Inner automorphism of $\mathfrak{sl}(2,k)$, $\operatorname{char}(k)=0$ and adjoint action If we have $sl(2,k)$, char $k = 0$, with standard basis $(x,y,h)$ and inner automorphism $\sigma = \exp(\operatorname{ad}x)\exp(-\operatorname{ad}y)\exp(\operatorname{ad}x)$. How can we show that $\sigma(x) = -y, \sigma(y) = -x$ and $\sigma(h) = -h$?
 A: I prove the relation  $\sigma(X)=-Y$; the others are similar. The idea is to use the definition of the operator $\exp(\operatorname{ad}M)$ with
$$\exp(\operatorname{ad}M)(R):=\sum_{k=0}^\infty \frac{(\operatorname{ad}M)^k(R)}{k!}$$
for nilpotent $M$'s and matrices $R$ s.t. $\exp(\operatorname{ad}M)(R)$ reduces to a finite sum (and so it is well defined). We have defined
$$(\operatorname{ad}M)^k(R):=(\operatorname{ad}M)((\operatorname{ad}M)^{k-1}(R))$$ for all $k\geq 2$ and
$$(\operatorname{ad}M)^0(R):=R.$$


*

*Proof of $\sigma(X)=-Y$
Denoting by $(H,X,Y)$ the standard basis of $sl(2,\mathbb K)$ ($ch(\mathbb K=0)$, then it is easy to prove that
$$X^2=Y^2=0$$
and 
$$[X,Y]=H,$$
$$[H,Y]=-2Y,$$
$$[H,X]=-2X.$$


*

*Long way: brute force
Let us compute $\sigma(X)$; we start with the first operator arriving at
$$\exp(\operatorname{ad}X)(X)=X+\operatorname{ad}X(X)+\frac{1}{2!}(\operatorname{ad}X)^2(X)+\dots=X,$$
as $\operatorname{ad}X(X)=X^2-X^2=0$ (which implies the vanishing of the higher compositions). We are left with
$$\exp(\operatorname{-ad}Y)(X)=X+(\operatorname{-ad}Y)(X)+\frac{1}{2!}(\operatorname{-ad}Y)^2(X)+\dots;$$
as 
$$\operatorname{-ad}Y(X):=-YX+XY=-[Y,X]=H,$$ 
$$(\operatorname{-ad}Y)^2(X)=
(\operatorname{-ad}Y)(H)=-YH+HY=-[Y,H]=-2Y$$
and so $(\operatorname{-ad}Y)^n(X)=0$ for all $n\geq 3$, we arrive at
$$\exp(\operatorname{-ad}Y)(X)=X+H-Y.$$
To finish the proof we need to compute
$$\exp(\operatorname{ad}X)(X+H-Y);$$
as 
$$\operatorname{ad}X(X+H-Y)=[X,X]+[X,H]-[X,Y]=-2X-H,$$
$$(\operatorname{ad}X)^2(X+H-Y)=-2[X,X]-[X,H]=2X,$$
then, using $(\operatorname{ad}X)^n(X+H-Y)=0$ for all $n\geq 3$ we arrive at
$$\sigma(X)=X+H-Y-2X-H+X=-Y,$$
as stated.


*

*Short way: nilpotency
As $X$ and $Y$ are nilpotent, it is easy to see that $(\operatorname{ad}X)^n=(\operatorname{ad}Y)^n=0$ for all $n\geq 3$. Then
$$\exp(\operatorname{ad}X)=1+\operatorname{ad}X+\frac{1}{2!}(\operatorname{ad}X)^2,$$
and similarly for $\exp(\operatorname{ad}Y)$. Repeating the above lines, the thesis follows.
