Lie algebra generator that does not appear on rhs of the algebra
Is it possible that in a lie algebra one may have a generator that does not appear on any of the commutators?
If the above line is too physicisty for you, maybe I can try to translate to a more mathematicians question:
A lie algebra is a vector space $mathfrak{g}$ equipped with an commutator map from $mathfrak{g}timesmathfrak{g}longrightarrowmathfrak{g}$ that satisfies the Jacobi identity. Is it possible that the range of this map to be a proper subset of $mathfrak{g}$?
I hope the last question conveys the same message. Any reference or link will be of help too.
abstract-algebra lie-algebras mathematical-physics
add a comment |
Is it possible that in a lie algebra one may have a generator that does not appear on any of the commutators?
If the above line is too physicisty for you, maybe I can try to translate to a more mathematicians question:
A lie algebra is a vector space $mathfrak{g}$ equipped with an commutator map from $mathfrak{g}timesmathfrak{g}longrightarrowmathfrak{g}$ that satisfies the Jacobi identity. Is it possible that the range of this map to be a proper subset of $mathfrak{g}$?
I hope the last question conveys the same message. Any reference or link will be of help too.
abstract-algebra lie-algebras mathematical-physics
1
If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$.
– Torsten Schoeneberg
Jan 4 at 17:34
@TorstenSchoeneberg I was about to post that as an answer. Why don't you do that?
– José Carlos Santos
Jan 4 at 17:34
@JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now.
– Torsten Schoeneberg
Jan 4 at 17:55
add a comment |
Is it possible that in a lie algebra one may have a generator that does not appear on any of the commutators?
If the above line is too physicisty for you, maybe I can try to translate to a more mathematicians question:
A lie algebra is a vector space $mathfrak{g}$ equipped with an commutator map from $mathfrak{g}timesmathfrak{g}longrightarrowmathfrak{g}$ that satisfies the Jacobi identity. Is it possible that the range of this map to be a proper subset of $mathfrak{g}$?
I hope the last question conveys the same message. Any reference or link will be of help too.
abstract-algebra lie-algebras mathematical-physics
Is it possible that in a lie algebra one may have a generator that does not appear on any of the commutators?
If the above line is too physicisty for you, maybe I can try to translate to a more mathematicians question:
A lie algebra is a vector space $mathfrak{g}$ equipped with an commutator map from $mathfrak{g}timesmathfrak{g}longrightarrowmathfrak{g}$ that satisfies the Jacobi identity. Is it possible that the range of this map to be a proper subset of $mathfrak{g}$?
I hope the last question conveys the same message. Any reference or link will be of help too.
abstract-algebra lie-algebras mathematical-physics
abstract-algebra lie-algebras mathematical-physics
edited Jan 4 at 17:36
José Carlos Santos
152k22123225
152k22123225
asked Jan 4 at 17:27
AriAri
1385
1385
1
If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$.
– Torsten Schoeneberg
Jan 4 at 17:34
@TorstenSchoeneberg I was about to post that as an answer. Why don't you do that?
– José Carlos Santos
Jan 4 at 17:34
@JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now.
– Torsten Schoeneberg
Jan 4 at 17:55
add a comment |
1
If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$.
– Torsten Schoeneberg
Jan 4 at 17:34
@TorstenSchoeneberg I was about to post that as an answer. Why don't you do that?
– José Carlos Santos
Jan 4 at 17:34
@JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now.
– Torsten Schoeneberg
Jan 4 at 17:55
1
1
If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$.
– Torsten Schoeneberg
Jan 4 at 17:34
If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$.
– Torsten Schoeneberg
Jan 4 at 17:34
@TorstenSchoeneberg I was about to post that as an answer. Why don't you do that?
– José Carlos Santos
Jan 4 at 17:34
@TorstenSchoeneberg I was about to post that as an answer. Why don't you do that?
– José Carlos Santos
Jan 4 at 17:34
@JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now.
– Torsten Schoeneberg
Jan 4 at 17:55
@JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now.
