A Principled(?) Way to Determine a Lie Algebra Automorphism from a Dynkin Diagram Automorphism (and invariant subalgebra)

Question

I am wondering about how to write out an explicit form of a Lie Algebra automorphism from an automorphism of the associated Dynkin Diagram of the root system. Within this question are some other subquestions that may help clarify the matter.

Let me give an example to illustrate my question in depth. Consider the Dynkin Diagram $A_{2n-1}$ (which I am not skilled enough in LaTeX to typeset, but whose image can be found here). It corresponds to the root system coming from $SU_{2n}$, which has a lie algebra given by skew-Hermitian matrices: $$\begin{bmatrix} it_1 & a_{1~2} & ... & a_{1~2n} \\ -\bar a_{1~2} & it_2 & ... & a_{2~2n} \\ \vdots & & & \vdots \\ -\bar a_{1~2n} &... & -\bar a_{2n-1~2n} & it_{2n} \end{bmatrix}.$$ The maximal torus is given by the diagonal matrices with positive roots given by $t_i- t_j$ for $i<j$ acting on the root spaces given by $a_{i~j}$ and $a_{j i}(=-\bar a_{ij})$ and we can choose a set of simple roots as being $t_i-t_{i+1}$ with $1\leq i\leq 2n$.

There is an involution of the Dynkin Diagram given by "flipping" the order of the chain. This corresponds to switching the roots by $t_i-t_{i+1} \leftrightarrow t_{2n-i}-t_{2n+1-i}$ leaving invariant $t_n-t_{n+1}$. Now I would like to lift this to some explicit automorphism of $\mathfrak{su}_{2n}$. The seemingly natural thing would be to look for some automorphism that swaps the order of the diagonal and then multiplies it by -1. One could consider, for example, for any matrix $A\in\mathfrak{su}_{2n}$ sending $A\mapsto BAB^{-1}$ for $$B=\begin{bmatrix} 0 & ... & 0 & i \\ 0 & ... &i & 0 \\ \vdots & & & \vdots \\ i & 0 & ... & 0 \end{bmatrix}.$$

However, although this matrix does what we desire on the roots, it also acts nontrivially on the root spaces, sending $a_{ij}\mapsto -\bar a_{2n+1-j~2n+1-i}$.

Edit for above: I did not choose B to act properly to get the desired result, see the answer below for the correct form.

So, some questions raised by this exercise are:

1. To lift an automorphism of the Dynkin Diagram to an automorphism of the Lie Algebra, what do we desire of its action on the root spaces and the Lie Algebra at large? How do we know we've done it "properly?" (Compare with this question)
2. Do we always given an automorphism by conjugation by some matrix? Are we always searching for this matrix? Or, put another way, why should one think to find the matrix B above (if that's even the correct automorphism)?
3. This question mentions that on $\mathfrak{sl}_{2n}$ the automorphism on $A_n$ corresponds to taking the negative transpose. As I understand it, $\mathfrak{sl}_{2n}$ is the complexification of $\mathfrak{su}_{2n}$. Are these two automorphisms compatible? Can we understand one in terms of the other?
4. Finally, we might wonder about the invariant subspace of the automorphism and its Dynkin Diagram. We note that the sum of the two roots that get swapped are invariant under the action, and it seems plausible to believe that these would become new roots for the invariant Lie Algebra. Is this a true belief, and why or why not?

Other questions that consider a similar topic (aside from the ones already linked) are here and here.

Thank you, have a nice day.

Answer 1

Denote by $e_{ij}$ the basis matrix with $1$ in row $i$, column $j$ and all other entries zero.

