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The problem is stated as follows. Let $X$ be a differential vector field on $M$ and let $f: M \rightarrow N$ be an isometry. Show that the pushforward $Y=df(X)$ is a Killing field on $N$ if and only if $X$ is a Killing field on $M$.

The "only if" part is trivial once you've proved the "if" direction because $X$ is the pushforward of $Y=df(X)$ by $f^{-1}$.

I am stuck here. I am trying to prove this using the definition that a Killing field is a vector field whose flows are isometries. I'm using the notation that if $\varphi: (-\epsilon, \epsilon) \times M \rightarrow M$ is the flow of $X$ and $\psi: (-\epsilon, \epsilon) \times N \rightarrow N$ is the flow of $Y=df(X)$ (I know I'm making some minor simplifying assumptions here) then, for fixed $t_{0} \in (-\epsilon,\epsilon)$ and $p_{0}\in M$ the map

$\varphi_{t_0}(p): M \rightarrow M$ is a a diffemorphism and

$\varphi_{p_0}(t): (-\epsilon, \epsilon) \rightarrow M$ is the integral curve of $X$ starting from $p_{0}$.

Analogous statements hold for $\psi$.

Also assuming $X$ is a killing field $$\varphi_{t_0}(p): M \rightarrow M$$ is an isometry for each $t_{0}$.

I am trying to use the fact that if $\gamma: (-\epsilon, \epsilon)\rightarrow M$ is the integral curve of $X$ starting at $\p$, then $f \circ \gamma$ is the integral curve of $Y=df(X)$ starting from $f(p)$ and the fact that the flows of $X$ are isometries but I can't seem to write it down properly.

For example, for fixed $q_0 \in N$, and writing $\hat{q}_0=f^{-1}(q)$ we have

$\psi_{q_0}(t)=f(\varphi_{\hat{q}_0}(t))$

but I can't translate this into getting that $\psi_{t_{0}}(q):N \rightarrow N$ is an isometry for each $t_{0}$.

Tuo
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    Why not use other formulaic characterization instead? That is, $X$ is killing iff $\langle \nabla_Z X, Y\rangle + \langle \nabla_Y X, Z\rangle = 0$ for all $Y,Z$. – Jason DeVito - on hiatus Aug 31 '18 at 13:26
  • I thought about it, but I don't see a relationship between the pushforward and the connection $\nabla$. – Tuo Aug 31 '18 at 15:42
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    The connection is that $f_\ast(\nabla_X^M Y) = \nabla_{f_\ast X}^N (f_\ast Y)$. To prove this, note that its equivalent to proving $\nabla_X^M Y = (f_\ast)^{-1}\left(\nabla^N_{f_\ast X} (f_\ast Y)\right)$. To establish this equality, use the Koszul formula. (I'll won't be in my office Tuesday, but will on Wednesday if you wanna chat.) – Jason DeVito - on hiatus Sep 02 '18 at 03:01
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    It's Killing, not killing. The name comes from a surname, not the act of killing. – tomasz Sep 02 '18 at 23:06
  • @tomasz: I knew that - simple typo above. But thanks! – Jason DeVito - on hiatus Sep 03 '18 at 02:59
  • @JasonDeVito Jason, given that this relationship between the connection and its pushforward hold, the problem is trivial from there. This seems simpler than the solution you've given below but I don't quite see how to hammer the relationship out of the Koszul formula. I'll probably come ask you about it on Wednesday. – Tuo Sep 03 '18 at 02:59
  • @JasonDeVito I thought tomasz was talking about my typos. – Tuo Sep 03 '18 at 03:00
  • @JasonDeVito: Yeah, as OP said, I meant that as a comment to OP, not your comment. – tomasz Sep 03 '18 at 18:01
  • @tomasz: I assumed you were referring to my first comment, when I mistakenly didn't capitalize Killing. In any event, thanks! – Jason DeVito - on hiatus Sep 04 '18 at 01:42

1 Answers1

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Here's a proof using the definition of Killing fields in terms of flows being isometries. I'll use notation as you did in your post. For a point $m\in M$, I will use the notation $\phi_m$ for the curve $\phi(\cdot, m):(-\epsilon,\epsilon)\rightarrow M$ given by mapping $(t,m)$ to $(t, f(m))$.

Lemma: For any $m\in M$, $\psi_{f(m)} = f\circ \phi_m$.

Proof: Let $m\in M$. We will show that both $\psi_{f(m)}$ and $f\circ \phi_m$ are the integral curves of the vector field $Y$. By uniqueness of these integral curves, we must then have $\psi_{f(m)} = f\circ \phi_m$.

At time $t = 0$, we have $\psi_{f(m)}(0) = f(m)$ and $f(\phi_m(0)) = f(m)$, so they agree when $t = 0$.

Now, we also have $\frac{d}{dt}|_{t=0} \psi_{f(m)} = Y_{f(m)}$ because $\psi$ is the flow of $df(X) = Y$. On the other hand by the chain rule, we have $\frac{d}{dt}|_{t=0}(f\circ \phi_m) = df\left( \frac{d}{dt}|_{t=0} \phi_m\right) = df(X_m) = Y_{f(m)}$. $\square$

(By the way, the above lemma is does not use the Riemannian structures anywhere - it works for general manifolds!)

With this lemma established, we note that it really means that $\psi(t,f(m)) = f(\phi(t,m))$ for any $(t,m)\in (-\epsilon, \epsilon)\times M$. This in turn implies that $\psi_t = f\circ \phi_t \circ f^{-1}$.

Now, letting $g_N$ denote the metric on $N$, and similarly for $g_M$, we have \begin{align*} \psi_{t}^\ast g_N &= (f\circ \phi_t\circ f^{-1})^\ast g_N\\ &= (f^{-1})^\ast \phi_t^\ast f^\ast g_N\\ &= (f^{-1})^\ast \phi_t^\ast g_M\\ &= (f^{-1})^\ast g_M \\ &= g_N.\end{align*}

Jason DeVito - on hiatus
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