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I went about it this way:

$\xi$ an $n$-rank vector bundle over a topological space $M$ is orientable.

$\iff$ its top exterior power $\bigwedge^n\xi$ is trivial

$\iff$ the 1-frame bundle $V_1(\bigwedge^n\xi)$ admits a section

$\iff$ the obstruction class $w_1(\bigwedge^n\xi)=\mathfrak o_1(\bigwedge^n\xi)\in H^1(M,\pi_0(V_1(\mathbb R)))$ vanishes

Is the argument okay so far?

If yes, I have proven $\xi$ is orientable $\iff w_1(\bigwedge^n\xi)=0$. Can one help me prove $w_1(\bigwedge^n\xi)=w_1(\xi)$?

If no, then how does one argue?

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up vote 6 down vote accepted

First we need a few preliminary facts:

Let $G_n$ be the infinite Grassmann manifold, i.e., the set of $n$-dimensional subspaces in $\mathbb R^\infty$

We know $G_n=V_n(\mathbb R^\infty)/O(n)$. Let $\widetilde G_n$ be its double cover $V_n(\mathbb R^\infty)/SO(n)$. Let $\gamma_n$ be the canonical bundle over $G_n$ and $\widetilde\gamma_n$ be its pullback under the covering projection.

  1. By the homotopy exact sequences for fiber bundles $SO(n)\hookrightarrow V_n(\mathbb R^\infty)\rightarrow \widetilde G_n$ and $O(n)\hookrightarrow V_n(\mathbb R^\infty)\rightarrow G_n$ we see that:$$\pi_1(\widetilde G_n)\cong\pi_0(SO(n))=0$$and$$\pi_1(G_n)\cong\pi_0(O(n))=\mathbb Z_2$$since $V_k(\mathbb R^\infty)$ is contractible.

  2. Also, as a consequence: $H_1(G_n)=\mathbb Z_2$, whence $H_1(G_n,\mathbb Z_2)\cong H_1(G_n)\otimes\mathbb Z_2=\mathbb Z_2$, implying $H^1(G_n,\mathbb Z_2)\cong\text{Hom}(H_1(G_n,\mathbb Z_2),\mathbb Z_2)=\mathbb Z_2$ (Since $\mathbb Z_2$ is a field, it is elementary homological algebra that the $\text{Ext}$ and the $\text{Tor}$ groups in the Universal Coefficient Theorem vanish)

  3. The canonical map $\pi_1(X)\rightarrow H_1(X)$ is a surjection and is by above, an isomorphism when $\pi_1(X)=\mathbb Z_2$

Now we prove:

As always, we have a map $f:M\rightarrow G_n$ such that $f^*\gamma_n=\xi$, where $\gamma_n$ is the canonical bundle over $G_n$ (the fibre over a subspace is the subspace itself)

Suppose $w_1(\xi)=0$

Then, $f^*w_1(\gamma_n)=0$ (naturality of Steifel-Whitney classes)

$\Rightarrow f^*:H^1(G_n,\mathbb Z_2)\rightarrow H^1(M,\mathbb Z_2)$ is zero (because $w_1(\gamma_n)$ generates $H^1(G_n,\mathbb Z_2)$)

$\Rightarrow f_*:H_1(M,\mathbb Z_2)\rightarrow H_1(G_n,\mathbb Z_2)$ is zero (because $H^1(-,\mathbb Z_2)=\text{Hom}(H_1(-,\mathbb Z_2),\mathbb Z_2)$ and the zero map on dual modules induces the zero map on the modules themselves)

$\Rightarrow f_*:H_1(M)\rightarrow H_1(G_n)(=\mathbb Z_2)$ is zero (because $H_1(-,\mathbb Z_2)=H_1(-)\otimes\mathbb Z_2$ and a non-zero map $M\rightarrow \mathbb Z_2$ of modules induces a non-zero module map $M\otimes\mathbb Z_2\rightarrow\mathbb Z_2\otimes\mathbb Z_2$)

Now the surjection from $\pi_1$ to $H_1$ (fact 3 above) of a space and $f_*$ result in a commutative diagram $\require{AMScd}$ \begin{CD} \pi_1(M) @>{h}>> H_1(M);\\ @VVV @VVV \\ \pi_1(G_n) @>{h}>> H_1(G_n); \end{CD}

This coupled with the $\pi_1$ to $H_1$ map being an isomorphism in case of $X=G_n$ (fact 3) implies that $f_*:\pi_1(M)\rightarrow\pi_1(G_n)$ is the zero map.

So by the lifting criterion $f$ lifts to $\tilde f:M\rightarrow \widetilde G_n$, so that now $\xi=f^*\gamma_n=\tilde f^*\widetilde\gamma_n$.

Since the $\gamma_n$ is orientable, so is its pullback $\xi$.

To prove the converse,

If $\xi$ is orientable, then there is a map $f:M\rightarrow\widetilde G_n$ such that $f^*\widetilde\gamma_n=\xi$.

We certainly have $w_1(\widetilde\gamma_n)=0$ (has to be, because $\pi_1(G_n)=0$, whence $H_1$ vanishes and so it does with coefficients in $\mathbb Z_2$, implying that $H^1(\widetilde G_n;\mathbb Z_2)=0$ we see that $w_1(\xi)=f^*w_1(\widetilde\gamma_n)=0$)$\quad\square$

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