Tannaka duality for semisimple groups Planned maintenance scheduled April 23, 2019 at 23:30...
Tannaka duality for semisimple groups
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
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$begingroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcal{C}$ equipped with a fiber functor $F: mathcal{C} to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcal{C} cong Rep(G)$. 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 $mathcal{C}$ 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 2 hours ago
leibnewtzleibnewtz
55428
$endgroup$
add a comment |
$begingroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcal{C}$ equipped with a fiber functor $F: mathcal{C} to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcal{C} cong Rep(G)$. 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 $mathcal{C}$ 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 2 hours ago
leibnewtzleibnewtz
55428
$endgroup$
add a comment |
$begingroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcal{C}$ equipped with a fiber functor $F: mathcal{C} to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcal{C} cong Rep(G)$. 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 $mathcal{C}$ 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 2 hours ago
leibnewtzleibnewtz
55428
$endgroup$
Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcal{C}$ equipped with a fiber functor $F: mathcal{C} to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcal{C} cong Rep(G)$. 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 $mathcal{C}$ 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 2 hours ago
leibnewtzleibnewtz
55428
asked 2 hours ago
leibnewtzleibnewtz
55428
asked 2 hours ago
leibnewtzleibnewtz
55428
asked 2 hours ago
leibnewtzleibnewtz
55428
asked 2 hours ago
leibnewtzleibnewtz
55428
55428
add a comment |
add a comment |
2 Answers
2
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 1 hour ago
M MuegerM Mueger
1635
$endgroup$
$begingroup$
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
51 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins 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 31 mins ago
Will SawinWill Sawin
68.7k7140285
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
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 1 hour ago
M MuegerM Mueger
1635
$endgroup$
$begingroup$
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
51 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins 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 1 hour ago
M MuegerM Mueger
1635
$endgroup$
$begingroup$
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
51 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins 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 1 hour ago
M MuegerM Mueger
1635
$endgroup$
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 1 hour ago
M MuegerM Mueger
1635
answered 1 hour ago
M MuegerM Mueger
1635
answered 1 hour ago
M MuegerM Mueger
1635
answered 1 hour ago
M MuegerM Mueger
1635
1635
$begingroup$
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
51 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins 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
51 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins 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
51 mins 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
51 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
47 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins ago
$begingroup$
Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
$endgroup$
– Will Sawin
30 mins 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 31 mins ago
Will SawinWill Sawin
68.7k7140285
$endgroup$
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 31 mins ago
Will SawinWill Sawin
68.7k7140285
$endgroup$
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 31 mins ago
Will SawinWill Sawin
68.7k7140285
$endgroup$
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 31 mins ago
Will SawinWill Sawin
68.7k7140285
answered 31 mins ago
Will SawinWill Sawin
68.7k7140285
answered 31 mins ago
Will SawinWill Sawin
68.7k7140285
answered 31 mins ago
Will SawinWill Sawin
68.7k7140285
68.7k7140285
add a comment |
add a comment |
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