Let $\mathfrak g = \mathfrak{sl}(2, \mathbb{C})$. Let $\gamma \in \operatorname{Aut}(\mathfrak{g})$. How to show that $\gamma$ is conjugation by some $u \in \mathrm{SL}(2, \mathbb C)$?


2 Answers 2


The following uses the theory of Jordan decomposition in finite dimensional semisimple complex Lie algebras (namely in $\mathfrak{sl}_2(\mathbb{C})$) and some basic knowledge about the finite dimensional representations of $\mathfrak{sl}_2(\mathbb{C})$. There is probably a much shorter and easier way to do this, but it’s the best I’m coming up with right now.

Let $(e,h,f)$ be the usual standard basis of $\mathfrak{g} := \mathfrak{sl}_2(\mathbb{C})$, i.e. $$ e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad h = \begin{pmatrix} 1 & \phantom{-}0 \\ 0 & -1 \end{pmatrix}, \quad f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. $$ Because $h$ is diagonalizable as a matrix it is a semisimple element of $\mathfrak{g}$. As $\gamma$ is a Lie algebra automorphism we find that $\gamma(h)$ is also a semisimple element of $\mathfrak{g}$.

Because $\gamma(h)$ is a semisimple of $\mathfrak{g}$ it follows that $\gamma(h)$ is diagonalizable as a matrix, with eigenvalues $\lambda$ and $-\lambda$ (because $\gamma(h)$ is traceless). We actually have $\{\lambda, -\lambda\} = \{1,-1\}$:

The natural representation $\rho \colon \mathfrak{g} \to \mathfrak{gl}(\mathbb{C}^2)$ is an irreducible representation of $\mathfrak{sl}_2(\mathbb{C})$. As $\gamma \colon \mathfrak{g} \to \mathfrak{g}$ is an Lie algebra automorphism it follows that the compositon $\rho \circ \gamma \colon \mathfrak{g} \to \mathfrak{gl}(\mathbb{C}^2)$ is also an irreducible representation (both $\rho$ and $\rho \circ \gamma$ have precisely the same subrepresentations). By the usual classification of finite dimensional irreducible representations of $\mathfrak{sl}_2(\mathbb{C})$ we find that $\mathbb{C}^2$ decomposes into the $h$-weight spaces $1$ and $-1$ with respect to $\rho \circ \gamma$. But these weight spaces are precisely the eigenspaces of $\gamma(h)$, and the weights $1$ and $-1$ are precisely the eigenvalues of $\gamma(h)$.

As $\gamma(h)$ is diagonalizable with eigenvalues $1$ and $-1$ there exists some $u \in \mathrm{GL}_2(\mathbb{C})$ with $$ u \gamma(h) u^{-1} = \begin{pmatrix} 1 & \phantom{-}0 \\ 0 & -1 \end{pmatrix} = h. $$ We can assume that $u \in \mathrm{SL}_2(\mathbb{C})$, i.e. $\det u = 1$. Otherwise we could choose some $\mu \in \mathbb{C} \setminus \{0\}$ with $\mu^2 = \det u$ and replace $u$ with $\mu^{-1} u$.

Now consider the Lie algebra automorphism $\beta \colon \mathfrak{g} \to \mathfrak{g}$ with $\beta(x) = u x u^{-1}$. We already have $\beta(\gamma(h)) = h$. It remains to show the following:

Lemma: Let $\alpha \colon \mathfrak{g} \to \mathfrak{g}$ be a Lie algebra automorphism with $\alpha(h) = h$. Then there exists $v \in \mathrm{SL}_2(\mathbb{C})$ with $\alpha(x) = v x v^{-1}$ for all $x \in \mathfrak{g}$.

From this it then follows that $\beta(\gamma(x)) = vxv^{-1}$, and thus $$ \gamma(x) = \beta^{-1}(vxv^{-1}) = u^{-1} v x v^{-1} u = (u^{-1} v) x (u^{-1} v)^{-1} $$ with $u^{-1} v \in \mathrm{SL}_2(\mathbb{C})$.

To prove the lemma notice that $$ [h,\alpha(e)] = [\alpha(h),\alpha(e)] = \alpha([h,e]) = 2\alpha(e), $$ so $\alpha(e)$ is an eigenvalues of $\operatorname{ad}(h)$ with eigenvalue $2$. But the corresponding eigenspace is spanned by $e$, so $\alpha(e) = \lambda e$ for some $\lambda \in \mathbb{C}$. Similarly $\alpha(f) = \lambda' f$ for some $\lambda' \in \mathbb{C}$. Because $$ h = \alpha(h) = \alpha([e,f]) = [\alpha(e), \alpha(f)] = \lambda \lambda' [e,f] = \lambda \lambda' h $$ we have $\lambda \lambda' = 1$, so $\lambda$ is non-zero with $\lambda' = \lambda^{-1}$. So $\alpha$ is given by $\alpha(e) = \lambda e$, $\alpha(h) = h$ and $\alpha(f) = \lambda^{-1} f$. Choosing $\mu \in \mathbb{C}$ with $\mu^2 = \lambda$ set $$ v = \begin{pmatrix} \mu & 0 \\ 0 & \mu^{-1} \end{pmatrix}. $$ Then a straightforward calculation shows that $v e v ^{-1} = \lambda e$, $v h v^{-1} = h$ and $v f v^{-1} = \lambda^{-1} f$.

  • $\begingroup$ very nice! thanks! $\endgroup$
    – Ronald
    Jan 30, 2016 at 4:04

The above answer is very nice and elementary. The following is a proof that uses some machinery:

Let $G$ be a semisimple linear algebraic Lie group and $\operatorname{Ad}\colon G \to \operatorname{Aut}(\mathfrak{g})$ the adjoint representation on the Lie algebra $\mathfrak{g}$ of $G$. Then $\operatorname{Ad}(G)$ consists exactly of those automorphisms of $\mathfrak{g}$ that act by conjugation, as in your question. These are also called inner automorphisms.

Theorem 1 [Thm 14.1, Linear Algebraic Groups, Humphreys]: $\operatorname{Ad}(G) = \operatorname{Aut}^\circ(\mathfrak{g})$.

Theorem 2 [Prop D.40, Representation Theory, Fulton-Harris]: The outer automorphism group $\operatorname{Aut}(\mathfrak{g})/\operatorname{Aut}^\circ(\mathfrak{g})$ is isomorphic to the automorphism group of the Dynkin diagram.

In the case of $G = \operatorname{SL}(2,\mathbb{C})$ and $\mathfrak{g}= \mathfrak{sl}(2,\mathbb{C})$, the Dynkin diagram consists of a single node, since the Weyl group consists of a single reflection (+ the identity). Therefore there are no outer automorphisms and $\operatorname{Ad}(\operatorname{SL}(2,\mathbb{C})) = \operatorname{Aut}(\mathfrak{sl}(2,\mathbb{C}))$.


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