Homework 3 Solutions

Homework 3 Solutions Differential Topology (math876 - Spring2006 Søren Kold Hansen Problem 1: Exercise 3.2 p. 246 in [MT]. Let {ɛ 1,..., ɛ n } be the basis of Alt 1 (R n dual to the standard basis {e 1,..., e n } of R n and let be the star operator w.r.t. vol = ɛ 1... ɛ n Alt n (R n as defined in Exercise 2.9. Then for ω Ω p (U, define ω by applying on Alt p (R n pointwise, i.e. ( ω(x = (ω(x (1 for x U. Here ω : U Alt p (R n is a smooth map, and ω : U Alt n p (R n is just a map. Let W p (U be the space of maps U Alt p (R n. Then W p (U is a real vector space like Ω p (U and Ω p (U is a subspace of W p (U. In fact W p (U is a W 0 (U module exactly in the same way as Ω p (U is a Ω 0 (U module. Namely, an element f W 0 (U is simply a function f : U R, and if ω W p (U we let fω W p (U be the element given by (fω(x = f(xω(x, x U. Note that we can actually define : W p (U W n p (U by (1 thus extending our on Ω p (U. The first job is to prove that ω is smooth if ω is smooth. To do that we first prove some other facts about. First we note that since the Hodge star operator on Alt p (R n is R-linear, we have that : Ω p (U Ω n p (U is Ω 0 (U-linear, i.e. it is R-linear and (fω = f ω (2 for f Ω 0 (U and ω Ω p (U. The last claim simply follows by ( (fω(x = ((fω(x = (f(xω(x = f(x (ω(x = f(x( ω(x = (f ω(x for each x U. By the definition (1 and the results in Exercise 2.9 in [MT] we immediately get that (dx 1... dx p = dx p+1... dx n (3 and more generally that for all σ S(p, n p. ( dx σ(1... dx σ(p = sign(σdxσ(p+1... dx σ(n (4 Before we continue, let us introduce some compact notation. For a shuffle σ S(p, n p we write dx σ = dx σ(1... dx σ(p and dx σ = dx σ(p+1... dx σ(n and similarly ɛ σ = ɛ σ(1... ɛ σ(p and ɛ σ = ɛ σ(p+1... ɛ σ(n for their values. Now if ω Ω p (U we have ω = σ S(p,n p f σdx σ for f σ Ω 0 (U and then ω = σ S(p,n p sign(σf σdx σ by the Ω 0 (U-linearity of : Ω p (U Ω n p (U and (4. Thus ω Ω n p (U. 1

By (1 and Exercise 2.9 we also conclude that = ( 1 p(n p on Ω p (U (and actually on W p (U. Let d : Ω p (U Ω p 1 (U be as defined in [MT] Exercise 3.2, i.e. d = ( 1 np+n 1 d. (Note that this operator is not defined on the full space W p (U. Then d d = λ d d = λ d d = 0 for some signs λ, λ {±1} (which we don t need to specify in this argument. Finally let us prove the formula d (fdx 1... dx p = ( 1 j dx 1... dx x ˆ j... dx p (5 j and more generally the formula d ( fdx σ(1... dx σ(p = ( 1 ν dx σ(1... dx x ˆ σ(ν... dx σ(p (6 σ(ν for f Ω 0 (U and σ S(p, n p. By (2 and (4 we have d (fdx σ = sign(σd (fdx σ = sign(σdf dx σ = sign(σ = sign(σ Thus, by Ω 0 (U-linearity of x σ(ν dx σ(ν dx σ(p+1... dx σ(n x j dx j dx σ d (fdx σ = x σ(ν sign(σ( 1 np+n 1 ( dx σ(ν dx σ so left is to prove that sign(σ( 1 np+n 1 ( dx σ(ν dx σ = ( 1 ν dx σ(1... or equivalently by (1 that sign(σ( 1 np+n 1 ( ɛ σ(ν ɛ σ = ( 1 ν ɛ σ(1... ˆ dx σ(ν... dx σ(p ɛ σ(ν ˆ... ɛ σ(p, (7 where is Hodge s star operator on Alt n p+1 (R n w.r.t. vol = ɛ 1... ɛ n. Let ν {1, 2,..., p} and put ω = ɛ σ(ν ɛ σ. We have ω Ω p 1 (U. Write ω = τ S(p 1,n p+1 r τɛ τ, r τ R. By the definition of Hodge s star operator, see Exercise 2.9, we have r τ vol = ω, ɛ τ = ω ɛ τ. 2

