440-3 Geometry/Topology: Cohomology Northwestern University Solution of Homework 5

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1 440-3 Geometry/Topology: Cohomology Northwestern University Solution of Homework 5 You can use freely the following simple results from complex analysis (which can be proved in the same way as the case of one complex variable). Recall that CP n = (C n+ \{0})/C, where C acts by scalar multiplication. Let (z 0,..., z n ) be the coordinates on C n+, and write z j = x j + iy j, 0 j n, with (x 0,..., y n ) real coordinates on C n+. We shall use complex derivatives = 2 ( x j i y j ), = 2 ( x j + i y j and complex -forms dz j = dx j + idy j, dz j = dx j idy j. Then for any smooth function f we have df = j=0 f dz j + j=0 f dz j. ), We also have that We will write z 2 = n j=0 z j 2. z k = δ jk, z k = 0. ) On C n+ \{0} let Prove that the complex 2-form g jk = log z 2. z k ω = i g jk dz j dz k, on C n+ \{0} is real (i.e. ω = ω), closed and invariant under the C action. Denote by ω FS the closed real 2-form it induces on CP n. Solution. First of all we claim that g jk = g kj, i.e. that at each point the matrix g jk is Hermitian. This is clear, because log z 2 is a real-valued function and ω = i z k = z k. Therefore g kj dz j dz k = i g kj dz k dz j = ω,

2 which means that ω is real (and after switching to real coordinates, it becomes a usual real 2-form). To show that ω is closed, we compute dω = i (dg jk ) dz j dz k = i log z 2 dz l dz j dz k z l z k j,k j,k,l + i log z 2 dz l dz j dz k = 0, z l z k j,k,l because partial derivatives commute. Let us now compute g jk = ( ) zk z 2 = z 2 δ jk z k z j z 4. If λ C and F λ : C n+ \{0} C n+ \{0} is given by F λ (z) = λz then F λ ω = i λ 2 (g jk F λ )dz j dz k = ω. 2) In the chart U 0 = {[z 0 : : z n ] CP n z 0 0} = C n with coordinates w j = z j /z 0, j =,..., n, show that ω n FS = n! ( + w 2 ) n+ idw dw idw n dw n. Solution. On the subset of C n+ \{0} where z 0 0, we have and log z 2 = log z log( + w 2 ), It follows that on U 0 we can write ω FS = i log z 0 2 = z 0 2 z 0 2 z k z 0 4 = 0. j,k= ( + w 2 )δ jk w k w j ( + w 2 ) 2 dw j dw k. Let us write h jk = ( + w 2 )δ jk w k w j ( + w 2 ) 2, 2

3 which is a Hermitian n n matrix at each point. If we apply a unitary change of coordinates, we may diagonalize h jk, so that in these new coordinates ζ j we have ω FS = i λ j dζ j dζ j, with λ j R. Taking the n th wedge product of this form, we get n ωfs n = n! idζ dζ idζ n dζ n, j= λ j j= and switching back to the original coordinates, ω n FS = n! det(h jk )idw dw idw n dw n, and we only need to compute the determinant of h jk. This equals det(h jk ) = ( + w 2 ) 2n det(( + w 2 )δ jk w k w j ). The Hermitian matrix w kw j represents the rank linear transformation w 2 which is the projection onto the complex line spanned by the nonzero complex number w, and therefore the matrix w k w j has the eigenvalue w 2 with multiplicity and the eigenvalue 0 with multiplicity n. Since (+ w 2 )δ jk is a multiple of the identity (which remains such in all coordinate systems, for example in coordinates which diagonalize w k w j ), we see that the matrix ( + w 2 )δ jk w k w j has eigenvalue with multiplicity and eigenvalue ( + w 2 ) with multiplicity n. It follows that det(( + w 2 )δ jk w k w j ) = ( + w 2 ) n, and so as required. det(h jk ) = ( + w 2 ) n+, 3) Again using the chart U 0, compute CP n ω n FS = (2π) n. 3

4 Deduce from this that [ω k FS ] 0 in H2k (CP n ) for all 0 k n, and show that therefore [ω k FS ] is a generator of H2k (CP n ). Solution. Note that idw j dw j = 2dx j dy j, so that idw dw idw n dw n = 2 n dµ, where dµ is the usual Lebesgue measure in C n. Since CP n \U 0 CP n has measure zero, we see that ωfs n = 2 n n! CP n C n ( + w 2 dµ(w). ) n+ We write C n = C n C, with w = (w, w n ), and first integrate in the variable w n. We have C n ( + w 2 dµ(w) = ) n+ C n C ( + w 2 + w n 2 ) n+ dµ(w n)dµ(w ). Use polar coordinates in C and write the innermost integral as 2π 0 so that r ( + w 2 + r 2 dr = π ) n+ C n ( + w 2 ) n+ dµ(w) = π n which can be iterated to get and so 0 ( + w 2 + s) n+ ds = π n( + w 2 ) n, C n ( + w 2 ) n dµ(w ), πn C n ( + w 2 dµ(w) = ) n+ n!, CP n ω n FS = 2 n n! πn n! = (2π)n. If we had [ωfs k ] = 0 in H2k (CP n ) for some 0 k n, then by taking the cup product with [ω n k FS ] we would get that also [ωn FS ] = 0 in H2n (CP n ). By Stokes s Theorem this would imply that ωfs n = 0, CP n a contradiction. Since we computed in Homework 2 that H 2k (CP n ) = R for 0 k n, it follows that any nonzero element in H 2k (CP n ) is a generator, 4

5 so in particular this holds for [ω k FS ]. 4) Use the Lefschetz fixed point theorem and show that for any even n, every smooth map F : CP n CP n has a fixed point. Solution. Since [ω FS ] is a generator of H 2 (CP n ), we have that F [ω FS ] = λ[ω FS ] for some λ R. Then F [ωfs k ] = [(F ω FS ) k ] = λ k [ωfs k ]. But these are generators for the cohomology of CP n (which vanishes in odd degree), and so by definition the Lefschetz number of F is L(F ) = + λ + λ λ n. If λ = then clearly L(F ) 0. If λ, then L(F ) = (λ n+ )/(λ ). If this was zero, then we would have that λ n+ =. If n is even, this is impossible, since λ is real. Therefore, if n is even we conclude that L(F ) 0. Thanks to the Lefschetz fixed point theorem, F has a fixed point. 5) For any odd n, find a smooth map F : CP n CP n without fixed points. Solution. We define F : CP 2n+ CP 2n+ by [z 0 : : z 2n+ ] [z : z 0 : z 3 : z 2 : : z 2n+ : z 2n ]. This is clearly a smooth map. If [z 0 : : z 2n+ ] was a fixed point for F, then we would have z 0 = λz, z = λz 0, for some λ C, and so z 0 = λ 2 z 0, which forces z 0 = z = 0. By the same token, z j = 0 for all j, a contradiction. 5