3.1. Endoscopic groups for $SL(2)$

As an example, we determine the unramified endoscopic groups of $G = SL(2)$. The character group $X^*$ can be identified with $\mathbb {Z}$, where $n\in \mathbb {Z}$ is identified with the character on the diagonal torus given by

$$\begin{pmatrix} t&0\\ 0&t^{-1} \end{pmatrix} \mapsto t^n.$$

The set $\Phi$ can be identified with the subset $\{\pm 2\}$ of $\mathbb {Z}$:

$$\begin{pmatrix} t&0\\ 0&t^{-1} \end{pmatrix} \mapsto t^{\pm 2}.$$

The cocharacter group $X_*$ is also identified with $\mathbb {Z}$, where $n\in \mathbb {Z}$ is identified with

$$t\mapsto \begin{pmatrix} t^n&0\\ 0&t^{-n} \end{pmatrix}.$$

Under this identification $\Phi ^{\vee } = \{\pm 1\}$. Since the group is split, $\sigma = 1$.

We get an unramified endoscopic group by selecting $s\in \operatorname {Hom}(X_*,\mathbb {C}^\times )\cong \mathbb {C}^\times$ and $w\in W(\Phi )$.

$$\begin{equation} \begin{array}{lll} \Phi _H^{\vee } = \{\alpha {\ \vert \ }s(\alpha )=1\} &= \{n\in \{\pm 1\}{\ \vert \ }s^n = 1\} \\&= \text{if}\ \ (s = 1)\ \ \text{then}\ \ \Phi _G^{\vee }\ \ \text{else}\ \ \emptyset . \end{array} \tag{3.0.1} \end{equation}$$

We consider two cases, according as $w$ is nontrivial or trivial. If $w$ is the nontrivial reflection, then $\sigma _H = w$ acts by negation on $\mathbb {Z}$. Thus,

$$(\sigma _H(s) = s)\ \ \implies \ \ ( s^{-1} = s)\ \ \implies (s = \pm 1).$$

If $s=1$, then $\sigma _H$ does not fix a set of simple roots as required. So $s=-1$ and $\Phi _H^{\vee } = \emptyset$. Thus, the root data of $H$ is

$$(\mathbb {Z},\mathbb {Z},\emptyset ,\emptyset ,w)$$

This determines $H$ up to isomorphism as $H=U_E(1)$, a $1$-dimensional torus split by an unramified quadratic extension $E/F$.

If $w$ is trivial, then there are two further cases, according as $\Phi _H$ is empty or not:

• The endoscopic group $\mathbb {G}_m$ has root data $$(\mathbb {Z},\mathbb {Z},\emptyset ,\emptyset ,1).$$

• The endoscopic group $H=SL(2)$ has root data $$(\mathbb {Z},\mathbb {Z},\{\pm 2\},\{\pm 1\},1).$$

In summary, the three unramified endoscopic groups of $SL(2)$ are $U_E(1)$, $\mathbb {G}_m$, and $SL(2)$ itself.

3.2. Endoscopic groups for $PGL(2)$

As a second complete example, we determine the endoscopic groups of $PGL(2)$. The group $PGL(2)$ is dual to $SL(2)$ in the sense that the coroots of one group can be identified with the roots of the other group. The root data for $PGL(2)$ is

$$(\mathbb {Z},\mathbb {Z},\{\pm 1\},\{\pm 2\},1).$$

When the Weyl group element is trivial, then the calculation is almost identical to the calculation for $SL(2)$. We find that there are again two cases, according as $\Phi _H$ is empty or not:

• The endoscopic group $\mathbb {G}_m$ has root data $$(\mathbb {Z},\mathbb {Z},\emptyset ,\emptyset ,1).$$

• The endoscopic group $H=PGL(2)$ has root data $$(\mathbb {Z},\mathbb {Z},\{\pm 1\},\{\pm 2\},1).$$

When the Weyl group element $w$ is nontrivial, then $s\in \{\pm 1\}$, as in the $SL(2)$ calculation.

$$\begin{equation} \Phi _H^{\vee } = \{\alpha {\ \vert \ }s(\alpha )=1\} = \{n \in \{\pm 2\}{\ \vert \ }s^n = 1\} = \Phi _G^{\vee }. \tag{3.0.2} \end{equation}$$

From this, we see that picking $w$ to be nontrivial is incompatible with the requirement that $\sigma _H=w$ must fix a set of simple roots. Thus, there are no endoscopic groups with $w$ nontrivial.

