# 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!

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$(\bigstar)$.

Edit ( September 22, 2016)
I forgot to mention the interesting fact that $T_vE$ and $\pi^*E$ are canonically isomorphic as vector bundles on $E$, so that we have a canonical exact sequence on $E$: $$0\to \pi^*E\to TE\stackrel {T\pi}{\to} \pi^*TM\to 0$$The equality $T_vE=\pi^*E$ ultimately rests on the fact that the tangent space to a vector space $V$ at any point $v\in V$ is canonically isomorphic to that vector space : $T_vV=V$.

• 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, 2012 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. Dec 5, 2012 at 21:17
• Thank you! I still need to think some more about the second point. Do you happen to know a reference by any chance?
– Dave
Dec 5, 2012 at 22:26
• Dear Dave, you might find what you want in Spivak's A Comprehensive Introduction to Differential Geometry, Vol. 1 , to which I have no access right now. Don't forget to check the exercises. Dec 5, 2012 at 22:47
• I will take a look. Thank you very much.
– Dave
Dec 5, 2012 at 23:13

[Note to moderators: This is mostly a comment on the previous answer.]

A reference for the short exact sequence

$$0 \to \pi^*E\to TE \to \pi^* TM \to 0$$

is [1, Vol. I, Chapter 3, Exercise 29 (page 103)]. Its restriction $0 \to E \to TE_{|M} \to TM \to 0$ splits canonically, irrespective of any compactness assumption on the base, since $\pi\colon E\to M$ is split by the zero section. So we always have a canonical isomorphism of vector bundles:

$$TE_{|M} \cong TM \oplus E$$

As $\pi$ is a homotopy equivalence, it follows that the unrestricted sequence also splits, so that we have a (non-canonical) isomorphism

$$TE \cong \pi^*TM \oplus \pi^*E$$

A reference for this argument is [2, Kapitel IX, Satz 6.9 (page 365)].

[1] Michael Spivak, A Comprehensive Introduction to Differential Geometry
[2] Tammo tom Dieck, Topologie

• Existence of a splitting also follows from the perhaps more basic observation that the space of such splittings is an affine space modelled on the space of sections of $Hom(TM, E)$, and hence is not empty. Aug 16, 2020 at 13:59
• Why is the splitting of the unrestricted sequence not canonical? I would have thought that the projection $TE \rightarrow VE$ gives a left split and both projection onto the vertical subbundle of $TE$ and the identification $VE \simeq \pi^*E$ are canonical.
– GFR
May 25, 2023 at 10:45

Just make a comment what have shown above we can use this exact sequence to compute differential of vector bundle morphism :

Let $$E\to M$$ , $$F\to M$$ be 2 smooth vector bundles, and $$f : E\to F$$ be a smooth vector bundle morphism.We know that we have exact sequence

$$0\to E\to TE|_M\to TM\to 0\\0\to F\to TF|_M\to TM\to 0$$ and bundle morphism $$f, df|_M, id$$ provide a morphism between these two complex, the second square commute since $$\pi_E = \pi_F \circ f$$, therefore, $$d\pi_E = d\pi_F \circ df$$, the first square commute as $$E\to TE|_M$$ simply as the inclusion of zero section and $$df$$ will maps zero to zero as it's $$\Bbb{R}-$$linear.

Therefore the map $$df|M : TE|_M \to TF|_M$$ under the identication above is just the map $$df|_M:TM\oplus E \to TM\oplus F\\(v,e)\mapsto (v,f(e))$$