2
$\begingroup$

What is known about existence of quasi-invariant smooth function with some eigencharacter on Lie algebra of a reductive Lie group? Consider reductive Lie group $G$ and its Lie algebra $\mathfrak{g}$. We have the conjugate action of $G$ on $\mathfrak{g}$: $g\cdot X=gXg^{-1}$. The following classical result claims the existence of $G$-invariant bump function on $ \mathfrak{g} $ for invariant enighborhood near semisimple point:

For semisimple point $X\in\mathfrak{g}$ and an invariant open neighborhood $\Omega$ of $X$, then there exists a $G$-invariant smooth function $f\in C^\infty(\mathfrak{g})^G$ such that
(1) $0\leq f\leq1$, $f=1$ in an neighborhood of $X$
(2) $\operatorname{supp}(f)\subseteq\Omega$

Suppose now $ \eta:G\to\mathbb{C}^\times $ is a unitary character of $ G $. We call a function $f:G\to\mathbb{C}$ is quasi-invariant with eigencharacter $\eta$ if $ f(g\cdot X)=\eta(g)f(X) $ for $g\in G,X\in\mathfrak{g}$.

Can we construct a quasi-invariant smooth function on $\mathfrak{g}$ with eigencharacter $\eta$ still satisfies the above conditions?

$\endgroup$
8
  • $\begingroup$ A unitary character with target $\mathbb{R}$ can only take values $\pm 1$. If $G$ is connected, such chracter will be trivial. $\endgroup$ Commented Feb 1 at 14:03
  • $\begingroup$ @Nandor Yes, what I'm really concerned about is the non connected group $ G=\operatorname{GL}_n $ $\endgroup$ Commented Feb 1 at 16:01
  • $\begingroup$ You cannot expect positivity, then. But other than that, you can take a $G^0$ invariant function $f_0$ and then set $f(X)=f_0(X)-f_0(\alpha X)$, where $\alpha$ is the non-trivial element in $G/G^0$. The worst that can happen is, then $f=0$. That would entail an automatic invariance of $G^0$ invariant functions under $\alpha$. $\endgroup$ Commented Feb 1 at 18:47
  • $\begingroup$ Surely you want your character to take values in the group $\mathbb R^\times$, not $\mathbb R$? $\endgroup$ Commented Feb 1 at 20:22
  • $\begingroup$ @Nandor The function $f$ you defined does not seem to be an quasi-invariant function with eigencharacter. $\endgroup$ Commented Feb 2 at 14:37

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.