Let $v \in T_p \mathbb{R}^n$. Assume that $A : T_p \mathbb{R}^n \to T_p \mathbb{R}^n$ is an antisymmetric endomorphism. Then is it true that there exists a Killing vector field $V \in \chi(\mathbb{R}^n)$ sucht that $V|_p = v$ and $\nabla V|p = A$?
It should follow from the fact that skew symmetric matrices are the Lie algebra of rotations, right?
(I know that this is probably a trivial question, but I have never studied Lie groups theory..)
Since $A$ is anti symmetric so there is a path $c(t)=\exp\ tA\in O(n)$.
Clearly, $c(t)$ is family of isometries on $\mathbb{R}^n$ so that we have a Killing field $$ X(x)=\frac{d}{dt}\bigg|_{t=0}\ c(t)\cdot x $$ By replacing origin, we can assume that $X(x)=v$.
Here $$ w(X)=\frac{d}{ds}\ X(x+sw) = \frac{d}{ds}\frac{d}{dt}\ c(t)\cdot (x+sw) = Aw $$