Local form of a fundamental vector field

Let $M$ be a smooth manifold of dimension $2l$ on which acts a lie group $G$. Let $X$ be a element in the lie algebra $\mathfrak{g}$ of $G$, we associate to it the vector field $X_M$ defined by $X_M(m)= \frac{d}{dt}\biggr\vert_{t=0} e^{-tX}.m$. Let $p$ be a zero of the vector field $X_M$.

In the page 6 of the article [1], the authors say that we can find some local coordinates $x_1,...,x_{2l}$ around $p$ such that the vector field $X_M$ is linearized: $$X_M= a_1 \left( x_2 \frac{\partial}{\partial x_1} -x_1 \frac{\partial}{\partial x_2} \right)+\ldots+a_l\left(x_{2l} \frac{\partial}{\partial x_{2l-1}}-x_{2l-1}\frac{\partial}{\partial x_{2l}}\right),$$ $a_1,..,a_l \in \mathbb{R}.$

How can we prove that $X_M$ has the above form locally (around p)?

References

[1] Michèle Vergne, "Cohomologie équivariante et théorème de Stokes" (rédigé par Sylvie Paycha) (French) Analysis on Lie groups and representation theory. Proceedings of the summer school, Kénitra, France, 1999, Séminaires et Congrès 7, Paris: Société Mathématique de France (ISBN 2-85629-142-2/pbk), pp. 1-43 (2003), MR2038647, Zbl 1045.57021.

Solution 1:

If $M$ admits a $G$-invariant Riemannian metric (which is the case if $G$ is compact) then the exponential map $T_pM\to M$ is locally a diffeomorphism and it is $G$-equivariant. So we just need to understand the action of $G$ on $T_pM$. That action is linear, and preserves the inner product on $T_pM$, so $X$ acts as an element of $so(T_pM)$. Finally, any element of $so(\mathbb R^n)$ can be put to a block-diagonal form (with blocks of size $2\times 2$) by a suitable choice of an orthonormal basis.