Let $G$ be a quasi-split connected reductive group over a perfect field $F$, with Borel subgroup and maximal torus $B \supseteq T$ defined over $F$. Assume everything splits over a Galois extension $L/F$, with $\Gamma = \textrm{Gal}(L/F)$. The choice of $B$ gives us a set of simple roots $\Delta \subseteq \Phi(G,T)$, and for each $\alpha \in \Delta$, there exists an isomorphism of algebraic groups $x\_{\alpha}$ of $\mathbf{G}\_a$ onto a closed, one dimensional subgroup $U\_{\alpha}$ of $R\_u(B)$, such that 


$$t x\_{\alpha}(a)t^{-1} = x\_{\alpha}(\alpha(t)a)$$


for all $t \in T, a \in \mathbf{G}\_a$. Such a map $x\_{\alpha}$ is called a **root vector** for $\alpha$.


If $x\_{\alpha}'$ is another isomorphism onto $U\_{\alpha}$ with the same property, then there is a unique $0 \neq c \in \mathbf{G}\_a$ such that $x\_{\alpha}(a) = x'\_{\alpha}(ca)$.


$\Gamma$ acts on $\Delta$ by $\alpha^{\sigma} = \sigma \alpha \sigma^{-1}$. If $x\_{\alpha}$ is a root vector for $\alpha$, then $x\_{\alpha}^{\sigma} := \sigma x\_{\alpha} \sigma^{-1}$ is one for $\alpha^{\sigma}$.


What I would like to do is the following:



> 
> **Goal**: *Choose for each $\alpha \in \Delta$ a root vector $x\_{\alpha}$ such that $x\_{\alpha}^{\sigma} = x\_{\alpha^{\sigma}}$ for all $\alpha \in \Delta, \sigma \in \Gamma$.* 
> 
> 
> 


This does not seem possible, since for a fixed $\alpha \in \Delta$, and $\sigma \neq \tau \in \Gamma$ we can have $\alpha^{\sigma} = \alpha^{\tau}$, and there is no guarantee that $x\_{\alpha}^{\sigma} = x\_{\alpha}^{\tau}$ for any choice of root vector $x\_{\alpha}$ for $\alpha$.


The next best thing seems to be to, for each orbit of $\Delta$ under the action of $\Gamma$, choose an $\alpha$ in the orbit, and replace $\Gamma$ by some quotient $\textrm{Gal}(F\_{\alpha}/F)$ for some appropriate Galois extension $F\_{\alpha}$ of $F$. In *Eisenstein Series and Automorphic L-Functions* by F. Shahidi, such a field $F\_{\alpha}$ is mentioned, but not defined, and it is called the **splitting field of $x\_{\alpha}$ over $F$.** 


My question is, how should this field $F\_{\alpha}$ be defined, and how close can we get to achieving the Goal mentioned above?



Splitting field of a root vector $x_{\alpha}: \mathbf{G}_a \rightarrow U_{\alpha}$
This is explained by the Borel-Tits relative structure theory for connected reductive groups over arbitrary fields. In particular, there is no need to assume $F$ is perfect. The explanation below is long when written out, but the underlying principles are rather natural and clean.
Moreover, we will find a collection of root vectors satisfying the stronger requested Galois-equivariance property as near the start of the question. I will use Roman letters to denote roots, so "$a$" rather than "$\alpha$". 


Consider a relative root $a$ in a chosen basis for the relative root system (it corresponds to a Galois orbit in a basis of the absolute root system such that the members of the orbit have nontrivial restriction to a maximal split torus, as we review below). The $a$-root space may have massive dimension as an $F$-vector space, but we will show that it is naturally a **line** over a finite separable extension $F\_a$ of $F$. The construction of this enhanced linear structure over such an extension will make clear the sense in which $F\_a/F$ and that $F\_a$-linear structure refining the given $F$-linear structure is canonical for non-multipliable $a$ (and nearly canonical 
for multipliable $a$). This will use in a crucial way the quasi-split hypothesis. 


