Prove that $\gamma$ is constant iff $\gamma′(t) = 0$ for all $t \in J$ if there is always an interval containing $t$ in which $ \gamma$ is const

Question

I am trying to prove the following lemma.

Let $M$ be a smooth manifold

(a) Let $J$ be an open interval and $ \gamma: J \to M$ a smooth curve. Suppose for every $t \in J$ there is an open interval $J_t \subseteq J$ with $t \in J_t$, on which $\gamma$ is constant. Prove that $\gamma $ is constant.

(b) Prove that $\gamma$ is constant if and only if $\gamma′(t) = 0$ for all $t \in J$.

I have already proven (a), but I am having trouble with (b)

This is what I've done

$(\implies)$ If $\gamma $ is constant, then for every $f \in C^{\infty}(M)$, $\gamma'(t)f=\frac{d}{dt}(f \circ \gamma)=0$. Since $f \circ \gamma: J \subseteq \Bbb \to \Bbb R$ is a constant function, it follows from ordinary calculus that $f \circ \gamma(t)=0 \implies$ $\gamma'(t)f=0$. So $\gamma'(t)$ is the zero vector field on J.

$(\impliedby)$ I am stuck here. I think I have to take a chart, so that I can write the vector field in the basis $\{\left. \frac{\partial}{\partial x^1}\right|_{\gamma(t)}, \dots, \left. \frac{\partial}{\partial x^n}\right|_{\gamma(t)} \}$. (Here, $n=\text{dim}M$.)

$$\gamma'(t)=\sum_{i+1}^{n}c_i(t)\left. \tfrac{\partial}{\partial x^i}\right|_{\gamma(t)}=0$$

From $\gamma'(t)=0$, .I guess it follows that $c_i(t)=0$ for $i=1,...,n$

My questions:

1. Is $(\implies)$ correct?

2. How do I prove $(\impliedby)$? Somehow I have to use part (a) there

---

Edit:

Following the hints:

The third bullet in the answer below is proven like this:

The $a^i:J' \to \Bbb R$. Since $0=\gamma'(t)=\frac{d a^i(t)}{dt}\frac{\partial}{\partial x^i}|_{\gamma(t)} \implies \frac{d a^i(t)}{dt}=0 \implies a^i(t)=$const for all $t \in J'$ , $i \in 1,...,n$. So we have that $\varphi \circ \gamma(t)=(a^1(t),...,a^n(t))=const$, for all $t \in J'$. Since $\varphi$ is a diffeomorphism,we can compose with $\varphi^{-1}$ to get: $\gamma(t)=\varphi^{-1}(t)\circ\varphi(t)\gamma(t)=\varphi^{-1}(const)=$const2 another constant, so we have that $\gamma$ is constant for all $t \in J'$.

But now putting together the result of the first and third bullet below we have show that for every $t_0 \in J$ there exists an interval $J'\subset S , t \in J'$ where $\gamma$ is constant. So now I can apply part (a) of my post to conclude that $\gamma$ is constant in all of $J$

Answer 1

Your argument for $\implies$ looks fine.

For $\impliedby$, I agree that we should use part (a), and I also like the fact that you're thinking about how $\gamma$ looks in a chart. Here's a way to proceed.

• Pick any $t_0 \in J$, and let $(U, \varphi)$ be a coordinate chart on $M$ containing the point $\gamma(t_0)$. Using the continuity of $\gamma$, you can argue that there exists an open interval $J ' \subset J$ containing $t_0$ such that $\gamma(J') \subset U$.
• Now suppose that $\varphi \circ \gamma (t) = (a^1(t), \dots, a^n(t))$ is the coordinate representation of $\gamma$ with respect to the chart $(U, \varphi)$, valid for $t \in J'$. You should be able to write down an expression for $\gamma'(t)$ in terms of the tangent vectors $\frac{\partial}{\partial x^1}|_{\gamma(t)},...,\frac{\partial}{\partial x^n}|_{\gamma(t)}$ and the time derivatives of $a^1(t), \dots, a^n(t)$. (This expression would be valid for $t \in J'$.)
• Having derived an expression for $\gamma'(t)$, you should be able to use calculus to show if $\gamma'(t)$ is zero for all $t$, then the functions $a^1(t), \dots, a^n(t)$ must all be constant functions.