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Notes on the K-Theory of Complex Grassmannians

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Notes on the K-Theory of Complex Grassmannians Notes on the K-theory of Complex Grassmannians Lenny Ng 18.979 May 1997 We wish to study the K-theory of the Grassmannian G(k, n) of complex k-planes in Cn. The Chern character gives a ring isomorphism ch : K(G(k, n)) ⊗ Q −→ H∗(G(k, n), Q) between the rationalized K-theory of G(k, n) and the rational cohomology ring of G(k, n). The CW-complex structure of G(k, n) (see, e.g., [3]) immediately implies that, as an additive n (k) n group, K(G(k, n)) ⊗ Q = Q is a free group on k generators. Our goal is to make this description more concrete. ¡ ¢ Note that G(k, n)= U(n)/(U(k) × U(n − k)) = St(k, n)/U(k) is the quotient by U(k) of the complex Stiefel manifold St(k, n) of partial orthonormal k-frames in Cn. Thus any finite-dimensional representation ρ : U(k) −→ GL(W ) of U(k) gives rise to the natural vector bundle St(k, n) ×ρ W −→ G(k, n), where St(k, n) −→ G(k, n) is viewed as a principal U(k)- bundle. This vector bundle, which we will denote by Φ(ρ), is most concretely visualized as a quotient by U(k) of the trivial U(k)-equivariant vector bundle St(k, n) × W(ρ) −→ St(k, n). (Here (ρ) denotes the action of U(k) on W by ρ.) This trivial vector bundle is simply the pullback under the natural map St(k, n) −→ G(k, n) of Φ(ρ). It is an easy exercise to verify that given two representations ρ1, ρ2 of U(k), Φ(ρ1 ⊗ ρ2) = Φ(ρ1) ⊗ Φ(ρ2). Hence the map ρ → Φ(ρ) gives a semiring homomorphism from representations of U(k) to vector bundles over G(k, n), which extends by universality to a ring homomorphism Φ : R(U(k)) ⊗ Q −→ K(G(k, n)) ⊗ Q from the rationalized representation (or character) ring of U(k) to the rationalized K-theory of G(k, n). Using techniques not understood by me, Φ can be shown to be surjective. We thus obtain a very concrete method by which elements of K(G(k, n)) can be constructed. It is then of interest to compute ker(Φ). We first need to understand the representation ring of U(n). For a full account, see [2]; we will give an abridged answer here. A maximal torus in U(k) is given by T (k) = iθ1 iθk −1 −1 {diag(e ,...,e )}. The representation ring of T (k) is easily seen to be Z[x1, x1 ,...,xk, xk ], e1 ek where the monomial x1 ··· xk corresponds to the one-dimensional representation for which 1 1 k the j-th factor of T (k)=(S ) acts on C via the standard weight ej representation; in other words, diag(eiθ1 ,...,eiθk ) · z = ei(e1θ1+···+ekθk)z. The Weyl group of U(k) is the full symmetric group Sk on k elements, and thus the ratio- nalized representation ring of U(k) is −1 −1 Sk −1 R(U(k)) ⊗ Q = Q[x1, x1 ,...,xk, xk ] = Q[σ1, . , σk, σk ], the subgroup of finite Laurent series in x1,...,xk which is invariant under permutations of the indeterminates. Here σi is the i-th elementary symmetric polynomial in x1,...,xk. Thus, given an element f = f(x1,...,xk) of R(U(k)) ⊗ Q, that is, a symmetric fi- nite Laurent series in x1,...,xk with rational coefficients, we obtain an element Φ(f) of K(G(k, n)) ⊗ Q. To better understand the structure of Φ(f), we look at its pullback bundle over St(k, n)/T (k). Note that the projection St(k, n) −→ G(k, n) factors into π St(k, n) −→ St(k, n)/T (k) −→ G(k, n), where π is defined to be the second projection. There is a map Φ˜ : R(T (k)) ⊗ Q −→ K(St(k, n)/T (k)) ⊗ Q constructed along the same lines as Φ, and a map π∗ : R(U(k))⊗Q −→ R(T (k))⊗Q defined canonically by restriction. Then an element f of the representation ring of U(k) naturally gives rise to two K-classes over St(k, n)/T (k), namely the bundle Φ(˜ π∗(f)) and the pullback bundle π∗(Φ(f)). Lemma 1 Φ(˜ π∗(f)) = π∗(Φ(f)). Proof. Straightforward; left as an exercise. 