Let $\mathfrak g$ be a complex simple Lie algebra. Fix a Cartan subalgebra $\mathfrak h$ of $\mathfrak g$, let $\Delta$ denote the corresponding root system. Pick a partial order on $\mathfrak h$, which induces a positive root system $\Delta^+$, fundamental Weyl chamber, etc.

I wish a reference for the following result, which must be well known. My proof is case-by-case and a bit long.

For all but finitely many finite dimensional irreducible representations $(\pi, V)$ of $\mathfrak g$, there is a weight $\mu$ of $\pi$ lying in the fundamental Weyl chamber.

Recall that $\mu\in\mathfrak h^*$ is called a weight of $\pi$ if $V(\mu):=\{v\in V: \pi(X)\cdot v=\mu(X)\, v\quad\forall \, X\in\mathfrak h\}$ is non-zero. Furthermore, the fundamental Weyl chamber is given by elements $\mu\in\mathfrak h^*$ satisfying $\langle \mu,\alpha\rangle >0$ for all $\alpha\in\Delta^+$.

The proof is immediately reduced to consider irreducible representations of $\mathfrak g$ with highest weight lying in a face of the fundamental Weyl chamber. Furthermore, it reduces to check irreducible representations with highest weight a multiple of a fundamental weight.

It is easy to see that the result is no longer valid when $\mathfrak g$ is assumed semisimple in place of simple.

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    $\begingroup$ Do I understand correctly that the issue here is that you define the Weyl chamber to be the open chamber where $\langle \ , \alpha \rangle > 0$ for all positive roots $\alpha$? $\endgroup$ May 7, 2019 at 14:39
  • $\begingroup$ @David E Speyer: yes, the Weyl chamber is assumed open. $\endgroup$
    – emiliocba
    May 7, 2019 at 18:51

3 Answers 3


Here's maybe another (more conceptual?) way to think about it.

First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.)

The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this.

Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.

Therefore $2\rho - 2\rho_i = \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha= c\omega_i$ for some integer $c > 0$. But one of the $\alpha \in \Phi^+\setminus \Phi_i^+$ is the highest root $\theta$ of $\Phi$, which when written $\theta=\sum_{j=1}^{n}a_j\alpha_j$ satisfies $a_j \geq 1$. So there is some multiple $k$ of $2\rho - 2\rho_i$ for which $k(2\rho - 2\rho_i) - 2\rho$ is a nonnegative sum of simple roots. This is to say the strictly dominant weight $2\rho$ appears in $V^{kc\omega_i}$. And then for any $m\geq kc$, the strictly dominant weight $2\rho + (m-kc)\omega_i$ appears in $V^{m\omega_i}$, completing the proof.

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    $\begingroup$ But I wonder what is the significance of having a weight strictly in the dominant chamber? $\endgroup$ May 7, 2019 at 18:07

The problem with your highlighted formulation is that it's wrong as stated, unless for example you require that an "irreducible" representation be finite dimensional or have an integral highest weight. For example, Jantzen's older 1979 Lecture Notes in Mathematics 750 (Springer-Verlag), his Habilitationsschrift (in German) and my more recent 2008 AMS graduate textbook have clear treatments of the irreducible highest weight representtions $L(\lambda)$ for an arbitrary linear functional $\lambda$ in the dual of a Cartan subalgebra. Notation varies in these sources from what you use, as it does in other sources such as Dixmier's 1974 book (in French, but translated into English) and the 1976 BGG article in a Russian journal, also translated into English.

With the above-mentioned restrictions on $\lambda$, you assert that there are only finitely many weights available on the walls of a Weyl chamber. But this seems harder to pin down in the literature without making the restrictions explicit. Can you at least give the details in a typical rank 2 case? As David Speyer's comment indicates, there is some skepticism. (I'm trying to think just in characteristic 0 terms, though most of the tricky problems arise now in prime characteristic.)

  • $\begingroup$ Sorry for the misunderstanding. Yes, I consider only finite-dimensional representations of $\mathfrak g$. $\endgroup$
    – emiliocba
    May 7, 2019 at 18:56

I don't know about a reference, but here is a short uniform proof. I assume that all of your representations are finite dimensional.

Let $\alpha_1$, ..., $\alpha_n$ be the simple roots and let $\omega_1$, ..., $\omega_n$ be the dual weights, so $\langle \alpha_i, \omega_j \rangle = \delta_{ij}$. So, if $\mu$ is an integral positive weight, then $\mu = \sum c_j \omega_j$ for $c_j \in \mathbb{Z}_{\geq 0}$.

Let $A_{ij}$ be the Cartan matrix. Let $\Gamma$ be the Dynkin diagram, so this is a graph with vertices $1$, $2$, ..., $n$ and an edge $(i,j)$ if $A_{ij} \neq 0$.

For a positive root $\mu$, let $a_r(\mu) = \{ j : c_j \leq r \}$. We partially order the integral positive weights as follows: We put $\mu \leq \nu$ if there is some index $q$ for which $a_q(\mu) > a_q(\nu)$ and $a_r(\mu) = a_r(\nu)$ for $0 \leq r < q$.

Fix your representation $V$ and let $\mu = \sum c_j \omega_j$ be minimal among the integral positive weights of $V$, with respect to the above order.

Key lemma: For every edge $(i,j)$ of $\Gamma$, we have $c_i - c_j \leq 2$.

Proof: Suppose for the sake of contradiction that $c_i \geq c_j + 3$. Since $c_i = \langle \alpha_i, \mu \rangle > 0$, the weight $\mu' := \mu - \alpha_i$ is also a weight of $V$. We will verify that $\mu'>\mu$. We have $\mu' = \sum c'_k \omega_k$ where $c'_i = c_i-2$ and $c'_k = c_k + (-A_{ik})$ for $k \neq i$.

Put $q = c_j$. I claim that $a_q(\mu') > a_q(\mu)$ and $a_r(\mu') \geq a_r(\mu)$ for $r < q$. Since $c'_i = c_i-2 \geq c_j+1=r+1$, the change between $c'_i$ and $c_i$ won't change the value of $a_q$ for $q \leq r$. Each coefficient other than the $i$-th gets larger going from $\mu$ to $\mu'$, so all the $a_q$ get weakly larger. Moreover, $c_j=r$ and $c'_j = r + (-A_{ij}) > r$, so $a_r$ gets strictly larger. $\square$

Let $\delta$ be the diameter of the graph $\Gamma$. (Here is where we use that the Lie algebra is simple, so $\Gamma$ is connected.) So, if $\mu = \sum c_j \omega_j$ is the above minimal weight, and one of the $c_j$ are $0$, then all of the $c_j$ are bounded by $2 \delta$. In particular, there are only finitely many choices for $\mu$.

Remark Experimentally, every representation seems to have a weight with $c_i - c_j \leq 1$, but it isn't always the minimum in the order I've defined, and I don't want to work hard enough to get this better bound.

  • $\begingroup$ Just confused: by “your representation $V$”, you mean one assumed to have no weight in the open Weyl chamber? $\endgroup$ May 7, 2019 at 17:52
  • $\begingroup$ @FrancoisZiegler Right. $\endgroup$ May 7, 2019 at 17:52

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