Understanding the proof of Cartier duality
up vote
0
down vote
favorite
I'm trying to understand the above proof of Cartier Duality.
The step I don't understand is the following
It says
$$ phipsi(a) = ((phi otimes psi) circ Delta(a) = (phi otimes psi)( a otimes a) $$ for all $f,g$
Why does it implies $Delta(a) = a otimes a$?
It may possible that $operatorname{ker}(phi otimes psi)$ is not trivial for all $phi$ and $psi$.
Here the $Delta$ is the co-multiplication map of the Hopf algebra $A$.
algebraic-geometry commutative-algebra hopf-algebras group-schemes
asked Nov 22 at 4:46
grok
607410
add a comment |
up vote
0
down vote
favorite
I'm trying to understand the above proof of Cartier Duality.
The step I don't understand is the following
It says
$$ phipsi(a) = ((phi otimes psi) circ Delta(a) = (phi otimes psi)( a otimes a) $$ for all $f,g$
Why does it implies $Delta(a) = a otimes a$?
It may possible that $operatorname{ker}(phi otimes psi)$ is not trivial for all $phi$ and $psi$.
Here the $Delta$ is the co-multiplication map of the Hopf algebra $A$.
algebraic-geometry commutative-algebra hopf-algebras group-schemes
asked Nov 22 at 4:46
grok
607410
Somehow it seems transparent to prove that $G(R) cong Hom(G^{check{}}, mathbb{G}_m)$, and then use the statement for $G^{check{}}$. Atleast that's what they is done in Algebraic Geometry - II Mumford Oda.
– random123
Nov 22 at 7:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to understand the above proof of Cartier Duality.
The step I don't understand is the following
It says
$$ phipsi(a) = ((phi otimes psi) circ Delta(a) = (phi otimes psi)( a otimes a) $$ for all $f,g$
Why does it implies $Delta(a) = a otimes a$?
It may possible that $operatorname{ker}(phi otimes psi)$ is not trivial for all $phi$ and $psi$.
Here the $Delta$ is the co-multiplication map of the Hopf algebra $A$.
algebraic-geometry commutative-algebra hopf-algebras group-schemes
asked Nov 22 at 4:46
grok
607410
I'm trying to understand the above proof of Cartier Duality.
The step I don't understand is the following
It says
$$ phipsi(a) = ((phi otimes psi) circ Delta(a) = (phi otimes psi)( a otimes a) $$ for all $f,g$
Why does it implies $Delta(a) = a otimes a$?
It may possible that $operatorname{ker}(phi otimes psi)$ is not trivial for all $phi$ and $psi$.
Here the $Delta$ is the co-multiplication map of the Hopf algebra $A$.
algebraic-geometry commutative-algebra hopf-algebras group-schemes
algebraic-geometry commutative-algebra hopf-algebras group-schemes
asked Nov 22 at 4:46
grok
607410
asked Nov 22 at 4:46
grok
607410
asked Nov 22 at 4:46
grok
607410
asked Nov 22 at 4:46
grok
607410
asked Nov 22 at 4:46
grok
607410
607410
Somehow it seems transparent to prove that $G(R) cong Hom(G^{check{}}, mathbb{G}_m)$, and then use the statement for $G^{check{}}$. Atleast that's what they is done in Algebraic Geometry - II Mumford Oda.
– random123
Nov 22 at 7:02
add a comment |
Somehow it seems transparent to prove that $G(R) cong Hom(G^{check{}}, mathbb{G}_m)$, and then use the statement for $G^{check{}}$. Atleast that's what they is done in Algebraic Geometry - II Mumford Oda.
– random123
Nov 22 at 7:02
Somehow it seems transparent to prove that $G(R) cong Hom(G^{check{}}, mathbb{G}_m)$, and then use the statement for $G^{check{}}$. Atleast that's what they is done in Algebraic Geometry - II Mumford Oda.
– random123
Nov 22 at 7:02
Somehow it seems transparent to prove that $G(R) cong Hom(G^{check{}}, mathbb{G}_m)$, and then use the statement for $G^{check{}}$. Atleast that's what they is done in Algebraic Geometry - II Mumford Oda.
– random123
Nov 22 at 7:02
add a comment |
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',
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%2f3008748%2funderstanding-the-proof-of-cartier-duality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3008748%2funderstanding-the-proof-of-cartier-duality%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
Somehow it seems transparent to prove that $G(R) cong Hom(G^{check{}}, mathbb{G}_m)$, and then use the statement for $G^{check{}}$. Atleast that's what they is done in Algebraic Geometry - II Mumford Oda.
– random123
Nov 22 at 7:02