The stifle-Whitney classes of a tensor product

[samiraahansaz]

Homework Statement

Let ξ^m and ηⁿ be vector bundles over a paracompact base space. Show that the stifle-Whitney classes of the tensor product ξ^m ⊗ ηⁿ (or of the isomorphic bundle Hom (ξ^m, ηⁿ) can be computed as follows. If the fiber dimensions m and n are both 1 then:
w₁ (ξ¹ ⊗ η¹) = w₁(ξ¹) + w₁(η¹)
More generally there is a universal formula of the form:
w(ξ^m ⊗ ηⁿ) = P_m,n(w₁(ξ^m),…,w_m(ξ^m), w₁(ηⁿ),…,w_n(ηⁿ))
Where the polynomial P_m,n in m+n variables can be characterized as minutes t₁,…, t_m and if σʹ₁,…, σʹ_n are the elementary symmetric functions of tʹ₁,…., tʹ_n
P_m,n (σ₁,…, σ_m , σʹ₁,…, σʹ_n) = ∏^m_i=1∏ⁿ_j=1(1+t_i+ tʹ_j)
(Help: The chohomology of G_m × G_n can be computed by the künneth theorem. The formula for w (ξ^m × ηⁿ) can be verified first in the special case when ξ^m and ηⁿ are Whitney sums of line bundles).

Homework Equations

The Attempt at a Solution

 

[lippyka]

Since ξm and ηn are vector bundles over a paracompact base space, by the Künneth Theorem, we know that the cohomology of ξm × ηn can be computed by the Künneth theorem. So, since the stifle-Whitney classes of the tensor product ξm ⊗ ηn (or of the isomorphic bundle Hom (ξm, ηn) can be computed as follows, we know that the cohomology of ξm × ηn can be used to compute the stifle-Whitney classes of the tensor product ξm ⊗ ηn (or of the isomorphic bundle Hom (ξm, ηn). If the fiber dimensions m and n are both 1 then:w1 (ξ1 ⊗ η1) = w1(ξ1) + w1(η1)This can be seen by using the Künneth formula for the cohomology of ξ1 × η1. Since the Künneth formula is given by∗(X×Y) ≅ H∗(X) ⊗ H∗(Y),we can see that the stifle-Whitney class of the tensor product ξ1 ⊗ η1 is equal to the sum of the stifle-Whitney classes of ξ1 and η1.More generally there is a universal formula of the form:w(ξm ⊗ ηn) = Pm,n(w1(ξm),…,wm(ξm), w1(ηn),…,wn(ηn))Where the polynomial Pm,n in m+n variables can be characterized as minutes t1,…, tm and if σʹ1,…, σʹn are the elementary symmetric functions of tʹ1,…., tʹnPm,n (σ1,…, σm , σʹ1,…, σʹn) = ∏mi=1∏nj=1(1+ti+