super commutation of matrices
$begingroup$
Let $M_{p|q}(mathbb{C}) = M_{p|q}(mathbb{C})_0 oplus M_{p|q}(mathbb{C})_1$ be the super algebra of all $(p+q) times (p+q)$ matrices.
Let $A, B in M_{p|q}(mathbb{C})$ (not necessarily homogeneous), If anybody can help me with finding the necessary and sufficient condition for $A$ and $B$ to super commute, that would be very helpful to me.
Super commutator is defined as $[X,Y] = XY - (-1)^{|X||Y|}XY ,forall, X,Y in M_{p|q}(mathbb{C})_0 sqcup M_{p|q}(mathbb{C})_1$ and extend this definition bilinearly to full $M_{p|q}(mathbb{C})$. $|X| = text{homogeneous degree of X}$
Thanks for your thoughts.
Have a good day.
matrices matrix-equations superalgebra lie-superalgebras
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
$endgroup$
add a comment |
$begingroup$
Let $M_{p|q}(mathbb{C}) = M_{p|q}(mathbb{C})_0 oplus M_{p|q}(mathbb{C})_1$ be the super algebra of all $(p+q) times (p+q)$ matrices.
Let $A, B in M_{p|q}(mathbb{C})$ (not necessarily homogeneous), If anybody can help me with finding the necessary and sufficient condition for $A$ and $B$ to super commute, that would be very helpful to me.
Super commutator is defined as $[X,Y] = XY - (-1)^{|X||Y|}XY ,forall, X,Y in M_{p|q}(mathbb{C})_0 sqcup M_{p|q}(mathbb{C})_1$ and extend this definition bilinearly to full $M_{p|q}(mathbb{C})$. $|X| = text{homogeneous degree of X}$
Thanks for your thoughts.
Have a good day.
matrices matrix-equations superalgebra lie-superalgebras
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
$endgroup$
add a comment |
$begingroup$
Let $M_{p|q}(mathbb{C}) = M_{p|q}(mathbb{C})_0 oplus M_{p|q}(mathbb{C})_1$ be the super algebra of all $(p+q) times (p+q)$ matrices.
Let $A, B in M_{p|q}(mathbb{C})$ (not necessarily homogeneous), If anybody can help me with finding the necessary and sufficient condition for $A$ and $B$ to super commute, that would be very helpful to me.
Super commutator is defined as $[X,Y] = XY - (-1)^{|X||Y|}XY ,forall, X,Y in M_{p|q}(mathbb{C})_0 sqcup M_{p|q}(mathbb{C})_1$ and extend this definition bilinearly to full $M_{p|q}(mathbb{C})$. $|X| = text{homogeneous degree of X}$
Thanks for your thoughts.
Have a good day.
matrices matrix-equations superalgebra lie-superalgebras
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
$endgroup$
Let $M_{p|q}(mathbb{C}) = M_{p|q}(mathbb{C})_0 oplus M_{p|q}(mathbb{C})_1$ be the super algebra of all $(p+q) times (p+q)$ matrices.
Let $A, B in M_{p|q}(mathbb{C})$ (not necessarily homogeneous), If anybody can help me with finding the necessary and sufficient condition for $A$ and $B$ to super commute, that would be very helpful to me.
Super commutator is defined as $[X,Y] = XY - (-1)^{|X||Y|}XY ,forall, X,Y in M_{p|q}(mathbb{C})_0 sqcup M_{p|q}(mathbb{C})_1$ and extend this definition bilinearly to full $M_{p|q}(mathbb{C})$. $|X| = text{homogeneous degree of X}$
Thanks for your thoughts.
