Regarding a proof that $T(M\times N) \cong TM\times TN$

Question

I want to ask about the answer in this link:

Tangent Bundle of Product Manifold

How is the identification $T_{(x,y)}(M\times N)=T_xM\oplus T_yN$ used in (*) ?

And in writing $T(M\times N)$ and $TM\oplus TN$ this way as sets (lines 4-6), are we also using that $T_xM=R^m$ and $T_yN=R^n$?

Why is $\phi$ a diffeomorphism? And is the last conclusion that ${\phi}^{\sim}$ a diffeomorphis a sort of a theorem?

Thanks a lot

Answer 1

Perhaps it would be better to think about it this way: from the definition of tangent bundle we have $$T(M\times N) = \{(z,u): z\in M\times N\text{ and } u\in T_z(M\times N)\}.$$ Now, clearly $z\in M\times N$ can be decomposed into $z=(x,y)$ where $x\in M$ and $y\in N$. But how can we decompose $u\in T_z(M\times N)$ into vectors from the tangent spaces of $M$ and $N$? That is where the (∗) assumption comes in: since $T_{(x,y)}(M\times N)\cong T_xM\oplus T_yN$, we can say $u=(v,w)$ where $v\in T_xM$ and $w\in T_yM$. Thus the tangent bundle of the product manifold can be rewritten as $$T(M\times N) = \{(x,y,v,w): x\in M, y\in N, v\in T_xM,\text{ and } w\in T_yN\}.$$

And in writing $T(M\times N)$ and $TM\oplus TN$ this way as sets (lines 4-6), are we also using that $T_xM=R^m$ and $T_yN=R^n$?

In the text of the linked answer, yes. It is how they can say the tangent spaces of the product manifold $T_z(M\times N)$ can be identified with $\mathbb{R}^{m+n}$. You don't really need this fact (yet) in the way that I've written the sets.

Why is $\phi$ a diffeomorphism? And is the last conclusion that $\tilde{\phi}$ a diffeomorphism a sort of a theorem?

I'm not surprised this step was confusing, because this was the most imprecise part of the linked answer. The goal here is to show that the "switching map" that puts the sets $T(M\times N)$ and $T(M)\times T(M)$ in bijection is actually a diffeomorphism with respect to their manifold structure.

Let's change notation a bit and consider the map $F: U\times V\times\mathbb{R}^{m+n} \to (U\times \mathbb{R}^m)\times (V\times \mathbb{R}^n)$ (where $U\subset \mathbb{R}^m$ and $V\subset \mathbb{R}^n$ are open) defined as $$(x,y,v,w) \mapsto (x,v,y,w).$$ This is just an invertible linear map defined on some open subsets of Euclidean space, and hence a diffeomorphism in the Euclidean sense (i.e., it is a smooth bijection of open sets with smooth inverse). But don't let the choice of variables $x,y,v,w$ fool you--these are elements of Euclidean space for the moment. We aren't on the manifold yet.

To obtain a map defined (locally) on the manifold $T(M\times N)$, we need to go through coordinate charts. We can take a chart neigborhood of $(x,y,u,v)\in T(M\times N)$ which takes us to $U\times V\times \mathbb{R}^{n+m}$, where $U$ and $V$ are open as above. Call this $\Phi$. Then we can apply $F$ to switch the coordinates around, so we are in the Cartesian product of $U\times\mathbb{R}^m$ and $V\times \mathbb{R}^n$. But these now look like the product of images of chart neighborhoods of $TM$ and $TN$, which we denote $\phi$ and $\psi$ respectively. So we can go through the map $\phi^{-1}\times\psi^{-1}$ to get from $(U\times\mathbb{R}^m) \times (V\times \mathbb{R}^n)$ to $TM\times TN$. In total, our map is defined locally as $((\phi^{-1}\times\psi^{-1})\circ F\circ \Phi)$. This map is a diffeomorphism because $F$ is a diffeomorphism. By doing this procedure on every coordinate chart of $T(M\times N)$, we can define a map $\tilde F:T(M\times N)\to TM\times TN$ defined on the entire manifold (the only worry is when two coordinate charts of $T(M\times N)$ overlap, since this could result in definitional ambiguity, but the transition maps between charts being diffeomorphisms will guarantee there is no ill-definedness).

Now, we have a map $\tilde F$ which is a bijection and, by construction, a local diffeomorphism. We can upgrade this to a global diffeomorphism by the inverse function theorem for manifolds: indeed, since $\tilde F$ is a bijection there is an inverse $(\tilde F)^{-1}: TM\times TN\to T(M\times N)$. If $\tilde F$ is a local diffeomorphism at $p$, then the inverse function theorem says $(\tilde F)^{-1}$ is differentiable at $F(p)$. Since every point $q\in TM\times TN$ can be written as $q=F(p)$ for some $p$, we see that $(\tilde F)^{-1}$ is differentiable everywhere and the result follows. (Note also this argument applies to any bijective local diffeomorphism of manifolds, so yes, this is a theorem).

Some parting remarks: the linked answer has a small blunder in the first line when they assume $M$ and $N$ live in $\mathbb{R}^m$ and $\mathbb{R}^n$ respectively. Since they later identify the tangent spaces of $M$ and $N$ with these same sets, I assume they mean for $M$ to be $m$-dimensional and likewise for $N$ to be $n$-dimensional. But one cannot generally embed an $m$-dimensional manifold into $\mathbb{R}^m$ (see the Whitney Embedding Theorem).

Also, this is generally a messy way to analyze these objects. Once you've learned a little about vector bundles (I presume you haven't seen them yet), you should try understanding this problem from the bundle perspective.