Coordinate superalgebra
$begingroup$
Lie algebra: Let $G$ be a semisimple, simply connected linear algebraic group with a fixed Borel subgroup $B$. Let $P$ be a parabolic subgroup containing $B$. Let $lambda$ be a dominant integral weight. It follows from the Borel-Weil-Bott theorem that the homogeneous coordinate ring $A_lambda(G/P)$ is a sum of highest weight representations. Namely,
begin{align*}
A_lambda(G/P) = oplus_{n in mathbb{N}} V(nlambda).
end{align*}
Lie superalgebra: Do we have an analogue result of this in the super case? More precisely, let $G$ be a simply connected analytic supergroup with $mathfrak{g} = Lie(G)$ (a basic classical Lie superalgebra) and a Borel subsupergroup $B$. Let $P$ be a subsupergroup containing $B$. I know that there is an analogue of Borel-Weil-Bott theorem for supercase, but I wonder if there is a result for the coordinate ring as above. I guess we might need to assume $lambda$ is a typical weight.
algebraic-geometry representation-theory lie-groups lie-superalgebras
asked Jan 1 at 3:18
NongAmNongAm
12318
$endgroup$
add a comment |
$begingroup$
Lie algebra: Let $G$ be a semisimple, simply connected linear algebraic group with a fixed Borel subgroup $B$. Let $P$ be a parabolic subgroup containing $B$. Let $lambda$ be a dominant integral weight. It follows from the Borel-Weil-Bott theorem that the homogeneous coordinate ring $A_lambda(G/P)$ is a sum of highest weight representations. Namely,
begin{align*}
A_lambda(G/P) = oplus_{n in mathbb{N}} V(nlambda).
end{align*}
Lie superalgebra: Do we have an analogue result of this in the super case? More precisely, let $G$ be a simply connected analytic supergroup with $mathfrak{g} = Lie(G)$ (a basic classical Lie superalgebra) and a Borel subsupergroup $B$. Let $P$ be a subsupergroup containing $B$. I know that there is an analogue of Borel-Weil-Bott theorem for supercase, but I wonder if there is a result for the coordinate ring as above. I guess we might need to assume $lambda$ is a typical weight.
algebraic-geometry representation-theory lie-groups lie-superalgebras
asked Jan 1 at 3:18
NongAmNongAm
12318
$endgroup$
add a comment |
$begingroup$
Lie algebra: Let $G$ be a semisimple, simply connected linear algebraic group with a fixed Borel subgroup $B$. Let $P$ be a parabolic subgroup containing $B$. Let $lambda$ be a dominant integral weight. It follows from the Borel-Weil-Bott theorem that the homogeneous coordinate ring $A_lambda(G/P)$ is a sum of highest weight representations. Namely,
begin{align*}
A_lambda(G/P) = oplus_{n in mathbb{N}} V(nlambda).
end{align*}
Lie superalgebra: Do we have an analogue result of this in the super case? More precisely, let $G$ be a simply connected analytic supergroup with $mathfrak{g} = Lie(G)$ (a basic classical Lie superalgebra) and a Borel subsupergroup $B$. Let $P$ be a subsupergroup containing $B$. I know that there is an analogue of Borel-Weil-Bott theorem for supercase, but I wonder if there is a result for the coordinate ring as above. I guess we might need to assume $lambda$ is a typical weight.
algebraic-geometry representation-theory lie-groups lie-superalgebras
asked Jan 1 at 3:18
NongAmNongAm
12318
$endgroup$
Lie algebra: Let $G$ be a semisimple, simply connected linear algebraic group with a fixed Borel subgroup $B$. Let $P$ be a parabolic subgroup containing $B$. Let $lambda$ be a dominant integral weight. It follows from the Borel-Weil-Bott theorem that the homogeneous coordinate ring $A_lambda(G/P)$ is a sum of highest weight representations. Namely,
begin{align*}
A_lambda(G/P) = oplus_{n in mathbb{N}} V(nlambda).
end{align*}
Lie superalgebra: Do we have an analogue result of this in the super case? More precisely, let $G$ be a simply connected analytic supergroup with $mathfrak{g} = Lie(G)$ (a basic classical Lie superalgebra) and a Borel subsupergroup $B$. Let $P$ be a subsupergroup containing $B$. I know that there is an analogue of Borel-Weil-Bott theorem for supercase, but I wonder if there is a result for the coordinate ring as above. I guess we might need to assume $lambda$ is a typical weight.
algebraic-geometry representation-theory lie-groups lie-superalgebras
algebraic-geometry representation-theory lie-groups lie-superalgebras
asked Jan 1 at 3:18
NongAmNongAm
12318
asked Jan 1 at 3:18
NongAmNongAm
12318
asked Jan 1 at 3:18
NongAmNongAm
12318
asked Jan 1 at 3:18
NongAmNongAm
12318
asked Jan 1 at 3:18
NongAmNongAm
12318
12318
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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%2fmath.stackexchange.com%2fquestions%2f3058200%2fcoordinate-superalgebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f3058200%2fcoordinate-superalgebra%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
