General commutators of derivations of the exterior algebra
$begingroup$
Let $M$ be a smooth manifold and let $Omega(M)$ be the exterior algebra of smooth differential forms over $M$.
The $mathbb R$-linear map $D:Omega(M)rightarrowOmega(M)$ is a derivation of the exterior algebra of degree $rinmathbb Z$ if it maps $k$-forms to $k+r$ forms and it satisfies the Leibniz-rule $$D(omegawedgeeta)=Domegawedgeeta+omegawedge Deta.$$
This map is instead an antiderivation if it satisfies the anti-Leibniz rule $$D(omegawedgeeta)=Domegawedgeeta+(-1)^komegawedge Deta,$$ where $k$ is the degree of $omega$.
It is a known fact that if $D,D^prime$ are antiderivations of odd degrees, then the anticommutator $${D,D^prime}=DD^prime+D^prime D$$ is a derivation of degree $deg D+deg D^prime$. For example this is often used to prove Cartan's magic formula.
I am looking for a more general version of this statement. Ideally, given any derivation or anti-derivation $D$ and derivation or antiderivation $D^prime$, some kind of "graded commutator" $$ [D,D^prime ]_pm=DD^prime pm D^prime D $$ should be definable, which should result in either a derivation or antiderivation, whose degree is the sum of the degrees of $D$ and $D^prime$ .
I have not been able to find one such relation. In fact I have tried to prove that for antiderivations of arbitrary degree $DD^prime -(-1)^{deg D deg D^prime}D^prime D$ is a derivation, but I couldn't cancel enough terms in the general case.
I have also looked up "graded commutator" and "graded super Lie algebra" on Wikipedia, however it seems to me that case is different, as a graded derivation if defined with the anti-Leibniz rule $D(omegaeta)=(Domega)eta+(-1)^{kdeg D}omega (Deta)$, which is slightly different from our anti-Leibniz rule here.
Question:
How can I unify the treatment of derivations and antiderivations of the exterior algebra to (hopefully) turn them into a graded super Lie algebra.
In particular, given any two derivations or antiderivations (including the case of one being one, the other being the other), how do I define a (graded) commutator that will produce a derivation or antiderivation?
abstract-algebra differential-geometry exterior-algebra lie-superalgebras
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth manifold and let $Omega(M)$ be the exterior algebra of smooth differential forms over $M$.
The $mathbb R$-linear map $D:Omega(M)rightarrowOmega(M)$ is a derivation of the exterior algebra of degree $rinmathbb Z$ if it maps $k$-forms to $k+r$ forms and it satisfies the Leibniz-rule $$D(omegawedgeeta)=Domegawedgeeta+omegawedge Deta.$$
This map is instead an antiderivation if it satisfies the anti-Leibniz rule $$D(omegawedgeeta)=Domegawedgeeta+(-1)^komegawedge Deta,$$ where $k$ is the degree of $omega$.
It is a known fact that if $D,D^prime$ are antiderivations of odd degrees, then the anticommutator $${D,D^prime}=DD^prime+D^prime D$$ is a derivation of degree $deg D+deg D^prime$. For example this is often used to prove Cartan's magic formula.
I am looking for a more general version of this statement. Ideally, given any derivation or anti-derivation $D$ and derivation or antiderivation $D^prime$, some kind of "graded commutator" $$ [D,D^prime ]_pm=DD^prime pm D^prime D $$ should be definable, which should result in either a derivation or antiderivation, whose degree is the sum of the degrees of $D$ and $D^prime$ .
I have not been able to find one such relation. In fact I have tried to prove that for antiderivations of arbitrary degree $DD^prime -(-1)^{deg D deg D^prime}D^prime D$ is a derivation, but I couldn't cancel enough terms in the general case.
I have also looked up "graded commutator" and "graded super Lie algebra" on Wikipedia, however it seems to me that case is different, as a graded derivation if defined with the anti-Leibniz rule $D(omegaeta)=(Domega)eta+(-1)^{kdeg D}omega (Deta)$, which is slightly different from our anti-Leibniz rule here.
Question:
How can I unify the treatment of derivations and antiderivations of the exterior algebra to (hopefully) turn them into a graded super Lie algebra.
In particular, given any two derivations or antiderivations (including the case of one being one, the other being the other), how do I define a (graded) commutator that will produce a derivation or antiderivation?
abstract-algebra differential-geometry exterior-algebra lie-superalgebras
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
$endgroup$
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– Travis
Dec 17 '18 at 16:30
$begingroup$
@Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:"
$endgroup$
– jgon
Dec 17 '18 at 18:15
$begingroup$
That said, OP why do you want this particular definition of antiderivation?
$endgroup$
– jgon
Dec 17 '18 at 18:18
add a comment |
$begingroup$
Let $M$ be a smooth manifold and let $Omega(M)$ be the exterior algebra of smooth differential forms over $M$.
The $mathbb R$-linear map $D:Omega(M)rightarrowOmega(M)$ is a derivation of the exterior algebra of degree $rinmathbb Z$ if it maps $k$-forms to $k+r$ forms and it satisfies the Leibniz-rule $$D(omegawedgeeta)=Domegawedgeeta+omegawedge Deta.$$
This map is instead an antiderivation if it satisfies the anti-Leibniz rule $$D(omegawedgeeta)=Domegawedgeeta+(-1)^komegawedge Deta,$$ where $k$ is the degree of $omega$.
