# Tangent manifold $D(M_1 \times M_2)$ is canonically diffeomorphic to the product $DM_1 \times DM_2$.

Let $M \subset \mathbb{R}^A$ and $M_2 \subset \mathbb{R}^B$ be smooth manifolds. How do I see that the tangent manifold $D(M_1 \times M_2)$ is canonically diffeomorphic to the product $DM_1 \times DM_2$?

Note that a function $x \mapsto (f_1(x), f_2(x))$ from $M$ to $M_1 \times M_2$ is smooth if and only if both $f_1 : M \to M_1$ and $f_2: M \to M_2$ are smooth.

## 1 Answer

By definition, $M_1\times M_2\subset\mathbb R^A\times\mathbb R^B=\mathbb R^{A+B}$. Now the projections $p_1$ and $p_2$ from $\mathbb R^A\times\mathbb R^B$ are linear maps and hence smooth, so they restrict to smooth maps $p_1:M_1\times M_2\to M_1$ and $p_2:M_1\times M_2\to M_2$. Hence the derivatives are smooth maps $D(p_1):D(M_1\times M_2)\to DM_1$ and likewise for $D(p_2)$ so taking them as components, you get a smooth map $(D(p_1),D(p_2)):D(M_1\times M_2)\to DM_1\times DM_2$.

To see that this is indeed a diffeomorphism, you observe that $DM_1\subset\mathbb R^A\times\mathbb R^A=\mathbb R^{2A}$ has the form $\{(x,v):x\in M_1,v\in D_xM_1\}$. (I hope, this is the notation you are used to, I just mean that $v$ is tangent to $M_1$ in the point $x$.) Doing this similarly for $M_2$, you see that there is an evidently smooth map $DM_1\times DM_2\to \mathbb R^{A+B}\times\mathbb R^{A+B}$ mapping $((x,v),(y,w))$ to $((x,y),(v,w))$. A moment of thought shows that this has values in $D(M_1\times M_2)$ and clearly it defines an inverse to $(D(p_1),D(p_2))$.