Tangent bundle of a trivial bundle

I was asking myself if the tangent bundle of a trivial bundle $$\mathcal{P}=M\times V$$ with fiber $$\pi: \mathcal{P}\to M$$ (actually it would be a principal trivial bundle, but I think it doesn't matter for the question), where $$V$$ is a vector space (where a certain group $$G$$ acts), has same fibers of the tangent bundle over $$M$$, so if this holds:

$$T_p\mathcal{P}=T_xM,$$

where $$p\in \mathcal{P}$$ such that $$\pi(p)=x\in M$$. And therefore $$\bigcup_{p\in \mathcal P} \{p\}\times T_p \mathcal P = V \times \bigcup_{x\in M}\{x\}\times T_x M .$$

*edit: of course (according to last equation too) $$T_p\mathcal{P}=T_xM\oplus V,$$

• First, it is not "the same", at best, they can be canonically isomorphic. And second $T_p\mathcal{P}$ has the same dimension as $\mathcal{P}$, and $T_xM$ has the same dimension as $M$, so they can not possibly be isomorphic, unless $V$ is trivial. What you have is $T_p\mathcal{P}\simeq T_xM\oplus V$, this might extend to a bundle isomorphism, but you'll have to check carefully if the isomorphism can be made canonical globally. – Conifold Mar 4 '19 at 11:04
• Surely you know it is not good to use all caps? This has been the norm on the internet for a quarter century or more. If you didn't know, well, let me be the first to welcome you to the internet! – rschwieb Mar 4 '19 at 13:42
• Thanks for reporting the error. I immediately corrected it, otherwise last equation did not hold. I also thought about the different dimensions, but I don't know out turn this around: I am reading on a paper that that if I have $\sigma : M\to\mathcal P$ a trivializing section, then I can pull back algebra valued differential forms $\textbf A_{\mu} \in \Omega^1(\mathcal P, \mathfrak g)$ to $\textbf A_{\mu} \in \Omega^1(M, \mathfrak g)$. – Bellem Mar 4 '19 at 13:53
• Since they are sections on a certain tensor product bundle I would say that if that holds, then $\sigma ^* \mathcal P \times \mathfrak g \otimes T^*\mathcal P \simeq M \times \mathfrak g \otimes T^*M$ must be holding as well and hence, since a pull-back bundle "shares same fibers" of the bundle, follows what I asked. – Bellem Mar 4 '19 at 13:54
• I explain better what I mean: since in general $\Omega^1(M, V)=\Gamma(M\times V \otimes T^*M)$ then a pull-back section is a section on the pull-back bundle (right?) which then shares same fibers (right?). So they should share same fibers, but apparently they do not, so I was trying to figure that out... – Bellem Mar 4 '19 at 14:18

In general for a bundle of smooth manifolds $$N\to E \to M$$ with projection $$\pi$$, locally $$E$$ looks like $$U_x\times N$$ for $$U_x$$ an neighbourhood of $$x\in M$$, and wrt to this chart the tangent space at a point $$(x,y)$$ looks like $$T_{(x,y)}E\cong T_xM\times T_yN$$. There is even a short exact sequence of bundles
$$ker(D\pi) \to TE \to TM$$
and choice of connection on $$E$$ corresponds to a choice of splitting for this sequence, and hence a direct sum decomposition of $$TE$$ into "vertical" and "horizontal" vectors.
Now let $$M^n$$ be a smooth manifold, $$TM$$ its tangent bundle. At any point $$(x, v)$$ in $$TM$$ the tangent bundle looks like $$T_{(x,v)}\cong T_pM\times T_vT_pM\cong \mathbb{R}^{2n}$$, so the fibres of $$TTM\to TM$$ have twice the dimension as the fibres of $$TM \to M$$. In fact if we assume as in your question that our tangent bundle is trivial, then there is a global isomorphism
$$T(TM) \cong TM \times \mathbb{R}^{2n}$$