Thus r τ = 0 except if {τ(1,..., τ(p 1} = {σ(1,..., σ(ν 1, σ(ν + 1,..., σ(p}. For this τ we have ω ɛ τ = ɛ σ(ν ɛ σ ɛ σ(1... ɛ σ(ν 1 ɛ σ(ν+1... ɛ σ(p = ( 1 n p+ν 1 ɛ σ ɛ σ = ( 1 n p+ν 1 ( 1 p(n p ɛ σ(1... ɛ σ(n = ( 1 n p+ν 1 ( 1 p(n p sign(σvol, where we proved the last identity in our solution to Exercise 2.9. Thus since p 2 + p is even. Thus which is equivalent to (7. r τ = ( 1 n p+ν 1 ( 1 p(n p sign(σ = ( 1 np+n 1 ( 1 ν sign(σ ω = ( 1 np+n 1 ( 1 ν sign(σɛ σ(1... Problem 2: Exercise 3.3 pp. 246 247 in [MT]. ɛ σ(ν ˆ... ɛ σ(p The Laplace operator : Ω p (U Ω p (U (one for each p is defined by = d d + d d (8 where d the exterior differential and d is defined in Exercise 3.2 p. 246 in [MT] (see also above. Let us first prove that (f = (9 x 2 j for f Ω 0 (U. We have df = n x j dx j and hence by (6 d df = ( d dx j = x j ( Since d is zero on Ω n (U we have that d (f = 0 and hence that (f = d df. Next we claim that x 2 j. (fdx σ = (fdx σ (10 for p > 0, σ S(p, n p and f Ω 0 (U, where the notation is as in the solution to Exercise 3.2 above. By (6, linearity of d and Lemmas 3.3 and 3.4 in [MT] we have dd (fdx σ = = ( ( 1 ν d dx σ(1... x σ(ν ( ( 1 ν d dx σ(1... x σ(ν dx ˆ σ(ν... dx σ(p ˆ dx σ(ν... dx σ(p 3

= = ( 1 ν ( 1 ν + ( 1 ν dx j dx σ(1... dx x j x ˆ σ(ν... dx σ(p σ(ν dx σ(j dx σ(1... dx x σ(j x ˆ σ(ν... dx σ(p σ(ν 2 f x 2 σ(ν = ( 1 p 1 ( 1 ν x 2 σ(ν dx σ. dx σ(ν dx σ(1... ˆ dx σ(ν... dx σ(p dx σ(1... dx x σ(j x ˆ σ(ν... dx σ(p dx σ(j σ(ν To calculate d d(fdx σ we first use Lemmas 3.3 and 3.4 in [MT] to get d(fdx σ = df dx σ = If σ = id we then get by (6 that d d(fdx σ = ( 1 p = ( 1 p +( 1 p ( 1 p+1 x j dx j dx σ = ( d dx 1... dx p dx j x σ(j x σ(j dx σ(j dx σ. ( 1 ν dx 1... dx x ν x ˆ ν... dx p dx j j dx x 2 id j where ( 1 p ( 1 p+1 = 1. Adding the thus calculated d d(fdx id and dd (fdx id gives (10 (using (9. The general case is more demanding, the problem being that we only have (6 for σ S(p, n p i.e. for σ(1 < σ(2 <... < σ(p. Thus we will follow another strategy, namely let us give a proof of (10 using results form Exercise (2.11 in [MT]. First observe that if ω is a constant p-form then ω is a constant (n p-form by (1. Also observe that for any integers i 1,..., i p {1,..., n} (with repetitions allowed and not necessarily ordered we have ( ɛ i1... ɛ ip is zero if ik = i l for some k l and is equal to λɛ j1... ɛ jn p for some λ {±1} if the i k s are mutually different, where {j 1,..., j n p } = {1,..., n}\{i 1,..., i p }. Moreover, for any integers i 1,..., i p and any f Ω 0 (U we have d ( fdx i1... dx ip = df dxi1... dx ip. Namely, both sides are zero if i k = i l for some k l and if that is not the case then dx i1... dx ip = λdx j1... dx jp for some λ {±1}, where {j 1,..., j p } = {i 1,..., i p } with 1 j 1 < j 2 <... < j p n. Thus d ( ( ( fdx i1... dx ip = d fλdxj1... dx jp = λd fdxj1... dx jp = λdf dx j1... dx jp = df dx i1... dx ip 4

