Mixing
Cartan-Killing metric and Lie-groups
$begingroup$
We defined the Cartan-Killing metric of a Lie-Group $G$ as $$g_{ab}equiv C_{acd}C_{bdc},$$
where $C_{abc}$ are the structure constants of the Lie-algebra $mathfrak{g}$. According to my professor it is possible to show that
$$operatorname{tr}(A_aA_b)=g_{ab},$$
where $A_i$ denote the generators in the adjoint representation.
I'm honestly having a really hard time on trying to show that. As far as I understand we have
$$A_a=left.frac{partial tau_A(g)}{partial alpha_a}right|_{g=e},$$
where $tau_A$ is the adjoint representation and $e$ the unit-element of $G$. I know that one can write $|A_a|_{bc}=C_{abc}$. This would then imply that
$$g_{ab}=C_{acd}C_{bdc}= sum_{c,d=1}^3|A_a|_{cd}|A_{b}|_{dc}.$$
The problem is that I don't really see what this has to do with the trace of the two matrices $A_a$ and $A_b$.
lie-groups lie-algebras
asked Feb 2 at 15:04
SitoSito
19417
$endgroup$
add a comment |
$begingroup$
We defined the Cartan-Killing metric of a Lie-Group $G$ as $$g_{ab}equiv C_{acd}C_{bdc},$$
where $C_{abc}$ are the structure constants of the Lie-algebra $mathfrak{g}$. According to my professor it is possible to show that
$$operatorname{tr}(A_aA_b)=g_{ab},$$
where $A_i$ denote the generators in the adjoint representation.
I'm honestly having a really hard time on trying to show that. As far as I understand we have
$$A_a=left.frac{partial tau_A(g)}{partial alpha_a}right|_{g=e},$$
where $tau_A$ is the adjoint representation and $e$ the unit-element of $G$. I know that one can write $|A_a|_{bc}=C_{abc}$. This would then imply that
$$g_{ab}=C_{acd}C_{bdc}= sum_{c,d=1}^3|A_a|_{cd}|A_{b}|_{dc}.$$
The problem is that I don't really see what this has to do with the trace of the two matrices $A_a$ and $A_b$.
lie-groups lie-algebras
asked Feb 2 at 15:04
SitoSito
19417
$endgroup$
1
$begingroup$
Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra.
$endgroup$
– Moishe Kohan
Feb 2 at 15:11
$begingroup$
@MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it...
$endgroup$
– Sito
Feb 2 at 15:15
$begingroup$
@MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more..
$endgroup$
– Sito
Feb 2 at 18:51
add a comment |
$begingroup$
We defined the Cartan-Killing metric of a Lie-Group $G$ as $$g_{ab}equiv C_{acd}C_{bdc},$$
where $C_{abc}$ are the structure constants of the Lie-algebra $mathfrak{g}$. According to my professor it is possible to show that
$$operatorname{tr}(A_aA_b)=g_{ab},$$
where $A_i$ denote the generators in the adjoint representation.
I'm honestly having a really hard time on trying to show that. As far as I understand we have
$$A_a=left.frac{partial tau_A(g)}{partial alpha_a}right|_{g=e},$$
where $tau_A$ is the adjoint representation and $e$ the unit-element of $G$. I know that one can write $|A_a|_{bc}=C_{abc}$. This would then imply that
$$g_{ab}=C_{acd}C_{bdc}= sum_{c,d=1}^3|A_a|_{cd}|A_{b}|_{dc}.$$
The problem is that I don't really see what this has to do with the trace of the two matrices $A_a$ and $A_b$.
lie-groups lie-algebras
asked Feb 2 at 15:04
SitoSito
19417
$endgroup$
We defined the Cartan-Killing metric of a Lie-Group $G$ as $$g_{ab}equiv C_{acd}C_{bdc},$$
where $C_{abc}$ are the structure constants of the Lie-algebra $mathfrak{g}$. According to my professor it is possible to show that
$$operatorname{tr}(A_aA_b)=g_{ab},$$
where $A_i$ denote the generators in the adjoint representation.