2 Thus we obtain the following diagram: ∗ St(k, n) × W(f) −→ Φ(˜ π (f)) −→ Φ(f) ↓ ↓ ↓ π St(k, n) −→ St(k, n)/T (k) −→ G(k, n) Recall that any representation of T (k) splits into a direct sum of one-dimensional e1 ek representations, of the form of the form x1 ··· xk , where, as before, we will use xj to denote the weight 1 representation associated to the j-th factor of T (k), and the exponents denote tensor product powers. Corresponding to x1,...,xk are line bundles L1,...,Lk over St(k, n)/T (k). The following result is nearly immediate. m1 mk Proposition 2 Let f(x1,...,xk) = am1···mk x1 ··· xk be an element of R(U(k)) ⊗ Q. Then ∗ P ⊗m1 ⊗mk π (Φ(f)) = am1···mk L1 ··· Lk . X 2 Proof. Use Lemma 1 and the fact that Φ˜ is a ring homomorphism. 2 A bit more notation is needed at this point. Let S∗ be the dual to the tautological ∗ k-plane bundle over G(k, n), and let γ1, . , γk be the Chern roots of S , so that the equation ∗ ∗ (1 + γ1) ··· (1 + γk)=1+ c1(S )+ ··· + ck(S ) is formally satisfied. Any symmetric function of γ1, . , γk can be written in terms of their ∗ ∗ elementary symmetric functions c1(S ),...,ck(S ), and thus should be interpreted as an element of H∗(G(k, n)) ⊗ Q. ∗ Lemma 3 Φ(x1 + ··· + xk)= S . →֒ (Proof. Note that σ1 = x1 + ··· + xk corresponds to the standard representation U(k GL(k, C). This argument was done in class for k = 1, n = 2. Essentially, lift Φ(σ1) to a U(k)-equivariant bundle Φ(˜ σ1) over St(k, n), tensor it with the “tautological bundle” over St(k, n) (i.e., the pullback of the tautological bundle over G(k, n)), and find a U(k)- equivariant nowhere vanishing section of this tensor product. 2 We are now in a position to compute the kernel of Φ using the Chern character iso- morphism. The following proposition is the crucial result in this direction. Proposition 4 Suppose f ∈ R(U(k)) ⊗ Q. Then ch(Φ(f)) = f(eγ1 ,...,eγk ). Proof. The following lemma will be useful. Lemma 5 (Borel-Hirzebruch [1]) If P −→ B is a principal G-bundle, and T is a max- imal torus in G, then the bundle projection factors as P −→ P/T −→ B, and the induced map H∗(B) −→ H∗(P/T ) is an injection; in fact, H∗(B) = H∗(P/T )W , where W is the Weyl group of G. In our case, this lemma implies that π∗ : H∗(G(k, n)) −→ H∗(St(k, n)/T (k)) is an injection. m1 mk Let f(x1,...,xk) = am1···mk x1 ··· xk , as before. By naturality of the Chern character and Proposition 2, P ∗ ∗ m1 mk π ch(Φ(f)) = ch(π Φ(f)) = am1···mk (chL1) ··· (chLk) m1c1(L1) mkck(Lk) = X am1···mk e ··· e = fX(ec1(L1),...,ec1(Lk)). ∗ ∗ But Lemma 3 implies that π (c(S )) = (1+c1(L1)) ··· (1+c1(Lk)), where c denotes the total ∗ Chern class, and so c1(L1),...,c1(Lk) are formally the image under π of the Chern roots ∗ γ1, . , γk. Since π is injective as a map on cohomology, the proposition follows. 2 We now apply this proposition to the simplest case, the case of G(1, n)= CP n−1. 3 Corollary 6 If k = 1, then j −1 i ker(Φ) = {f = ajx ∈ Q[x, x ] : j aj = 0,i = 0,...,n − 1} j j X X = {f : 0= f(1) = f ′(1) = ··· = f (n−1)(1)}. In terms of χ(f) : U(1) −→ C, the character of f, this corollary says that χ(f) vanishes to order n at 1 ∈ U(1). These conditions on f have an intuitive explanation. The condition that j aj = 0 says simply that the “fibers” of Φ(f) have dimension 0. For the next condition, jaj = 0, Pj recall that K(S2) = Z ⊕ Z, where one factor comes from the dimension of a vector bundle P over S2, and the other comes from the winding number of the clutching function given by this vector bundle. Then j jaj gives the winding number corresponding to the K-class Φ(f) restricted to S2 = CP 1 ⊂ CP n−1. The higher conditions (i ≥ 2) correspond to higher P winding numbers. Proof of Corollary 6. The kernel of Φ is precisely the set of f so that f(eγ1 ) = ∗ ∗ n−1 n ch(Φ(f)) = 0. But γ1 = c1(S ) generates H (CP ) with (γ1) = 0; therefore, if f = j j ajx , then n− 1 (γ )i P f(eγ1 )= a ejγ1 = jia 1 . j j i! j i=0 Ã j ! X X X The corollary follows. 2 References [1] Borel and Hirzebruch, Ann. Math., 1954? [2] T. Br¨ocker and T. tom Dieck, Representations of Compact Lie Groups (Springer-Verlag, New York, 1985). [3] J. W. Milnor and J. D. Stasheff, Characteristic Classes (Princeton University Press, Princeton, 1974). 4.
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