In summary, the two endoscopic groups of $PGL(2)$ are $\mathbb {G}_m$ and $PGL(2)$ itself.

3.3. Elliptic Endoscopic groups

The origin of the term elliptic is the following. We will see below that each Cartan subgroup of $H$ is isomorphic to a Cartan subgroup of $G$. (Here and elsewhere, when we speak of an isomorphic between algebraic groups defined over $F$, we mean an isomorphism over $F$.) The condition on $H$ for it to be elliptic is precisely the condition that is needed for some Cartan subgroup of $H$ to be isomorphic to an elliptic Cartan subgroup of $G$.

3.4. An exercise: elliptic endoscopic groups of unitary groups

This exercise is a calculation of the elliptic unramified endoscopic groups of $U(n,J)$. We assume that $J$ is a cross-diagonal matrix with units along the cross-diagonal as in Section 1.6.1. We give a few facts about the endoscopic groups of $U(n,J)$ and leave it as an exercise to fill in the details.

Let $T = \{\operatorname {diag}(t_1,\mathinner {\ldotp \ldotp \ldotp },t_n)\}$ be the group of diagonal $n$ by $n$ matrices. The character group $X^*$ can be identified with $\mathbb {Z}^n$ in such a way that the character

$$\operatorname {diag}(t_1,\mathinner {\ldotp \ldotp \ldotp },t_n) \mapsto t_1^{k_1}\cdots t_n^{k_n}$$

is identified with $(k_1,\mathinner {\ldotp \ldotp \ldotp },k_n)$.

The cocharacter group can be identified with $\mathbb {Z}^n$ in such a way that the cocharacter

$$t\mapsto \operatorname {diag}(t^{k_1},\mathinner {\ldotp \ldotp \ldotp },t^{k_n})$$

is identified with $(k_1,\mathinner {\ldotp \ldotp \ldotp },k_n)$.

Let $e_i$ be the basis vector of $\mathbb {Z}^n$ whose $j$-th entry is Kronecker $\delta _{ij}$. The set of roots can be identified with

$$\Phi = \{e_i-e_j {\ \vert \ }i\ne j\}.$$

The set of coroots $\Phi ^{\vee }$ can be identified with the set of roots $\Phi$ under the isomorphism $X_* \cong \mathbb {Z}^n \cong X^*$.

We may identify $\operatorname {Hom}(X_*,\mathbb {C}^\times )$ with $\operatorname {Hom}(\mathbb {Z}^n,\mathbb {C}^\times )= (\mathbb {C}^\times )^n$. Thus, we take the element $s$ in the definition of endoscopic group to have the form $s=(s_1,\mathinner {\ldotp \ldotp \ldotp },s_n)\in (\mathbb {C}^\times )^n$. The element $\sigma = \sigma _G$ acts on characters and cocharacters by

$$\sigma (k_1,\mathinner {\ldotp \ldotp \ldotp },k_n) = (-k_n,\mathinner {\ldotp \ldotp \ldotp },-k_1).$$

Let $I = \{1,\mathinner {\ldotp \ldotp \ldotp },n\}$. Show that if $H$ is an elliptic unramified endoscopic group, then there is a partition

$$I = I_1\coprod I_2$$

with $s_i = 1$ for $i\in I_1$ and $s_i = -1$ otherwise. The elliptic endoscopic group is a product of two smaller unitary groups $H = U(n_1)\times U(n_2)$, where $n_i = \#I_i$, for $i=1,2$.