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Suppose $\mathfrak{g}$ is a semisimple Lie algebra over $\mathbb{C}$, and $\tilde{G}$ is the unique connected, simply connected Lie group whose Lie algebra is $\mathfrak{g}$. Let $C$ be any discrete subgroup of $Z(\tilde{G})$, and let $G=\tilde{G}/C$. Let $\mathfrak{h}\subset\mathfrak{g}$ be a Cartan subalgebra, and suppose we fix an ordering of the roots, $R$, of the Cartan decomposition of $\mathfrak{g}$ with respect to $\mathfrak{h}$ by choosing some real linear functional $l:\Lambda_R\to\mathbb{R}$ on the root lattice $\Lambda_R$. Let $\lambda$ be any element in the intersection of the weight lattice and the Weyl chamber defined by $l$, so that there is a unique irreducible representation of $\mathfrak{g}$ with highest weight $\lambda$, which we denote $\Gamma_{\lambda}$.

Now, I understand that representations of $G$ correspond to representations of $\tilde{G}$ which vanish on $C$, but the proof Lemma $23.12$ of "Representation Theory: A First Course" by Fulton and Harris says that $\Gamma_{\lambda}$ is a representation of $G$ if and only if $g\cdot v=v$ for all $g\in C$, where $v$ is a highest weight vector.

How can I conclude that $g$ acts trivially on all of $\Gamma_{\lambda}$ if I only know that it acts trivially on $v$?

I understand that $\Gamma_{\lambda}$ is "generated by $v$" in the sense that $\mathfrak{g}\cdot v=\Gamma_{\lambda}$, but I'm not too sure how to show the following implication:

$$g\cdot v=v\quad\Longrightarrow\quad g\cdot(X\cdot v)= X\cdot v$$

for any $X\in\mathfrak{g}$. Thanks in advance for any help here.

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1 Answer 1

up vote 1 down vote accepted

It's easier to work with groups than Lie algebras here because understanding how group actions commute is easy.

Since $\Gamma_{\lambda}$ is irreducible, for $v$ the highest weight vector, $$\text{span} \{G \cdot v\} = \Gamma_{\lambda}.$$

So, it suffices to show that if every $g \in C$ fixes $v$, then it fixes $G\cdot v$, pointwise.

This follows from the fact that $C \subset Z(G)$: for any $g \in G, h \in C$ $$h\cdot v = v \Rightarrow h\cdot g\cdot v = g \cdot h \cdot v = g \cdot v.$$

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