Mathematics Asked on January 5, 2022

Suppose $M,N$ are manifolds, and consider the product $Mtimes N$.

From this answer, I know that:

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

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

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!)

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

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 $T(M times N) to T(M) times T(N)$

Answered by Brozovic on January 5, 2022

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