I'm honestly having a really hard time on trying to show that. As far as I understand we have
$$A_a=left.frac{partial tau_A(g)}{partial alpha_a}right|_{g=e},$$
where $tau_A$ is the adjoint representation and $e$ the unit-element of $G$. I know that one can write $|A_a|_{bc}=C_{abc}$. This would then imply that
$$g_{ab}=C_{acd}C_{bdc}= sum_{c,d=1}^3|A_a|_{cd}|A_{b}|_{dc}.$$
The problem is that I don't really see what this has to do with the trace of the two matrices $A_a$ and $A_b$.
lie-groups lie-algebras
lie-groups lie-algebras
asked Feb 2 at 15:04
SitoSito
19417
asked Feb 2 at 15:04
SitoSito
19417
asked Feb 2 at 15:04
SitoSito
19417
asked Feb 2 at 15:04
SitoSito
19417
asked Feb 2 at 15:04
SitoSito
19417
19417
1
$begingroup$
Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra.
$endgroup$
– Moishe Kohan
Feb 2 at 15:11
$begingroup$
@MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it...
$endgroup$
– Sito
Feb 2 at 15:15
$begingroup$
@MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more..
$endgroup$
– Sito
Feb 2 at 18:51
add a comment |
1
$begingroup$
Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra.
$endgroup$
– Moishe Kohan
Feb 2 at 15:11
$begingroup$
@MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it...
$endgroup$
– Sito
Feb 2 at 15:15
$begingroup$
@MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more..
$endgroup$
– Sito
Feb 2 at 18:51
1
1
$begingroup$
Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra.
$endgroup$
– Moishe Kohan
Feb 2 at 15:11
$begingroup$
Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra.
$endgroup$
– Moishe Kohan
Feb 2 at 15:11
$begingroup$
@MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it...
$endgroup$
– Sito
Feb 2 at 15:15
$begingroup$
@MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it...
$endgroup$
– Sito
Feb 2 at 15:15
$begingroup$
@MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more..
$endgroup$
– Sito
Feb 2 at 18:51
$begingroup$
@MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more..
$endgroup$
– Sito
Feb 2 at 18:51
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I will write a solution using the notation for structural constants that I prefer:
If $e_1,...,e_N$ is a basis of the Lie algebra ${mathfrak g}$, then
$$
[e_i,e_j]=sum_{l} c^l_{ij} e_l.
$$
Now, $ad(e_i)circ ad(e_j)(x)= [e_i, [e_j, x]]$ for $xin {mathfrak g}$. If $x=e_k$ then
$$
[e_i, [e_j, e_k]] = [e_i, sum_{l} c^l_{jk} e_l]= sum_{l,m} c^l_{jk} c^m_{il} e_m.
$$
Thus, the linear map $ad(e_i)circ ad(e_j)$ sends
$$
e_kmapsto sum_{l,m} c^l_{jk} c^m_{il} e_m
$$
Computing the trace of this map means setting $k=m$ and taking the sum over $k=1,...N$:
$$
tr(ad(e_i)circ ad(e_j))= sum_{l,k} c^l_{jk} c^k_{il}.
$$
That's your professor's formula.
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
$endgroup$
add a comment |
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$begingroup$
I will write a solution using the notation for structural constants that I prefer:
If $e_1,...,e_N$ is a basis of the Lie algebra ${mathfrak g}$, then
$$
[e_i,e_j]=sum_{l} c^l_{ij} e_l.
$$
Now, $ad(e_i)circ ad(e_j)(x)= [e_i, [e_j, x]]$ for $xin {mathfrak g}$. If $x=e_k$ then
$$
[e_i, [e_j, e_k]] = [e_i, sum_{l} c^l_{jk} e_l]= sum_{l,m} c^l_{jk} c^m_{il} e_m.
$$
Thus, the linear map $ad(e_i)circ ad(e_j)$ sends
$$
e_kmapsto sum_{l,m} c^l_{jk} c^m_{il} e_m
$$
Computing the trace of this map means setting $k=m$ and taking the sum over $k=1,...N$:
$$
tr(ad(e_i)circ ad(e_j))= sum_{l,k} c^l_{jk} c^k_{il}.
$$
That's your professor's formula.
