Suppose $M,N$ are manifolds, and consider the product $M\times N$. From this answer, I know that:

$T_{(m,n)}(M \times N) \cong T_m M \oplus T_n N $

Can we conclude that $T(M\times N) \cong T(M) \oplus T(N)$

  • $\begingroup$ Consider a bundle which is locally $\mathbb{C}\oplus\mathbb{C}$. From this information, it is not enough to conclude that the total bundle is a sum of two trivial bundles. $\endgroup$ – Michael Burr Jun 24 '15 at 11:28
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    $\begingroup$ Yes, but it takes more work than just identifying that the two vector bundles are isomorphic at each point. You need to show that the global map $T(M\times N) \to T(M) \oplus T(N)$ that you've implicitly defined is actually an isomorphism of smooth vector bundles. $\endgroup$ – Alex G. Jun 24 '15 at 11:28
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    $\begingroup$ To be (constructively) pedantic, the correct assertion is that if $\pi_{1}:M \times N \to M$ and $\pi_{2}:M \times N \to N$ are the projections, then$$T(M \times N) \simeq \pi_{1}^{*} TM \oplus \pi_{2}^{*} TN.$$In order to take a direct sum of vector bundles on the right, the base spaces have to be the same. (And to re-iterate the two previous comments, the isomorphism of bundles is not implied by the fibrewise isomorphism, though if the fibrewise isomorphism is "sufficiently independent of bundle chart" then "it's clear the bundles are in fact isomorphic".) $\endgroup$ – Andrew D. Hwang Jun 24 '15 at 11:43
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    $\begingroup$ So if I wanted to prove this, I would need to construct a homeomorphism $h : T(M \times N) \to \pi_1^{\ast} TM \oplus \pi_2^{\ast} TN$, which takes each fiber from the LHS to the corresponding fiber on the RHS by a linear isomorphism? $\endgroup$ – Mark B Jun 24 '15 at 11:53
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    $\begingroup$ Yes, but you already know such a map! The fibers above $(x, y)$ in the two bundles are $T_{(x, y)}(M\times N)$ and $T_xM \oplus T_yN$ respectively. We have a linear isomorphism from the first to the second. Thus, these isomorphisms all collectively give you a function $T(M\times N) \to \pi_1^*TM \oplus \pi_2^*TN$. Just show that this function is smooth and you'll be done (smoothness of its inverse is automatic). Checking smoothness is easy too, just figure out what the local coordinates of each bundle are. $\endgroup$ – Alex G. Jun 24 '15 at 12:11

Let, $M \subset \Bbb R^m , N \subset \Bbb R^n$ (i.e. considered as subsets of the Euclidean spaces of respective dimensions to start off with!)

$T(M \times N)=\{((x,y),(v,w))\in M \times N \times \Bbb R^{n+m}: (v,w)\in T_{(x,y)}(M \times N)\}=\{(x,y,v,w)\in M \times N \times \Bbb R^{n+m}: (v,w)\in T_{(x,y)}(M \times N)\} \dots (*)$ and $TM \oplus TN=\{(x,v,y,w)\in M \times \Bbb R^m \times N \times \Bbb R^n : v \in T_xM,w\in T_y N\}$

Note that we can write $(*)$ due to the identification of $T_{(x,y)}(M \times N)=T_x M \oplus T_yN$

Let's make out intentions clear, we want to "just switch $y$ and $v$" .

Now take open sets $U,V$ in $\Bbb R^m,\Bbb R^n$ (respectively) containing $x \in X$ and $y \in Y$ respectively.Note that $U \times V$ is again an open set in $\Bbb R^{m+n}$. Then look at the "switching map" here, i.e. $$\phi: T(U \times V) \to T(U) \times T(V)$$ $$(x,y,v,w) \to (x,v,y,w)$$

Again note that, $T(U \times V)=U \times V \times \Bbb R^{n+m}$ and $T(U) \times T(V)= U \times \Bbb R^n \times V \times \Bbb R^m$ , hence the switching here makes perfect sense and in fact a diffeomorphism! ( Since there is no local dilemma!)

Hence this map $\phi$ extends the "switching map" locally and hence $\tilde{\phi}: T(M \times N) \to T(M) \times T(N)$ defined by, $(x,y,v,w) \mapsto (x,v,y,w)$ is a local diffeomorphism. It is clear that it is a bijection. Thus bijection + local diffeomorphism $\implies$ that $\tilde{\phi}$ defines a diffeomorphism from $T(M \times N) \to T(M) \times T(N)$


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