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By an application of Gromov's $h$-principle, an open manifold $W$ admits a symplectic structure precisely if $TW$ admits an almost complex structure. In the case you are considering it does, as $TW$ i …

The condition for having a $Pin^+$-structure is the vanishing of $w_2$, and for having a $Pin^-$-structure is the vanishing of $w_2 +w_1^2$, for manifolds of any dimension. This is because the Lie gro …

I don't know a reference, but you can proceed as follows. By the splitting principle, it suffices to give the formula for vector bundles which are sums of complex line bundles, and we may as well then …

I may have miscalculated, but writing $e(\nu)$ for the Euler class of the normal bundle to such an immersion I find that $$\int_{\mathbb{CP}^3} e(\nu) c^2 = 2,$$ which appears to contradict 1). To se …

No: for example, there is no 1-manifold with Stiefel--Whitney number for $w_1$ equal to 1.

No, because the Wu formulae express $\mathrm{Sq}^j(w_i)$ in terms of $w_k$'s, so if the $x_i$ you choose don't satisfy this formula, they cannot possibly arise as Stiefel--Whitney classes.

No, because there are not enough of them. The lowest Steenrod operation $\mathcal{P}^1$ raises degree by $2(q-1) = 4 \tfrac{q-1}{2}$, so $\mathcal{P}^1(p_1)$ has degree $4(\tfrac{q-1}{2}+1)$ and so yo …

I will explain that as long as $n>2$ the real vector bundles $\Omega^{even}$ and $\Omega^{odd}$ over $M$ are isomorphic. If $n>2$ then $dim(\Omega^{even}) = dim(\Omega^{odd}) = 2^{n-1} > n$ and so $\O …