Why is the L group of an even unitary group what it is?

Question

It is fairly easy to see from the formalism of the L group that the L group of a quasisplit unitary group will be a nontrivial semidirect product of $GL_n(\mathbb{C})$ (for the appropriate value of $n$), and a group of order two. When $n$ is odd this determines the isomorphism class uniquely. When $n$ is even, it does not. Indeed, an outer automorphism is of the form $g \mapsto A{}^tg^{-1}A^{-1}$ for some $A \in GL_n(\mathbb C).$ It is of order two if and only if $A$ is either symmetric or skew-symmetric. It seems to me that in the literature the $L$ group of a quasisplit unitary group is consistently defined using a skew-symmetric $A.$ Can anyone explain why this is correct?

Answer 1

There are three separate issues here: exactly which unitary or special unitary groups you're thinking about (in effect: exactly which non-degenerate hermitian spaces), how to define $L$-groups in general, and how that definition works out in the quasi-split case.

In general, if $G$ is a pinned split connected reductive group over a field $k$ and it has based root datum $(R, \Delta)$ arising from the pinning then via the notion of pinned automorphism we have $${\rm{Aut}}(G_{k_s}) = G^{\rm{ad}}(k_s) \ltimes {\rm{Aut}}(R, \Delta)$$ compatible with the action of ${\rm{Gal}}(k_s/k)$. In this way, the pointed set ${\rm{H}}^1(k, {\rm{Aut}}(R,\Delta))$ of conjugacy classes of actions of ${\rm{Gal}}(k_s/k)$ on $(R, \Delta)$ is in natural bijection with the set of $k$-isomorphism classes of quasi-split $k$-forms of $G$, and every $k$-form of $G$ has a unique quasi-split inner form.

For example, if $G = {\rm{SL}}_n$ for $n>1$ then it is simply connected with diagram A$_{n-1}$, so the split form is the only quasi-split form when $n=2$ (as A$_1$ has no nontrivial automorphisms) whereas otherwise the diagram automorphism group has order 2 and so the set of non-split quasi-split $k$-forms of ${\rm{SL}}_n$ is classified up to isomorphism by the set of quadratic Galois extensions $k'/k$ up to isomorphism. The link is that the group attached to $k'/k$ becomes split over $k'$. But what is it explicitly? That is where you preferred (special) unitary groups come in.

So now assume $n\ge 3$ and consider the hermitian space $({k'}^n, h_n)$ for a quadratic etale $k$-algebra $k'$ (i.e., a quadratic Galois field extension, or $k \times k$) with unique non-trivial automorphism denoted $z \mapsto \overline{z}$ and the non-degenerate hermitian form $h_n$ defined by $$h_{2m}(\vec{x}, \vec{y}) = \sum_{j=1}^{m}(x_j \overline{y}_{m+j}+x_{m+j}\overline{y}_j)$$ for $n = 2m$ and $$h_{2m+1} = x_0 \overline{y}_0 + h_{2m}((x_1,\dots,x_{2m}),(y_1,\dots,y_{2m}))$$ for $n=2m+1$. There are very many other kinds of rank-$n$ non-degenerate hermitian spaces over $k$ in general.

If $k' = k \times k$ then ${\rm{SU}}(k'/k, h_n)$ turns out to be ${\rm{SL}}_n$ by another name; see Exercise 3 at http://math.stanford.edu/~conrad/252Page/homework/hmwk7.pdf (where ${\rm{SU}}(h)$ for hermitian spaces $(V',h)$ over quadratic etale algebras $k'/k$ are described inside the Weil restriction ${\rm{R}}_{k'/k}({\rm{SL}}(V'))$). I claim that for $k'/k$ a quadratic Galois field extension, the $k$-group ${\rm{SU}}(k'/k, h_n)$ is the unique non-split quasi-split form of ${\rm{SL}}_n$ that splits over $k'$.

