Mixing
Finding quotients of Lie algebras $L$ with their centre $Z(L)$ and $[L,L]$
$begingroup$
I am currently trying to find the quotient Lie algebra of $L=gl(2,mathbb{C}),sl(2,mathbb{C}),u(2,mathbb{C})$ and $b(3,mathbb{C})$, when quotiented with both their centre $Z(L)$ and also $[L,L]$.
For $gl(2,mathbb{C})$ I believe I have found the centre to be $Z(L)=lambda I$ where $I$ is the $2 times 2$ identity matrix, and $[L,L]=sl(2,mathbb{C})$.
Now my problems seem to stem from not completely understanding quotients, as I am unsure as to what the quotient $gl(2,mathbb{C})/sl(2,mathbb{C})$ is isomorphic to. I think that maybe $gl(2,mathbb{C})/lambda I$ is isomorphic to $sl(2,mathbb{C})$ by considering the element of trace $0$ from each coset.
I think I am fine with $sl(2,mathbb{C})$, then for $b(2,mathbb{C})$ the $2 times 2$ upper triangular matrices, I believe that the centre is again $lambda I$ and $[L,L]=langle e_{12} rangle$, but again I cannot figure out what either quotient is in this case.
The same with $u(3,mathbb{C})$, the set of strictly upper triangular $3 times 3$ matrices. I believe $Z(L)=langle e_{13} rangle = [L,L]$ although I am not 100% sure on this, I have no idea where to start on thinking about the quotient.
Any help would be greatly appreciated thanks :)
lie-algebras
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
$endgroup$
add a comment |
$begingroup$
I am currently trying to find the quotient Lie algebra of $L=gl(2,mathbb{C}),sl(2,mathbb{C}),u(2,mathbb{C})$ and $b(3,mathbb{C})$, when quotiented with both their centre $Z(L)$ and also $[L,L]$.
For $gl(2,mathbb{C})$ I believe I have found the centre to be $Z(L)=lambda I$ where $I$ is the $2 times 2$ identity matrix, and $[L,L]=sl(2,mathbb{C})$.
Now my problems seem to stem from not completely understanding quotients, as I am unsure as to what the quotient $gl(2,mathbb{C})/sl(2,mathbb{C})$ is isomorphic to. I think that maybe $gl(2,mathbb{C})/lambda I$ is isomorphic to $sl(2,mathbb{C})$ by considering the element of trace $0$ from each coset.
I think I am fine with $sl(2,mathbb{C})$, then for $b(2,mathbb{C})$ the $2 times 2$ upper triangular matrices, I believe that the centre is again $lambda I$ and $[L,L]=langle e_{12} rangle$, but again I cannot figure out what either quotient is in this case.
The same with $u(3,mathbb{C})$, the set of strictly upper triangular $3 times 3$ matrices. I believe $Z(L)=langle e_{13} rangle = [L,L]$ although I am not 100% sure on this, I have no idea where to start on thinking about the quotient.
Any help would be greatly appreciated thanks :)
lie-algebras
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
$endgroup$
$begingroup$
$u(3,Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=langle zrangle $, as you have found.
$endgroup$
– Dietrich Burde
Jan 24 at 19:47
add a comment |
$begingroup$
I am currently trying to find the quotient Lie algebra of $L=gl(2,mathbb{C}),sl(2,mathbb{C}),u(2,mathbb{C})$ and $b(3,mathbb{C})$, when quotiented with both their centre $Z(L)$ and also $[L,L]$.
For $gl(2,mathbb{C})$ I believe I have found the centre to be $Z(L)=lambda I$ where $I$ is the $2 times 2$ identity matrix, and $[L,L]=sl(2,mathbb{C})$.
Now my problems seem to stem from not completely understanding quotients, as I am unsure as to what the quotient $gl(2,mathbb{C})/sl(2,mathbb{C})$ is isomorphic to. I think that maybe $gl(2,mathbb{C})/lambda I$ is isomorphic to $sl(2,mathbb{C})$ by considering the element of trace $0$ from each coset.
I think I am fine with $sl(2,mathbb{C})$, then for $b(2,mathbb{C})$ the $2 times 2$ upper triangular matrices, I believe that the centre is again $lambda I$ and $[L,L]=langle e_{12} rangle$, but again I cannot figure out what either quotient is in this case.
The same with $u(3,mathbb{C})$, the set of strictly upper triangular $3 times 3$ matrices. I believe $Z(L)=langle e_{13} rangle = [L,L]$ although I am not 100% sure on this, I have no idea where to start on thinking about the quotient.
