Linear Killing fields in $\Bbb R^n$

Question

$X$ is called a Killing field if for each one parameter group induced form the flow $\psi_{t}:U\subset M \rightarrow M$ is an isometry. Show that a linear field on $\mathbb R^n$, defined by a matrix $A$ is a Killing field if and only if $A$ is anti-symmetric.

Let $\psi_{t}$ equals to $\exp tA$, I want to show that for any tangent vectors at $p$ $$\langle v,v\rangle=\langle d\psi_t(v),d\psi_t(v)\rangle\Leftrightarrow A+A^T=0$$

My attempt:

$\Leftarrow$: $\psi_t=\exp(tA)$ is an element of $O(n)$ by $(\exp(tA))^T=\exp(tA)^{-1}$, since orthogonal transformation preserves the norms, so it is an isometry.

$\Rightarrow$：No idea.

This is a problem selected from do Carmo's Riemannian Geometry, but I've met some trouble, can anyone else give me some hints? (Not necessary the whole solutions).

Thanks for your feedback!

Answer 1

${d\over{dt}}_{t=0}\langle v,v\rangle=\langle d\psi_t(v),d\psi_t(v)\rangle=\langle A(u),v\rangle+\langle u,A(v)\rangle=0$.

$\langle A(u),v\rangle+\langle u,\rangle=\langle A(u),v\rangle+\langle A^T(u),v\rangle=\langle (A+A^T)(u),v\rangle=0$ for every $u,v$ implies $A+A^T=0$.

Conversely, suppose that $A+A^T=0$, for every $u,v,{d\over{dt}}_{t=t_0}\langle exp(tA)u,exp(tA)v\rangle=$

$\langle A(expt_0A)u,exp(t_0A)v\rangle+\langle exp(t_0A)u,A(exp(t_0A)v\rangle=$

$\langle (A+A^T)(exp(t_0A)(u),exp(t_0A)(v)\rangle=0$ implies $f(t)=\langle exp(tA)u,exp(tA)v\rangle$ is constant and $f(t)=f(0)=\langle u,v\rangle.$

Answer 2

In any Riemannian manifold $(M,g)$, $X\in \mathfrak{X}(M)$ is Killing if and only if $\mathcal{L}_Xg = 0$, where $\mathcal{L}_X$ denotes Lie derivative (this is obvious from your definition since $\mathcal{L}_X$ is defined with flows). On the other hand, you have that $$\mathcal{L}_Xg = (\nabla X)+ (\nabla X)^\top,$$where $\nabla$ is the Levi-Civita connection of $g$. In the case where $M = \Bbb R^n$ and $g$ is the standard dot product, we have that $\nabla$ is the standard flat connection. If $X_x = Ax$ for all $x \in \Bbb R^n$, for some matrix $A$, the fact that the total derivative of a linear map is itself gives that $\nabla X = A$. Thus $\mathcal{L}_Xg = A+A^\top$, and this vanishes if and only if $A$ is skew-symmetric.