Defining the Lie bracket on a tensor product Lie algebra
$begingroup$
So, my question is the following:
Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.
vector-spaces lie-algebras
asked Aug 12 '18 at 16:28
G KG K
1198
$endgroup$
|
show 2 more comments
$begingroup$
So, my question is the following:
Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.
vector-spaces lie-algebras
asked Aug 12 '18 at 16:28
G KG K
1198
$endgroup$
$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13
$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13
$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20
1
$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28
$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30
|
show 2 more comments
$begingroup$
So, my question is the following:
Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.
vector-spaces lie-algebras
asked Aug 12 '18 at 16:28
G KG K
1198
$endgroup$
So, my question is the following:
Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.
vector-spaces lie-algebras
vector-spaces lie-algebras
asked Aug 12 '18 at 16:28
G KG K
1198
asked Aug 12 '18 at 16:28
G KG K
1198
asked Aug 12 '18 at 16:28
G KG K
1198
asked Aug 12 '18 at 16:28
G KG K
1198
asked Aug 12 '18 at 16:28
G KG K
1198
1198
$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13
$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13
$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20
1
$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28
$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30
|
show 2 more comments
$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13
$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13
$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20
1
$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28
$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30
$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13
$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13
$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13
$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13
$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20
$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20
1
1
$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28
$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28
$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30
$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
$$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
$endgroup$
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%2f2880510%2fdefining-the-lie-bracket-on-a-tensor-product-lie-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
$begingroup$
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
$$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
$endgroup$
add a comment |
$begingroup$
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
$$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
$endgroup$
add a comment |
$begingroup$
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
$$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
$endgroup$
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
$$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
answered Dec 27 '18 at 15:55
moutheticsmouthetics
50137
50137
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%2f2880510%2fdefining-the-lie-bracket-on-a-tensor-product-lie-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
$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13
$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13
$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20
1
$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28
$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30