# Question regarding Ehresmann connection

Assume we have a fiber bundle $$F\to E\stackrel \pi\to B$$.

In the wikipedia article it is stated that $$E$$ the vertical bundle $$V=\ker d\pi$$ consisting of vectors along the fibers is canonically defined while the horizontal bundle of vectors along the base is not.
Specifying a horizintal subspace of $$TE$$ is then called an Ehresmann connection on $$E$$.

But isn't $$\pi^* TB$$ a canonical subbundle of $$TE$$ which consists of vectors along $$B$$?

## 2 Answers

No! There is no reason why $$\pi^*TB$$ is a subbundle of $$TE$$ -- this is precisely why you need a connexion.

Recall $$\pi^*TB$$ is actually constructed as $$E\times_B TB$$: $$\pi^*TB=\{(e,(p,v))\in E\times TB\mid \pi(e)=\operatorname{proj}_{TB\to B}(p,v)=p\}$$ where $$\operatorname{proj}_{TB\to B}$$ is the usual projection $$TB\to B$$.

• Then why is it claimed differently here? map.mpim-bonn.mpg.de/Tangent_bundles_of_bundles_(Ex) – Peter Jun 6 at 8:59
• A connexion $\nabla$ is precisely a choice of splitting $TE=H\oplus V$ with $H\cong \pi^*TB$. You don't have a canonical choice of $H$. – user10354138 Jun 6 at 9:03
• So independently of the connection, there always is an isomorphism $H\to \pi^* TB$? Specifying such an isomorphism does then define a unique connection? – Peter Jun 6 at 9:36

By definition of the pullback, $$\pi^*B=\{(x,y,v):y\in\pi^{-1}(x), v\in T_xB\}$$ therefore it is not a subbundle of $$TE$$.