Show that every derivation of the non-abelian two-dimensional Lie algebra is inner

Question

For any $2$-dimensional non-abelian Lie algebra $g$, there exists a basis $a,b$ such that $[a,b]=a$. Now I want to prove that any derivation of $L$ is inner. My proof: $D[x,y]=[Dx,y]+[x,Dy]= [k_1 a+k_2b,b]+[a,k_3a+k_4b]= k_1a+k_4a=ka$, for some $k\in F$.

Now, $ka=[a,kb]=ad_a(l)$, for some $l\in g$.

Is my proof correct?

Answer 1

You have shown that for every $x,y \in \mathfrak{g}$ there exists som $l \in \mathfrak{g}$ with $D[x,y] = \mathrm{ad}_a(l)$. What you need to show is that there exists some $x \in \mathfrak{g}$ with $D(y) = \mathrm{ad}_x(y)$ for every $y \in \mathfrak{g}$.

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There is probably a short and elegant solution, but this should work too:

With respect to the basis $(a,b)$ of $\mathfrak{g}$ we have $$ \mathrm{ad}_a = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \quad\text{and}\quad \mathrm{ad}_b = \begin{pmatrix} -1 & 0 \\ \phantom{-}0 & 0 \end{pmatrix}. $$ (We are a bit sloppy and identify the endomorphisms of $\mathfrak{g}$ with the corresponding matrices with respect to our basis.) Thus the space of inner automorphisms is given by $$ I = \left\{ \begin{pmatrix} c & d \\ 0 & 0 \end{pmatrix} \,\middle|\, c,d \in K, \right\} $$ where $K$ denotes the ground field. Let $D \colon \mathfrak{g} \to \mathfrak{g}$ be a dervation with $$ D = \begin{pmatrix} c_1 & c_2 \\ c_3 & c_4 \end{pmatrix}. $$ Notice that $$ c_1 a + c_3 b = D(a) = D([a,b]) = [D(a),b] + [a,D(b)] = c_1 a + c_4 b; $$ comparing both sides gives $c_3 = c_4 =: c$. Now $$ D' := D - c_2 \mathrm{ad}_a + c_1 \mathrm{ad}_b $$ is a derivation of $\mathfrak{g}$ with $$ D' = \begin{pmatrix} 0 & 0 \\ c & c \end{pmatrix}. $$ Because $$ cb = D'(a) = D'([a,b]) = [D'(a),b] + [a,D'(b)] = [cb,b] + [a,cb] = ca $$ we find that $c = 0$. So $D$ is already of the form $$ D = \begin{pmatrix} c_1 & c_2 \\ 0 & 0 \end{pmatrix}, $$ i.e. $D$ is an inner derivation.

Answer 2

The result follows indeed by an explicit computation of all derivations of the $2$-dimensional. nonabelian Lie algebra, as demonstrated in the answer of Jendrik. I think this is a perfect solution.

Later one could use more general results, by noting that this Lie algebra is the affine Lie algebra $\mathfrak{aff}(\mathbb{R}^1)$, and using the well-known result that all affine Lie algebras $\mathfrak{aff}(\mathbb{R}^n)$ are complete, i.e., have zero center and all derivations are inner.