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 2 hours ago
leibnewtzleibnewtz
55428
$endgroup$
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 2 hours ago
leibnewtzleibnewtz
55428
$endgroup$
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 2 hours ago
leibnewtzleibnewtz
55428
$endgroup$
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 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
56 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 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
35 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 36 mins ago
Will SawinWill Sawin
68.7k7140285
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328495%2ftannaka-duality-for-semisimple-groups%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
56 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 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
35 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
56 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 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
35 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
56 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 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
35 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
56 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 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
35 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
56 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
56 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 mins ago
$begingroup$
I think so. But I’m more into topological groups...
$endgroup$
– M Mueger
52 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
35 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
35 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 36 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 36 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 36 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 36 mins ago
Will SawinWill Sawin
68.7k7140285
answered 36 mins ago
Will SawinWill Sawin
68.7k7140285
answered 36 mins ago
Will SawinWill Sawin
68.7k7140285
answered 36 mins ago
Will SawinWill Sawin
68.7k7140285
68.7k7140285
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328495%2ftannaka-duality-for-semisimple-groups%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
