Tannaka duality for semisimple groups Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?Tannaka formalism and the étale fundamental groupIs there a ``path'' between any two fiber functors over the same field in Tannakian formalism?Counter example in Tannaka reconstruction?Recovering classical Tannaka duality from Lurie's version for geometric stacksTannaka DualityCan one explain Tannaka-Krein duality for a finite-group to … a computer ? (How to make input for reconstruction to be finite datum?)Tannakian Formalism for the Quaternions and Dihedral GroupTannakian theory for Lie algebrasIs it possible to reconstruct a finitely generated group from its category of representations?
Tannaka duality for semisimple groups
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?Tannaka formalism and the étale fundamental groupIs there a ``path'' between any two fiber functors over the same field in Tannakian formalism?Counter example in Tannaka reconstruction?Recovering classical Tannaka duality from Lurie's version for geometric stacksTannaka DualityCan one explain Tannaka-Krein duality for a finite-group to … a computer ? (How to make input for reconstruction to be finite datum?)Tannakian Formalism for the Quaternions and Dihedral GroupTannakian theory for Lie algebrasIs it possible to reconstruct a finitely generated group from its category of representations?
$begingroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. This means that any such category is associated with a root datum.
Is there a version of this reconstruction theorem that will tell us when a category $mathcalC$ is the category of finite dimensional representations of a semisimple group? I would like to be able to associate with a Tannakian category a root system, and not just a root datum.
ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
$endgroup$
add a comment |
$begingroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. This means that any such category is associated with a root datum.
Is there a version of this reconstruction theorem that will tell us when a category $mathcalC$ is the category of finite dimensional representations of a semisimple group? I would like to be able to associate with a Tannakian category a root system, and not just a root datum.
ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
$endgroup$
$begingroup$
The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories.
$endgroup$
– anon
2 days ago
$begingroup$
To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348
$endgroup$
– anon
2 days ago
$begingroup$
Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it).
$endgroup$
– LSpice
2 days ago
$begingroup$
L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$.
$endgroup$
– anon
2 days ago
add a comment |
$begingroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. This means that any such category is associated with a root datum.
Is there a version of this reconstruction theorem that will tell us when a category $mathcalC$ is the category of finite dimensional representations of a semisimple group? I would like to be able to associate with a Tannakian category a root system, and not just a root datum.
ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
$endgroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. This means that any such category is associated with a root datum.
Is there a version of this reconstruction theorem that will tell us when a category $mathcalC$ is the category of finite dimensional representations of a semisimple group? I would like to be able to associate with a Tannakian category a root system, and not just a root datum.
ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category
ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
asked Apr 19 at 22:39
leibnewtzleibnewtz
58429
58429
$begingroup$
The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories.
$endgroup$
– anon
2 days ago
$begingroup$
To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348
$endgroup$
– anon
2 days ago
$begingroup$
Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it).
$endgroup$
– LSpice
2 days ago
$begingroup$
L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$.
$endgroup$
– anon
2 days ago
add a comment |
$begingroup$
The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories.
$endgroup$
– anon
2 days ago
$begingroup$
To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348
$endgroup$
– anon
2 days ago
$begingroup$
Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it).
$endgroup$
– LSpice
2 days ago
$begingroup$
L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$.
$endgroup$
– anon
2 days ago
$begingroup$
The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories.
$endgroup$
– anon
2 days ago
$begingroup$
The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories.
$endgroup$
– anon
2 days ago
$begingroup$
To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348
$endgroup$
– anon
2 days ago
$begingroup$
To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348
$endgroup$
– anon
2 days ago
$begingroup$
Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it).
$endgroup$
– LSpice
2 days ago
$begingroup$
Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it).
$endgroup$
– LSpice
2 days ago
$begingroup$
L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$.
$endgroup$
– anon
2 days ago
$begingroup$
L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$.
$endgroup$
– anon
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In order for $mathcal C$ to come from an algebraic group rather than a pro-algebraic one, you want $mathcal C$ to be finitely generated. And for semisimplicity, you want the group to have finite center. The center can be read off from the category. Cf. my paper “On the center of a compact group”, Intern. Math. Res. Notes. 2004:51, 2751-2756 (2004) or math.CT/0312257.
answered Apr 19 at 23:42
M MuegerM Mueger
36017
$endgroup$
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
1
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
1
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
Another criterion is that there should be only finitely many objects of bounded dimension. This condition might be easy to check in practice from abstract finiteness theorems. The proof is that, if the group is not semi simple, you can take any 1-dimensional character of the identity component and induce up to the main group. Because there are infinitely many characters, infinitely many representations.
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
$endgroup$
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
1
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
I've decided to turn my comments into an answer.
(a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories.
(b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne.
