How to understand the Galois *-action on a Dynkin diagram

Choose $y in L$ with $sigma(y)=-y$. Let's concretely look at the two examples case 1: $H = pmatrix{0&0&0&1\0&0&1&0\0&1&0&0\1&0&0&0}$ and case 2: $H = pmatrix{1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1}$.

Part 1: Seeing the standard Galois action $alpha mapsto sigma cdot alpha$ on roots

Method 1: With the Lie algebra

Case 1: The Lie algebra of $G(k)$ is

$$mathfrak g = { X in M_4(L): Tr(X)=0 text{ and } XH + H X^* =0} = {X in M_4(L): Tr(X) = 0 text{ and } x_{ij} = -sigma(x_{5-j,5-i})}$$

 $$= { pmatrix{a+by&c+dy&e+fy&gy\h+iy&j-by&ky&-e+fy\l+my&ny&-j-by&-c+dy\oy&-l+my&-h+iy&-a+by} : a, ..., o in k }.$$

Surely its scalar extension $mathfrak g_K$ is (isomorphic to) the split Lie algebra $mathfrak{sl}_4(L) = {X in M_4(L): Tr(X)=0}$, but let us look closely how Galois operates on that; namely, it does not just act on the entries of the matrix (when written as above), but rather as

$$sigma_{mathfrak g_L} := sigma otimes id_{mathfrak g} qquad qquad (1)$$

on $mathfrak g_L:= L otimes_k mathfrak g$. That is, e.g. the matrix $$pmatrix{0&1&0&0\0&0&0&0\0&0&0&0\0&0&0&0} in mathfrak g_L$$ "correctly viewed" as an element of the tensor product $L otimes_k mathfrak g$ is

$$=frac12 otimes underbrace{pmatrix{0&1&0&0\0&0&0&0\0&0&0&-1\0&0&0&0}}_{in mathfrak g} + frac{1}{2y} otimes underbrace{pmatrix{0&y&0&0\0&0&0&0\0&0&0&y\0&0&0&0}}_{in mathfrak g}$$

so that $$sigma(pmatrix{0&1&0&0\0&0&0&0\0&0&0&0\0&0&0&0})= frac12 otimes pmatrix{0&1&0&0\0&0&0&0\0&0&0&-1\0&0&0&0} + frac{1}{2color{red}{cdot(-y)}} otimes pmatrix{0&y&0&0\0&0&0&0\0&0&0&y\0&0&0&0} = pmatrix{0&0&0&0\0&0&0&0\0&0&0&-1\0&0&0&0} $$

from which one gets $sigma (pmatrix{0&x&0&0\0&0&0&0\0&0&0&0\0&0&0&0}) = pmatrix{0&0&0&0\0&0&0&0\0&0&0&-sigma(x)\0&0&0&0}$ for all $x in L$. Likewise, for example,

$$ sigma(underbrace{pmatrix{1&0&0&0\0&-1&0&0\0&0&0&0\0&0&0&0}}_{=:H_{alpha_1}})= underbrace{pmatrix{0&0&0&0\0&0&0&0\0&0&1&0\0&0&0&-1}}_{=:H_{alpha_3}})$$

whereas $pmatrix{0&0&0&0\0&1&0&0\0&0&-1&0\0&0&0&0}) in mathfrak g$ and hence is actually fixed by $sigma$.

Now in this case, indeed the diagonal matrices in $mathfrak g$ form a maximal toral subalgebra (a.k.a. Cartan subalgebra) which contain a maximal split toral subalgebra (namely, the two-dimensional $mathfrak s := { pmatrix{a&0&0&0\0&j&0&0\0&0&-j&0\0&0&0&-a} : a,j in k }$), so we can use the diagonal as above to see how Galois acts on the roots. Turns out that $sigma$ maps the root space $mathfrak g_{L, alpha_1}$ onto the root space $mathfrak g_{L, alpha_3}$ (I call $alpha_i = eta_i - eta_{i-1}$ the standard roots of the diagonal in $mathfrak{sl}_4(L)$); and, accordingly, interchanges the coroots $H_{alpha_1}$ and $H_{alpha_3}$; in short, now there are many ways to define the action of Galois on the roots via $sigma cdot alpha_1 := alpha_3$ and $sigma cdot alpha_2 =alpha_2$. The actual technical formula to do it is

$$sigma cdot alpha := sigma circ alpha circ sigma_{mathfrak g_L}^{-1}$$

where the first $sigma$ is just the normal Galois on $L$, but the second (inner) $sigma_{mathfrak g_L}$ is defined via $(1)$ and in particular is not just Galois action on entries, so that this normal Galois action $alpha mapsto sigma cdot alpha$ is not trivial.

