Is the root cone is contained in the weight cone?

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Let $G$ be a semisimple algebraic group over a field $k$. Let $A_0$ be a maximal split torus of $G$, and $mathfrak a_0 = operatornameHom(X(A_0), mathbb R)$ the real Lie algebra of $A_0$ with dual $mathfrak a_0^ast = X(A_0) otimes mathbb R$.

Let $Delta subset mathfrak a_0^ast$ be a set of simple roots of $A_0$ in $G$. Let $Delta^vee subset mathfrak a_0$ be the coroots for the corresponding root system, and let $hatDelta^vee$ the coweights (the dual basis to $Delta$).

Let $$A = v in mathfrak a_0^ast : langle v, alpha^vee rangle > 0 textrm for all $alpha^vee in Delta^vee$$$

$$B = v in mathfrak a_0^ast : langle v, varpi_alpha^vee rangle > 0 textrm for all varpi_alpha^vee in hatDelta^vee $$

Is it true that $A subseteq B$?

This is claimed in line (2.2) of these notes on Langlands classification. However, I can't seem to prove this and am beginning to think it is not true.

representation-theory lie-groups lie-algebras root-systems

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asked Jul 22 at 23:49

D_S

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Let $G$ be a semisimple algebraic group over a field $k$. Let $A_0$ be a maximal split torus of $G$, and $mathfrak a_0 = operatornameHom(X(A_0), mathbb R)$ the real Lie algebra of $A_0$ with dual $mathfrak a_0^ast = X(A_0) otimes mathbb R$.

Let $Delta subset mathfrak a_0^ast$ be a set of simple roots of $A_0$ in $G$. Let $Delta^vee subset mathfrak a_0$ be the coroots for the corresponding root system, and let $hatDelta^vee$ the coweights (the dual basis to $Delta$).

Let $$A = v in mathfrak a_0^ast : langle v, alpha^vee rangle > 0 textrm for all $alpha^vee in Delta^vee$$$

$$B = v in mathfrak a_0^ast : langle v, varpi_alpha^vee rangle > 0 textrm for all varpi_alpha^vee in hatDelta^vee $$

Is it true that $A subseteq B$?

This is claimed in line (2.2) of these notes on Langlands classification. However, I can't seem to prove this and am beginning to think it is not true.

representation-theory lie-groups lie-algebras root-systems

share|cite|improve this question

asked Jul 22 at 23:49

D_S

12.8k51550

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $G$ be a semisimple algebraic group over a field $k$. Let $A_0$ be a maximal split torus of $G$, and $mathfrak a_0 = operatornameHom(X(A_0), mathbb R)$ the real Lie algebra of $A_0$ with dual $mathfrak a_0^ast = X(A_0) otimes mathbb R$.

Let $Delta subset mathfrak a_0^ast$ be a set of simple roots of $A_0$ in $G$. Let $Delta^vee subset mathfrak a_0$ be the coroots for the corresponding root system, and let $hatDelta^vee$ the coweights (the dual basis to $Delta$).

Let $$A = v in mathfrak a_0^ast : langle v, alpha^vee rangle > 0 textrm for all $alpha^vee in Delta^vee$$$

$$B = v in mathfrak a_0^ast : langle v, varpi_alpha^vee rangle > 0 textrm for all varpi_alpha^vee in hatDelta^vee $$

Is it true that $A subseteq B$?

This is claimed in line (2.2) of these notes on Langlands classification. However, I can't seem to prove this and am beginning to think it is not true.

representation-theory lie-groups lie-algebras root-systems

share|cite|improve this question

asked Jul 22 at 23:49

D_S

12.8k51550

Let $G$ be a semisimple algebraic group over a field $k$. Let $A_0$ be a maximal split torus of $G$, and $mathfrak a_0 = operatornameHom(X(A_0), mathbb R)$ the real Lie algebra of $A_0$ with dual $mathfrak a_0^ast = X(A_0) otimes mathbb R$.

Let $Delta subset mathfrak a_0^ast$ be a set of simple roots of $A_0$ in $G$. Let $Delta^vee subset mathfrak a_0$ be the coroots for the corresponding root system, and let $hatDelta^vee$ the coweights (the dual basis to $Delta$).

Let $$A = v in mathfrak a_0^ast : langle v, alpha^vee rangle > 0 textrm for all $alpha^vee in Delta^vee$$$

$$B = v in mathfrak a_0^ast : langle v, varpi_alpha^vee rangle > 0 textrm for all varpi_alpha^vee in hatDelta^vee $$

Is it true that $A subseteq B$?

This is claimed in line (2.2) of these notes on Langlands classification. However, I can't seem to prove this and am beginning to think it is not true.

representation-theory lie-groups lie-algebras root-systems

share|cite|improve this question

asked Jul 22 at 23:49

D_S

12.8k51550

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 23:49

D_S

12.8k51550

asked Jul 22 at 23:49

D_S

12.8k51550

asked Jul 22 at 23:49

D_S

12.8k51550

12.8k51550

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This statement appears to be true. The elements of $A$ are non-negative linear combinations of the fundamental weights, and elements of $B$ are non-negative linear combinations of simple roots. Formula $(2.1)$ implies that the fundamental weights are non-negative linear combinations of simple roots.

Draw a picture of $A$ and $B$ for a couple of rank 2 root systems and you'll see what is going on.

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answered Jul 25 at 16:59

David Hill

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1 Answer
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1 Answer
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active

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up vote
1
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This statement appears to be true. The elements of $A$ are non-negative linear combinations of the fundamental weights, and elements of $B$ are non-negative linear combinations of simple roots. Formula $(2.1)$ implies that the fundamental weights are non-negative linear combinations of simple roots.

Draw a picture of $A$ and $B$ for a couple of rank 2 root systems and you'll see what is going on.

share|cite|improve this answer

answered Jul 25 at 16:59

David Hill

7,9311618

add a comment | 

up vote
1
down vote

This statement appears to be true. The elements of $A$ are non-negative linear combinations of the fundamental weights, and elements of $B$ are non-negative linear combinations of simple roots. Formula $(2.1)$ implies that the fundamental weights are non-negative linear combinations of simple roots.

Draw a picture of $A$ and $B$ for a couple of rank 2 root systems and you'll see what is going on.

share|cite|improve this answer

answered Jul 25 at 16:59

David Hill

7,9311618

add a comment | 

up vote
1
down vote

up vote
1
down vote

This statement appears to be true. The elements of $A$ are non-negative linear combinations of the fundamental weights, and elements of $B$ are non-negative linear combinations of simple roots. Formula $(2.1)$ implies that the fundamental weights are non-negative linear combinations of simple roots.

Draw a picture of $A$ and $B$ for a couple of rank 2 root systems and you'll see what is going on.

share|cite|improve this answer

answered Jul 25 at 16:59

David Hill

7,9311618

This statement appears to be true. The elements of $A$ are non-negative linear combinations of the fundamental weights, and elements of $B$ are non-negative linear combinations of simple roots. Formula $(2.1)$ implies that the fundamental weights are non-negative linear combinations of simple roots.

Draw a picture of $A$ and $B$ for a couple of rank 2 root systems and you'll see what is going on.

share|cite|improve this answer

answered Jul 25 at 16:59

David Hill

7,9311618

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 16:59

David Hill

7,9311618

answered Jul 25 at 16:59

David Hill

7,9311618

answered Jul 25 at 16:59

David Hill

7,9311618

7,9311618

add a comment | 

add a comment | 

 

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