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David E Speyer
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I don't know about a reference, but here is a short uniform proof. I assume that all of your representations have integral highest weights. (This follows fromare finite dimensionalitydimensional.)

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. (Of course, we need the set of integral positive weights to be nonempty.)

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.

I don't know about a reference, but here is a short uniform proof. I assume that all of your representations have integral highest weights. (This follows from finite dimensionality.)

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. (Of course, we need the set of integral positive weights to be nonempty.)

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.

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.

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David E Speyer
  • 156.2k
  • 14
  • 420
  • 763

I don't know about a reference, but here is a short uniform proof. I assume that all of your representations have integral highest weights. (This follows from finite dimensionality.)

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. (Of course, we need the set of integral positive weights to be nonempty.)

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$$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.

I don't know about a reference, but here is a short uniform proof. I assume that all of your representations have integral highest weights. (This follows from finite dimensionality.)

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. (Of course, we need the set of integral positive weights to be nonempty.)

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 $\mu$.

I don't know about a reference, but here is a short uniform proof. I assume that all of your representations have integral highest weights. (This follows from finite dimensionality.)

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. (Of course, we need the set of integral positive weights to be nonempty.)

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.

Source Link
David E Speyer
  • 156.2k
  • 14
  • 420
  • 763

I don't know about a reference, but here is a short uniform proof. I assume that all of your representations have integral highest weights. (This follows from finite dimensionality.)

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. (Of course, we need the set of integral positive weights to be nonempty.)

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 $\mu$.