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What is weight lattice modulo coroot lattice?
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In Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov, the $S$ matrix (expression 3.3.7) is expressed in the form $vert P/kQ^vee vert^-1/2times(cdots)$, where $kin mathbbN$ is the shifted level, $P$ is the weight lattice and $Q^vee$ is the coroot lattice. This expression seems to suggest that $Q^vee$ can be interpreted as a sublattice of $P$, but this is giving me a confusion because $Psubset mathfrakh^*$ while $Q^veesubsetmathfrakh$. What is the natural way to embed $Q^vee$ inside $P$?
For instance, in case of $G_2$, let's say $alpha$ and $beta$ are simple short and long roots, respectively. The fundamental weights are then $w_1 = 2alpha+beta$, $w_2 = 3alpha+2beta$. Hence the weight lattice in this case is the same as the root lattice; $P=Q$. Now the coroots are $alpha^vee = alpha$, $beta^vee = frac13beta$, and they don't sit inside $P$.
Should I just assume that $Q^vee$ is the sublattice of $P$ generated by long roots, or is there a better way to interpret $vert P/Q^vee vert$?
lie-algebras root-systems quantum-groups lattices-in-lie-groups
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
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up vote
3
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In Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov, the $S$ matrix (expression 3.3.7) is expressed in the form $vert P/kQ^vee vert^-1/2times(cdots)$, where $kin mathbbN$ is the shifted level, $P$ is the weight lattice and $Q^vee$ is the coroot lattice. This expression seems to suggest that $Q^vee$ can be interpreted as a sublattice of $P$, but this is giving me a confusion because $Psubset mathfrakh^*$ while $Q^veesubsetmathfrakh$. What is the natural way to embed $Q^vee$ inside $P$?
For instance, in case of $G_2$, let's say $alpha$ and $beta$ are simple short and long roots, respectively. The fundamental weights are then $w_1 = 2alpha+beta$, $w_2 = 3alpha+2beta$. Hence the weight lattice in this case is the same as the root lattice; $P=Q$. Now the coroots are $alpha^vee = alpha$, $beta^vee = frac13beta$, and they don't sit inside $P$.
Should I just assume that $Q^vee$ is the sublattice of $P$ generated by long roots, or is there a better way to interpret $vert P/Q^vee vert$?
lie-algebras root-systems quantum-groups lattices-in-lie-groups
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov, the $S$ matrix (expression 3.3.7) is expressed in the form $vert P/kQ^vee vert^-1/2times(cdots)$, where $kin mathbbN$ is the shifted level, $P$ is the weight lattice and $Q^vee$ is the coroot lattice. This expression seems to suggest that $Q^vee$ can be interpreted as a sublattice of $P$, but this is giving me a confusion because $Psubset mathfrakh^*$ while $Q^veesubsetmathfrakh$. What is the natural way to embed $Q^vee$ inside $P$?
For instance, in case of $G_2$, let's say $alpha$ and $beta$ are simple short and long roots, respectively. The fundamental weights are then $w_1 = 2alpha+beta$, $w_2 = 3alpha+2beta$. Hence the weight lattice in this case is the same as the root lattice; $P=Q$. Now the coroots are $alpha^vee = alpha$, $beta^vee = frac13beta$, and they don't sit inside $P$.
Should I just assume that $Q^vee$ is the sublattice of $P$ generated by long roots, or is there a better way to interpret $vert P/Q^vee vert$?
lie-algebras root-systems quantum-groups lattices-in-lie-groups
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
In Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov, the $S$ matrix (expression 3.3.7) is expressed in the form $vert P/kQ^vee vert^-1/2times(cdots)$, where $kin mathbbN$ is the shifted level, $P$ is the weight lattice and $Q^vee$ is the coroot lattice. This expression seems to suggest that $Q^vee$ can be interpreted as a sublattice of $P$, but this is giving me a confusion because $Psubset mathfrakh^*$ while $Q^veesubsetmathfrakh$. What is the natural way to embed $Q^vee$ inside $P$?
For instance, in case of $G_2$, let's say $alpha$ and $beta$ are simple short and long roots, respectively. The fundamental weights are then $w_1 = 2alpha+beta$, $w_2 = 3alpha+2beta$. Hence the weight lattice in this case is the same as the root lattice; $P=Q$. Now the coroots are $alpha^vee = alpha$, $beta^vee = frac13beta$, and they don't sit inside $P$.
