Second Stiefel-Whitney Class of a Five Manifold There is a unique rank 4 nontrivial orientable vector bundle over the 2-torus, denote this by $p:E\rightarrow T^2$. Denote the associated sphere bundle by $S(E)$. Then since $S(E)$ is orientable, the first Stiefel-Whitney class of the five manifold $S(E)$ is trivial. By Wu's formula, it's not hard to see that the second Stiefel-Whitney class $w_2(S(E))$ equals the second Wu class. But I have trouble in finding the second Wu class. Can anyone help me this out? Thanks!
 A: I'm assuming that your end goal is the computation of $w_2(TS(E))$, rather than that of the Wu class.  If so, here's how I'd proceed.  First, I'll abuse notation and let $p:S(E)\rightarrow \mathbb{T}^2$ denote the projection map.
First note that $\pi_1(\mathbb{T}^2)\cong\mathbb{Z}^2$ acts trivially on $H^\ast(S^3)$.  This follows because the bundle is orientable and the only non-zero cohomology group of $S^3$ is in degree $3$, and is generated by a choice of orientation.  Thus, in particular, the Serre spectral sequence machinery works easily (no extension problems!), and we get $H^\ast(S(E)) \cong H^\ast(S^3\times \mathbb{T}^2)$.  We also see from the edge homomorphism, that $p^\ast:H^2(T^2)\rightarrow H^2(S(E))$ is an isomorphism.  This implies that $H^\ast(S(E);\mathbb{Z}_2)\cong H^\ast(S^3\times \mathbb{T}^2; \mathbb{Z}_2)$ and that the induced map on $H^2$ with $\mathbb{Z}_2$ coefficieints is also an isomorphism.
Now, we use a nice trick.  I claim that $TS(E)\oplus 1$ is isomorphic to $p^\ast(T \mathbb{T}^2)\oplus p^\ast(E)$.
Believing this, it follows that $w(TS(E)\oplus 1) = p^\ast(w(T\mathbb{T}^2))\cup p^\ast (w(E))$.  But $\mathbb{T}^2$ is parallelizable, so this reduces to $w(TS(E)) = p^\ast(w(E))$.  In particular, since $w_1(E) = w_k(E) = 0$ for $k>2$ while $w_2(E)$ is a generator of $H^2(\mathbb{T}^2;\mathbb{Z}_2)$, we conclude that $w(TS(E))$ is only nonzero in degree $2$ (and degree $0$), where it is the unique nonzero element.
So, why does the nice trick work?  Well, it works for any oriented sphere bundle $S$ which is the sphere bundle of a vector bundle $S = S(E)$.  The idea is to pick Riemannian metrics on the total space $E$ and on the base for which the projection is a Riemannian submersion.  This allows you to canonically decompose $T_{(b,v)}E$ into three pieces: vectors tangent to $B$, those tangent to the sphere in $p^{-1}(b)$, and those vectors in $p^{-1}(b)$ pointing radially inward/outward.  (The inward/outward is there the $\oplus 1$ comes from.)