It is a known fact that if $D,D^prime$ are antiderivations of odd degrees, then the anticommutator $${D,D^prime}=DD^prime+D^prime D$$ is a derivation of degree $deg D+deg D^prime$. For example this is often used to prove Cartan's magic formula.
I am looking for a more general version of this statement. Ideally, given any derivation or anti-derivation $D$ and derivation or antiderivation $D^prime$, some kind of "graded commutator" $$ [D,D^prime ]_pm=DD^prime pm D^prime D $$ should be definable, which should result in either a derivation or antiderivation, whose degree is the sum of the degrees of $D$ and $D^prime$ .
I have not been able to find one such relation. In fact I have tried to prove that for antiderivations of arbitrary degree $DD^prime -(-1)^{deg D deg D^prime}D^prime D$ is a derivation, but I couldn't cancel enough terms in the general case.
I have also looked up "graded commutator" and "graded super Lie algebra" on Wikipedia, however it seems to me that case is different, as a graded derivation if defined with the anti-Leibniz rule $D(omegaeta)=(Domega)eta+(-1)^{kdeg D}omega (Deta)$, which is slightly different from our anti-Leibniz rule here.
Question:
How can I unify the treatment of derivations and antiderivations of the exterior algebra to (hopefully) turn them into a graded super Lie algebra.
In particular, given any two derivations or antiderivations (including the case of one being one, the other being the other), how do I define a (graded) commutator that will produce a derivation or antiderivation?
abstract-algebra differential-geometry exterior-algebra lie-superalgebras
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
$endgroup$
Let $M$ be a smooth manifold and let $Omega(M)$ be the exterior algebra of smooth differential forms over $M$.
The $mathbb R$-linear map $D:Omega(M)rightarrowOmega(M)$ is a derivation of the exterior algebra of degree $rinmathbb Z$ if it maps $k$-forms to $k+r$ forms and it satisfies the Leibniz-rule $$D(omegawedgeeta)=Domegawedgeeta+omegawedge Deta.$$
This map is instead an antiderivation if it satisfies the anti-Leibniz rule $$D(omegawedgeeta)=Domegawedgeeta+(-1)^komegawedge Deta,$$ where $k$ is the degree of $omega$.
It is a known fact that if $D,D^prime$ are antiderivations of odd degrees, then the anticommutator $${D,D^prime}=DD^prime+D^prime D$$ is a derivation of degree $deg D+deg D^prime$. For example this is often used to prove Cartan's magic formula.
I am looking for a more general version of this statement. Ideally, given any derivation or anti-derivation $D$ and derivation or antiderivation $D^prime$, some kind of "graded commutator" $$ [D,D^prime ]_pm=DD^prime pm D^prime D $$ should be definable, which should result in either a derivation or antiderivation, whose degree is the sum of the degrees of $D$ and $D^prime$ .
I have not been able to find one such relation. In fact I have tried to prove that for antiderivations of arbitrary degree $DD^prime -(-1)^{deg D deg D^prime}D^prime D$ is a derivation, but I couldn't cancel enough terms in the general case.
I have also looked up "graded commutator" and "graded super Lie algebra" on Wikipedia, however it seems to me that case is different, as a graded derivation if defined with the anti-Leibniz rule $D(omegaeta)=(Domega)eta+(-1)^{kdeg D}omega (Deta)$, which is slightly different from our anti-Leibniz rule here.
Question:
How can I unify the treatment of derivations and antiderivations of the exterior algebra to (hopefully) turn them into a graded super Lie algebra.
In particular, given any two derivations or antiderivations (including the case of one being one, the other being the other), how do I define a (graded) commutator that will produce a derivation or antiderivation?
abstract-algebra differential-geometry exterior-algebra lie-superalgebras
abstract-algebra differential-geometry exterior-algebra lie-superalgebras
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
asked Dec 17 '18 at 16:04
Bence RacskóBence Racskó
3,418823
3,418823
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– Travis
Dec 17 '18 at 16:30
$begingroup$
@Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:"
$endgroup$
– jgon
Dec 17 '18 at 18:15
$begingroup$
That said, OP why do you want this particular definition of antiderivation?
$endgroup$
– jgon
Dec 17 '18 at 18:18
add a comment |
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– Travis
Dec 17 '18 at 16:30
$begingroup$
@Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:"
$endgroup$
– jgon
Dec 17 '18 at 18:15
$begingroup$
That said, OP why do you want this particular definition of antiderivation?
$endgroup$
– jgon
Dec 17 '18 at 18:18
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– Travis
Dec 17 '18 at 16:30
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– Travis
Dec 17 '18 at 16:30
$begingroup$
@Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:"
$endgroup$
– jgon
Dec 17 '18 at 18:15
$begingroup$
@Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:"
$endgroup$
– jgon
Dec 17 '18 at 18:15
$begingroup$
That said, OP why do you want this particular definition of antiderivation?
$endgroup$
– jgon
Dec 17 '18 at 18:18
$begingroup$
That said, OP why do you want this particular definition of antiderivation?
$endgroup$
– jgon
Dec 17 '18 at 18:18
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%2f3044116%2fgeneral-commutators-of-derivations-of-the-exterior-algebra%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%2f3044116%2fgeneral-commutators-of-derivations-of-the-exterior-algebra%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
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– Travis
Dec 17 '18 at 16:30
$begingroup$
@Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:"
$endgroup$
– jgon
Dec 17 '18 at 18:15
$begingroup$
That said, OP why do you want this particular definition of antiderivation?
$endgroup$
– jgon
Dec 17 '18 at 18:18