Let $L/k$ be a (Galois) quadratic field extension, and let $\sigma \in \operatorname{Gal}(L/k)$ be the nontrivial automorphism. Let $h$ be a Hermitian form on $L^{4}$, and let $G = \operatorname{SU}_{4}$ be the special unitary group associated to $h$. Concretely, if $H$ is the matrix of $h$ in some fixed basis, then the $k$-points of $G$ are
$$G(k) = \{ X \in \operatorname{SL}_{4}(L) : X^* H X = H \}$$
After extension to $L$, $G$ becomes isomorphic to $\operatorname{SL}_{4}$.
$$G_L \cong \operatorname{SL}_{4}$$
In particular,
$$G(L) \cong \operatorname{SL}_{4}(L)$$
Let $T \subset G$ be the diagonal subgroup, which is a maximal torus. $T$ is not split over $k$, but is split over $L$. On $L$-points, $T(L)$ corresponds to the usual diagonal subgroup in $\operatorname{SL}_{4}(L)$. For $i=1, 2, 3, 4$, let
$$\eta_i:T(L) \to L^\times$$
be the character which picks off the $i$th diagonal entry. The (absolute) roots for $G$ are
$$
\Phi = \{ \eta_i – \eta_j : i \neq j \}
$$
which is a root system of type $A_3$. A set of simple roots is
$$
\Pi = \{ \eta_1 – \eta_2, \eta_2 – \eta_3, \eta_3 – \eta_4 \}
$$
and the associated Dynkin diagram $A_3$ has the root $\eta_2 – \eta_3$ corresponding to the middle vertex.
I have been unable to understand any sources which describe the $*$-action of the Galois group $\operatorname{Gal}(L/k)$ on the Dynkin diagram. Possibly this should instead focus on the absolute Galois group $\operatorname{Gal}(k^{\operatorname{sep}}/k$), I am not sure.
There is not that much that can possibly happen here, since the automorphism group $\operatorname{Aut}(D)$ of the Dynkin diagram is just $\mathbb{Z}/2\mathbb{Z}$, with the nontrivial element given by reflection across the middle node. So either $\sigma$ acts as this nontrivial automorphism of $D$, or it acts trivially on $D$. However, I can't find any definition concrete enough to tell me which it should be in this particular case.
One definition I've been told is that the $*$-action here should be given by taking a character $\eta$ and then
$$
\sigma \cdot \eta = \sigma \circ \eta \circ \sigma^{-1}
$$
If this definition is correct, then the action is trivial, since conjugation commutes with picking off matrix entries. Is this definition correct? If not, how else to think about this?
Best Answer
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 $$\sigma(\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0})= \frac12 \otimes \pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} + \frac{1}{2\color{red}{\cdot(-y)}} \otimes \pmatrix{0&y&0&0\\0&0&0&0\\0&0&0&y\\0&0&0&0} = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} $$
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.