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On a double covering there is a differential form $\omega$ arises by the pullback of a differential form under the projection iff it is the pullback of $\omega$ under the map $i$, where $i$ is the map induced from the involution of orientation, $\omega$ itself. Why is this the case?

I ask this question because I am really stuck reading this document.

The step is on page 148 last paragraph. Thanks in advance.

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1 Answer 1

If $\omega$ is a pullback, say $\omega=\pi^*\alpha$, then you can show that $\omega$ is $i$-invariant just from functoriality: $$i^*\omega=i^*\pi^*\alpha=(\pi \circ i)^*\alpha=\pi^*\alpha=\omega \, .$$

Conversely, since $\pi$ is a covering map, it has local inverses, so you can always locally define a form $(\pi^{-1})^*\omega$. The problem is, different choices of inverse might lead to different forms. If you write everything out carefully, you can use the same kind of functoriality argument to see that if $\omega$ is $i$-invariant, all these different forms will be equal, and so $(\pi^{-1})^*\omega$ will be globally well-defined.

EDIT for a lot more detail on the hard direction:

We have a double cover $\pi:\tilde M \to M$, its corresponding involution $i$, and an $i$-invariant form $\omega$ on $\tilde M$. We would like to show that $\omega$ is the pullback of some form $\alpha$ on $M$.

Let $\mathcal{C}$ be a trivializing open cover of $M$. We will define $\alpha$ on $M$ by defining it on each $U \in \mathcal{C}$, and then showing that these definitions all agree on the intersections.

Suppose $U \subset M$ is open and trivializes $\pi$, so $\pi^{-1}(U)$ consists of two disjoint copies of $U$. That is, we have disjoint embeddings $p:U \to \tilde M$, $q:U \to \tilde M$ which locally invert $\pi$ and thus satisfy $i \circ p=q$, $i \circ q=p$. Moreover, by the same kind of functoriality calculation we did before, $$ p^* \omega = (i \circ q)^* \omega = q^*i^*\omega = q^* \omega \, .\\ $$ So there is a unique form $\alpha_U=p^*\omega=q^* \omega$ that can be constructed in this way. Moreover, $$ \omega|_{p(U)}=(p\circ\pi)^*\omega|_{p(U)}=\pi^*p^*\omega=\pi^*\alpha_U \\ \omega|_{q(U)}=(q\circ\pi)^*\omega|_{q(U)}=\pi^*q^*\omega=\pi^*\alpha_U $$ since $p \circ \pi$ and $q \circ \pi$ are the identity maps on $p(U),q(U)$. So $\omega|_{\pi^{-1}(U)}=\pi^*\alpha_U$.

Finally, it is clear from the construction that for $W \subset U$, $\alpha_W=\alpha_U|_W$. So, in particular, for any $U,V \in \mathcal{C}$, we have $$ \alpha_U|_{U \cap V}=\alpha_{U \cap V}=\alpha_V|_{U \cap V} $$ and thus the $\alpha_U$ are all restrictions of the same global form $\alpha$. Since $\omega$ is everywhere locally a pullback of $\alpha$, it is globally a pullback of $\alpha$ as desired.

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Thanks for that response unfortunately I am really hung up on the details of this local inversion argument. Can you go into more detail? Thanks in advance! –  Coherentsheaf Aug 26 '12 at 10:14
    
@Coherentsheaf: I added a bunch more detail. Now that I notice your user name, I'll also suggest that you could think of this as a sheaf-theoretic argument: we're showing the existence of a global section that has the desired properties by seeing that there's no cohomological barrier to its existence... –  Micah Aug 26 '12 at 15:18

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