# Prove that the weight space of the highest weight is always one dimensional for an irreducible representation.

Prove that the weight space of the highest weight is always one dimensional for an irreducible representation, where the weight has the usual meaning in representation theory. If H is a operator and Hφ = M (h)φ, M is called the weight.

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Is your Lie algebra (semi-)simple? I have not heard of weights being used in another context, and there $h$ is restricted to come from a maximal toral subalgebra $H$. –  Jyrki Lahtonen Apr 4 '12 at 19:36
And in the context of finite dimensional representations of a semisimple Lie algebra this follows from the properties of so called standard cyclic modules. See chapter 20 of Humphreys. –  Jyrki Lahtonen Apr 4 '12 at 19:44

Well in general, I believe, it is not true. As for finite irreps of connected solvable Lie groups, it is just the statement of the Lie Theorem:

Theorem (S. Lie): Every irreducible finite dimensional representation of connected solvable Lie group in a complex vector space is of dimension 1.

(I present here the proof that can be found in the book by D.P. Zelobenko "Compact Lie Groups and Their Representations".)

We will need a couple of definitions beforehand

Def. 1: A group $G$ is called solvable if $G^{(p)}=\{e\}$ for some integer $p$ and $G^{(p)}= [G^{(p-1)},G^{(p-1)}]$, where we write $[]$ instead of Lie derivative.

Proof: The minimal number $p$, s.t. $G^{(p)}=\{e\}$ we call rank of the group. If $p=1$ then the theorem is true. We proceed further by using the method of mathematical induction in $p$.

Let $V$ be the carrier space of the representation $T$ of $G$ and let $V_0$ be an invariant subspace w.r.t. $T$. The group $G'=[G,G]$ is solvable and connected, because the $G$ is. The rank of $G'$ is strictly smaller than the rank of $G$. Due to our induction hypothesis we can assume that $dim V_0=1$. Therefore $$T_h\xi_0=\lambda(h)\xi_0,$$ where $h\in G'$, $\xi_0\in V_0$. $G'$ is a normal subgroup of $G$ and therefore $hg=g(g^{-1}hg)=g\tilde{h}$. Now consider the action of $T_h$ on elements of an orbit $T_g\xi_0=\xi_g$. We have $$T_h\xi_g=T_hT_g\xi_0=T_gT_{\tilde{h}}\xi_0=\lambda(\tilde{h})\xi_g.$$ Therefore every vector $\xi_g$ is an eigenvector of $T_h$ with an eigenvalue $\lambda_g(h)=\lambda(g^{-1}hg)$.

In the next step we will essentially use the fact that $V$ is finite-dimensional and $G$ is connected. Since $dim V$ is finite then $\lambda_g(h)$ as a function of $g$ can take values only from a finite set $\Lambda=\{\mu(h)\}$ - the set of all 'weights' of the representation $T_h$. Since $G$ is connected, we have that $\lambda_g$ is a continuous function of $g$. From that reasoning we conclude that $\lambda_g$ can not depend on $g$ and therefore $\lambda(g^{-1}hg)=\lambda(h)$ must hold for all $g\in G$. Let $$V_{\lambda}={\xi\in V, T_h\xi=\lambda(h)\xi}$$ for some fixed $\lambda=\lambda(h)$. $V_{\lambda}$ is invariant subspace of $V$ w.r.t. all group $G$. Hence $V_{\lambda}=V$. If $h$ is a commutator, i.e. $h=aba^{-1}b^{-1}$, then $det T_h=1$ and therefore $$\lambda(h)^n=1, \: n=dim V,$$ for all commutators $h\in G'$ and therefore for all elements of $G'$. Since $G'$ is connected and $\lambda(h)$ is continuous we conclude that $\lambda(h)=1$. Hence, $T_h$ is equal to identity operator in $V$ and the representation $T_g$ is abelian. But then due to Schur Lemma $dimV=1$, which finishes the proof.

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OK. How do you prove it then? –  ramanujan_dirac Apr 4 '12 at 9:09