Any help would be greatly appreciated thanks :)
lie-algebras
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
$endgroup$
I am currently trying to find the quotient Lie algebra of $L=gl(2,mathbb{C}),sl(2,mathbb{C}),u(2,mathbb{C})$ and $b(3,mathbb{C})$, when quotiented with both their centre $Z(L)$ and also $[L,L]$.
For $gl(2,mathbb{C})$ I believe I have found the centre to be $Z(L)=lambda I$ where $I$ is the $2 times 2$ identity matrix, and $[L,L]=sl(2,mathbb{C})$.
Now my problems seem to stem from not completely understanding quotients, as I am unsure as to what the quotient $gl(2,mathbb{C})/sl(2,mathbb{C})$ is isomorphic to. I think that maybe $gl(2,mathbb{C})/lambda I$ is isomorphic to $sl(2,mathbb{C})$ by considering the element of trace $0$ from each coset.
I think I am fine with $sl(2,mathbb{C})$, then for $b(2,mathbb{C})$ the $2 times 2$ upper triangular matrices, I believe that the centre is again $lambda I$ and $[L,L]=langle e_{12} rangle$, but again I cannot figure out what either quotient is in this case.
The same with $u(3,mathbb{C})$, the set of strictly upper triangular $3 times 3$ matrices. I believe $Z(L)=langle e_{13} rangle = [L,L]$ although I am not 100% sure on this, I have no idea where to start on thinking about the quotient.
Any help would be greatly appreciated thanks :)
lie-algebras
lie-algebras
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
asked Jan 23 at 21:36
UsernameInvalidUsernameInvalid
676
676
$begingroup$
$u(3,Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=langle zrangle $, as you have found.
$endgroup$
– Dietrich Burde
Jan 24 at 19:47
add a comment |
$begingroup$
$u(3,Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=langle zrangle $, as you have found.
$endgroup$
– Dietrich Burde
Jan 24 at 19:47
$begingroup$
$u(3,Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=langle zrangle $, as you have found.
$endgroup$
– Dietrich Burde
Jan 24 at 19:47
$begingroup$
$u(3,Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=langle zrangle $, as you have found.
$endgroup$
– Dietrich Burde
Jan 24 at 19:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The first quotient is
$$
mathfrak{gl}_n(K)/mathfrak{sl}_n(K)cong K,
$$
because $mathfrak{sl}_n(K)$ is a $1$-codimensional ideal. And the only $1$-dimensional Lie algebra over $K$ is $K$ itself. In the question, $n=2$ and $K=Bbb C$, but it is true in general.
Since $L=mathfrak{gl}_n(Bbb C)$ is reductive, we have $mathfrak{gl}_n(Bbb C)cong [L,L]oplus Z(L)$, with the ideals $[L,L]cong mathfrak{sl}_n(Bbb C)$ and $Z(L)cong Bbb C$. For the quotients, these are first of all quotient vector spaces. So if you have
$$
W=Uoplus V,
$$
what is $W/U$ and $W/V$? For the Heisenberg Lie algebra $L=mathfrak{n}_3(Bbb C)$, we indeed have $[L,L]=Z(L)=langle zrangle$, which is $1$-dimensional. The quotient $L/Z(L)$ is an abelian Lie algebra of dimension $2$ over $Bbb C$, hence isomorphic to the Lie algebra $Bbb C^2$.
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
add a comment |
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1 Answer
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$begingroup$
The first quotient is
$$
mathfrak{gl}_n(K)/mathfrak{sl}_n(K)cong K,
$$
because $mathfrak{sl}_n(K)$ is a $1$-codimensional ideal. And the only $1$-dimensional Lie algebra over $K$ is $K$ itself. In the question, $n=2$ and $K=Bbb C$, but it is true in general.
Since $L=mathfrak{gl}_n(Bbb C)$ is reductive, we have $mathfrak{gl}_n(Bbb C)cong [L,L]oplus Z(L)$, with the ideals $[L,L]cong mathfrak{sl}_n(Bbb C)$ and $Z(L)cong Bbb C$. For the quotients, these are first of all quotient vector spaces. So if you have
$$
W=Uoplus V,
$$
what is $W/U$ and $W/V$? For the Heisenberg Lie algebra $L=mathfrak{n}_3(Bbb C)$, we indeed have $[L,L]=Z(L)=langle zrangle$, which is $1$-dimensional. The quotient $L/Z(L)$ is an abelian Lie algebra of dimension $2$ over $Bbb C$, hence isomorphic to the Lie algebra $Bbb C^2$.
