2

Let $M$ be a smooth manifold.

Let $P\rightarrow M, Q\rightarrow M$ and $R\rightarrow M$ be vector bundles over $M$ which combine to form a short exact sequence of vector bundles

$$0\rightarrow P\rightarrow Q\rightarrow R\rightarrow 0.$$

Question : Does this short exact sequence of vector bundles induce short exact sequence of vector space of sections? Is $$0\rightarrow \Gamma(M,P)\rightarrow \Gamma(M,Q)\rightarrow \Gamma(M,R)\rightarrow 0$$ an exact sequence?

Given a smooth map $s:M\rightarrow R$, we can always define a map $M\rightarrow Q$, but there is no good reason for this to be a smooth map. I am looking for a simple example where this is not true.

Praphulla Koushik
  • 925
  • 1
    This looks a lot like the long exact sequence on cohomology. $H^1(M,P)=0$ will guarantee surjectivity to $\Gamma(M,R)$. Since you're working in the smooth category, the sheaf of smooth sections of a vector bundle is a fine sheaf and therefore has vanishing $H^k$ for $k\ge 1$. So, yes, what you want does hold. – Ted Shifrin Oct 11 '20 at 04:19
  • @Ted: is it clear that we can do homological algebra on vector bundles? Vector bundles don't form an abelian category. If you're instead doing homological algebra on sheaves, you'd first need to show that a short exact sequence of vector bundles induces a short exact sequence of sheaves. I guess this isn't so bad though. – Qiaochu Yuan Oct 11 '20 at 04:20
  • 1
    @Qiaochu, yes, I think it works out because we're certainly surjective on sections on sufficiently small open sets. – Ted Shifrin Oct 11 '20 at 04:24
  • @TedShifrin can you please say in 1 more sentence what does it mean by “it works out because we're certainly surjective on sections on sufficiently small open sets”. I guessed something, just want to confirm :) – Praphulla Koushik Oct 11 '20 at 05:07
  • Even in the holomorphic category, given $p\in U\in M$, a surjection of vector bundles should give rise to an open set $p\subset V\subset U$ so that $\Gamma(U,E)\to\Gamma(V,F)$ is surjective. (I think probably you can take $V=U$, in fact, when both $E$ and $F$ trivialize over $U$.) – Ted Shifrin Oct 11 '20 at 19:15
  • More generally, if $P\to Q\to R$ is an exact sequence of smooth vector bundles over $M$ and $U\subset M$ is open, then $P(U)\to Q(U)\to R(U)$ is an exact sequence of $C^\infty(U)$-modules. I proved this here. – Elías Guisado Villalgordo Jul 26 '23 at 14:33

1 Answers1

4

Yes. Every short exact sequence of vector bundles splits; for example, if we pick a Riemannian metric on the bundles we can split them by taking an orthogonal complement. So WLOG every short exact sequence has the form

$$0 \to P \to P \oplus R \to R \to 0$$

and then we get a split exact sequence of spaces of sections

$$0 \to \Gamma(M, P) \to \Gamma(M, P) \oplus \Gamma(M, R) \to \Gamma(M, R) \to 0.$$

Qiaochu Yuan
  • 419,620