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?


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 $w_1(\xi)=f^*w_1(\widetilde\gamma_n)=0$)$\quad\square$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.