3

I am working through Lee's Introduction to Smooth Manifolds and trying to show the following:

Let $M_{1}$ and $M_{2}$ be smooth manifolds of dimensions $n_{1}$ and $n_{2}$ respectively and let $\pi_{j}: M_{1}\times M_{2}\rightarrow M_{j}$ be the projection onto the $M_{j}$ factor for $j = 1, 2$. For any $p = (p_{1}, p_{2})\in M_{1}\times M_{2}$, the map $$\alpha: T_{p}(M_{1}\times M_{2})\rightarrow T_{p_{1}}M_{1}\oplus T_{p_{2}}M_{2}$$ defined by $$\alpha(v) = (d(\pi_{1})_{p}(v), d(\pi_{2})_{p}(v))$$ is an isomorphism.

I am aware of the standard approach to solve this problem (shown, for instance, here and here) in which one constructs the inverse map $\beta$ and shows that $\alpha\circ\beta$ is the identity map on $T_{p_{1}}M_{1}\oplus T_{p_{2}}M_{2}$. However, I'm curious whether the following (less elegant) approach is correct wherein I directly show that $\alpha$ is injective.

Approach:

Let $v\in T_{p}(M_{1}\times M_{2})$ such that $\alpha(v) = 0$ and let $g\in C^{\infty}(M_{1}\times M_{2})$. The idea is to construct a function $h\in C^{\infty}(M_{1}\times M_{2})$ that locally agrees with $g$ and ''depends only on one coordinate up to first order''.

For $i = 1, 2$, let $(B_{i}, \varphi_{i})$ be a regular coordinate neighborhood of $p_{i}$. Then for $(q_{1}, q_{2})\in \overline{B_{1}\times B_{2}}$ we may use Taylor's theorem to obtain $$g(q_{1}, q_{2}) = g(p_{1}, p_{2})+\sum_{i_{1} = 1}^{n_{1}}a_{1, i_{1}}(\varphi_{1, i_{1}}(q_{1})-\varphi_{1, i_{1}}(p_{1}))+\sum_{i_{2} = 1}^{n_{2}}a_{2, i_{2}}(\varphi_{2, i_{2}}(q_{2})-\varphi_{2, i_{2}}(p_{2}))+\text{second order terms},$$ where $\{a_{1, i_{1}}\}_{i_{1} = 1}^{n_{1}}$ and $\{a_{2, i_{2}}\}_{i_{2} = 1}^{n_{2}}$ are real numbers. Here, a typical second order term looks like $$(\varphi_{1, i_{1}}(q_{1})-\varphi_{1, i_{1}}(p_{1}))(\varphi_{1, i_{2}}(q_{2})-\varphi_{1, i_{2}}(p_{2}))\int_{0}^{1}(1-t)f(\varphi_{1}(q_{1}), \varphi_{2}(q_{2}))\:\text{d}t$$ for some smooth function $f$ defined on a neighborhood of $(\varphi_{1}\times\varphi_{2})(\overline{B_{1}\times B_{2}})$. For $i = 1, 2$, by extending the coordinate chart $\varphi_{i}$ from $\overline{B_{i}}$ to a smooth function on $M_{i}$, we may extend the above local representation of $g$ to a smooth function $h$ defined on $M_{1}\times M_{2}$. Then $v(g) = v(h)$ and the latter vanishes since $\alpha(v) = 0$.

Could someone provide feedback about whether this approach is correct? Any help is greatly appreaciated.

Karthik Kannan
  • 1,600

1 Answers1

2

Your idea of proving injectivity of $\alpha$ is perfectly fine and your proof works modulo explaining some details near the end. A clearer way to understand what your proof is doing to note the following fact:

A vector $v \in T_p X$ is zero if and only if $v(x_i)=0$ for all $i$, where the $x_i$ are local coordinates in a neighbourhood of $p$.

(I'm using the fact $v$ can be canonically evaluated on smooth functions defined only in a neighbourhood of $p$.)

Now in the case of $X=M_1 \times M_2$, and with your notation, local coordinates about $p$ are given by

$$ \varphi_{1,1} \circ \pi_1, \ldots, \varphi_{1,n_1} \circ \pi_1, \varphi_{2,1} \circ \pi_2, \ldots, \varphi_{2,n_2} \circ \pi_2,$$ and we have $$ v(\varphi_{1,i} \circ \pi_1)=(d(\pi_1)_pv)(\varphi_i)=0,$$ $$ v(\varphi_{2,i} \circ \pi_2)=(d(\pi_2)_pv)(\varphi_i)=0,$$ using that $\alpha(v)=0$.

A Epelde
  • 639