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
add a comment |
$begingroup$
The first quotient is
$$
mathfrak{gl}_n(K)/mathfrak{sl}_n(K)cong K,
$$
because $mathfrak{sl}_n(K)$ is a $1$-codimensional ideal. And the only $1$-dimensional Lie algebra over $K$ is $K$ itself. In the question, $n=2$ and $K=Bbb C$, but it is true in general.
Since $L=mathfrak{gl}_n(Bbb C)$ is reductive, we have $mathfrak{gl}_n(Bbb C)cong [L,L]oplus Z(L)$, with the ideals $[L,L]cong mathfrak{sl}_n(Bbb C)$ and $Z(L)cong Bbb C$. For the quotients, these are first of all quotient vector spaces. So if you have
$$
W=Uoplus V,
$$
what is $W/U$ and $W/V$? For the Heisenberg Lie algebra $L=mathfrak{n}_3(Bbb C)$, we indeed have $[L,L]=Z(L)=langle zrangle$, which is $1$-dimensional. The quotient $L/Z(L)$ is an abelian Lie algebra of dimension $2$ over $Bbb C$, hence isomorphic to the Lie algebra $Bbb C^2$.
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
add a comment |
$begingroup$
The first quotient is
$$
mathfrak{gl}_n(K)/mathfrak{sl}_n(K)cong K,
$$
because $mathfrak{sl}_n(K)$ is a $1$-codimensional ideal. And the only $1$-dimensional Lie algebra over $K$ is $K$ itself. In the question, $n=2$ and $K=Bbb C$, but it is true in general.
Since $L=mathfrak{gl}_n(Bbb C)$ is reductive, we have $mathfrak{gl}_n(Bbb C)cong [L,L]oplus Z(L)$, with the ideals $[L,L]cong mathfrak{sl}_n(Bbb C)$ and $Z(L)cong Bbb C$. For the quotients, these are first of all quotient vector spaces. So if you have
$$
W=Uoplus V,
$$
what is $W/U$ and $W/V$? For the Heisenberg Lie algebra $L=mathfrak{n}_3(Bbb C)$, we indeed have $[L,L]=Z(L)=langle zrangle$, which is $1$-dimensional. The quotient $L/Z(L)$ is an abelian Lie algebra of dimension $2$ over $Bbb C$, hence isomorphic to the Lie algebra $Bbb C^2$.
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
The first quotient is
$$
mathfrak{gl}_n(K)/mathfrak{sl}_n(K)cong K,
$$
because $mathfrak{sl}_n(K)$ is a $1$-codimensional ideal. And the only $1$-dimensional Lie algebra over $K$ is $K$ itself. In the question, $n=2$ and $K=Bbb C$, but it is true in general.
Since $L=mathfrak{gl}_n(Bbb C)$ is reductive, we have $mathfrak{gl}_n(Bbb C)cong [L,L]oplus Z(L)$, with the ideals $[L,L]cong mathfrak{sl}_n(Bbb C)$ and $Z(L)cong Bbb C$. For the quotients, these are first of all quotient vector spaces. So if you have
$$
W=Uoplus V,
$$
what is $W/U$ and $W/V$? For the Heisenberg Lie algebra $L=mathfrak{n}_3(Bbb C)$, we indeed have $[L,L]=Z(L)=langle zrangle$, which is $1$-dimensional. The quotient $L/Z(L)$ is an abelian Lie algebra of dimension $2$ over $Bbb C$, hence isomorphic to the Lie algebra $Bbb C^2$.
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
edited Jan 23 at 21:51
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
answered Jan 23 at 21:40
Dietrich BurdeDietrich Burde
80.6k647104
80.6k647104
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
add a comment |
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
I wanted to check that $[L,L]$ is not just ${ [x,y], x,y in L }$ but the additive group generated by all the linear combination of the $[x,y]$, in the same way as the commutator subgroup is built from a group
$endgroup$
– reuns
Jan 25 at 6:13
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
$begingroup$
@reuns I see. Yes, this is the linear span of all $[x,y]$, you are right. It is an interesting question to ask which simple Lie algebras have the property that every element is already itself some commutator, not just a linear combination of commutators (but already answered at MSE).
$endgroup$
– Dietrich Burde
Jan 25 at 9:04
add a comment |
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$begingroup$
$u(3,Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=langle zrangle $, as you have found.
$endgroup$
– Dietrich Burde
Jan 24 at 19:47