First, we fix what seems to be a misconception: the correct definition of $F\_a$ will show that it is generally *not* a Galois extension of $F$. Rather, it is a finite separable extension of $F$, and the Galois orbit of absolute roots in question is not naturally indexed by the Galois group for $F\_a$ over $F$ (there is no natural base point for the orbit, and no such Galois group when $F\_a/F$ is not Galois) but rather is naturally indexed by the set of $F$-embeddings of
$F\_a$ into a fixed separable closure $F\_s$ (sanity check: ${\rm{Gal}}(F\_s/F)$ does
act naturally and transitively on that set, regardless of whether or not $F\_a/F$ is Galois!). 


We are going to postpone the quasi-split hypothesis for as long as possible, to identify most clearly where exactly this assumption is truly essential. For an arbitrary connected reductive $F$-group $G$ (not yet assumed to be quasi-split, to clarify ideas), let $S$ be a maximal split $F$-torus. The Borel-Tits relative structure theory over arbitrary fields ensures that:


(i) $S$ is contained in some minimal parabolic $F$-subgroup $P$ of $G$,


(ii) all minimal parabolic $F$-subgroups of $G$ are $G(F)$-conjugate to each other (so by (i), if $G$ is quasi-split then $S$ is contained
in a Borel $F$-subgroup), 


(iii) all pairs $(S, P)$ with $S \subset P$ are $G(F)$-conjugate to each other, and $S$ is non-central if and only if there exists a proper
parabolic $F$-subgroup,


(iv) $S' := (S \cap \mathscr{D}(G))^0\_{\rm{red}}$ is a maximal split $F$-torus in $\mathscr{D}(G)$, the map $\Phi(G,S) \rightarrow {\rm{X}}(S')$ via restriction is an injection into ${\rm{X}}(S') - \{0\}$
and constitutes a (possibly non-reduced!) root system spanning
${\rm{X}}(S')\_{\mathbf{Q}}$, 


(v) the positive systems of roots in $\Phi(G,S)$ are in bijective correspondence with the set of minimal parabolic $F$-subgroups $P \subset G$ containing $S$ via $P \mapsto \Phi(P,S)$ (the set of non-trivial $S$-weight occurring in ${\rm{Lie}}(P)$). 


Pick a minimal parabolic $F$-subgroup $P \supset S$, and a maximal $F$-torus $T \subset P$ containing $S$ (so $T$ is also maximal as an $F$-torus in $G$). Let ${}\_F\Delta$ be the basis of $\Phi(G,S)$ corresponding to the positive system of roots $\Phi(P, S)$. By the split theory applied over $F\_s$, we can pick a Borel $F\_s$-subgroup $B \subset G\_{F\_s}$ contained in $P\_{F\_s}$ and containing $T\_{F\_s}$. (We are
ultimately interested in the case that $G$ is quasi-split, so $B = P\_{F\_s}$, but we do not make that assumption just yet.) 


Let $\Delta$ be the basis of the absolute root system $\Phi = \Phi(G\_{F\_s}, T\_{F\_s})$ corresponding to the positive system of roots
$\Phi(B, T\_{F\_s})$, and 
let $\Delta\_0 \subset \Delta$ be the set of absolute roots whose restriction to $S\_{F\_s}$ is *trivial*. The restriction map
${\rm{X}}(T\_{F\_s}) \rightarrow {\rm{X}}(S\_{F\_s}) = {\rm{X}}(S)$ clearly carries $\Delta - \Delta\_0$ into $\Phi(G\_{F\_s}, T\_{F\_s})$, and since
$B \subset P\_{F\_s}$ it is immediate from the definitions that $\Delta - \Delta\_0$ is even carried into $\Phi(B, T\_{F\_s})$; with some
more work one shows that it lands inside ${}\_F\Delta$. 


