Multiplying tangent vectors by positive numbers (Gullemin-Pollack 1.8.2)

Question

This is a question from Differential Topology by Guillemin and Pollack:

Let $g$ be a smooth, everywhere-positive function on $X$. Check that the multiplication map $T(X)\rightarrow T(X)$, $(x,v)\mapsto (x,g(x)v)$ is smooth.

Here $X$ is a manifold embedded in some $\mathbb{R}^n$, and $T(X)$ is its tangent bundle, as a subset of $\mathbb{R}^n\times\mathbb{R}^n$.

I believe I have proved this by working locally, and showing this is smooth, since locally it is really just multiplication. But I have a feeling I'm under-thinking it, because that seems way too easy, and I haven't used the "everywhere-positive" part. Can someone explain why we need to assume $g(x)$ is everywhere positive?

Answer 1

Working locally is fine. However, you can simplify the proof using the following ingredients.

1. The authors define smoothness for maps $f : X \to \mathbb R^m$ defined on any $X \subset \mathbb R^n$. They do not explicity define smoothness for maps $f : X \to Y$ with $Y \subset \mathbb R^m$, but it is clear that such a map is called smooth if composed map $X \xrightarrow{f} Y \hookrightarrow \mathbb R^m$ is smooth.

2. As a consequence, if $X \subset Y \subset \mathbb R^m$, then the inclusion map $X \hookrightarrow Y$ is smooth because the the composed map $X \hookrightarrow Y \hookrightarrow \mathbb R^m$ is smooth as the restriction of the identity on $\mathbb R^m$.

3. The product $f \times g : X \times X' \to Y \times Y'$ of smooth maps $f : X \to Y$ and $g : X' \to Y'$ is smooth. [Exercise 14 on p.7]

4. If $f : X \to Y$ is smooth, then $\bar f : X \to \operatorname{graph} f, \bar f(x) = (x, f(x))$, is smooth (even a diffeomorphism). [Exercise 17 on p.7]

Let us analyze your map $G : TX \to TX, G(x,v) = (x,g(x)v)$, where $X \subset \mathbb R^n$. We define a map

$$\phi : TX \hookrightarrow X \times \mathbb R^n \xrightarrow{\bar g \times id} \operatorname{graph} g \times \mathbb R^n \hookrightarrow \mathbb R^n \times \mathbb R \times \mathbb R^n \xrightarrow{id \times \mu} \mathbb R^n \times \mathbb R^n$$ with multiplation $\mu : \mathbb R \times \mathbb R^n \to \mathbb R^n, \mu(t,x) = tx$.

This map is smooth by 2. - 4. But $\phi$ is nothing else than the composition $$TX \xrightarrow{G} TM \hookrightarrow \mathbb R^n \times \mathbb R^n$$ so that $G$ is smooth by 1.

You also see that this is true for each smooth $g$.