(c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –
answered 2 days ago
anonanon
1212
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In order for $mathcal C$ to come from an algebraic group rather than a pro-algebraic one, you want $mathcal C$ to be finitely generated. And for semisimplicity, you want the group to have finite center. The center can be read off from the category. Cf. my paper “On the center of a compact group”, Intern. Math. Res. Notes. 2004:51, 2751-2756 (2004) or math.CT/0312257.
answered Apr 19 at 23:42
M MuegerM Mueger
36017
$endgroup$
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
1
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
1
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
In order for $mathcal C$ to come from an algebraic group rather than a pro-algebraic one, you want $mathcal C$ to be finitely generated. And for semisimplicity, you want the group to have finite center. The center can be read off from the category. Cf. my paper “On the center of a compact group”, Intern. Math. Res. Notes. 2004:51, 2751-2756 (2004) or math.CT/0312257.
answered Apr 19 at 23:42
M MuegerM Mueger
36017
$endgroup$
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
1
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
1
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
In order for $mathcal C$ to come from an algebraic group rather than a pro-algebraic one, you want $mathcal C$ to be finitely generated. And for semisimplicity, you want the group to have finite center. The center can be read off from the category. Cf. my paper “On the center of a compact group”, Intern. Math. Res. Notes. 2004:51, 2751-2756 (2004) or math.CT/0312257.
answered Apr 19 at 23:42
M MuegerM Mueger
36017
$endgroup$
In order for $mathcal C$ to come from an algebraic group rather than a pro-algebraic one, you want $mathcal C$ to be finitely generated. And for semisimplicity, you want the group to have finite center. The center can be read off from the category. Cf. my paper “On the center of a compact group”, Intern. Math. Res. Notes. 2004:51, 2751-2756 (2004) or math.CT/0312257.
answered Apr 19 at 23:42
M MuegerM Mueger
36017
answered Apr 19 at 23:42
M MuegerM Mueger
36017
answered Apr 19 at 23:42
M MuegerM Mueger
36017
answered Apr 19 at 23:42
M MuegerM Mueger
36017
36017
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
1
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
1
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
1
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
1
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct?
$endgroup$
– leibnewtz
Apr 19 at 23:54
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
Apr 19 at 23:58
1
1
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
Apr 20 at 0:15
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
$begingroup$
@Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.).
$endgroup$
– M Mueger
2 days ago
1
1
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
$begingroup$
@M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A otimes B = I$.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
Another criterion is that there should be only finitely many objects of bounded dimension. This condition might be easy to check in practice from abstract finiteness theorems. The proof is that, if the group is not semi simple, you can take any 1-dimensional character of the identity component and induce up to the main group. Because there are infinitely many characters, infinitely many representations.
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
$endgroup$
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
1
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
Another criterion is that there should be only finitely many objects of bounded dimension. This condition might be easy to check in practice from abstract finiteness theorems. The proof is that, if the group is not semi simple, you can take any 1-dimensional character of the identity component and induce up to the main group. Because there are infinitely many characters, infinitely many representations.
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
$endgroup$
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
1
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
Another criterion is that there should be only finitely many objects of bounded dimension. This condition might be easy to check in practice from abstract finiteness theorems. The proof is that, if the group is not semi simple, you can take any 1-dimensional character of the identity component and induce up to the main group. Because there are infinitely many characters, infinitely many representations.
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
$endgroup$
Another criterion is that there should be only finitely many objects of bounded dimension. This condition might be easy to check in practice from abstract finiteness theorems. The proof is that, if the group is not semi simple, you can take any 1-dimensional character of the identity component and induce up to the main group. Because there are infinitely many characters, infinitely many representations.
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
answered Apr 20 at 0:14
Will SawinWill Sawin
68.8k7140285
68.8k7140285
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
1
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
1
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
$begingroup$
Could you say something about what bounded dimension means?
$endgroup$
– leibnewtz
2 days ago
1
1
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
$begingroup$
@leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor.
$endgroup$
– Will Sawin
2 days ago
add a comment |
$begingroup$
I've decided to turn my comments into an answer.
(a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories.
(b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne.
(c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –
answered 2 days ago
anonanon
1212
$endgroup$
add a comment |
$begingroup$
I've decided to turn my comments into an answer.
(a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories.
(b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne.
(c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –
answered 2 days ago
anonanon
1212
$endgroup$
add a comment |
$begingroup$
I've decided to turn my comments into an answer.
(a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories.
(b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne.
(c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –
answered 2 days ago
anonanon
1212
$endgroup$
I've decided to turn my comments into an answer.
(a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories.
(b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne.
(c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –
answered 2 days ago
anonanon
1212
edited 2 days ago
answered 2 days ago
anonanon
1212
answered 2 days ago
anonanon
1212
answered 2 days ago
anonanon
1212
1212
add a comment |
add a comment |
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The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $Dto Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories.
$endgroup$
– anon
2 days ago
$begingroup$
To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348
$endgroup$
– anon
2 days ago
$begingroup$
Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it).
$endgroup$
– LSpice
2 days ago
$begingroup$
L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$.
$endgroup$
– anon
2 days ago