The Galois group $\Gamma\_F = {\rm{Gal}}(F\_s/F)$ acts naturally (on the left) on the geometric character lattice $X = {\rm{X}}(T\_{F\_s})$, and as such it certainly preserves $\Phi$ (since $T$ is an $F$-torus and the absolute root system encodes the weight spaces for non-trivial weights for the $F\_s$-scalar extension of the $F$-linear $T$-action on ${\rm{Lie}}(G)$ over $F$. For each $\gamma \in \Gamma\_F$, $\gamma(\Delta)$ is a basis for $\Phi$. It is exactly in the quasi-split case that we can find a Borel $F$-subgroup, and then pick $T$ inside that, so then the corresponding $\Delta$ would be $\Gamma\_F$-stable. But we want to do better for now, so we confront the possibility that $\gamma(\Delta) \ne \Delta$ for some $\gamma$, perhaps no matter what $T$ and $\Delta$ we may try. This is handled by Tits' so-called $\ast$-action of $\Gamma\_F$ on $\Delta$, as we now review.


Recall that the Weyl group of any root system acts simply transitively on the set of bases for the root system (since the set of bases is in natural bijective correspondence with the set of Weyl chambers). But for the finite etale $F$-group $W(G,T) = N\_G(T)/T$ we know from the structure theory of split reductive groups (such as over $F\_s$) that the natural map
$$W(G,T)(F\_s) = N\_{G(F\_s)}(T\_{F\_s})/T(F\_s) \rightarrow W(\Phi)$$
is an isomorphism, so there is a unique $w\_{\gamma} \in W(G,T)(F\_s)$ such that $w\_{\gamma}.\gamma(\Delta) = \Delta$. One checks from the uniqueness aspect that $w\_{\gamma'} \gamma'(w\_{\gamma}) = w\_{\gamma'\gamma}$
(using the natural $\Gamma\_F$-action on $W(G,T)(F\_s)$), and that $w\_{\gamma} = 1$ for $\gamma$ in some open subgroup of $\Gamma\_F$, so
$$(\gamma, a) \mapsto \gamma \ast a := w\_{\gamma}(\gamma(a))$$
is a continuous left action of $\Gamma\_F$ on $\Delta$. In the quasi-split case, if we had chosen $T$ inside a Borel $F$-subgroup $B$ then for the corresponding $\Delta$ we would have $w\_{\gamma} = 1$ for all $\gamma$, so this $\ast$-action would agree with the effect of the natural $\Gamma\_F$-action on $\Phi$ that preserves the positive system of roots $\Phi(B\_{F\_s}, T\_{F\_s})$ and hence preserves its set $\Delta$ of simple elements. That establishes the contact with the quasi-split case, and in what follows we always work with the $\ast$-action to avoid a quasi-split hypothesis.


Here finally is the real substance of the matter. From the definitions one can check that the natural map
$$\pi: \Delta - \Delta\_0 \rightarrow {}\_F\Delta$$
introduced above is *invariant* under the $\ast$-action on $\Gamma\_F$, so $\pi$ carries each $\Gamma\_F$-orbit to a single point in ${}\_F\Delta$. 
(Really we have a map $\pi:\Delta \rightarrow {}\_F \Delta \cup \{0\}$,
and $\Delta\_0$ is the fiber over $\{0\}$ by definition.) 
Much more serious (and deeper) is the fact that $\pi$ is surjective
with fibers consisting of *precisely* the $\Gamma\_F$-orbits. So to each simple positive relative root $a$ (i.e., element of ${}\_F\Delta$) there is a $\Gamma\_F$-orbit of simple positive absolute roots $a'$ (i.e., elements of $\Delta$) such that $a'|\_{S\_{F\_s}} = a$ via the identification of ${\rm{X}}(S\_{F\_s})$ with ${\rm{X}}(S)$.


