We know that in general if M is a manifold provided with an action of a lie group G, then for X in the Lie algebra of G , we define the vector field $X_M$ by
$X_M(m) = \frac{d}{dt}_{|t=0}e^{-tX}.m $.
I've come across this statement wich I don't know how to prove it:
Let V be a euclidean vector space of dimension n with basis $e_i$, $1 \leq i \leq n$. We denote by $x_i$ the coordinate functions on V, and $\partial_i $ the direvation in the direction of the vector $e_i$. Suppose that V is provided with an action of a Lie group G.
Why this is true:
If $X \in \mathfrak{g}$, then the vector field $X_V$ is given by :
$X_V= - \sum_{i < j} (Xe_i,e_j)(x_i \partial_j - x_j \partial_i)$ ?
Thanks!
It isn't true. For example, take $V=\mathbb{R}^2$, $G=\operatorname{GL}(2,\mathbb{R})$, and $X = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\in\mathfrak{g}$. Then $$ X_V(x_1,x_2) = \frac{d}{dt}\Big\vert_{t=0}\begin{pmatrix} e^{-t} & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix} = \frac{d}{dt}\Big\vert_{t=0}\begin{pmatrix}e^{-t}x_1 \\ x_2\end{pmatrix} = -x_1\partial_1, $$ but according to your formula it would be $$ X_V(x_1,x_2) = -(Xe_1,e_1)(x_1\partial_2-x_2\partial_1) = -(x_1\partial_2-x_2\partial_1). $$ Perhaps the statement you read has some additional conditions on the group $G$?