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
$endgroup$
add a comment |
$begingroup$
I will write a solution using the notation for structural constants that I prefer:
If $e_1,...,e_N$ is a basis of the Lie algebra ${mathfrak g}$, then
$$
[e_i,e_j]=sum_{l} c^l_{ij} e_l.
$$
Now, $ad(e_i)circ ad(e_j)(x)= [e_i, [e_j, x]]$ for $xin {mathfrak g}$. If $x=e_k$ then
$$
[e_i, [e_j, e_k]] = [e_i, sum_{l} c^l_{jk} e_l]= sum_{l,m} c^l_{jk} c^m_{il} e_m.
$$
Thus, the linear map $ad(e_i)circ ad(e_j)$ sends
$$
e_kmapsto sum_{l,m} c^l_{jk} c^m_{il} e_m
$$
Computing the trace of this map means setting $k=m$ and taking the sum over $k=1,...N$:
$$
tr(ad(e_i)circ ad(e_j))= sum_{l,k} c^l_{jk} c^k_{il}.
$$
That's your professor's formula.
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
$endgroup$
add a comment |
$begingroup$
I will write a solution using the notation for structural constants that I prefer:
If $e_1,...,e_N$ is a basis of the Lie algebra ${mathfrak g}$, then
$$
[e_i,e_j]=sum_{l} c^l_{ij} e_l.
$$
Now, $ad(e_i)circ ad(e_j)(x)= [e_i, [e_j, x]]$ for $xin {mathfrak g}$. If $x=e_k$ then
$$
[e_i, [e_j, e_k]] = [e_i, sum_{l} c^l_{jk} e_l]= sum_{l,m} c^l_{jk} c^m_{il} e_m.
$$
Thus, the linear map $ad(e_i)circ ad(e_j)$ sends
$$
e_kmapsto sum_{l,m} c^l_{jk} c^m_{il} e_m
$$
Computing the trace of this map means setting $k=m$ and taking the sum over $k=1,...N$:
$$
tr(ad(e_i)circ ad(e_j))= sum_{l,k} c^l_{jk} c^k_{il}.
$$
That's your professor's formula.
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
$endgroup$
I will write a solution using the notation for structural constants that I prefer:
If $e_1,...,e_N$ is a basis of the Lie algebra ${mathfrak g}$, then
$$
[e_i,e_j]=sum_{l} c^l_{ij} e_l.
$$
Now, $ad(e_i)circ ad(e_j)(x)= [e_i, [e_j, x]]$ for $xin {mathfrak g}$. If $x=e_k$ then
$$
[e_i, [e_j, e_k]] = [e_i, sum_{l} c^l_{jk} e_l]= sum_{l,m} c^l_{jk} c^m_{il} e_m.
$$
Thus, the linear map $ad(e_i)circ ad(e_j)$ sends
$$
e_kmapsto sum_{l,m} c^l_{jk} c^m_{il} e_m
$$
Computing the trace of this map means setting $k=m$ and taking the sum over $k=1,...N$:
$$
tr(ad(e_i)circ ad(e_j))= sum_{l,k} c^l_{jk} c^k_{il}.
$$
That's your professor's formula.
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
answered Feb 3 at 2:11
Moishe KohanMoishe Kohan
48.7k344111
48.7k344111
add a comment |
add a comment |
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1
$begingroup$
Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra.
$endgroup$
– Moishe Kohan
Feb 2 at 15:11
$begingroup$
@MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it...
$endgroup$
– Sito
Feb 2 at 15:15
$begingroup$
@MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more..
$endgroup$
– Sito
Feb 2 at 18:51