We can start cooking with gas. For each $a \in \Phi(G,S)$ (such as an element of ${}\_F\Delta$, but can be more general for a moment) we have the associated codimension-1 $F$-subtorus $S\_a = (\ker a)^0\_{\rm{red}} \subset S$, and $Z\_G(S\_a)$ is a connected reductive $F$-group containing $T$ and $S$ in which $S\_a$ is central but $S$ is not since ${\rm{Lie}}(Z\_G(S\_a)) = {\rm{Lie}}(G)^{S\_a}$ contains the nonzero $a$-weight space for $S$ in ${\rm{Lie}}(G)$. In case $G$ is quasi-split,
so $P$ as above is a Borel $F$-subgroup, so is $P \cap Z\_G(S\_a)$
(as Borel subgroups meet centralizers of their subtori in Borel subgroups
for general smooth connected affine groups over fields). Thus,
if $G$ is quasi-split then so is $Z\_G(S\_a)$ and hence its derived group.


The $F$-group $G\_a := \mathscr{D}(Z\_G(S\_a))$ is connected semisimple containing the 1-dimensional isogeny complement
$S'\_a := (S \cap G\_a)^0\_{\rm{red}}$ to $S\_a$ in $S$ as a maximal split $F$-torus, and all absolute roots restricting to $a\_{F\_s}$ on $S\_{F\_s}$
have their weight spaces in ${\rm{Lie}}(G\_{F\_s}) = {\rm{Lie}}(G)\_{F\_s}$ supported inside ${\rm{Lie}}(G\_a)\_{F\_s}$! Also, 
$T \cap G\_a$ is a maximal $F$-torus in $G\_a$ (since $G\_a$ is normal in
$Z\_G(S\_a)$) and is an isogeny complement in $T$ to the maximal central $F$-torus in $Z\_G(S\_a)$. The absolute weight spaces and root groups for $(Z\_G(S\_a), T)$ coincide with those for $(G\_a, T \cap G\_a)$ since
passage to the derived group never affects such notions.


Thus, for any questions about the root groups or root spaces over $F\_s$ associated to absolute roots extending $a\_{F\_s}$, we lose nothing by passing to $(G\_a, S'\_a, T \cap G\_a)$ 
(also preserving the quasi-split property if $G$ is quasi-split,
as seen above) and we *gain* the property of $G$ being connected semisimple with $F$-rank equal to 1. We apply the preceding with $a \in {}\_F\Delta$ and are interested in the root groups and root spaces attached to the $\Gamma\_F$-orbit $\pi^{-1}(a) \subset \Delta$. This brings us to the case of $F$-rank 1 with the singleton $\{a\}$ a basis for the rank-1 (possibly non-reduced!) relative root system. (One has to make to small argument involving the insensitivity of the relative root system upon passage to the derived group, applied to $Z\_G(S\_a)$ and its derived group, to ensure that $a|\_{S'\_a}$ really is a basis for the relative root system for $(G\_a, S'\_a)$. This argument is easy, so we omit it here.) 


Passage to the simply connected central cover has no effect on the relative root system, nor on the quasi-split property (if it holds), nor on the absolute root system, nor on the absolute root groups or their root spaces or even the relative root spaces, since the Lie algebra of the (possibly inseparable) central isogeny is contained in the Lie algebra of any maximal $F$-torus. Thus, for the purpose of the question posed we may and do assume $G$ is not only connected semisimple with $F$-rank equal to 1 but is also *simply connected*. 


