How is the trace of the adjoint of Lie algebra elements defined? [duplicate]Coordinate-Free Definition of Trace.Proving the trace of a transformation is independent of the basis chosenOrthonormal basis of Cartan subalgebra relative to Killing formMetric over a Lie algebra $mathfraku(n)$Vector Space of Lie AlgebraAn $mathrmAd$-invariant inner product that agrees with the traceHow to visualise the Killing form of a Lie algebraDifferent definitions of Casimir elementScalar product on Lie algebra of compact Lie groupCommutator of Lie sub-algebraKilling form and trace form on a simple Lie algebraWhy is the adjoint representation in SO?
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How is the trace of the adjoint of Lie algebra elements defined? [duplicate]
Coordinate-Free Definition of Trace.Proving the trace of a transformation is independent of the basis chosenOrthonormal basis of Cartan subalgebra relative to Killing formMetric over a Lie algebra $mathfraku(n)$Vector Space of Lie AlgebraAn $mathrmAd$-invariant inner product that agrees with the traceHow to visualise the Killing form of a Lie algebraDifferent definitions of Casimir elementScalar product on Lie algebra of compact Lie groupCommutator of Lie sub-algebraKilling form and trace form on a simple Lie algebraWhy is the adjoint representation in SO?
$begingroup$
This question already has an answer here:
Coordinate-Free Definition of Trace.
1 answer
Consider an element $Xinmathfrak g$ of some Lie algebra $mathfrak g$.
I understand that $mathfrak g$ can be represented via its action on other elements of the same algebra, as $operatornamead(X)Yequiv[X,Y]$, so that $operatornamead(X)inoperatornameGL(mathfrak g)$.
One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $operatornamead(X)$.
However, I usually don't see any direct mention of the inner product with respect to which these things are defined.
To properly define what something such as $operatornamead(X)_ij$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $X_i$ the elements of some basis for the algebra)?
Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?
representation-theory lie-algebras inner-product-space trace
asked Mar 29 at 17:07
glSglS
790521
$endgroup$
marked as duplicate by glS, Community♦ Mar 29 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 3 more comments
$begingroup$
This question already has an answer here:
Coordinate-Free Definition of Trace.
1 answer
Consider an element $Xinmathfrak g$ of some Lie algebra $mathfrak g$.
I understand that $mathfrak g$ can be represented via its action on other elements of the same algebra, as $operatornamead(X)Yequiv[X,Y]$, so that $operatornamead(X)inoperatornameGL(mathfrak g)$.
One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $operatornamead(X)$.
However, I usually don't see any direct mention of the inner product with respect to which these things are defined.
To properly define what something such as $operatornamead(X)_ij$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $X_i$ the elements of some basis for the algebra)?
Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?
representation-theory lie-algebras inner-product-space trace
asked Mar 29 at 17:07
glSglS
790521
$endgroup$
marked as duplicate by glS, Community♦ Mar 29 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
Trace of a matrix is independent of the basis.
$endgroup$
– Moishe Kohan
Mar 29 at 17:15
3
$begingroup$
One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384.
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:16
$begingroup$
@MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace
$endgroup$
– glS
Mar 29 at 17:17
2
$begingroup$
Surely $mathfrakg$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it?
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:27
1
$begingroup$
I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^-1)=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/…
$endgroup$
– Moishe Kohan
Mar 29 at 18:18
|
show 3 more comments
$begingroup$
This question already has an answer here:
Coordinate-Free Definition of Trace.
1 answer
Consider an element $Xinmathfrak g$ of some Lie algebra $mathfrak g$.
I understand that $mathfrak g$ can be represented via its action on other elements of the same algebra, as $operatornamead(X)Yequiv[X,Y]$, so that $operatornamead(X)inoperatornameGL(mathfrak g)$.
One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $operatornamead(X)$.
However, I usually don't see any direct mention of the inner product with respect to which these things are defined.
To properly define what something such as $operatornamead(X)_ij$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $X_i$ the elements of some basis for the algebra)?
Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?
representation-theory lie-algebras inner-product-space trace
asked Mar 29 at 17:07
glSglS
790521
$endgroup$
This question already has an answer here:
Coordinate-Free Definition of Trace.
1 answer
Consider an element $Xinmathfrak g$ of some Lie algebra $mathfrak g$.
I understand that $mathfrak g$ can be represented via its action on other elements of the same algebra, as $operatornamead(X)Yequiv[X,Y]$, so that $operatornamead(X)inoperatornameGL(mathfrak g)$.
One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $operatornamead(X)$.
However, I usually don't see any direct mention of the inner product with respect to which these things are defined.
To properly define what something such as $operatornamead(X)_ij$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $X_i$ the elements of some basis for the algebra)?
Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?
This question already has an answer here:
Coordinate-Free Definition of Trace.
1 answer
representation-theory lie-algebras inner-product-space trace
representation-theory lie-algebras inner-product-space trace
asked Mar 29 at 17:07
glSglS
790521
asked Mar 29 at 17:07
glSglS
790521
asked Mar 29 at 17:07
glSglS
790521
asked Mar 29 at 17:07
glSglS
790521
asked Mar 29 at 17:07
glSglS
790521
790521
marked as duplicate by glS, Community♦ Mar 29 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by glS, Community♦ Mar 29 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
Trace of a matrix is independent of the basis.
$endgroup$
– Moishe Kohan
Mar 29 at 17:15
3
$begingroup$
One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384.
