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:


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) $T_p\mathcal{P}=T_xM\oplus V,$

  • $\begingroup$ 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. $\endgroup$ – Conifold Mar 4 '19 at 11:04
  • $\begingroup$ 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! $\endgroup$ – rschwieb Mar 4 '19 at 13:42
  • $\begingroup$ 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 $\textbf A_{\mu} \in \Omega^1(M, \mathfrak g)$. $\endgroup$ – Bellem Mar 4 '19 at 13:53
  • $\begingroup$ 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. $\endgroup$ – Bellem Mar 4 '19 at 13:54
  • $\begingroup$ 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... $\endgroup$ – 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

$$ T(TM) \cong TM \times \mathbb{R}^{2n} $$

  • $\begingroup$ Thank you very much, I still have problems with what I wrote in the comment section under my question. If you could take a look I would be glad! $\endgroup$ – Bellem Mar 4 '19 at 14:13

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