Now, at last, the mythic (and canonical!) "field of definition" $F\_a$ in the question (which will be finite separable over $F$) emerges from the shadows. This rests on the following fundamental result (the proof of which is part of the same circle of ideas that prove the Theorem stated in my answer to [Automorphism of restriction of scalars](https://mathoverflow.net/questions/255974/automorphism-of-restriction-of-scalars)):


**Theorem**. *If $G$ is a nontrivial connected semisimple group over a field $F$ and it is simply connected then $G \simeq {\rm{R}}\_{F'/F}(G')$ for a nonzero finite etale $F$-algebra $F'$ and a smooth affine $F'$-group $G'$ whose fiber over each factor field of $F'$ is connected semisimple, simply connected, and **absolutely simple**. Moreover, the pair $(F'/F, G')$ is unique up to unique isomorphism in the sense that if $(F''/F, G'')$ is a second such pair then any $F$-isomorphism
${\rm{R}}\_{F''/F}(G'') \simeq {\rm{R}}\_{F'/F}(G')$ arises from a unique pair $(\alpha, \varphi)$ consisting of an $F$-algebra isomorphism $\alpha:F' \simeq F''$ and a group isomorphism $\varphi:G'\simeq G''$ over $\alpha$.*


This result is basically just an exercise in Galois descent and the fact that the root datum of a split simply connected group is uniquely the direct product (or direct sum?) of root data for its irreducible components. A reference for a proof is the same as for the Theorem I stated in my answer at the link above, namely Proposition A.5.14 in the book *Pseudo-reducitve Groups*. 


Coming back to the case of interest, we use the preceding Theorem to describe our connected semisimple $G$ that is simply connected with $F$-rank 1 as ${\rm{R}}\_{F'/F}(G')$ for some nonzero finite etale $F$-algebra $F'$ and $F'$-group $G'$ as in the Theorem. Letting $\{F'\_i\}$ be the factor fields of $F'$ and $G'\_i$ the $F'\_i$-fiber of $G'$, we have
$$G = {\rm{R}}\_{F'/F}(G') \simeq \prod\_i {\rm{R}}\_{F'\_i/F'}(G'\_i)$$
where each $G'\_i$ is a connected semisimple $F'\_i$-group that is *absolutely simple* and simply connected. Maximal split $F$-tori in such a product have the unique form $\prod S\_i$
where $S\_i$ is the maximal $F$-split torus in ${\rm{R}}\_{F'\_i/F}(S'\_i)$ for a maximal split $F'\_i$-torus $S'\_i$ in $G'\_i$ (see Proposition A.5.15(2) in *Pseudo-reductive Groups* for a proof in a somewhat broader setting); note that $\dim S\_i = \dim S'\_i$. Thus, exactly one of the factors ${\rm{R}}\_{F'\_i/F}(G'\_i)$ has $F$-rank 1 (with the corresponding $G'\_i$ of $F'\_i$-rank 1) and the others are $F$-anisotropic.


Likewise, $T = {\rm{R}}\_{F'/F}(T')$ for a unique maximal $F'$-torus $T' \subset G'$ and moreover $T' \supset S'$. We have $T' = \coprod T'\_i$ for a unique maximal $F'\_i$-torus $T'\_i \subset G'\_i$, and $S'\_i \subset T'\_i$. Thus, $T = \prod {\rm{R}}\_{F'\_i/F}(T'\_i)$. Note in particular
that all $T\_{F\_s}$-weight spaces in ${\rm{Lie}}(G\_{F\_s})$ that extend $a\_{F\_s}$ are supported inside the $F\_s$-fiber of the unique
$F$-isotropic factor ${\rm{R}}\_{F'\_i/F}(G'\_i)$, and that the corresponding
root groups over $F\_s$ are also contained in the $F\_s$-fiber of that direct factor.


By Galois descent one also shows rather generally that if
$k'$ is a nonzero finite etale algebra over a field $k$ and $H'$ is a smooth affine $k'$-group with connected fibers over the factor fields of $k'$ then $Q' \mapsto {\rm{R}}\_{k'/k}(Q')$ is a bijective correspondence between the set of parabolic $k'$-subgroups of $H'$ and the set of parabolic $k$-subgroups of ${\rm{R}}\_{k'/k}(H')$, inclusion-preserving in
both directions. Thus, $P = {\rm{R}}\_{F'/F}(P')$
for a unique *minimal* parabolic $F'$-subgroup $P' \subset G'$
that necessarily contains $T'$, and $P'$ is a Borel $F'$-subgroup if and only if $P$ is a Borel $F$-subgroup. Note 
that $P' = \coprod P'\_i$ for a parabolic $F'\_i$-subgroup $P'\_i \subset G'\_i$ that is clearly minimal and contains $S'\_i$ for each $i$.