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:16
$begingroup$
@MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace
$endgroup$
– glS
Mar 29 at 17:17
2
$begingroup$
Surely $mathfrakg$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it?
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:27
1
$begingroup$
I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^-1)=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/…
$endgroup$
– Moishe Kohan
Mar 29 at 18:18
|
show 3 more comments
3
$begingroup$
Trace of a matrix is independent of the basis.
$endgroup$
– Moishe Kohan
Mar 29 at 17:15
3
$begingroup$
One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384.
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:16
$begingroup$
@MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace
$endgroup$
– glS
Mar 29 at 17:17
2
$begingroup$
Surely $mathfrakg$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it?
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:27
1
$begingroup$
I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^-1)=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/…
$endgroup$
– Moishe Kohan
Mar 29 at 18:18
3
3
$begingroup$
Trace of a matrix is independent of the basis.
$endgroup$
– Moishe Kohan
Mar 29 at 17:15
$begingroup$
Trace of a matrix is independent of the basis.
$endgroup$
– Moishe Kohan
Mar 29 at 17:15
3
3
$begingroup$
One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384.
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:16
$begingroup$
One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384.
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:16
$begingroup$
@MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace
$endgroup$
– glS
Mar 29 at 17:17
$begingroup$
@MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace
$endgroup$
– glS
Mar 29 at 17:17
2
2
$begingroup$
Surely $mathfrakg$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it?
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:27
$begingroup$
Surely $mathfrakg$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it?
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:27
1
1
$begingroup$
I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^-1)=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/…
$endgroup$
– Moishe Kohan
Mar 29 at 18:18
$begingroup$
I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^-1)=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/…
$endgroup$
– Moishe Kohan
Mar 29 at 18:18
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Consider $mathfraksl_2$ with basis $e,f,h$. You have $$ad(e)(e)=0,;;;ad(e)(f)=h,;;;ad(e)(h)=-2e$$
so the matrix of $ad(e)$ in this basis is
$$
beginpmatrix0&0&-2\0&0&0\0&1&0endpmatrix
$$
which has trace 0.
You can check that the matrix for $ad(h)$ is
$$
beginpmatrix2&0&0\0&-2&0\0&0&0endpmatrix
$$
which again has trace 0. Can you do $ad(f)$?
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
$endgroup$
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
1
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $mathfraksl_2$ with basis $e,f,h$. You have $$ad(e)(e)=0,;;;ad(e)(f)=h,;;;ad(e)(h)=-2e$$
so the matrix of $ad(e)$ in this basis is
$$
beginpmatrix0&0&-2\0&0&0\0&1&0endpmatrix
$$
which has trace 0.
You can check that the matrix for $ad(h)$ is
$$
beginpmatrix2&0&0\0&-2&0\0&0&0endpmatrix
$$
which again has trace 0. Can you do $ad(f)$?
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
$endgroup$
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
1
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
add a comment |
$begingroup$
Consider $mathfraksl_2$ with basis $e,f,h$. You have $$ad(e)(e)=0,;;;ad(e)(f)=h,;;;ad(e)(h)=-2e$$
so the matrix of $ad(e)$ in this basis is
$$
beginpmatrix0&0&-2\0&0&0\0&1&0endpmatrix
$$
which has trace 0.
You can check that the matrix for $ad(h)$ is
$$
beginpmatrix2&0&0\0&-2&0\0&0&0endpmatrix
$$
which again has trace 0. Can you do $ad(f)$?
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
$endgroup$
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
1
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
add a comment |
$begingroup$
Consider $mathfraksl_2$ with basis $e,f,h$. You have $$ad(e)(e)=0,;;;ad(e)(f)=h,;;;ad(e)(h)=-2e$$
so the matrix of $ad(e)$ in this basis is
$$
beginpmatrix0&0&-2\0&0&0\0&1&0endpmatrix
$$
which has trace 0.
You can check that the matrix for $ad(h)$ is
$$
beginpmatrix2&0&0\0&-2&0\0&0&0endpmatrix
$$
which again has trace 0. Can you do $ad(f)$?
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
$endgroup$
Consider $mathfraksl_2$ with basis $e,f,h$. You have $$ad(e)(e)=0,;;;ad(e)(f)=h,;;;ad(e)(h)=-2e$$
so the matrix of $ad(e)$ in this basis is
$$
beginpmatrix0&0&-2\0&0&0\0&1&0endpmatrix
$$
which has trace 0.
You can check that the matrix for $ad(h)$ is
$$
beginpmatrix2&0&0\0&-2&0\0&0&0endpmatrix
$$
which again has trace 0. Can you do $ad(f)$?
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
answered Mar 29 at 17:24
David HillDavid Hill
9,5461619
9,5461619
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
1
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
add a comment |
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
1
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
$begingroup$
You also need to know that the trace is independent of a choice of basis.
$endgroup$
– Qiaochu Yuan
Mar 29 at 17:35
1
1
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
$begingroup$
That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan
$endgroup$
– David Hill
Mar 29 at 17:57
add a comment |
3
$begingroup$
Trace of a matrix is independent of the basis.
$endgroup$
– Moishe Kohan
Mar 29 at 17:15
3
$begingroup$
One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384.
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:16
$begingroup$
@MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace
$endgroup$
– glS
Mar 29 at 17:17
2
$begingroup$
Surely $mathfrakg$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it?
$endgroup$
– Torsten Schoeneberg
Mar 29 at 17:27
1
$begingroup$
I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^-1)=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/…
$endgroup$
– Moishe Kohan
Mar 29 at 18:18