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My lecture notes on differential geometry read the following (without proof):

For $M$ a manifold, let $TM = \bigcup_{p \in M} T_p M$ be the (disjoint) union of all its tangent spaces. Then, there exists a unique topology and smooth structure on $TM$ making it into a smooth manifold such that any section $X:M \rightarrow TM$ of the canonical projection map is smooth if and only if for all smooth functions $f$ on $M$, the function $Xf$ is smooth.

I am not quite sure as to how to approach this (I am talking about the uniqueness part, the usual topology and smooth structure evidently imply the desired equivalence). I found a related post without an answer here: Why is the manifold structure on the tangent bundle unique? It seems that in this post, OP also assumes the projection map to be continuous, whereas my lecture notes do not.

Any help would be appreciated.

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This is two separate results tied in one. I'm not going to write all the details here since they're quite long and besides, their complete proof were all written in great details in Lee's $\textit{Introduction to Smooth Manifolds, 2nd ed}$ . The first one, $TM$ as smooth manifold, is proved in $\textbf{Proposition 3.18.}$, and the uniqueness follows from a lemma used to proved the proposition (Lemma 1.35).

The second is about characterization of smooth vector fields $X : M \to TM$, proved in $\textbf{Proposition 8.14}$., says that a vector field $$ X : M \to TM \text{ is smooth } \Leftrightarrow \forall f \in C^{\infty}(M), \text{ the function }Xf \text{ is smooth on }M \Leftrightarrow \text{ For every open subset } U\subseteq M \text{ and } \forall f \in C^{\infty}(U) \text{ then function } Xf \text{ is smooth on } U. $$ Good Luck.

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  • $\begingroup$ My problem is this: we want to use Lemma 1.35 to prove uniqueness, so following the notation in Lee, we want to prove that given the equivalence in Proposition 8.14, all bijections of type $\tilde{\phi_{\alpha}}$ are actually charts (i.e. diffeomorphisms). By using the equivalence, it follows that coordinate vector fields are smooth, so that $(\tilde{\phi_{\alpha}})^{-1}$ is smooth. But there is nothing we can say about $(\tilde{\phi_{\alpha}})$ itself. What am I missing? $\endgroup$ – Thomas Bakx Sep 9 '18 at 16:29
  • $\begingroup$ @ThomasBakx I don't really know what do you meant. First, once Lee's proved that $TM$ is a smooth manifold (Proposition 3.18), then by Lemma 1.35, the topology and smooth structure are unique. Then with this topology and smooth structure Proposition 8.14 says that any section $X : M \to TM$ is smooth iff $Xf$ is smooth function for any $f$. The uniqueness does not follows from Proposition 8.14. $\endgroup$ – Sou Sep 9 '18 at 17:15
  • $\begingroup$ It's important to be precise here. Proposition 3.18 says (using Lemma 1.35) that $TM$ has a unique smooth manifold structure such that the bijections $\tilde{\phi_{\alpha}}$ are smooth charts. Of course $TM$ as a set can be given any manifold structure you want it to have, but this one makes the canonical trivializations induced by charts into diffeomorphisms, which is desirable. So, I repeat, the real question is whether given the equivalence as stated in Proposition 8.14 (or in my original post), the canonical trivializations are indeed diffeomorphisms. And this I do not see. $\endgroup$ – Thomas Bakx Sep 9 '18 at 18:04
  • $\begingroup$ @ThomasBakx Even without the equivalence, $\phi : \pi^{-1}(U) \to \varphi(U) \times \Bbb{R}^n$ is indeed diffeomorphism by definition of smooth structure on $TM$. $\endgroup$ – Sou Sep 9 '18 at 18:34
  • $\begingroup$ You do not understand my question. We want to put a smooth structure on $TM$ such that the equivalence holds. That means that it is not defined yet. Hence, using only the equivalence, we need to show that the equivalence implies the smoothness of the maps $\tilde{\phi_{\alpha}}$ and their inverses. This does not seem to be possible, as we are only given a characterization of smooth maps into the manifold (which gives us smoothness of the inverses) but no smoothness of maps out of the manifold. $\endgroup$ – Thomas Bakx Sep 9 '18 at 18:42
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Since it seems that awarding the bounty isn't going to do much, I'll post my own (partial) answer anyways:

By Proposition 10.24 in Introduction to Smooth Manifolds (Lee), we can deduce the following:

Theorem: The natural topology and smooth structure on the set $TM = \cup_{p \in M} T_p M$ are unique when we require the following:

(i) $\pi: TM \rightarrow M$ is a smooth vector bundle, where $\pi$ is the canonical projection map $(p,v_p) \mapsto p$.

(ii) For $U \subset M$ open and $X: U \rightarrow TM$ an arbitrary local section, if $Xf \in C^{\infty}(U)$ for all $f \in C^{\infty}(U)$ then $X$ is smooth.

Proof: Indeed, it follows from property (ii) that local coordinate vector fields are smooth, so that we can apply Proposition 10.24.

Note that we did require the manifold structure of $TM$ to be of 'bundle' type. Intuitively, it would seem that we want the zero section to be a smooth embedding of $M$ into $TM$. Then, knowing the topology on the fibers as well and requiring a local 'product structure' seems to determine everything. As I pointed out in one of my comments above, it is not likely that we can achieve any such result without smoothness requirements on some map which goes out of $TM$, and the only canonical one we have is the projection. Perhaps condition (i) can be weakened to something like '$\pi$ is a smooth submersion'?

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  • $\begingroup$ If you assume condition (ii) and that $\pi$ is a smooth submersion, then you can characterise the smooth maps $N\to TM$ as those which locally factor through smooth sections. But knowing all smooth maps into a manifold characterises its topology and smooth structure. $\endgroup$ – Ben Sep 13 '18 at 8:19
  • $\begingroup$ Can you be more precise about that last statement and provide a proof of it? Because that's exactly the type of result I was after when I posted this question. Indeed, smooth vector fields (maps into $TM$) are fully characterized by the fact that they map smooth functions to smooth functions. If you can turn this comment into an answer it will receive the bounty. $\endgroup$ – Thomas Bakx Sep 13 '18 at 19:14
  • $\begingroup$ I don't know when I will find the time to do that, but I'll come back to this. $\endgroup$ – Ben Sep 13 '18 at 19:32
  • $\begingroup$ Hmm, there is some gap in the argument I had in mind and I don't quite have the time to try to fix it, but here is the idea. I thought from these two assumptions I could prove something like this: A map $g\colon N\to TM$ from any smooth manifold $N$ is smooth if and only if the composite $\pi\circ g\colon N\to M$ is smooth and if for every smooth function $M\to\mathbb R$, the induced map $N\to\mathbb R$ (using that tangent vectors are derivations) is smooth. But right now I can only show this for those maps $g$ such that $\pi\circ g$ is a local diffeomorphism, which is probably not enough. $\endgroup$ – Ben Sep 18 '18 at 16:37
  • $\begingroup$ Anyway, if we had this statement, then it would follow that the smooth structure is unique since it shows that the identity map $TM\to TM$ is smooth even if we put a-priori different smooth structures on the two copies of $TM$; by symmetry, it would be a diffeomorphism. $\endgroup$ – Ben Sep 18 '18 at 16:40

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