Hence, for our purposes we can focus entirely on the factor with $F$-rank 1, which means that we can assume $G = {\rm{R}}\_{F'/F}(G')$ where
$F'/F$ is a finite separable extension *field* and $G'$ is a connected semisimple $F'$-group that is *absolutely simple* and simply connected with $F'$-rank 1. We have $S \subset {\rm{R}}\_{F'/F}(S')$
for a unique $F$-split torus $S' \subset G'$ with dimension 1, 
and it is easy to check by inspection of Lie algebras (and using the good behavior of Lie algebras with respect to Weil restriction) that $\Phi(G',S') = \Phi(G,S)$ (equal to $\{\pm a\}$ or $\{\pm a, \pm 2a\}$). 


Finally, the payoff: we claim that the sought-after field $F\_a$
is equal to $F'$ or to a quadratic Galois extension of $F'$. The reason is due to the quasi-split hypothesis that we shall now use. The point is that although it is hopeless to describe the possibilities for the absolutely simple $G'$ over $F'$ 
with $F'$-rank 1 in general, in the quasi-split case 
the possibilities are very explicit:


**Proposition.** *If $H$ is a connected semisimple group over a field $k$ and it is **absolutely simple**, simply connected, and quasi-split with $k$-rank equal to $1$ then $H$ is either $k$-isomorphic to ${\rm{SL}}\_2$ or $H$ is the quasi-split special unitary group ${\rm{SU}}\_3(K/k)$ associated to a quadratic Galois extension $K/k$. In the latter case, $K/k$ is determined by $G$ up to $k$-isomorphism.*


*Proof*. If the absolute rank $n$ of $H$ is equal to 1 then the 1-dimensional maximal split tori are in fact maximal $k$-tori, so $H$ would be split and thus the conclusion clear. We therefore may assume $n \ge 2$. Let $H\_0$ denote the split $k$-form of $H$, so $H$ is quasi-split form on $H\_0$. The diagram $\Delta$ of $H\_0$ is connected by absolute simplicity, and has $n$ vertices.


The automorphism group of the based root datum for $H\_0$ coincides with the automorphism group of $\Delta$ since $H\_0$ is simply connected. Thus, the quasi-split $k$-forms of $H\_0$ are classified up to isomorphism by the pointed set ${\rm{H}}^1(k, {\rm{Aut}}(\Delta))$ of conjugacy classes of continuous homomorphism ${\rm{Gal}}(k\_s/k) \rightarrow {\rm{Aut}}(\Delta)$; the dictionary goes via inner-twisting by pinned automorphisms. This pointed set is trivial unless $\Delta$ admits non-trivial automorphisms, so by the classification of reduced and irreducible root systems the only possibilities we need to consider for $\Delta$ are A$\_n$ ($n \ge 2$), D$\_n$ $(n \ge 4)$, and E$\_6$. 


