Property of order 2 automorphism of simple Lie algebra If $\sigma$ is an automorphism of simple Lie algebra, of order 2, such that we fixed : $\sigma (x_\beta)=-y_\beta$ for each simple root of a given base $\Delta$, is it also necessarily true that $\sigma (x_\gamma)=-y_\gamma$ for nonsimple root $\gamma$ ? (Humphreys Lie algebra book, exercise 14.4)
And this is what I tried :
I first looked for a counter-example with $sl(3,\mathbb{C})$, 6 roots, a base contains 2 simple roots and I called the third (positive) one (with height 2) $\alpha_3$ : but after some calculations I found that : $\sigma (x_{{\alpha}_3})=-y_{{\alpha}_3}$ . $sl(3,\mathbb{C})$ was not a counter-example.
So I decided to try to prove that the property is true :
we know that $[L_{\alpha}L_{\beta}]=L_{\alpha+\beta}$. Suppose $\alpha$ and $\beta$ are two simple roots, and call $\gamma={\alpha+\beta}$. Let's choose $x_\gamma=[x_{\alpha}x_{\beta}]$. If we prove the property for $x_\gamma$, we're done by induction on height of the root.
$\sigma(x_\gamma)=[\sigma(x_\alpha)\sigma(x_\beta)]=[-y_{\alpha}-y_\beta]=[y_{\alpha}y_\beta]$
We know that $y_\gamma=t[y_{\alpha}y_\beta]$ for some t. But is $t$ necessarily $-1$ ? I'm stuck here. Thank you for help.
 A: Too long for a comment, and can maybe be upgraded to an answer: When we do the calculations in $\mathfrak{sl}_3$, and call
$$x_\alpha := \pmatrix{0&1&0\\0&0&0\\0&0&0}, \qquad x_\beta :=\pmatrix{0&0&0\\0&0&1\\0&0&0}$$
$$y_\alpha := \pmatrix{0&0&0\\1&0&0\\0&0&0}, \qquad y_\beta :=\pmatrix{0&0&0\\0&0&0\\0&1&0}$$
$$h_\alpha := \pmatrix{1&0&0\\0&-1&0\\0&0&0}=[x_\alpha, y_\alpha], \qquad h_\beta :=\pmatrix{0&0&0\\0&1&0\\0&0&-1} =[x_\beta, y_\beta]$$
then the issue is that $[x_\alpha, x_\beta]= \pmatrix{0&0&1\\0&0&0\\0&0&0}$ but $[y_\alpha, y_\beta] = \pmatrix{0&0&0\\0&0&0\\\color{red}{-1}&0&0}$, consequently
$[[x_\alpha, x_\beta],[y_\alpha, y_\beta]]= \pmatrix{\color{red}{-1}&0&0\\0&0&0\\0&0&\color{red}{1}}$ which does not act on $[x_\alpha, x_\beta]$ as $h_{\alpha+\beta}$ (which should be $=h_\alpha + h_\beta$) usually should. So if you want things to be consistent, and you want to call $x_\gamma= [x_\alpha, x_\beta]$ (in this order), then you should call $y_\gamma = \color{red}{-}[y_\alpha, y_\beta] = [y_\beta, y_\alpha]$.
(Too make things more annoying, some sources would call $y_\alpha$ and $y_\beta$ the negatives of my $y_\alpha$ and $y_\beta$ above. But then you get the same sign issue just with now an unexpected $+$ instead of a $-$.)
You say you checked what the order two automorphism $A \mapsto -A^T$ (negative transpose) does. It looks good if we follow the above definitions.
A: The point is you can choose $x_\gamma$ wherever you want in $L_\gamma$  ($=\{x\in L| [hx]=\gamma(h)x$ for all $h \in H\}$) for some nonsimple root $\gamma$. But there exists, for this $x_\gamma$, a unique $y_\gamma \in L_{-\gamma}$ such that $\{x_\gamma,y_\gamma,[x_{\gamma}y_\gamma]\}$ will form a $sl_2$-triple (Humphreys, prop. 8.4).
In your example, if we are choosing $x_\gamma=[x_{\alpha}y_\beta]$ or $x_\gamma=[x_{\beta}y_\gamma]$ we don't have a counter-example because $\sigma(x_\gamma)=-y_\gamma $ in both cases.
However, if we take : $$x_\gamma = \pmatrix{0&0&2\\0&0&0\\0&0&0}$$ then $y_\gamma =\pmatrix{0&0&0\\0&0&0\\\frac{1}{2}&0&0}$ so that $[{x_\gamma}y_\gamma]=h_\alpha+h_\beta$ (with your notation) and $\{x_\gamma,y_\gamma,h_\alpha+h_\beta\}$ is a $sl_2-$triple
But here : $\sigma(x_\gamma)=\sigma(2[{x_\alpha}x_\beta])=2[{y_\alpha}y_\beta]\ne -y_\gamma$
This indeed seems to be a counter-example !