where we used Lemma 3.4 in [MT]. Let us in the following not destinguish between a constant form and its value in our notation (exactly as when we write ɛ j for dx j. Now to the proof of (10. Let F ν : Alt p (R n Alt p+1 (R n and F ν : Alt p+1 (R n Alt p (R n, ν Alt 1 (R n, be the linear maps from Exercise 2.11 in [MT]. Claim 1 We have and d d(fdx σ = dd (fdx σ = k=p+1 k=p+1 x σ(k x σ(j F ɛ σ(j F ɛσ(k (ɛ σ x σ(k x σ(j F ɛσ(k F ɛ σ(j (ɛ σ x 2 σ(j x 2 σ(j F ɛ σ(j F ɛσ(j (ɛ σ (11 F ɛσ(j F ɛ σ(j (ɛ σ. (12 Moreover F ɛσ(j (ɛ σ = 0 for j {1,..., p} while F ɛ σ(j (ɛ σ = 0 for j {p + 1,..., n}. Proof of Claim 1 By (2, Lemma 3.3 in [MT] and the observations above Claim 1 we have ( d d(fdx σ = ( 1 n(p+1+n 1 d dx l dx σ x l l=1 = ( 1 n(p+1+n 1 k=p+1 = ( 1 n(p+1+n 1 = l=1 k=p+1 k=p+1 l=1 ( d [dx σ(k dx σ ] x σ(k ( dx l ( [dx σ(k dx σ ] x l x σ(k x l x σ(k F ɛ l F ɛσ(k (ɛ σ. Note that F ɛ l F ɛσ(k (ɛ σ = τ S(p,n p r τɛ τ, where r τ = F ɛ l F ɛσ(k (ɛ σ, ɛ τ = F ɛσ(k (ɛ σ, F ɛl (ɛ τ = ɛ σ(k ɛ σ, ɛ l ɛ τ. (13 In particular Fɛ l F ɛσ(k (ɛ σ = 0 unless l {σ(1,..., σ(p, σ(k} so we get (11. Similarly we get dd (fdx σ = F ɛl Fɛ x l x m (ɛ σ. m Here F ɛ m (ɛ σ = τ S(p 1,n p+1 r τɛ τ, where l=1 m=1 r τ = F ɛ m (ɛ σ, ɛ τ = ɛ σ, ɛ m ɛ τ, so F ɛ m (ɛ σ = 0 unless m {σ(1,..., σ(p} and in that case F ɛ m (ɛ σ = ɛ σ, ɛ m ɛ τ ɛ τ (14 5