The maximal $k$-tori in the Borel subgroups are the Weil restriction of ${\rm{GL}}\_1$ from the finite etale $k$-algebra $k'$ associated to the ${\rm{Gal}}(k\_s/k)$-action on $\Delta$: we associated to each Galois-orbit of vertices the subfield of $k\_s$ associated to the open subgroup fixing a choice of vertex in the orbit. In particular, if there are at least 2 Galois orbits then the $k$-rank of the associated quasi-split form is at least 2 since ${\rm{R}}\_{K/k}({\rm{GL}}\_1)$ contains ${\rm{GL}}\_1$ as a $k$-subgroup for any finite separable extension field $K/k$. But we assumed $H$ has $k$-rank equal to 1, so we only need to consider cases when ${\rm{Gal}}(k\_s/k)$ acts transitively on the set of vertices of $\Delta$. This can only happen when ${\rm{Aut}}(\Delta)$ acts transitively on $\Delta$. Inspection of the diagrams for
A$\_n$ ($n \ge 2$), D$\_n$ ($n \ge 4$), and E$\_6$ shows that the latter only happens for the A$\_2$ diagram. The quasi-split groups of type A with absolute rank $n \ge 2$ are exactly the special unitary groups ${\rm{SU}}\_{n+1}(K/k)$ associated to quadratic Galois extensions $K/k$ (and these $k$-groups have $k$-rank equal to $n-1$). The extension $K/k$ is unique up to isomorphism since it is the splitting field for the Galois action on the absolute diagram.


QED Proposition


We conclude that either (i) $G = {\rm{R}}\_{F'/F}({\rm{SL}}\_2)$
or (ii) $G = {\rm{R}}\_{F'/F}({\rm{SU}}\_3(F''/F'))$ for a quadratic Galois extension $F''/F'$. In the first case we will show $F'/F$ satisfies the desired properties to deserve being called $F\_a$, and in the second case likewise for $F''/F$. We emphasize that $F'$ is an *abstract* finite separable extension of $F$, not built inside $F\_s$! And of course $F'$ will generally *not* be Galois over $F$, as was noted near the start. The preceding method does give *intrinsic* meaning to $F'/F$ in terms of $a$ (even without reference to $P$). A slightly annoying feature is that we have not pinned down $F''/F'$ canonically, but only up to $F'$-isomorphism. 


Let's first consider case (i). We have $G = {\rm{R}}\_{F'/F}(G')$
where $G' \simeq {\rm{SL}}\_2$, but we do not choose such a latter isomorphism (as we want to keep everything canonical). The key point is that the geometric unipotent radical of the Borel $F'$-subgroup $P' \subset G'$ determined by $a$ descends to a unipotent smooth connected normal $F'$-subgroup $U' \subset P'$ that is split of dimension 1. This $U'$ is normalized by $S'$, with $S'$ acting through $a$ on the 1-dimensional Lie algebra $V'$ of $U'$ over $F'$. There is a unique $S'$-equivariant $F'$-isomorphism $f:U' \simeq \underline{V}'$ to the vector group over $F'$ associated to $V'$ such that ${\rm{Lie}}(f)$ is the identity map on $V'$. 


Thus, 
the analogous $F$-unipotent radical $U = {\rm{R}}\_{F'/F}(U')$ for ${\rm{R}}\_{F'/F}(P') = P'$ is identified with the vector group
$\underline{V}$ associated to the $F$-vector space $V$ underlying $V'$,
and this is compatible with the action of $S \subset {\rm{R}}\_{F'/F}(S') = T$. The $F\_s$-fiber $U\_{F\_s}$ is the unipotent radical of the Borel $F\_s$-subgroup $P\_{F\_s}$. The compatibility with Weil restriction with extension of the ground field and the canonical isomorphism of $F\_s$-algebras $F' \otimes\_F F\_s \simeq \prod\_{\sigma} F\_s$
via $y' \otimes t \mapsto (\sigma(y')t)$ (with $\sigma$ varying
through the set $\Sigma$ of $F$-embeddings of $F'$ into $F\_s$) identifies $T\_{F\_s}$ with $\prod\_{\sigma} (S' \otimes\_{F',\sigma} F\_s)$ and $U\_{F\_s}$ with $\prod\_{\sigma} (\underline{V}' \otimes\_{F',\sigma} F\_s)$.
Thus, if we pick an $F'$-basis of $V'$ then this defines
an $F'$-isomorphism $x':\underline{V}' \simeq \mathbf{G}\_{\rm{a}}$ 
so that $U\_{F\_s}$ is identified with $\mathbf{G}\_{\rm{a}}^{\Sigma}$
making the $\mathbf{G}\_{\rm{a}}$-factors exactly the $T\_{F\_s}$-root groups. This defines a canonical identification of
$\Sigma$ with the set of absolute roots $a'$ extending $a\_{F\_s}$;
another way to make that bijection is to identify $S'$ with $S\_{F'}$ via the map $S\_{F'} \rightarrow S'$ corresponding to the inclusion $S \hookrightarrow {\rm{R}}\_{F'/F}(S')$ and then identify ${\rm{R}}\_{F'/F}(S')\_{F\_s}$ with $\prod\_{\sigma} (S' \otimes\_{F',\sigma} F\_s)$, noting that each absolute root factors through the projection to exactly one of those factors $S' \otimes\_{F',\sigma} F\_s$.


