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Let $X$ be a complex manifold equipped with a smooth hermitian metric $h$. We can define a sub-fibration $B \to X$ of the tangent bundle $T_X$ by requiring that the fiber over a point be the unit ball in $T_X$ at that point, i.e.

$$ B_x = \{ \xi \in T_{X,x} \, | \, \| \xi \|_{h(x)} = 1 \}. $$

I want to see that if $X$ is compact, then the fibration $B$ is compact in the total space of $T_X$. This is just a global version of the usual fact that the unit ball in a normed finite dimensional vector space is compact.

I can prove this by using the (horrible) definition of $T_X$ as the disjoint union of the stalks $T_{X,x}$ modulo an equivalence relation. As each $B_x$ is compact, their disjoint union is compact in the disjoint union of the stalks by Tychonoff, and dividing by the equivalence relation is a continuous map so $B$ is compact in the total space of $T_X$.

Isn't there a nicer way to see this? I ask both because I really don't like the definition of the tangent space as a disjoint union of stalks, and because I have to prove a similar lemma for the relative tangent space associated to deformations of a manifold $X$. A similar proof works in that case, but it is quite dirty.

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I wonder if we can trick our way out of this by looking at the projectivized bundle $\mathbb P(T_X)$? The usual projective space is the unit ball $S^{2n+1}$ modulo $S^1$, so maybe the projectivized bundle is the unit ball fibration modulo $S^1 \times X$? Hmm... if we know that $\mathbb P^n$ is compact, then does it follow that $S^{2n+1}$ is compact? –  Gunnar Magnusson Feb 22 '11 at 17:16
why isn't it clear that $S^n$ is compact? –  Sean Tilson Feb 23 '11 at 4:16
It is, but to make the trick work one would have to deduce that from that $\mathbb P^n$ and $S^1$ are compact. –  Gunnar Magnusson Feb 23 '11 at 6:53
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1 Answer

Hint: Here is an outline of a possible proof.

1) Do the problem in the special case that the ball fibration is trivial, i.e., isomorphic to a product.

2) Convince yourself that the ball fibration is locally trivial.

3) Since the base is compact, you can find a finite open cover $\{U_i\}$ such that the restriction of the fibration to each $U_i$ is trivial. Now use the fact that a finite union of compact sets is compact.

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Isn't there a problem in 1), i.e. that $S^{2n+1} \times U$ is not compact in $\mathbb C^n \times U$ for $U$ open? We can get out of it by using the closures of relatively compact $V$ in $U$, but that still leaves the issue in 3) where we need to quotient by the equivalence relation induced by the transition functions. I have the impression that this is basically the same proof as I outlined, except we avoid applying Tychonoff. –  Gunnar Magnusson Feb 22 '11 at 18:10
Yes, you want to make sure to use $U_i$'s which have compact closure contained in some other trvializing open set. But since the base space is regular, this is no problem. As to whether it's the same as your argument: well, you said you wanted to avoid the definition of the tangent bundle as a disjoint union of stalks modulo an equivalence relation, and the argument I gave does seem to avoid this... –  Pete L. Clark Feb 22 '11 at 18:15
Indeed it does, thank you Pete. But what I'd ideally like is something like: "If $V \to X$ is a vector bundle over a compact base, and $B \to X$ is a sub-fibration where each fiber is compact, then $B$ is compact in $V$." Here we have more, because as you said, the ball fibration is locally trivial. Some dirty elementary topology can probably give what I need, but I wondered if there wasn't a clever way to do this before I broke out pages and pages of open coverings. –  Gunnar Magnusson Feb 22 '11 at 18:16
P.S.: I don't see where we're taking the quotient by anything in my argument. You will get the space you want as a finite union of compact subspaces, so it's compact. –  Pete L. Clark Feb 22 '11 at 18:17
@Gunnar: but are you really interested in non-locally trivial fibrations? That would seem to lie in the realm of algebraic topology, not manifold theory. If so, you should probably edit your question to reflect this. –  Pete L. Clark Feb 22 '11 at 18:19
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