Have a good day.
matrices matrix-equations superalgebra lie-superalgebras
matrices matrix-equations superalgebra lie-superalgebras
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
edited Dec 19 '18 at 5:11
GA316
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
asked Dec 17 '18 at 17:50
GA316GA316
2,6521232
2,6521232
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $$A~=~A_0+A_1qquadtext{and}qquad B~=~B_0+B_1,tag{1}$$ then the supercommutator
$$[A,B]~=~[A,B]_0+[A,B]_1 tag{2}$$
decomposes as
$$[A,B]_0~=~[A_0,B_0]+[A_1,B_1]qquadtext{and}qquad [A,B]_1~=~[A_0,B_1]+[A_0,B_1].tag{3}$$
In particular$$ Atext{ and }Btext{ supercommute}quadLeftrightarrowquad
[A,B]~=~0quadLeftrightarrowquad [A,B]_0~=~0~~wedge~~[A,B]_1 ~=~0,tag{4}$$
cf. OP's question.
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
$endgroup$
add a comment |
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%2f3044236%2fsuper-commutation-of-matrices%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $$A~=~A_0+A_1qquadtext{and}qquad B~=~B_0+B_1,tag{1}$$ then the supercommutator
$$[A,B]~=~[A,B]_0+[A,B]_1 tag{2}$$
decomposes as
$$[A,B]_0~=~[A_0,B_0]+[A_1,B_1]qquadtext{and}qquad [A,B]_1~=~[A_0,B_1]+[A_0,B_1].tag{3}$$
In particular$$ Atext{ and }Btext{ supercommute}quadLeftrightarrowquad
[A,B]~=~0quadLeftrightarrowquad [A,B]_0~=~0~~wedge~~[A,B]_1 ~=~0,tag{4}$$
cf. OP's question.
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
$endgroup$
add a comment |
$begingroup$
If $$A~=~A_0+A_1qquadtext{and}qquad B~=~B_0+B_1,tag{1}$$ then the supercommutator
$$[A,B]~=~[A,B]_0+[A,B]_1 tag{2}$$
decomposes as
$$[A,B]_0~=~[A_0,B_0]+[A_1,B_1]qquadtext{and}qquad [A,B]_1~=~[A_0,B_1]+[A_0,B_1].tag{3}$$
In particular$$ Atext{ and }Btext{ supercommute}quadLeftrightarrowquad
[A,B]~=~0quadLeftrightarrowquad [A,B]_0~=~0~~wedge~~[A,B]_1 ~=~0,tag{4}$$
cf. OP's question.
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
$endgroup$
add a comment |
$begingroup$
If $$A~=~A_0+A_1qquadtext{and}qquad B~=~B_0+B_1,tag{1}$$ then the supercommutator
$$[A,B]~=~[A,B]_0+[A,B]_1 tag{2}$$
decomposes as
$$[A,B]_0~=~[A_0,B_0]+[A_1,B_1]qquadtext{and}qquad [A,B]_1~=~[A_0,B_1]+[A_0,B_1].tag{3}$$
In particular$$ Atext{ and }Btext{ supercommute}quadLeftrightarrowquad
[A,B]~=~0quadLeftrightarrowquad [A,B]_0~=~0~~wedge~~[A,B]_1 ~=~0,tag{4}$$
cf. OP's question.
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
$endgroup$
If $$A~=~A_0+A_1qquadtext{and}qquad B~=~B_0+B_1,tag{1}$$ then the supercommutator
$$[A,B]~=~[A,B]_0+[A,B]_1 tag{2}$$
decomposes as
$$[A,B]_0~=~[A_0,B_0]+[A_1,B_1]qquadtext{and}qquad [A,B]_1~=~[A_0,B_1]+[A_0,B_1].tag{3}$$
In particular$$ Atext{ and }Btext{ supercommute}quadLeftrightarrowquad
[A,B]~=~0quadLeftrightarrowquad [A,B]_0~=~0~~wedge~~[A,B]_1 ~=~0,tag{4}$$
cf. OP's question.
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
answered Dec 30 '18 at 11:00
QmechanicQmechanic
4,91711855
4,91711855
add a comment |
add a comment |
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%2f3044236%2fsuper-commutation-of-matrices%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