By design, for any $a'$ and any $\gamma \in {\rm{Gal}}(F\_s/F)$, $x\_{\gamma(a')}$ is the scalar extension of $x\_{a'}$ through the $F$-automorphism $\gamma:F\_s \simeq F\_s$. Voila, so $F'$ has the desired property for $F\_a$ (and we see moreover that the true choice involved in picking those root vectors is a *single* choice, namely a basis of the canonical $F'$-line $V' = {\rm{Lie}}(U')$. That takes care of case (i)!


Next, we turn to case (ii): $G \simeq {\rm{R}}\_{F'/F}({\rm{SU}}\_3(F''/F'))$ for a quadratic Galois extension $F''/F'$ (unique up to $F'$-isomorphism). In this case the relative root system $\Phi(G,S)$ is BC$\_1$ (see Prop. 17.1.6(iii) with $n=3$ in Springer's book *Linear Algebraic Groups*); explicity, the $F'$-unipotent radical $U'$ of the Borel $F$-subgroup $P'$ is a Heisenberg group, with $S'$ acting through $2a$ on the Lie algebra of the 1-dimensional $F'$-split center and through $a$ on the 2-dimensional $F'$-split vector group quotient $U'/Z\_{U'}$. In particular, $a$ is a basis for ${\rm{X}}(S')$, which is to say that $a:S' \rightarrow {\rm{GL}}\_1$ (corresponding to $a: S \rightarrow {\rm{GL}}\_1$) is an *$F'$-isomorphism*. 


The root groups for the two absolute roots of $(G', T')$ corresponding to root lines in ${\rm{Lie}}(U'/Z\_{U'})\_{F'\_s}$ give a direct product decomposition of $(U'/Z\_{U'})\_{F'\_s}$ via the scalar extension of
the *$F'$-homomorphism* $U' \rightarrow U'/Z\_{U'}$. 
By explicit inspection of the construction of ${\rm{SU}}\_3(F''/F')$, we can identify $T'$ with ${\rm{R}}\_{F''/F'}(S'\_{F''}) \simeq {\rm{R}}\_{F''/F'}({\rm{GL}}\_1)$ (latter isomorphism via $a$!) respecting the inclusion of $S'$ into each such that the linear $T'$-action on the 2-dimensional $F'$-vector space ${\rm{Lie}}(U'/Z\_{U'})$ arises from a (necessarily unique) $F''$-vector space structure. Thus, 
for the $F''$-line $L := {\rm{Lie}}(U'/Z\_{U'})$ there is a unique
$T'$-equivariant isomorphism $U'/Z\_{U'} \simeq {\rm{R}}\_{F''/F'}(\underline{L})$ inducing the identity on Lie algebras, so an $F''$-basis of $L$ plays the same role as an $F'$-basis of $V'$ in case (i) to define root vectors of the desired type (thereby justifying that $F''$ deserves to be called $F\_a$ (with the annoyance that $F''$ is only unique up to $F'$-isomorphism and not shown to be truly canonical over $F$). 


