I'm relatively new to Lie Groups, and finding in a text the following statement, as it is supposed to be obvious. I would love some reference or context to either of these two (probably related) statements:
The context is $G$ a connected lie group with lie algebra $\mathfrak{g}$, $H$ closed connected subgroup with sub-lie-algebra $\mathfrak{h}$ of dimention $d$.
fixing $x\in \wedge^d \mathfrak h - \{0\}$, it is said that $\wedge^d Ad(g)x=x$ iff $g\in N(H)$ and $\det(Ad\:(g|_{\mathfrak h}))=1$
for some $\Gamma$ acting on $G$, if $\gamma(H\Gamma/\Gamma)=H\Gamma/\Gamma$ then $|\det(Ad\:(\gamma|_{\mathfrak h}))|=1$, and so $\wedge^d Ad(\gamma)x=\pm x$
the first one is somewhat related in my head to the formula of inner product of $e_1\wedge\ldots\wedge e_d$ with $\wedge^d Ad(g) e_1\wedge\ldots\wedge e_d$, and that it should be 1 if it's an eigenvector with eigenvalue $1$. I could not formulate it.
the second one looked like some argument of $\gamma$ preserving the volume and therefore the absolute value of the determinant, which basically means volume, must be 1. I could not formulate this either.
P.S:
Just to mention: I think I do understand the part about being in the normalizer, as $g\in N(H)$ implies $\phi_g (g)\in H$ for all $h\in H$and therefore $Ad(g)(x)\in \mathfrak{h}$ for all $x\in\mathfrak{h}$. Still, the determinant part is unclear.