for the unique τ S(p 1, n p+1 satisfying {τ(1,..., τ(p 1} = {σ(1,..., σ(p}\{m}. Thus F ɛl F ɛ m (ɛ σ = ɛ l F ɛ m (ɛ σ = 0 except if m {σ(1,..., σ(p} and l {m, σ(p + 1,..., σ(n} so we get (12. Finally F ɛσ(j (ɛ σ = ɛ σ(j ɛ σ = 0 for j {1,..., p} while we saw above that F ɛ σ(j (ɛ σ = 0 for j {p + 1,..., n}. This finalizes the proof of Claim 1. By Exercise 2.11 F ν F ν +F ν F ν is multiplication by ν 2 so by the last observations in Claim 1 we get = x 2 σ(j x 2 j F ɛ σ(j F ɛσ(j (ɛ σ x 2 σ(j (F ɛj F ɛj + F ɛj F ɛj (ɛ σ = (fdx σ, where (f is given by (9. Thus (10 follows if we can prove that k=p+1 F ɛσ(j F ɛ σ(j (ɛ σ ( F ɛσ(k Fɛ x σ(k x σ(j + Fɛ σ(j F ɛσ(k (ɛ σ = 0. σ(j By letting f run through the set of functions given by f kj (x 1,..., x n = x σ(k x σ(j, j {1,..., p} and k {p + 1,..., n}, we see that this identity is true for all smooth maps f if and only if the following Claim 2 is true. Claim 2 For j {1,..., p} and k {p + 1,..., n} we have (F ɛσ(k F ɛσ(j + F ɛσ(j F ɛσ(k (ɛ σ = 0. (15 Proof of Claim 2 By (14 we have so F ɛ σ(j (ɛ σ = ɛ σ, ɛ σ(j ɛ σ(1... Futher, by (13 we have F ɛσ(k F ɛ σ(j (ɛ σ = ( 1 j 1 ɛ σ(k ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p F ɛ σ(j F ɛσ(k (ɛ σ = ɛ σ(k ɛ σ, ɛ σ(j ɛ τ ɛ τ ɛ σ(j ˆ... ɛ σ(p. for the unique τ S(p, n p s.t. {σ(j, τ(1,..., τ(p} = {σ(1,..., σ(p, σ(k}. Thus (15 is equivalent to ɛ σ(k ɛ σ, ɛ σ(j ɛ τ ɛ τ = ( 1 j ɛ σ(k ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p (16 for that τ. But this is seen to be true the following way: First note that ɛ σ(k ɛ σ, ɛ σ(j ɛ τ ɛ τ = ( 1 j ɛ σ(j ɛ σ(k ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p, ɛ σ(j ɛ τ ɛ τ. 6

Secondly note that ɛ σ(k ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p = λɛ τ for a λ {±1} so ɛ σ(j ɛ σ(k ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p, ɛ σ(j ɛ τ ɛ τ = ɛ σ(j (λɛ τ, ɛ σ(j ɛ τ 1 λ ɛ σ(k ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p and ( 1 j times this is equal to the right-hand side of (16 finalizing the proof of Claim 2. At the same time this finalizes the proof of (10. A smooth p-form ω Ω p (U is said to be harmonic if (ω = 0. We claim that : Ω p (U Ω n p (U maps harmonic forms into harmonic forms. In fact we claim that and commute, that is ( ω = (ω (17 for all ω Ω p (U. This follows by a direct computation which is mainly an issue about signs: ( ω = (dd + d d( ω = ( 1 n 1 ( ( 1 n(n p d d ω + ( 1 n(n p+1 d d ω = ( 1 n 1 ( ( 1 n(n p+p(n p d dω + ( 1 n(n p+1 d d ω = ( 1 n 1 ( ( 1 n(n p+p(n p ( 1 (n (p+2(p+2 d dω + ( 1 n(n p+1 d d ω = ( 1 n 1 ( ( 1 n(n p+p(n p+(n (p+2(p+2 ( 1 n(p+1+n 1 d dω +( 1 n(n p+1 ( 1 np+n 1 dd ω = ( ( 1 n(n p+p(n p+(n (p+2(p+2+n(p+1 d dω + ( 1 n(n p+1+np dd ω = (ω. Let us give another proof that the Hodge star operator maps harmonic forms to harmonic forms. Thus assume that ω Ω p (U is harmonic. Write ω = σ S(p,n p f σdx σ and get by (2 and (4 that ω = σ S(p,n p sign(σf σdx σ. Since is linear we have by (10 that 0 = (ω = σ S(p,n p (f σ dx σ so (f σ = 0 for all σ S(p, n p. But then ( ω = sign(σ (f σ dx σ = 0. σ S(p,n p We use here that if σ S(p, n p, then there is a unique shuffle τ S(n p, p determined by τ(j = σ(p + j, j = 1, 2,..., n p, and dx σ = dx τ so (f σ dx σ = (f σ dx τ = (f σ dx τ = (f σ dx σ by (10. Before leaving this exercise, let us give a proof of the last claim in Exercise 2.11 in [MT] that is different from the one presented in my solutions to homework 2. The set up is as follows: 7

