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Consider the Lie algebra $gl(2, \mathbb{C})$. I already know that the centre of this is $\{\lambda I \mid \lambda \in \mathbb{C}\}$, where $I$ is the $2 \times 2$ identity matrix.

I'm struggling to calculate the quotient though. By definition, I can write:

$$gl(2, \mathbb{C})/Z(gl(2,\mathbb{C})) = \{x + \{\lambda I \mid \lambda \in \mathbb{C}\} \mid x \in gl(2, \mathbb{C})\}$$

I don't know how to proceed, but I'm tempted to continue by writing $$gl(2, \mathbb{C})/Z(gl(2,\mathbb{C})) = \{x + \lambda I \mid x \in gl(2, \mathbb{C}), \lambda \in \mathbb{C}\}$$ which would just be equal to $gl(2, \mathbb{C})$, which is obviously wrong.

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    Hint: Can you find a subalgebra which is a complement of the center? – 2017-01-26
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    @TobiasKildetoft What does complement mean in this context? Because the way I'm reading it, it just seems like the question repeated again. – 2017-01-26
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    A complement means a subspace intersecting the center trivially and whose sum with the center gives the full Lie algebra. – 2017-01-26
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    Hint 2: There's a named, well-known subalgebra with the required property. – 2017-01-26
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    @TobiasKildetoft It's not something we've covered in class, and no nice Lie algebras come to mind either. It seems like we want our set to be $\{(a_{ij}) \in M_2(\mathbb{C}) \mid a_{11} \neq a_{22}\}$ though. – 2017-01-26
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    @IrregularUser: Your set is not a complement of $\{\lambda I\}$ -- for example it contains both $({}^1_0\;{}^0_2)$ and $({}^2_0\;{}^0_3)$ whose difference is in the set of $\lambda I$s. A complement must be a set such that _every matrix_ in $gl(2)$ can be written _in exactly one way_ as a matrix from the center and a matrix from the complement. – 2017-01-26
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    @HenningMakholm So my set is too big... named Lie algebras that I have to use as ammunition here are $b(2, \mathbb{C}), u(2, \mathbb{C}), sp(2, \mathbb{C})$ and of course $sl(2, \mathbb{C})$ -- none of which seem to fit the bill. – 2017-01-26
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    @IrregularUser: One of the algebras you list _does_ fit the bill. Try for each of them to construct an argument why it doesn't fit; you ought to fail in one of the cases. – 2017-01-26
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    @HenningMakholm So $b(2, \mathbb{C})$ doesn't fit since $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}$ are in the same class, $u(2, \mathbb{C})$ doesn't fit because of the same reason. Ah! Looking carefully at $sp(2, \mathbb{C})$, we have matrices of the form $\begin{pmatrix} a & b \\ c & -a \end{pmatrix}$. Suppose $\begin{pmatrix} x & b \\ c & -x \end{pmatrix}$ was a different matrix. Then their difference can only be in the set of $\lambda I$s iff $a=x$, in which case they'd be the same matrix. – 2017-01-26
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    @IrregularUser: Correct. (Note that $Sp(2,F)$ and $SL(2,F)$ are the same group; it is the latter that generalizes your property to higher dimensions). – 2017-01-27
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    @HenningMakholm Ah, yes they are. Thanks for your help! (If you compile these comments as an answer, I'll be happy to accept it.) – 2017-01-27

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If we construct the quotient $\mathfrak{gl}(n,\mathbb C)/Z$ in the by-the-book way, the elements are cosets of $Z$ -- that is, each element is a set of matrices that differ from each other by adding or subtracting the same number to each diagonal element.

In search of a nicer representation of the quotient, let's see if we can appoint one of the matrices in each coset to be the canonical representative of the coset. Based on what characterizes each of the cosets, it seems that a natural choice for a representative would be the matrix where the sum of the diagonal elements is $0$ -- starting from a random matrix we can always achieve that by subtracting $\frac1n$ times the sum it already has from each diagonal element.

The sum of the diagonal elements is the trace of the matrix, so we're looking at the class of matrices with trace zero and wondering whether we can make a Lie algebra out of it.

If we paid attention in class, we will remember that the matrices with trace zero are exactly what make up $\mathfrak{sl}(n,\mathbb C)$ -- and it is now a simple task to check that $\mathfrak{sl}(n,\mathbb C)$ is indeed isomorphic to the quotient we're looking for.

The same holds for any field in place of $\mathbb C$, except when its characteristic divides $n$ (in which case $I$ has trace $0$, which ruins everything).

In the case $n=2$ we can also describe the quotient as $\mathfrak{sp}(n,\mathbb C)$, because $Sp(2,F)$ and $SL(2,F)$ are the same group for every field $F$.

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    Thanks for going to the effort of providing very good intuition here. – 2017-01-27