Simple coroots in a non-reduced root system do not form a base?
$begingroup$
Let $(V,R)$ be a not necessarily reduced root system, and $(V^{vee},R^{vee})$ the dual root system. A chamber of $R$ is a connected component of the complement in $V$ of the hyperplanes $H_{alpha} = { x in V : langle x, alpha^{vee} rangle =0}$. If $C$ is a chamber, the corresponding base $B = B(C)$ is the unique vector space basis $alpha_1, ... , alpha_l$ of roots such that
$$C = { x in V : langle x, alpha_i^{vee} rangle > 0 textrm{ for } 1 leq i leq l }$$
The elements of $B$ are usually called simple roots. I thought that the set $B^{vee}={alpha_1^{vee}, ... , alpha_l^{vee} rangle$ of "simple coroots" was always a base of the dual root system $R^{vee}$. Bourbaki, Lie Groups and Lie Algebras, Chapter VI, Section 10, seems to indicate otherwise:
Is it true that $B^{vee}$ is in general not a base of the root system $(V^{vee},R^{vee})$?
representation-theory lie-algebras root-systems
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
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$begingroup$
Let $(V,R)$ be a not necessarily reduced root system, and $(V^{vee},R^{vee})$ the dual root system. A chamber of $R$ is a connected component of the complement in $V$ of the hyperplanes $H_{alpha} = { x in V : langle x, alpha^{vee} rangle =0}$. If $C$ is a chamber, the corresponding base $B = B(C)$ is the unique vector space basis $alpha_1, ... , alpha_l$ of roots such that
$$C = { x in V : langle x, alpha_i^{vee} rangle > 0 textrm{ for } 1 leq i leq l }$$
The elements of $B$ are usually called simple roots. I thought that the set $B^{vee}={alpha_1^{vee}, ... , alpha_l^{vee} rangle$ of "simple coroots" was always a base of the dual root system $R^{vee}$. Bourbaki, Lie Groups and Lie Algebras, Chapter VI, Section 10, seems to indicate otherwise:
Is it true that $B^{vee}$ is in general not a base of the root system $(V^{vee},R^{vee})$?
representation-theory lie-algebras root-systems
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $(V,R)$ be a not necessarily reduced root system, and $(V^{vee},R^{vee})$ the dual root system. A chamber of $R$ is a connected component of the complement in $V$ of the hyperplanes $H_{alpha} = { x in V : langle x, alpha^{vee} rangle =0}$. If $C$ is a chamber, the corresponding base $B = B(C)$ is the unique vector space basis $alpha_1, ... , alpha_l$ of roots such that
$$C = { x in V : langle x, alpha_i^{vee} rangle > 0 textrm{ for } 1 leq i leq l }$$
The elements of $B$ are usually called simple roots. I thought that the set $B^{vee}={alpha_1^{vee}, ... , alpha_l^{vee} rangle$ of "simple coroots" was always a base of the dual root system $R^{vee}$. Bourbaki, Lie Groups and Lie Algebras, Chapter VI, Section 10, seems to indicate otherwise:
Is it true that $B^{vee}$ is in general not a base of the root system $(V^{vee},R^{vee})$?
representation-theory lie-algebras root-systems
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
$endgroup$
Let $(V,R)$ be a not necessarily reduced root system, and $(V^{vee},R^{vee})$ the dual root system. A chamber of $R$ is a connected component of the complement in $V$ of the hyperplanes $H_{alpha} = { x in V : langle x, alpha^{vee} rangle =0}$. If $C$ is a chamber, the corresponding base $B = B(C)$ is the unique vector space basis $alpha_1, ... , alpha_l$ of roots such that
$$C = { x in V : langle x, alpha_i^{vee} rangle > 0 textrm{ for } 1 leq i leq l }$$
The elements of $B$ are usually called simple roots. I thought that the set $B^{vee}={alpha_1^{vee}, ... , alpha_l^{vee} rangle$ of "simple coroots" was always a base of the dual root system $R^{vee}$. Bourbaki, Lie Groups and Lie Algebras, Chapter VI, Section 10, seems to indicate otherwise:
Is it true that $B^{vee}$ is in general not a base of the root system $(V^{vee},R^{vee})$?
