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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?2011-02-22
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    why isn't it clear that $S^n$ is compact?2011-02-23
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    It is, but to make the trick work one would have to deduce that from that $\mathbb P^n$ and $S^1$ are compact.2011-02-23

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