Vanishing of non top-order Chern classes

Question

Let $E \to B$ be a rank-$r$ complex vector bundle and denote by $c_1(E)$, $\ldots$, $c_r(E)$ its Chern classes. Then $c_r(E)$ is just the Euler class of the realization of $E$ as a real vector bundle $E_{\mathbb R}$: $c_r(E) = e(E_{\mathbb R})$. Hence, since $e(E_{\mathbb R})$ is the Poincaré dual of zero locus of any transversal to zero section of $E_{\mathbb R}$ then if $E$ posesses a nowhere vanishing section it will imply $c_r(E) = 0$. I think that I've seen in some book (but I don't remember in which) that if $E$ posesses $k$ nowhere vanishing and nowhere linearly dependent sections then $c_{r-k+1}(E)= 0$. Is it really true? If it is true, in which book I can find the proof of this statement that doesn't involve algebraic geometry (I think that I've seen something similar in a book on algebraic geometry)?

Answer 1

Assuming paracompactness, the presence of such sections gives a splitting $E = \theta^k \oplus \eta$, where $\theta$ is the trivial bundle and $\eta = \theta^\perp$. Hence $c(E) = c(\theta) c(\eta) = c(\eta)$ with $\eta$ a bundle of dimension $\leq r - k$.

Answer 2

This one is not a book, but it's worth a look:

http://www.math.ku.dk/~moller/students/mauricio.pdf

Answer 3

Let $\mathcal E$ be a $r$-bundle on $M^n$, and $\sigma_0,...,\sigma_{r-k}$ be general linearly independent global sections of $\mathcal E$. Then the degeneraci locus $\operatorname V(\sigma_0\wedge...\wedge\sigma_{r-k})$ of these sections (consisting of points where they are linearly dependent) should define a class in $H_{n-k}(M)$.

Then $k$-th Chern class is just Poincaré dual of it:

$$\widetilde{c}_k(\mathcal E)=[\operatorname V(\sigma_0\wedge...\wedge\sigma_{r-k})]^*.$$

If there is a collection of $r-k+1$ linearly independent sections, then there is even an open (in Zariski topology) non-empty subset of then, so $k$-th Chern class, calculated this way, is 0.

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To prove equivalence to the usual definition (and independence on choice of sections), let us suppose that $W \subset \Gamma(\mathcal E)$ is $m$-dimensional vector subspace of sections generating $\mathcal E$ and define

$$\phi: M \to Gr(m-r,W), p \mapsto \ker (W \to E|_p).$$

Then it is enough to check $\widetilde{c}_k(\mathcal E)=\phi^*(c_k(\gamma))$, but $c_k(\gamma)=\sigma_k^*$ is Poincaré dual to Schubert class, so by naturality of Poincaré duality it is equal to

$$\phi_*([V(\sigma_0\wedge...\wedge\sigma_{r-k})])=\sigma_k.$$

Denoting $U=\langle \sigma_0,...,\sigma_{r-k} \rangle$ vector space spanned by these sections,

$$\phi(V(\sigma_0\wedge...\wedge\sigma_{r-k}))=\{\Lambda \in Gr(m-r,W) : \Lambda \cap U \not= \emptyset\}=\Sigma_k(U).$$