representation-theory lie-algebras root-systems
representation-theory lie-algebras root-systems
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
asked Dec 8 '18 at 16:19
D_SD_S
13.5k51551
13.5k51551
add a comment |
add a comment |
1 Answer
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Let $(-,-)$ be a symmetric, positive definite Weyl group invariant form on $V$, so that $V$ identifies with $V^{vee}$ via $v mapsto (-,v)$. Then the coroot $alpha^{vee}$ identifies with $frac{2}{(alpha,alpha)}alpha$. It follows that if $alpha in B$ is such that $2alpha$ is a root, then $alpha^{vee}$ cannot be an element of a base of $R^{vee}$, since it is equal to $2 (2alpha)^{vee}$:
$$(2alpha)^{vee} = frac{2}{(2alpha,2alpha)} cdot 2alpha = frac{alpha}{(alpha,alpha)} = frac{1}{2} alpha^{vee}$$
answered Dec 8 '18 at 21:42
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$begingroup$
Let $(-,-)$ be a symmetric, positive definite Weyl group invariant form on $V$, so that $V$ identifies with $V^{vee}$ via $v mapsto (-,v)$. Then the coroot $alpha^{vee}$ identifies with $frac{2}{(alpha,alpha)}alpha$. It follows that if $alpha in B$ is such that $2alpha$ is a root, then $alpha^{vee}$ cannot be an element of a base of $R^{vee}$, since it is equal to $2 (2alpha)^{vee}$:
$$(2alpha)^{vee} = frac{2}{(2alpha,2alpha)} cdot 2alpha = frac{alpha}{(alpha,alpha)} = frac{1}{2} alpha^{vee}$$
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $(-,-)$ be a symmetric, positive definite Weyl group invariant form on $V$, so that $V$ identifies with $V^{vee}$ via $v mapsto (-,v)$. Then the coroot $alpha^{vee}$ identifies with $frac{2}{(alpha,alpha)}alpha$. It follows that if $alpha in B$ is such that $2alpha$ is a root, then $alpha^{vee}$ cannot be an element of a base of $R^{vee}$, since it is equal to $2 (2alpha)^{vee}$:
$$(2alpha)^{vee} = frac{2}{(2alpha,2alpha)} cdot 2alpha = frac{alpha}{(alpha,alpha)} = frac{1}{2} alpha^{vee}$$
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $(-,-)$ be a symmetric, positive definite Weyl group invariant form on $V$, so that $V$ identifies with $V^{vee}$ via $v mapsto (-,v)$. Then the coroot $alpha^{vee}$ identifies with $frac{2}{(alpha,alpha)}alpha$. It follows that if $alpha in B$ is such that $2alpha$ is a root, then $alpha^{vee}$ cannot be an element of a base of $R^{vee}$, since it is equal to $2 (2alpha)^{vee}$:
$$(2alpha)^{vee} = frac{2}{(2alpha,2alpha)} cdot 2alpha = frac{alpha}{(alpha,alpha)} = frac{1}{2} alpha^{vee}$$
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
$endgroup$
Let $(-,-)$ be a symmetric, positive definite Weyl group invariant form on $V$, so that $V$ identifies with $V^{vee}$ via $v mapsto (-,v)$. Then the coroot $alpha^{vee}$ identifies with $frac{2}{(alpha,alpha)}alpha$. It follows that if $alpha in B$ is such that $2alpha$ is a root, then $alpha^{vee}$ cannot be an element of a base of $R^{vee}$, since it is equal to $2 (2alpha)^{vee}$:
$$(2alpha)^{vee} = frac{2}{(2alpha,2alpha)} cdot 2alpha = frac{alpha}{(alpha,alpha)} = frac{1}{2} alpha^{vee}$$
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
answered Dec 8 '18 at 21:42
D_SD_S
13.5k51551
13.5k51551
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