If a change of basis preserves the Lie bracket, why is the automorphism group of a Lie algebra not the entire general linear group?

Question

According to Knapp (chapter 1, section 14), an automorphism of a Lie algebra is an invertible linear map $L$ that preserves brackets: $[L(X),L(Y)]=L[X,Y]$. Since this just means that the brackets must be given by the change of basis formula, I’m puzzled that he then goes on to say that $\operatorname{Aut}_{\mathbb R}\mathfrak g \subseteq \operatorname{GL}_\mathbb R (\mathfrak g)$. Why isn’t the set inclusion an equality?

Answer 1

What do you mean by “the brackets must be given by the change of basis formula”? If $\mathfrak{g}$ is a Lie algebra and you pick a basis $B=\{x_1,\ldots, x_n\}$ of $\mathfrak{g}$, then you can write $[x_i,x_j] = \sum_{k=1}^n c_{ij}^k x_k$, and the scalars $(c_{ij}^k)_{i,j,k \in \{1,2,\ldots,n\}}$ (called the “structure constants” with respect to the basis $B$) completely determine the Lie bracket.

If $L$ is an invertible linear map and $L(x_j) = \sum_{i=1}^n l_{ij}x_i$, we can write out the conditions on the matrix of $L$ given by the requirement that $[L(x),L(y)] = L([x,y])$ explicitly in terms of the structure constants:

$$[L(x_i),L(x_j)] = \left[\sum_{p=1}^n l_{pi}x_p,\sum_{q=1}^n l_{qj} x_q\right] = \sum_{p,q,r=1}^n l_{pi}l_{qj}c_{pq}^r x_r$$

On the other hand,

$$L([x_i,x_j]) = L\left(\sum_{k=1}^n c_{ij}^kx_k\right)= \sum_{k,r=1}^n l_{rk}c_{ij}^kx_r$$

Thus $L$ is an automorphism of $\mathfrak{g}$ precisely when

$$\sum_{k=1}^n l_{rk}c_{ij}^k = \sum_{p,q=1}^n l_{pi}l_{qj}c_{pq}^r.$$

This is a system of $n^3$ degree 2 equations in the entries of the matrix $(l_{ij})$.

Hopefully this convinces you that you the condition that $L([x,y]) = [L(x),L(y)]$ for all $x,y \in \mathfrak{g}$ is not automatically satisfied by any element of $\mathrm{GL}(\mathfrak{g})$, but to be completely concrete, take $\mathfrak{g} = \mathfrak{s}_2$ the two-dimensional non-abelian Lie algebra. This has basis $\{x_1,x_2\}$ and $[x_1,x_2]=x_2$ (and all other brackets of basis elements $0$ by the alternating property) so that the structure constants are zero except for $c_{12}^2$. Then if $L\colon \mathfrak{s}_2 \to \mathfrak{s}_2$ is a linear map with matrix $\left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$ we see that the entries of the matrix must satisfy

$$r=1: \quad l_{12} =0$$ $$r=2: \quad l_{22} = l_{11}l_{22}$$

Since $L \in \mathrm{GL}(\mathfrak{g})$, it follows that the matrix of $L$ has the form $\left(\begin{array}{cc} 1 & 0 \\ l_{21} & l_{22} \end{array}\right)$ (where $l_{22} \neq 0$ since we require $L$ to be invertible).

Answer 2

For a real abelian Lie algebra $L$ all bijective linear maps $L\colon L\rightarrow L$ satisfy $$ [L(x),L(y)]=0=L([x,y]). $$ Hence ${\rm Aut}(L)=GL_3(\Bbb R)$. However, in general such maps do not preserve the Lie bracket. For an easy example consider $\mathfrak{n}_3(\Bbb R)$, the $3$-dimensional Heisenberg Lie algebra. Then we have $$ {\rm Aut}(\mathfrak{n}_3(\Bbb R))=\left\{ \left. \begin{pmatrix} a & b & 0 \cr c & d & 0 \cr e & f & ad-bc \end{pmatrix}\,\right| ad-bc\neq 0\right\}, $$ which is different form $GL_3(\Bbb R)$. So the inclusion usually is not an equality.

Answer 3

Perhaps you will like a much shorter and easier version of the above answers.

Let $\mathfrak{g}$ be the non-abelian $2$-dimensional Lie algebra with underlying vector space $V=\Bbb R^2$ and basis $(x,y)$. We may assume that $[x,y]=x$. Let $L$ be the invertible linear map in $GL(V)$ given by $L(x)=y$ and $L(y)=x$. Then, if $L$ preserves Lie brackets, we have

$$L(x)=L([x,y])=[L(x),L(y)]=[y,x]=-[x,y]=-x,$$ which is not true. So $L$ doesn't preserve Lie brackets, and there is no equality between ${\rm Aut}(\mathfrak{g})$ and $GL(V)\cong GL_2(\Bbb R)$. Indeed $L\in GL(V)$, but not in ${\rm Aut}(\mathfrak{g})$ .

Thus you see that preserving Lie brackets is not a "change of basis" formula, but a strong condition on the linear map $L$.