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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 :)

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  • $u(3,\Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and $[x,y]=z$. So $Z(L)=[L,L]=\langle z\rangle $, as you have found. – Dietrich Burde Jan 24 '19 at 19:47

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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 $$ W=U\oplus V, $$ what is $W/U$ and $W/V$? For the Heisenberg Lie algebra $L=\mathfrak{n}_3(\Bbb C)$, we indeed have $[L,L]=Z(L)=\langle z\rangle$, 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 $\Bbb C^2$.

Dietrich Burde
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  • 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 – reuns Jan 25 '19 at 06:13
  • @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). – Dietrich Burde Jan 25 '19 at 09:04