <. If this complex computes the k th cohomology H k M of M, then {im V k } obey the Morse inequalities. One example: V k is a vector space whose basis is the set of critical points of inex k. This is Floer cohomology, an important mathematical tool. We now bring some geometry into the picture. We now put an arbitrary Riemannian metric on the manifol M, giving us a scalar prouct on the space of ifferential forms. ω 1, ω 2 = number. We also have an L 2 inner prouct: ω 1, ω 2 = ω 1, ω 2. We have : Ω p Ω p+1, an the ajoint : Ω p+1 Ω p is efine by the formula ω, α = ω, α for any p-form ω an any p + 1-form M

3 3 α. The Laplacian on forms is = + = + 2. We label p : Ω p Ω p. Hoge theory tells us the following. Given a class [ω] H p M note [ω 1 ] = [ω 2 ] iff ω 1 ω 2 = α for some p 1-form α. In each class, there is a unique form ω such that ω 2 = ω, ω is minimum. This form is the unique harmonic form in the class i.e. p ω = 0. For example, let M = S 1. Ientify functions on S 1 with 2π-perioic functions. H 0 S 1 = ker = {constant functions}. Note these are harmonic, because 2 constant = 0. Next, the one forms ω satisfy θ 2 ω = 0. We have to mo out by the image of on functions. Two one-forms are in the same class if an only if they integrate to the same number. So H 1 M is generate by θ. In fact, θ is harmonic, since 2 θ = 0. θ 2 The upshot is the following fact. Corollary 5. Hoge im ker p = im H p M = b p M. Motivation: in quantum physics, the eigenvalues of the Laplacian are energies. The energy is zero implies that the particle in the vacuum state. Important: what is the imension of vacuum space? Same as im ker p. We want to solve p ω = 0. The irect metho is very ifficult. However, the imension epens only on the topology, ue to Hoge theory. Sometimes it is ifficult to fin the b p of a very complicate manifol. The next best thing is to estimate b p using a Morse function, or a vector fiel, or a one-form. Witten s iea: given a smooth f : M R, the Witten ifferential is s = e sf e sf, where s R. Then 2 s = 0, s : Ω p Ω p+1. The Witten Laplacian is p s = s s + s s : Ω p Ω p. We have the following commutative iagram, so e sf is a quasi-isomorphism, an s yiels isomorphic cohomology groups.... Ω p Ω p+1 e sf e sf... Ω p s Ω p+1 So ker p s is isomorphic to H p s, so im ker p s = b p M.

4 4 We now compute Then s = s s + s s s ω = e sf e sf ω = + s f ω sω = + s f ω. s = + s f + s f + + s f + s f = + s {f, } + { f, } + s 2 f 2 = + s L f + L f + s 2 f 2 = + s L f + L f + s 2 f 2 The first orer erivative terms of L f an its ajoint are negatives of each other, so that L f + L f is a boune operator that we call B. Now, suppose that s is very large, an Then 0 = s ω s, ω s s ω s = 0. = ω s 2 + ω s 2 + s Bω s, ω s + s 2 f 2 ω s, ω s. The term Bω s, ω s is boune by a constant inepenent of s. Thus, ω s must ecay rapily away from the set where f = 0, ie. the critical points of f. So we hope that if f is a nice function, we can approximate it by functions efine only near the critical points. If f is a Morse function, ie. the critical points are nonegenerate, then f x = 0 implies that the quaratic part of the Taylor expansion is nonegenerate there. If this is the case, then, s can be approximate by a moel operator H. That is, the eigenvalues an eigenforms of the moel operator H will be close to the eigenvalues an eigenfunctions of s. We replace with the secon orer part with constant coefficients frozen at x. For example, on functions, = 1 g x i g ij g x j g ij x 2 x i x j. Next, replace B with B x. Finally, replace f 2 with its quaratic part at x. Theorem 6. For large s, s an H have the same number of eigenvalues in the interval [ 0, s 2/5]. Corollary 7. im ker s # eigenvalues of H in [ 0, s 2/5].

5 Restrict to a single critical point x. Choose coorinates so that x = 0, f x = m i=1 xi 2 + n i=m+1 xi 2, inex = m. Make the metric Eucliean near x. The moel operator is the irect sum of moel operators at each critical point. At the critical point x, the moel operator is H x = 2 x i 2 ± 2s [ x i, e i ] + s 2 x 2 i. i=1 These operators summe commute with one another. Note that 2 x i 2 + s 2 x i 2 has spectrum 2 + 4j s, j 0 with multiplicity one. The lowest eigenvalue is ψ 0 = exp sx 2 /2. Note that [x i, e i ] is x i e i x i e i, which acts by +1 if x i is in the form, 1 if x i is not in the form. So H can have something in the kernel, it can only have an m-form of the form exp s x i 2 x i... x m 2 for a critical point of inex m. Therefore, for large s, # eigenvalues of H in [ 0, s 2/5] = # of critical points of inex p = m p, so b p m p. 5