Case 2: Now the Lie algebra of $G(k)$ is

$$mathfrak g = { X in M_4(L): Tr(X)=0 text{ and } X = - X^* } = {X in M_4(L): Tr(X) = 0 text{ and } x_{ij} = -sigma(x_{j,i})}$$

 $$= { pmatrix{ay&b+cy&d+ey&f+gy\-b+cy&hy&i+jy&k+ly\-d+ey&-i+jy&my&n+oy\-f+gy&-k+ly&-n+oy&(-a-h-m)y} : a, ..., o in k }.$$

Again its scalar extension $mathfrak g_K$ is (isomorphic to) the split Lie algebra $mathfrak{sl}_4(L)$, but this time Galois operates on that via

$$sigma_{mathfrak g_L} : X mapsto -X^* (= -sigma(X)^{tr}).$$

This seems to immediately give us an action $sigma cdot alpha_i = -alpha_i$ ($i=1,2,3$) on the roots, however we have to be very careful here. The problem is that although again the diagonal matrices in $mathfrak g$ define a maximal toral subalgebra (CSA) $mathfrak t$, and certainly $mathfrak t_L$ is a split CSA in $mathfrak g_L$, this $mathfrak t$ does not necessarily contain a maximal split toral subalgebra of $mathfrak g$ (e.g. if $k$ is a $p$-adic field, one can show there are non-zero ad-diagonalisable elements in $mathfrak g$, but obviously none is in $mathfrak t$): and we need that to set up the proper Galois action on the roots.

But if we assume that $k=mathbb R$, $y=i$ (imaginary unit), then one can show that $mathfrak g$ has no non-zero split toral subalgebra at all, and we are good to go, and indeed $sigma$ operates on the entire root system as $-id$.

Method 2: With tori and algebraic geometry

(References: Galois action on character group and Galois action on the character group of a torus, but I find these a bit misleading because they focus on cases where the Galois action is just the one on points / matrix entries. Actually Definition of the unitary group via a cocycle on $operatorname{GL}_n$ shows better what's going on here.)

Case 1: Our group has a torus with $T(L) = { pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} : x_i in L, x_1 x_2 x_3 x_4 =1}$ which is not split in $G(k)$, rather we get

$$T(k) = { pmatrix{a&0&0&0\0&b&0&0\0&0&sigma(b)^{-1}&0\0&0&0&sigma(a)^{-1}} : a,b in L, ab in k}.$$

Accordingly, the Galois action on $T(L)$ is again not given by $sigma$ acting on entries, but rather via

$$sigma_{T(L)} : pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} mapsto pmatrix{sigma(x_4)^{-1}&0&0&0\0&sigma(x_3)^{-1}&0&0\0&0&sigma(x_2)^{-1}&0\0&0&0&sigma(x_1)^{-1}}.$$

Again we define the $sigma$-action on a character $chi : T(L) rightarrow L^times$ of the torus via

$$sigma cdot chi := sigma circ chi circ sigma_{T(L)}^{-1}$$

and again we can see that rather than being trivial, e.g. the character $chi_1 : pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} mapsto dfrac{x_1}{x_2}$ (i.e. the root $alpha_1$ written multiplicatively) gets sent to

$$ sigma cdot chi_1 : pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} mapsto dfrac{x_3}{x_4}$$

i.e. the root $alpha_3$ written multiplicatively, etc.

Case 2: (But again assuming that the maximal $k$-split torus of $G$ is trivial, e.g in the case $k=mathbb R$, but not for $p$-adic fields $k$.) Again we can choose $T(L) = { pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} : x_i in L, x_1 x_2 x_3 x_4 =1}$ but this time in $G(k)$ we get

$$T(k) = { pmatrix{a&0&0&0\0&b&0&0\0&0&c&0\0&0&0&d} : a,b,c,d in L, abcd =1, a^{-1} = sigma(a), b^{-1}=sigma(b), c^{-1}=sigma(c)}.$$

