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Algebraic Topology Mid-Semestral Examination

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Algebraic Topology Mid-Semestral Examination Algebraic Topology M.Math II Mid-Semestral Examination http://www.isibang.ac.in/∼statmath/oldqp/WWWW Midterm Solution (1). Decide whether the following statements are TRUE or FALSE. Justify your answers. State precisely the results that you use. (a) The pull back of a trivial bundle is trivial. (b) The tangent bundle of S2 splits as the Whitney sum of two line bundles. (c) !(γ1) = 1 + a. (d) There exists a vector bundle η such that γ1 ⊕ η is trivial. 2 3 (e) The manifolds RP × RP and S5 are cobordant. Solution: (a) True Justification: Let ' : E ! B × V be a trivialization of a vector bundle p : E ! B. Let f : X ! B be Id×' a continuous map. Then : f ∗E ⊆ X × E −! X × B × V ! X × V is a trivialization of f ∗E with f×Id '−1 inverse induced by the projection g1 : X × V ! X and the map g2 : X × V −! B × V −! E. (b)False Justification: Theorem 1 (Classification of line bundles) The association E ! !1(E) gives a bijection between iso- 1 morphism classes of real line bundles on B and H (B; Z2). Theorem 2 (Hairy Ball Theorem) There is no nonvanishing continuous tangent vector field on even dimensional n-spheres. 2 2 2 Suppose the tangent bundle of S , T S = L1 ⊕ L2, where L1 and L2 are line bundles over S . By the classification of line bundles, L1 and L2 are trivial bundles. This implies that there is a nonvanishing continuous tangent vector field on S2. This is a contradiction by Hairy Ball Theorem, and therefore our assumption is wrong. (c) True 1 Justification: Let γ1 be the canonical line bundle over RP . Then we have an obvious bundle map 1 1 E(γ1 ) −−−−! E(γ ) ? ? ?γ1 ? 1 (1) y 1 yγ 1 inclus 1 S1 = RP −−−−! RP ∗ 1 ∗ 1 1 Therefore i (!1(γ )) = !1(i (γ )) = !1(γ1 ) 6= 0, by the axiom of the Stiefel-Whitney classes. This shows 1 1 1 ∼ 1 1 1 that !1(γ ) = a is the generator of H (RP ; Z2) = Z2. Hence !(γ ) = !0(γ ) + !1(γ ) = 1 + a, since the bundle is one dimensional. (d) False 1 Justification: Suppose there exists a bundle η : E ! RP such that γ1 ⊕η is trivial. Since !(γ1) = 1+a, 1 1 1 −1 2 where a is the generator of H (RP ; Z2) and hence !(E) must be (1 + a) = 1 + a + a + ··· . Since i i 1 we are using Z2 coefficients. Thus !i(E) = a , which is non-zero in H (RP ; Z2)for all i. However, this contradicts the fact that !i(E) = 0 for i > dim E. (e) False Justification: Theorem 3 (Pontrjagin) If B is a smooth compact (n + 1)-dimensional manifold with boundary equal to M, then the Stiefel-Whitney numbers of M are all zero. Theorem 4 Two smooth closed n-manifolds belong to the same cobordism class if and only if all of their corresponding Stiefel-Whitney numbers are equal. 2 3 5 2 3 2 3 Claim: RP × RP and S are not cobordant. Since !1(RP × RP ) = !1(RP ) [ !0(RP ) = 3a [ 1 6= 0 1 2 3 1 2 2 3 in H (RP × RP ; Z2), where a is the generator of H (RP ; Z2) of RP × RP . So not all Stiefel-Whitney 2 3 numbers are 0. But S5 = @D6. So, by Theorem 3 and Theorem 4, RP × RP and S5 are not cobordant. (2). 9 (a). Determine the least integer k assuming that RP can be immersed in R9+k. Solution: Now suppose that a manifold M n of dimension n can be immersed in Rn+k. Recall that a n n+k n n+k function f : M ! R is an immersion if at every point x 2 M , Dfx : TxM ! R is injective. This means that the tangent bundle TM is an n-dimensional vector bundle, and the normal bundle ν n is a k-dimensional vector bundle. From the Whitney Duality theorem, !i(ν) = !i(M ). So the dual n 9 Stiefel-Whitney classes !i(M ) = 0 for i > k. For M = RP , 9 2 8 !(RP ) = 1 + a + a and 9 2 4 6 !(RP ) = 1 + a + a + a : 9 So if RP can be immersed in R9+k, then k must be atleast 6. n (b). Show that the tangent bundle of RP has a sub bundle of rank 1 if and only if n is odd. n n Solution: Since !(RP ) = (1 + a)n+1. If T (RP ) = ξ ⊕ η for a line bundle ξ, then we must have n !(ξ)!(η) = !(RP ) = (1 + a)n+1. !(ξ) and !(η) must be order 1 and n − 1 respectively(by the axioms of Stiefel-Whitney classes) so it must be the case that !(ξ) = 1 + a or 1. If n is even, then in the first case, n n n n n+1 n n !(η) = (1 + a) has !n(η) = n a = a , and in the later case !n(η) = n a = a . This contradicts the fact that for an (n − 1)-bundle such as η, !n(η) = 0. Thus such a splitting cannot happen. If n is n n odd, then !n(RP ) = 0. This implies that T (RP ) admits one global section by whose span defines a sub-bundle of dimension 1. n (c). Let n = 2k. Find all the Stiefel-Whitney numbers of RP . n Solution: !(RP ) = (1 + a)n+1 = (1 + a)(1 + a)n. Since we are working modulo 2 and n = 2k. We have n n (1+a)n = 1+an. So !(RP ) = (1+a)(1+a)n = 1+a+an +an+1 = 1+a+an. Thus !(RP ) = 1+a+an. n n n n n This implies !0(RP ) = 1, !1(RP ) = a, !n(RP ) = a and !i(RP ) = 0 8 i 6= 0; 1; n. So the Stiefel- n n n n n n n ∼ Whitney numbers h!n(RP ); [RP ]i and h!1(RP ) ; [RP ]i are the only non-zero in H (RP ; Z2) = Z2. All other Stiefel-Whitney numbers are zero. 2 1 1 k (d). Let k ≥ 1 be an integer and fk : S ! S be the power map z 7! z . For which values of k is the ∗ 1 bundle fk (γ1 ) trivial? 1 1 1 1 1 Solution: Let γ1 be the canonical line bundle over RP = S . Since fk : S ! S has degree k, then ∗ 1 1 1 1 ∗ ∗ 1 1 fk : H (S ; Z2) ! H (S ; Z2) such that fk (x) = kx. Suppose the pull back bundle p : fk (γ1 ) ! S is ∗ 1 ∗ 1 ∗ 1 trivial.This implies that 0 = !1(fk (γ1 )) = fk (!(γ1 )) = fk (a) = ka where a = !1(γ1 ) is the generator of 1 1 ∼ H (S ; Z2) = Z2. This shows that k ≡ 0 mod 2 3.
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