If you look at Example 7.1.10 (and Example 7.1.5) in the article "Reductive Groups Schemes" from the 2011 Luminy summer school on SGA3 (and set the base $S$ there to be ${\rm{Spec}}(k)$) you'll see a description for any $n>2$ of the Galois-twisting encoding exactly the unique non-split quasi-split $k$-form $G$ of ${\rm{SL}}_n$ that is split by $k'/k$, using the pinning associated to the upper-triangular Borel subgroup $B$ and the order-2 automorphism of ${\rm{SL}}_n$ given by $g \mapsto w({}^tg^{-1})w^{-1}$ where $w$ is the anti-diagonal matrix with alternating $1$'s and $-1$'s beginning with $1$ in the upper-right (and the role of transpose is to ensure that $B_{k'}$ is stable under the Galois descent, so the associated $k$-form really is quasi-split). From that explicit description of $G$ one describes $G(A)$ inside ${\rm{SL}}_n(k' \otimes_k A)$ for any $k$-algebra $A$ as fixed-points of a specific involution, and you can work out that this coincides exactly with the description of ${\rm{SU}}(k'/k, h_n) \subset {\rm{R}}_{k'/k}({\rm{SL}}_n)$ (on $A$-points, for any $A$).

So we have learned two things, for all $n > 2$ on equal footing: the groups ${\rm{SU}}(k'/k, h_n)$ for varying quadratic Galois $k'/k$ really are characterized in terms of the quasi-split property over $k$ and the split property over $k'/k$ as claimed, and we see exactly how they are made from ${\rm{SL}}_n$ via an explicit involutive automorphism over $k'$. We likewise conclude that ${\rm{U}}(k'/k, h_n)$ admits the same Galois-twisting description in terms of ${\rm{GL}}_n$ and $k'/k$ for each $n > 2$. This Galois-twisting description with an explicit involution is what is going to answer your actual question concerning the $L$-group for ${\rm{U}}(k'/k, h_n)$. But first we have to recall what exact is the definition of the $L$-group.

Strictly speaking, the uniform definition of the $L$-group treating all connected reductive groups $G$ over a local or global field $k$ on an equal footing involves a semi-direct product against the entire Weil group $W_k$ of $k$ for a specific action of $W_k$ on the connected reductive $\mathbf{C}$-group $^{L}G^0$ with the dual based root datum, so it has infinitely many connected components. But in practice one passes to the quotient by the finite-index open kernel of the $W_k$-action to collapse it down to a semi-direct product against a finite Galois group ${\rm{Gal}}(k'/k)$. It is the latter viewpoint you're thinking of, but really that is not the actual definition of the $L$-group (though in practice it works just as well). However you slice it, the real issue is: what is the $W_k$-action on ${}^LG^0(\mathbf{C})$? Or in our case of interest with $G = {\rm{U}}(k'/k, h_n)$ with $n \ge 3$ and $k'/k$ a quadratic Galois field extension, where it collapses down to an action of ${\rm{Gal}}(k'/k)$ on ${\rm{GL}}_n(\mathbf{C})$ (really the "dual"), what is the associated involution of ${\rm{GL}}_n(\mathbf{C})$? I claim that it is exactly the same recipe which we saw in the very description of $G$ as a quasi-split form of ${\rm{GL}}_n$ obtained by $k'/k$-twisting against the involution $g \mapsto w({}^tg^{-1})w^{-1}$ of ${\rm{GL}}_n$ over $k'$.

In general, the classification of connected reductive groups over fields involves Tits' notion of the $\ast$-action of the absolute Galois group on the based root datum, a notion that simplifies a lot in the quasi-split case (in that a certain intervention of the Weyl group becomes trivial): see section 12.2 of the course notes http://math.stanford.edu/~conrad/249BW16Page/handouts/249B_2016.pdf for the $\ast$-action and how it works out in the quasi-split case in terms of the explicit Galois-twisting that makes the group from the split form. Coming back to ${\rm{U}}(k'/k, h_n)$, for which we worked out the $k'/k$-twisting to get it from ${\rm{GL}}_n$ above, this gives exactly the recipe you wanted provided that we can sort out the role of the identification of ${\rm{GL}}_n$ as its own dual (to rigorously justify the correctness of how we describe the action on the $\mathbf{C}$-group ${}^L{\rm{GL}}_n$). Passing to the dual torus lattice and its dual basis of coroots (due to how ${}^LG^0$ is defined as a pinned $\mathbf{C}$-group), we get desired formula on the associated pinned "dual ${\rm{GL}}_n$" because the automorphism $f:g \mapsto w({}^tg^{-1})w^{-1}$ satisfies $${}^t (f({}^tg)) = {}^t(w({}^t({}^tg)^{-1})w^{-1}) = ({}^tw^{-1})({}^tg^{-1})({}^tw) = w({}^tg^{-1})w^{-1} = f(g);$$ the second-to-last equality uses that ${}^tw = (-1)^{n-1}w=w^{-1}$.