– Torsten Schoeneberg
Jan 4 at 17:55
add a comment |
1 Answer
1
active
oldest
votes
It happens quite often that the image of
$$ [ cdot, cdot]: mathfrak{g} times mathfrak{g} rightarrow mathfrak{g}$$
is a proper subset of $mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $lbrace 0 rbrace$ (but $mathfrak{g}$ can be any vector space you like).
Other examples are all nilpotent and even all solvable Lie algebras.
It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[mathfrak{g}, mathfrak{g}]$, whereas formally it should be something like "$langle[mathfrak{g}, mathfrak{g}]rangle$") is all of $mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.
Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)
As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $Bbb R$ or $Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $mathfrak{sl}_n(Bbb R)$ and $mathfrak{sl}_n(Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
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%2f3061865%2flie-algebra-generator-that-does-not-appear-on-rhs-of-the-algebra%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
It happens quite often that the image of
$$ [ cdot, cdot]: mathfrak{g} times mathfrak{g} rightarrow mathfrak{g}$$
is a proper subset of $mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $lbrace 0 rbrace$ (but $mathfrak{g}$ can be any vector space you like).
Other examples are all nilpotent and even all solvable Lie algebras.
It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[mathfrak{g}, mathfrak{g}]$, whereas formally it should be something like "$langle[mathfrak{g}, mathfrak{g}]rangle$") is all of $mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.
Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)
As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $Bbb R$ or $Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $mathfrak{sl}_n(Bbb R)$ and $mathfrak{sl}_n(Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
add a comment |
It happens quite often that the image of
$$ [ cdot, cdot]: mathfrak{g} times mathfrak{g} rightarrow mathfrak{g}$$
is a proper subset of $mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $lbrace 0 rbrace$ (but $mathfrak{g}$ can be any vector space you like).
Other examples are all nilpotent and even all solvable Lie algebras.
It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[mathfrak{g}, mathfrak{g}]$, whereas formally it should be something like "$langle[mathfrak{g}, mathfrak{g}]rangle$") is all of $mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.
Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)
As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $Bbb R$ or $Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $mathfrak{sl}_n(Bbb R)$ and $mathfrak{sl}_n(Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
add a comment |
It happens quite often that the image of
$$ [ cdot, cdot]: mathfrak{g} times mathfrak{g} rightarrow mathfrak{g}$$
is a proper subset of $mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $lbrace 0 rbrace$ (but $mathfrak{g}$ can be any vector space you like).
Other examples are all nilpotent and even all solvable Lie algebras.
It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[mathfrak{g}, mathfrak{g}]$, whereas formally it should be something like "$langle[mathfrak{g}, mathfrak{g}]rangle$") is all of $mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.
Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)
As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $Bbb R$ or $Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $mathfrak{sl}_n(Bbb R)$ and $mathfrak{sl}_n(Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384
It happens quite often that the image of
$$ [ cdot, cdot]: mathfrak{g} times mathfrak{g} rightarrow mathfrak{g}$$
is a proper subset of $mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $lbrace 0 rbrace$ (but $mathfrak{g}$ can be any vector space you like).
Other examples are all nilpotent and even all solvable Lie algebras.
It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[mathfrak{g}, mathfrak{g}]$, whereas formally it should be something like "$langle[mathfrak{g}, mathfrak{g}]rangle$") is all of $mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.
Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)
As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $Bbb R$ or $Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $mathfrak{sl}_n(Bbb R)$ and $mathfrak{sl}_n(Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384
edited Jan 4 at 21:44
answered Jan 4 at 17:53
Torsten SchoenebergTorsten Schoeneberg
3,8762833
3,8762833
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
add a comment |
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds).
– Tobias Kildetoft
Jan 4 at 19:06
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.
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%2f3061865%2flie-algebra-generator-that-does-not-appear-on-rhs-of-the-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
1
If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$.
– Torsten Schoeneberg
Jan 4 at 17:34
@TorstenSchoeneberg I was about to post that as an answer. Why don't you do that?
– José Carlos Santos
Jan 4 at 17:34
@JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now.
– Torsten Schoeneberg
Jan 4 at 17:55