# Splitting of the tangent bundle of a vector bundle

Let $\pi:E\to M$ be a rank $k$ vector bundle over the (compact) manifold $M$ and let $i:M\hookrightarrow E$ denote the zero section. I'm interested in a splitting of $i^*(TE)$, the restriction of the tangent bundle $TE$ to the zero section.

Intuitively I would guess that one could show the following:

$$i^*(TE)\cong TM\oplus E$$

Is this true? If so, how does the proof work?

Any details and references are appreciated!

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The morphism $\pi:E\to M$ (which is a submersion) induces a surjective tangent morphism $T\pi: TE\to \pi^*TM\to 0$ whose kernel is (by definition) the vertical tangent bundle $T_vE$ .
There results the exact sequence of bundles on E $$0\to T_vE\to TE\stackrel {T\pi}{\to} \pi^*TM\to 0$$ Pulling back that exact sequence to $M$ via the embedding $i$ yields the exact sequence of vector bundles on M: $$0\to E\to TE\mid M \to TM\to 0 \quad (\bigstar )$$ The hypothesis that $M$ is compact is irrelevant to what precedes.
However if $M$ is paracompact, the displayed sequence $(\bigstar )$ splits and you may write $$TE\mid M \cong E\oplus TM$$
Since however the splitting of $(\bigstar)$ is not canonical, I do not recommend this transformation of the preferable (because intrinsic) exact sequence$(\bigstar)$.
Thanks for this great answer! I'm a bit unclear on a few points: 1. How is the map $T\pi$ defined? 2. Why is $E$ the pullback of $T_vE$ under $i$? (3. Why is paracompactness needed for the sequence to split?) – Dave Dec 5 '12 at 17:29
$\:T\pi$ is the disjoint union for $e\in E$ of the tangent maps $D_e\pi: T_eE\to T_{\pi(e)}M\quad$ 2. This is a bit involved. A key ingredient is the *canonical* identification of the tangent space at any point $v$ of a vector space $V$ with that vector space: $T_v V=V\quad$ 3.You can show that any short exact sequence of vector bundles splits with the help of a Riemannian metric, and such a metric always exists on a paracompact manifold. – Georges Elencwajg Dec 5 '12 at 21:17