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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.

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    $\begingroup$ What if $\Bbb R^+$ acts on $\Bbb R^N$ by $t\cdot x = tx$? $\endgroup$ Commented Jan 3, 2022 at 20:06

1 Answer 1

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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.

If $G$ is not compact, the statement is in general not true.

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  • $\begingroup$ could you please elaborate further your answer @user8268. I didn't understand why does it suffices to understand the action of $G$ on $T_pM$, and why how can we construct the local coordinates $x_i$ ? $\endgroup$
    – Mira
    Commented Jan 19, 2022 at 17:48
  • $\begingroup$ you may have to tell me more about what you don's understand: 1. the exp map $T_pM\to M$ is a local diffeo (a diffeo of a neighourhood of $0\in T_pM$ to a neighbourhood of $p\in M$) and $G$-equivariant, therefore, it is enough to understand the action of $G$ (or of $X$) on $T_pM$; 2. the local coordinates on $T_pM$ are linear, given by a suitable orthonormal basis (chosen to simplify $X_{T_pM}$); they then get transferred to $M$ via the local diffeo $\endgroup$
    – user8268
    Commented Jan 19, 2022 at 19:22
  • $\begingroup$ could you please explain it if dimension of M is equal to 2. $\endgroup$
    – Mira
    Commented Jan 19, 2022 at 20:02
  • $\begingroup$ I understand that if choose a suitable orthonormal basis, the matrix which represents the action of $G$ on $T_pM$ is of the form \begin{pmatrix} 0 & - a \\ a & 0 \end{pmatrix} where $a \in \mathbb{R}$ $\endgroup$
    – Mira
    Commented Jan 19, 2022 at 20:06
  • $\begingroup$ Suppose $m$ is in a neighborhood of $p$, how can we prove that there exists coordinates $x_1, x_2$ such that $$X_M(m) =a(x_2 \frac{\partial}{\partial_{x_1}}\vert_m - x_1\frac{\partial}{\partial_{x_2}}\vert_m ).$$ $\endgroup$
    – Mira
    Commented Jan 19, 2022 at 20:10

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