Accordingly, the Galois action on $T(L)$ is again not given by $sigma$ acting on entries, but rather via

$$sigma_{T(L)} : pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} mapsto pmatrix{sigma(x_1)^{-1}&0&0&0\0&sigma(x_2)^{-1}&0&0\0&0&sigma(x_3)^{-1}&0\0&0&0&sigma(x_4)^{-1}}.$$

Again we define the $sigma$-action on a character $chi : T(L) rightarrow L^times$ of the torus via

$$sigma cdot chi := sigma circ chi circ sigma_{T(L)}^{-1}$$

and again we can see that rather than being trivial, e.g. the character $chi_1 : pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} mapsto dfrac{x_1}{x_2}$ (i.e. the root $alpha_1$ written multiplicatively) gets sent to

$$ sigma cdot chi_1 : pmatrix{x_1&0&0&0\0&x_2&0&0\0&0&x_3&0\0&0&0&x_4} mapsto dfrac{x_2}{x_1}$$

i.e. the root $color{red}{-}alpha_1$ written multiplicatively, etc.

Part 2: The $*$-action

(References: 6.2 in Borel-Tits, 2.3 in Tits' Classification article in the Boulder Proceedings, ...)

Now this can be done in various ways, using maximal parabolic subgroups and whatnot, but here is a way of doing it just on the root level:

Say we are given a root system $R$ in an Euclidean space $V$, and an action of a finite Galois group $Gamma$ via automorphisms of $R$.

Set $V_0 := {v in V: sum_{sigma in Gamma} sigma cdot v =0 }$. (To make things work, here it is crucial that in the above one had chosen a maximal Torus $T$ which contains a maximal $k$-split torus $S$: Because then and only then will $V/V_0$ be well-behaved enough and realise the $k$-relative roots as quotient ("folding") of the original root system $R$.)

A $Gamma$-linear order on $R$ (or $V$) is a choice of a positive half-space in $V$ such that for all $v in V setminus V_0$, we have $v$ positive $implies sigma cdot v$ positive for all $sigma in Gamma$. One can show that such $Gamma$-linear orders always exist.

A $Gamma$-basis for our root system $R$ is a set of simple roots among the ones positive for a $Gamma$-linear order. I.e. a basis consisting of roots which are positive for a $Gamma$-linear order.

Examples: In case 1, there are eight possible $Gamma$-bases: ${alpha_1, alpha_2, alpha_3}$ and its negative, ${alpha_1, -alpha_1-alpha_2-alpha_3, alpha_3}$ and its negative, ${alpha_1+alpha_2, -alpha_2, alpha_2+alpha_3}$ and its negative, and a fourth one I leave to you. Whereas e.g. ${alpha_2, alpha_3, -alpha_1-alpha_2-alpha_3}$ is a basis, but not a $Gamma$-basis of $R$.

In case 2, every basis of the root system is a $Gamma$-basis (because $V=V_0$, the extra condition for positive systems to be $Gamma$-linear is empty).

Now the $*$-action is defined as follows: For any basis $Delta$ of the root system, $sigma cdot Delta$ is again a basis. As is well-known, the Weyl group operates simply transitively on the set of bases of the root system; further, let's take a $Gamma$-basis for $Delta$. Then there is a Weyl group element $w_{sigma}$ (actually, a unique one) such that $w_{sigma}(sigma cdot Delta) = Delta$. Then we define the $*$-action as

$$sigma * alpha = w_{sigma} (sigma cdot alpha).$$

So to say, the $*$-action is the normal Galois action but "bent back" from a $Gamma$-basis into itself via the Weyl group.

Now in our examples, in case 1, the normal action already stabilises each $Gamma$-basis, and so the $*$-action is just the normal action. In case 2, the normal action actually sends each basis to its own negative, so as $w_sigma$ we have to take the longest element of the Weyl group, which again gives the non-trivial automorphism of the Dynkin diagram.

In both cases, the $*$-action of the non-trivial element of the Galois group is given by $alpha_1 mapsto alpha_3, alpha_2 mapsto alpha_2$.

Now you will notice that a lot of this seems like overkill. Indeed, in both cases, it seems as if one does not need $Gamma$-bases at all! Indeed, one can readily tell whether the "normal" Galois action acts through the Weyl group or not, and if not, then it necessarily gives the non-trivial diagram automorphism.

I am sure though that in more complicated examples (different root systems but more importantly, more intricate Galois groups!) this will make a difference. And then it is a crucial fact remarked in the above sources that the $*$-action, unlike the normal action, does not depend on many choices, in particular of the involved tori.