We have a real vector space V with inner product and an ONB {e 1,..., e n } with dual basis {ɛ 1,..., ɛ n } for Alt 1 (V. For ν Alt 1 (V we have a map F ν : Alt p (V Alt p+1 (V with adjoint F ν : Alt p+1 (V Alt p (V (w.r.t. the inner products on the spaces Alt k (V, k = 0, 1,..., n, induced from the inner product on V, see Exercise 2.5 in [MT]. We want to prove that the endomorphism F ν F ν + F ν F ν on Alt p (V is given by multiplication by ν 2. We first note that all we proved above concerning the maps F ɛk and F ɛ k remain true in this more general setting where R n with its standard inner product is replaced by the inner product space V. Thus the last part of Claim 1, the remarks around (13 and (14 and Claim 2 are all true in this general setting. We therefore have the following generalization of Claim 2: Claim 3 The identity holds for all l, m {1, 2,..., n} with l m. F ɛl F ɛ m + F ɛ m F ɛl = 0 Proof of Claim 3 By linearity we have to prove that ( Fɛl F ɛ m + F ɛ m F ɛl (ɛσ = 0 for all σ S(p, n p. Thus let σ S(p, n p and choose k, j {1, 2,..., n} such that l = σ(k and m = σ(j. Then k j. Since Claim 2 is valid in our present situation we only have to consider the cases where k, j {1, 2,..., p} or k, j {p + 1,..., n}. But if k, j {1, 2,..., p} then F ɛσ(k (ɛ σ = 0 by Claim 1 and F ɛσ(k F ɛ σ(j (ɛ σ = 0 by the remarks below (14 since k j. If k, j {p + 1,..., n} we have F ɛ σ(j (ɛ σ = 0 by Claim 1 and F ɛ σ(j F ɛσ(k (ɛ σ = 0 by the remarks around (13 since k j finalizing the proof of Claim 3. Write ν = n r jɛ j. Since ν F ν is linear and hence also ν F ν is linear we get F ν F ν + F ν F ν = ( r l r m Fɛl Fɛ m + Fɛ m F ɛl = l=1 m=1 l=1 r 2 l ( Fɛl F ɛ l + F ɛ l F ɛl, where the last equality follows by Claim 3. Left is to prove that F ɛl F ɛ l +F ɛ l F ɛl is the identity for all l. To this end, let σ S(p, n p and let us calculate (F ɛσ(j F ɛσ(j + F ɛσ(j F ɛσ(j (ɛ σ for all j. If j {1, 2,..., p} then F ɛσ(j (ɛ σ = 0. Moreover, following the remarks around (14 we get that F ɛ σ(j (ɛ σ = ɛ σ, ɛ σ(j ɛ τ ɛ τ for the unique τ S(p 1, n p+1 such that {τ(1,..., τ(p 1} = {σ(1,..., σ(p}\{σ(j}. Thus F ɛσ(j F ɛ σ(j (ɛ σ = ɛ σ, ɛ σ(j ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p ɛ σ(j ɛ σ(1... ɛ σ(j ˆ... ɛ σ(p 8

and this is simply equal to ɛ σ since ɛ σ, ɛ σ = 1. Next assume that j {p + 1,..., n}. Then F ɛ σ(j (ɛ σ = 0 by Claim 1. Moreover, by the remarks around (13 we have F ɛ σ(j F ɛσ(j (ɛ σ = τ S(p,n p r τɛ τ where r τ = ɛ σ(j ɛ σ, ɛ σ(j ɛ τ. Thus r τ = 0 for τ σ and r σ = 1 finalizing the proof. Problem 3: Exercise 4.1 p. 248 in [MT]. Problem 4: Exercise 4.4 p. 249 in [MT]. 9