Should I just assume that $Q^vee$ is the sublattice of $P$ generated by long roots, or is there a better way to interpret $vert P/Q^vee vert$?
lie-algebras root-systems quantum-groups lattices-in-lie-groups
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
edited Jul 27 at 5:14
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
asked Jul 24 at 18:31
Sunghyuk Park
1,444621
1,444621
add a comment |Â
add a comment |Â
1 Answer
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In Theorem 3.3.6 the authors of said book mention that $Q^vee$ is embedded into $mathfrak h^*$ via the form $langle , rangle$. Here $langle, rangle$ denotes an invariant bilinear form on $mathfrak g$ such that $langle alpha,alpharangle = 2$ for long roots (see the beginning of section 1.4).
In this way $kQ^vee$ is viewed as a sublattice of $P$. Actually, the authors state in the proof of Theorem 3.3.20 that $W^a = Wltimes kQ^vee$ acts on $P subseteq mathfrak h^*$; as such this means that $P/kQ^vee$ is the set of orbits of the action of $kQ^vee$ on $P$ (which is the same thing).
answered Jul 27 at 11:17
Claudius
3,7111515
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In Theorem 3.3.6 the authors of said book mention that $Q^vee$ is embedded into $mathfrak h^*$ via the form $langle , rangle$. Here $langle, rangle$ denotes an invariant bilinear form on $mathfrak g$ such that $langle alpha,alpharangle = 2$ for long roots (see the beginning of section 1.4).
In this way $kQ^vee$ is viewed as a sublattice of $P$. Actually, the authors state in the proof of Theorem 3.3.20 that $W^a = Wltimes kQ^vee$ acts on $P subseteq mathfrak h^*$; as such this means that $P/kQ^vee$ is the set of orbits of the action of $kQ^vee$ on $P$ (which is the same thing).
answered Jul 27 at 11:17
Claudius
3,7111515
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
add a comment |Â
up vote
1
down vote
accepted
In Theorem 3.3.6 the authors of said book mention that $Q^vee$ is embedded into $mathfrak h^*$ via the form $langle , rangle$. Here $langle, rangle$ denotes an invariant bilinear form on $mathfrak g$ such that $langle alpha,alpharangle = 2$ for long roots (see the beginning of section 1.4).
In this way $kQ^vee$ is viewed as a sublattice of $P$. Actually, the authors state in the proof of Theorem 3.3.20 that $W^a = Wltimes kQ^vee$ acts on $P subseteq mathfrak h^*$; as such this means that $P/kQ^vee$ is the set of orbits of the action of $kQ^vee$ on $P$ (which is the same thing).
answered Jul 27 at 11:17
Claudius
3,7111515
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In Theorem 3.3.6 the authors of said book mention that $Q^vee$ is embedded into $mathfrak h^*$ via the form $langle , rangle$. Here $langle, rangle$ denotes an invariant bilinear form on $mathfrak g$ such that $langle alpha,alpharangle = 2$ for long roots (see the beginning of section 1.4).
In this way $kQ^vee$ is viewed as a sublattice of $P$. Actually, the authors state in the proof of Theorem 3.3.20 that $W^a = Wltimes kQ^vee$ acts on $P subseteq mathfrak h^*$; as such this means that $P/kQ^vee$ is the set of orbits of the action of $kQ^vee$ on $P$ (which is the same thing).
answered Jul 27 at 11:17
Claudius
3,7111515
In Theorem 3.3.6 the authors of said book mention that $Q^vee$ is embedded into $mathfrak h^*$ via the form $langle , rangle$. Here $langle, rangle$ denotes an invariant bilinear form on $mathfrak g$ such that $langle alpha,alpharangle = 2$ for long roots (see the beginning of section 1.4).
In this way $kQ^vee$ is viewed as a sublattice of $P$. Actually, the authors state in the proof of Theorem 3.3.20 that $W^a = Wltimes kQ^vee$ acts on $P subseteq mathfrak h^*$; as such this means that $P/kQ^vee$ is the set of orbits of the action of $kQ^vee$ on $P$ (which is the same thing).
answered Jul 27 at 11:17
Claudius
3,7111515
answered Jul 27 at 11:17
Claudius
3,7111515
answered Jul 27 at 11:17
Claudius
3,7111515
answered Jul 27 at 11:17
Claudius
3,7111515
3,7111515
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
add a comment |Â
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $mathfrakg$, one normalized so that $langlelanglealpha,alpharanglerangle=2$ for short roots while the other normalized so that $langle alpha,alpha rangle$ for long roots. After all, $Q^vee$ is the sublattice of $Q$ generated by long roots, as I expected.
– Sunghyuk Park
Jul 27 at 17:25
add a comment |Â
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