Answer 2

Just to put what znt said into answer form: the key reference is Borel's Corvallis paper MR0546608 and the key point is addressed in 1.2 and 2.3. We can formulate the problem as follows. We have a canonical action of the Galois group on the based root datum of the L group but as our example demonstrates, knowing what an automorphism does to the based root datum doesn't actually pin it down precisely. Write $\psi_0(G)$ for the root datum. Then we have an element of $\psi_0(G)$ and we must choose a lifting of it to an element of $\operatorname{Aut} G.$ In other words, we must choose a splitting of the exact sequence $$1\to \operatorname{Inn}\ ^LG \to \operatorname{Aut}\ ^LG \to \operatorname{Aut} \psi_0(^LG) \to 1.$$ For each simple root $\alpha$ choose an element $x_\alpha$ of the corresponding root subgroup. Then for each element $\sigma$ of $\operatorname{Aut} \psi_0(G)$ there is a unique automorphism $\widetilde \sigma$ of $^LG$ which maps one's fixed maximal torus and Borel to themselves and permutes $\{ x_\alpha: \alpha \in \Delta\}.$ A splitting obtained in this way is called admissible. The $L$ group should be defined using an admissible splitting. Changing the choice of $x_\alpha$'s is equivalent to conjugating by the inner automorphism attached to an element of the torus, so it doesn't change the isomorphism class of the group.

Consider the case of a unitary group in four variables. It suffices to consider one symmetric and one skew symmetric option, so let's assume $A$ is $$ \begin{pmatrix}&&&1\\&&1&\\&\pm1 &&\\ \pm1 &&&\end{pmatrix} $$

Let $$ x_1(r_1)=\begin{pmatrix}1&r_1&&\\&1&&\\&&1&\\&&&1\end{pmatrix}, x_2(r_2)=\begin{pmatrix}1&&&\\&1&r_2&\\&&1&\\&&&1\end{pmatrix}, \text{ and } x_3(r_3)=\begin{pmatrix}1&&&\\&1&&\\&&1&r_3\\&&&1\end{pmatrix}. $$ Then $g\mapsto A\ ^tg^{-1} A^{-1}$ maps $x_1(r_1)$ to $x_3(-r_1),$ $x_3(r_3)$ to $x_1(-r_3)$ and $x_2(r_2)$ to $x_2(\mp r_2).$ Hence, if we want to choose three elements which are permuted by the action of our automorphism we must take $r_3 = -r_1$ and the sign must be $-.$

One may reasonably ask why sticking to ``admissible'' splittings is the correct thing to do. In this regard, I would just note that in either semidirect product $GL_4(\mathbb C) \rtimes \langle$involution$\rangle$ we have the group $(GL_2(\mathbb C) \times GL_2(\mathbb C)) \rtimes \langle$involution$\rangle,$ which is the $L$ group of $\operatorname{Res}GL_2$ (restriction of scalars). For either choice of $A$ above our involution will preserve the parabolic subgroup of $GL_4(\mathbb C)$ whose Levi is $^L\operatorname{Res}GL_2,$ but the action of $^L\operatorname{Res}GL_2$ on the unipotent radical is different in the two cases. So, results in the Langlands-Shahidi method which show that the $L$ function appearing in the constant term is the one attached to the representation we get from the admissible splitting as opposed to the inadmissible splitting make a persuasive case that the admissible one is indeed the right one.