In $\mathfrak{su}(2n)$ let $$ h_i=\sqrt{-1}\left(e_{ii}-e_{i+1,i+1}\right)\\ x_i=\left(e_{i,i+1}-e_{i+1,i}\right)\\ y_i=\sqrt{-1}\left(e_{i,i+1}+e_{i+1,i}\right) $$ for $1\le i<2n$. The $h_i$ are a basis (over $\mathbb R$) for a Cartan subalgebra. They're also a basis (over $\mathbb C$) for a Cartan subalgebra in the complexification $\mathfrak{sl}(2n)$. I'll let you verify the following brackets $$ \begin{matrix} [h_i,x_i]=2y_i&[h_i,y_i]=-2x_i\\ [h_{i-1},x_i]=[h_{i+1},x_i]=-y_i&[h_{i-1},y_i]=[h_{i+1},y_i]=x_i \end{matrix} $$ and $[h_j,x_i]=[h_j,y_i]=0$ if $|j-i|>1$. Letting $V_i=\operatorname{span}(x_i,y_i)$, the $V_i$ are $\operatorname{ad}(H)$-invariant subspaces where $H=\operatorname{span}(h_1,\ldots,h_{2n-1})$ is the Cartan subalgebra. In the complexification $\mathfrak{sl}(2n)$, the matrices $e_{i,i+1}$ are root vectors corresponding to simple positive roots, and the matrices $e_{i+1,i}$ are root vectors corresponding to simple negative roots. In $\mathfrak{su}(2n)$ the root vectors join in opposing pairs to form the invariant subspaces $V_i$.

Now let $\pi=\sum_{i=1}^{2n} \sqrt{-1}\left(e_{i,2n+1-i}\right)$. This is the matrix $B$ from the original post. It is $\sqrt{-1}$ times a permutation matrix that corresponds to the permutation $(1,2n)(2,2n-1)\ldots(n,n+1)$ that reverses the indices in the range $1,\ldots,2n$. We have $\det\pi=(-1)^n\sqrt{-1}^{2n}=1$ which is the reason for scaling the permutation by $\sqrt{-1}$. We also have $\pi^{-1}=-\pi$, and if $M\in M_{2n}(\mathbb C)$ then $\pi M \pi^{-1}$ consists of reversing the columns of $M$, and then reversing the rows of $M$ (or vice versa which gives the same result.)

Now we wish to use the automorphism $\sigma$ of $\mathfrak{su}(2n)$ given by $\sigma(x)=\pi\bar{x}\pi^{-1}$ where $\bar{x}$ is complex conjugation. This is an outer automorphism (for $n>1$) which is in the same coset as the outer automorphism complex conjugation. From the definition $$ \sigma(h_i)=\pi\left(-h_i\right)\pi^{-1}=-\sqrt{-1}\left(e_{2n+1-i,2n+1-i}-e_{2n-i,2n-i}\right)=h_{2n-i}\\ \sigma(x_i)=\pi x_i \pi^{-1}=\left(e_{2n+1-i,2n-i}-e_{2n-i,2n+1-i}\right)=-x_{2n-i}\\ \sigma(y_i)=\pi \left(-y_i\right) \pi^{-1}=-\sqrt{-1}\left(e_{2n+1-i,2n-i}+e_{2n-i,2n+1-i}\right)=-y_{2n-i}$$

So the effect of $\sigma$ is to reverse the order of the $h_i$ and the order of the invariant subspaces $V_i$ - corresponding to flipping the diagram. Note also that complex conjugation on $\mathfrak{su}(2n)$ is the restriction to $\mathfrak{su}(2n)$ from $\mathfrak{sl}(2n)$ of the outer automorphism $x\mapsto -x^{\intercal}$. Using this outer automorphism the same way on the complexification $\mathcal{sl}(2n)$ is necessary in order to reverse the positive root vectors $e_{i,i+1}$ to $\pm e_{2n-i,2n+1-i}$. Without it they map to $\pm e_{2n+1-i,2n-i}$ which are the negative root vectors, not positive ones. In $\mathfrak{su}(2n)$, if we drop complex conjugation from the definition of $\sigma$, we still get $h_i$ mapped to $-h_{2n-i}$, $x_i$ mapped the same and $y_i$ mapped to $y_{2n-i}$.