Naive question on characteristic classes of $Gr(k,n)$

Question

Let $Gr(k,n)$ be the Grassmann manifold of $k$-planes in $\mathbb{R}^n$ and $\gamma_k$ its tautological $k$-plane bundle. Is it obvious to see (or even true) that the Stiefel-Whitney classes $w_i(\gamma_k)$ are non-zero? ($ i \leq k$)

Answer 1

I will answer this question for the axiomatic definition of Stiefel-Whitney-Classes (in the end they are all the same classes).

First of all note that every $k$-dimension vector bundle over $X$ living in an $n$-dimensional trivial bundle can be written as a pullback of the tautological bundle over $Gr(k,n)$ (This is fairly standard, note here that this is not the classifying space since there are non-homotopic maps giving the same bundles)

Now the sumation axiom and the non triviality give us that $\gamma_1 \oplus \gamma_1$ is a bundle with non-trivial second stiefel whitney class over $\mathbb{R}P^2$. Now this bundle has a classifying map into $Gr(2,6)$. By naturality we get that the second Stiefel Whitney class of the tautoligcal bundle over $Gr(2,6)$ is non trivial.

With this procedure we can go further to prove that every Stiefel-Whitney class of the tautological bundles is non-zero. At certain points you will need the gauss map, which gives you an isomorphism between $Gr(k,n)$ and $Gr(n-k,n)$ and also that external products of vector bundles behave nicely and that you can compute them easily (This is künneth together with naturality).

For example you can use $\gamma_1 \hat{\oplus} \gamma_1$ as a bundle over $S^1 \oplus S^1$ to get that the second Stiefel-Whitney-Class of $Gr(2,4)$ is non-zero.