Mixing
$SU(2)$ and its representations
1
$begingroup$
Does the Lie-Algebra of $SU(2)$ always have 3 generators? Because I'm reading about different representations of $SU(2)$ as $2times2$, $3times3$, $4times4$ matrices etc. But I guess that even if we are looking at $7times7$ matrix representation of $SU(2)$ we only need $3$ such matrices corresponding to the $3$ generators right?
And if we have a representation by say $7times 7$ matrices, is it not possible by using similarity transformations to find an $3-$dimensional invariant subspace corresponding to these $3$ linear operators? Then what point is it to look for higher representations of $SU(2)$ then just $3times 3$ real matrices?
lie-algebras
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
asked Jan 5 at 3:35
HiggsinoHiggsino
595
$endgroup$
add a comment |
1
$begingroup$
Does the Lie-Algebra of $SU(2)$ always have 3 generators? Because I'm reading about different representations of $SU(2)$ as $2times2$, $3times3$, $4times4$ matrices etc. But I guess that even if we are looking at $7times7$ matrix representation of $SU(2)$ we only need $3$ such matrices corresponding to the $3$ generators right?
And if we have a representation by say $7times 7$ matrices, is it not possible by using similarity transformations to find an $3-$dimensional invariant subspace corresponding to these $3$ linear operators? Then what point is it to look for higher representations of $SU(2)$ then just $3times 3$ real matrices?
lie-algebras
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
asked Jan 5 at 3:35
HiggsinoHiggsino
595
$endgroup$
3
$begingroup$
If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular.
$endgroup$
– Torsten Schoeneberg
Jan 5 at 5:34
2
$begingroup$
And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: mathoverflow.net/q/153740/27465
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:06
2
$begingroup$
It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation.
$endgroup$
– Andreas Blass
Jan 6 at 2:58
$begingroup$
ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on.
$endgroup$
– Higgsino
Jan 6 at 22:38
1
$begingroup$
Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace.
$endgroup$
– Torsten Schoeneberg
Jan 7 at 21:10
add a comment |
1
1
1
1
$begingroup$
Does the Lie-Algebra of $SU(2)$ always have 3 generators? Because I'm reading about different representations of $SU(2)$ as $2times2$, $3times3$, $4times4$ matrices etc. But I guess that even if we are looking at $7times7$ matrix representation of $SU(2)$ we only need $3$ such matrices corresponding to the $3$ generators right?
And if we have a representation by say $7times 7$ matrices, is it not possible by using similarity transformations to find an $3-$dimensional invariant subspace corresponding to these $3$ linear operators? Then what point is it to look for higher representations of $SU(2)$ then just $3times 3$ real matrices?
lie-algebras
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
asked Jan 5 at 3:35
HiggsinoHiggsino
595
$endgroup$
Does the Lie-Algebra of $SU(2)$ always have 3 generators? Because I'm reading about different representations of $SU(2)$ as $2times2$, $3times3$, $4times4$ matrices etc. But I guess that even if we are looking at $7times7$ matrix representation of $SU(2)$ we only need $3$ such matrices corresponding to the $3$ generators right?
And if we have a representation by say $7times 7$ matrices, is it not possible by using similarity transformations to find an $3-$dimensional invariant subspace corresponding to these $3$ linear operators? Then what point is it to look for higher representations of $SU(2)$ then just $3times 3$ real matrices?
lie-algebras
lie-algebras
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
asked Jan 5 at 3:35
HiggsinoHiggsino
595
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
asked Jan 5 at 3:35
HiggsinoHiggsino
595
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
edited Jan 5 at 19:09
Torsten Schoeneberg
3,9562833
3,9562833
asked Jan 5 at 3:35
HiggsinoHiggsino
595
asked Jan 5 at 3:35
HiggsinoHiggsino
595
asked Jan 5 at 3:35
HiggsinoHiggsino
595
595
3
$begingroup$
If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular.
$endgroup$
– Torsten Schoeneberg
Jan 5 at 5:34
2
$begingroup$
And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: mathoverflow.net/q/153740/27465
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:06
2
$begingroup$
It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation.
$endgroup$
– Andreas Blass
Jan 6 at 2:58
$begingroup$
ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on.
$endgroup$
– Higgsino
Jan 6 at 22:38
1
$begingroup$
Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace.
$endgroup$
– Torsten Schoeneberg
Jan 7 at 21:10
add a comment |
3
$begingroup$
If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular.
$endgroup$
– Torsten Schoeneberg
Jan 5 at 5:34
2
$begingroup$
And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: mathoverflow.net/q/153740/27465
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:06
2
$begingroup$
It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation.
$endgroup$
– Andreas Blass
Jan 6 at 2:58
$begingroup$
ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on.
$endgroup$
– Higgsino
Jan 6 at 22:38
1
$begingroup$
Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace.
$endgroup$
– Torsten Schoeneberg
Jan 7 at 21:10
3
3
$begingroup$
If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular.
$endgroup$
– Torsten Schoeneberg
Jan 5 at 5:34
$begingroup$
If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular.
$endgroup$
– Torsten Schoeneberg
Jan 5 at 5:34
2
2
$begingroup$
And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: mathoverflow.net/q/153740/27465
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:06
$begingroup$
And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: mathoverflow.net/q/153740/27465
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:06
2
2
$begingroup$
It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation.
$endgroup$
– Andreas Blass
Jan 6 at 2:58
$begingroup$
It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation.
$endgroup$
– Andreas Blass
Jan 6 at 2:58
$begingroup$
ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on.
$endgroup$
– Higgsino
Jan 6 at 22:38
$begingroup$
ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on.
$endgroup$
– Higgsino
Jan 6 at 22:38
1
1
$begingroup$
Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace.
$endgroup$
– Torsten Schoeneberg
Jan 7 at 21:10
$begingroup$
Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace.
$endgroup$
– Torsten Schoeneberg
Jan 7 at 21:10
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
});
}
});
draft saved
draft discarded
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%2f3062380%2fsu2-and-its-representations%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
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3062380%2fsu2-and-its-representations%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
3
$begingroup$
If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular.
$endgroup$
– Torsten Schoeneberg
Jan 5 at 5:34
2
$begingroup$
And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: mathoverflow.net/q/153740/27465
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:06
2
$begingroup$
It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation.
$endgroup$
– Andreas Blass
Jan 6 at 2:58
$begingroup$
ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on.
$endgroup$
– Higgsino
Jan 6 at 22:38
1
$begingroup$
Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace.
$endgroup$
– Torsten Schoeneberg
Jan 7 at 21:10