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This is my first post on the stackexchange, so I'm sorry if its rambling.

I've been working through Lee's Introduction to Smooth Manifolds and I'm having some trouble with one of the exercises. Most of the earlier sections are understandable, and I can follow the proofs given in the text. Although working on the exercises are hard for me to evaluate the my proofs on my own.

I'm trying to prove that $TM\times TN$ is diffeomorphic to $T(M\times N)$, given that $M$ and $N$ are smooth manifolds. Since $M$ and $N$ are smooth their product manifolds $M\times N$ are smooth, and tangent bundles of those product manifolds will be smooth manifolds as well. So $TM\times TN$ and $T(M\times N)$ are both smooth manifolds, but I don't know how to show they are diffeomorphic.

Any advice would be appreciated.

user90242
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  • Which definition of the tangent space at a point of a manifold do you use? – Dan Rust Aug 13 '13 at 18:56
  • Lee defines the tangent space to be the set of derivations to a point on the manifold, where a derivation is a linear map from the set of smooth functions defined on the manifold, into the real numbers. – user90242 Aug 13 '13 at 19:06

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For each fixed $(p,q)\in M\times N$ you have an isomorphism $T_{(p,q)}(M\times N)\cong T_p M\times T_q N$. The isomorphism is merely $X\mapsto (\pi^1_\ast(X),\pi^2_\ast(X))$ where $\pi^i$ is the projection of $M\times N$ onto the $i$th coordinate.

So, now consider the map $T(M\times N)\to TM\times TN$ which, on fibers, is precisely the described above map. If we can prove that this is smooth, then evidently it is a bijective bundle map, and so a bundle isomorphism.

But, check in coordinates (the proper coordinates--think about how you can combine coordinates of $M$ and $N$ to get coordinates of $M\times N$) that this map has a very nice coordinate representation (it's constant).

Alex Youcis
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  • I like this approach, but unfortunately I haven't gotten to bundle isomorphisms in the text yet... – user90242 Aug 13 '13 at 19:09
  • @user90242 Those are just fancy trappings :) The map is still a diffeomorphism – Alex Youcis Aug 13 '13 at 19:10
  • It looks like one could represent coordinates of $M\times N$ by $(x^1 ,...,x^m ,y^1 ,..., y^n)$ where $x^i$ and $y^i$ are coordinate functions of $M$ and $N$ respectively. – user90242 Aug 13 '13 at 19:14
  • @user90242 Exactly! Now, see what those coordinates on M,N and MxN lift as coordinates of TMxTN and T(MxN) respectively. The coordinate representation should then be pretty easy :) – Alex Youcis Aug 13 '13 at 19:17
  • In my own attempts at the proof I tried writing smooth charts of $T(M\times N)$ as $(U\times V, \phi \times \psi)$ where $\phi \times \psi$ is the map defined by $\psi \times \phi (u^i \frac{\partial}{\partial x^i}|_p ,v^i \frac{\partial}{\partial y^i}|_p )=(x^1 (p),..., x^m (p), y^1 (p),..., y^n (p) ,u^1 ,...,u^m ,v^1 ,...,v^n)$. Was this along the lines of the coordinate representation you were describing? – user90242 Aug 13 '13 at 19:34
  • @user90242 That looks exactly right to me. So, what does the coordinate map look like in that chart (and the obvious chart on $TM\times TN$)? – Alex Youcis Aug 13 '13 at 19:36
  • $((x^i ,u^i),(y^i ,v^i))$ ? – user90242 Aug 13 '13 at 19:42
  • @user90242 Yes, the obvious coordinates send those to the same thing as your other chart! – Alex Youcis Aug 13 '13 at 19:42
  • Wonderful! Thank you Alex, you've been very helpful at explaining this. – user90242 Aug 13 '13 at 19:48
  • @user90242 No problem! Best of luck! – Alex Youcis Aug 13 '13 at 19:48