Everything is over an algebraically closed field of characteristic $\neq 2$. I had constructed a root datum for $\textrm{GSp}_4$ in a less than ideal way, and I had hoped to show that it was self dual. I wanted to know if my approach was salvegeable.

I came up with the following two lemmas to help me do it.

Lemma 1: Let $(X,R,X^{\wedge},R^{\wedge})$ be the root datum of a connected reductive group $G$. Then the root datum of $G \times \mathbf{G_m}$ is $(X \oplus \mathbb{Z}, R, X^{\wedge}\oplus \mathbb{Z},R^{\wedge})$ where a perfect pairing for $X \oplus \mathbb{Z}, X^{\wedge} \oplus \mathbb{Z}$ is extended in the obvious way from the one for $X,X^{\wedge}$, and the roots and coroots are embedded as $\alpha \mapsto (\alpha,0)$.

This is easy to see from the fact that if $T$ is a maximal torus of $G$, then $T \times \mathbf{G}_m$ is a maximal torus of $G \times \mathbf{G}_m$. One then looks at the action of this torus on the Lie algebra $\mathfrak g \oplus \mathbb{A}^1$.

The next lemma is a little more technical, but similarly straightforward to establish:

Lemma 2: Let $(X,R,X^{\wedge},R^{\wedge})$ be the root datum of a connected reductive group $G$. Let $N$ be a finite normal subgroup lying in the center of $G$. Then one can obtain the root datum of $G/N$ as follows:

Let $T$ be a maximal torus of $G$ with $X = X(T), X^{\wedge}= Y(T)$. Then $X(T/N)$ injects into $X(T)$ as a subgroup of finite index, $R$ is actually contained in $X(T/N)$, and $Y(T/N) = \textrm{Hom}(X(T/N),\mathbb{Z})$ can be identified with $Y(T) = \textrm{Hom}(X(T),\mathbb{Z})$ via $\gamma \mapsto \gamma|X(T/N)$.

Let $v_1, ... , v_n$ be a basis for $X(T)$ such that $x_1 = d_1v_1, ... , x_n = d_nv_n$ is a basis for $X(T/N)$ for some integers $1 \leq d_1 \mid \cdots \mid d_n$.

Let $w_1, ... , w_n$ be a basis for $Y(T)$ which is dual to $v_1, ... , v_n$. Define a pairing $X(T/N) \times Y(T/N) \rightarrow \mathbb{Z}$ by

$$\langle c_1x_1 + \cdots + c_nx_n, r_1w_1 + \cdots + r_nw_n \rangle_0 := \frac{1}{d_1}c_1r_1 + \cdots + \frac{1}{d_n} c_nr_n $$

Then this is a perfect pairing. Keep the roots $R \subseteq X(T/N)$, and for the coroots $R^{\wedge} \subseteq Y(T/N)$, replace any coroot $\gamma = r_1w_1 + \cdots + r_nw_n$ by the element $d_1r_1w_1 + \cdots + d_nr_nw_n$. Call the resulting set $R_1^{\wedge}$. Then with this perfect pairing $\langle - , - \rangle_0$ and the bijection $R \rightarrow R_1^{\wedge}$ induced by the bijection $R \rightarrow R^{\wedge}$,

$$(X(T/N),R,Y(T/N), R_1^{\wedge})$$

is the root datum of $G/N$.

To construct the root datum of $\textrm{GSp}_4$, I began with the root datum of $G = \textrm{Sp}_4$ with $X$ having basis $v_1,v_2$, $X^{\wedge}$ having basis $w_1,w_2$, and

$$R = \pm \{v_1 - v_2, 2v_2, v_1 + v_2, 2v_1 \}$$

$$R^{\wedge} = \pm \{w_1 - w_2, w_2,w_1+w_2,w_1 \}$$

Via $X$ we can choose a maximal torus for $\textrm{Sp}_4$ and identify it with $\textrm{Diag}(a,b)$ and identify the nontrivial element of the center as $-I := \textrm{Diag}(-1,-1)$. This is because the center of a connected, reductive group is the intersection of the kernels of all the roots. Now the definition I used for $\textrm{GSp}_4$ is

$$(\textrm{Sp}_4 \times \mathbf{G}_m)/N$$

where $N$ is the subgroup generated by $(-I,-1)$.

If you go through both of the lemmas, you will obtain the following $(X,\Phi,Y,\Phi^{\wedge})$ as the root datum for $\textrm{GSp}_4$: $X,Y$ are rank three free abelian groups with bases $x_1,x_2,x_3;y_1,y_2,y_3$ in obvious perfect pairing,

$$\Phi = \pm \{x_1,x_2,x_1+x_2,2x_1+x_2 \}$$

$$\Phi^{\wedge} = \pm \{ 2y_1-2y_2 - y_3,-y_1+2y_2 + y_3, 2y_2 + y_3,y_1 \}$$

where $x_1$ corresponds to the coroot $2y_1-2y_2 - y_3$ etc.

Supposedly $\textrm{GSp}_4$ is self dual, but that does not seem clear at all from the root datum I have obtained. Is it feasible to show that this root datum is self dual?


2 Answers 2


Define $A=\begin{pmatrix}1& 0\\0&1\\0&0\end{pmatrix},\ B=\begin{pmatrix}2&-1\\-2&2\\-1&1\end{pmatrix} $.

The matrices $A$ and $B$ consist of coefficents of simple roots and simple coroots, respectively. To be explicit, each column of $A$ corresponds to a simple root vector and consists of the coefficients of the simple root vector in the basis you presented above. Even more, the columns of $B$ (simple coroots) correspond to the columns of $A$ (the simple roots) such that ${}^TAB=({}^TA)B$ is a Cartan matrix. Define $C=\begin{pmatrix}-1&2\\2&-2\\1&-1 \end{pmatrix},\ D=\begin{pmatrix}0&1\\1&0\\0&0\end{pmatrix} $. The matrices $C$, respectively, $D$ are coefficients of the simple roots, respectively, simple coroots of the dual. Also, ${}^TC D={}^TAB$.

Define $\phi= \begin{pmatrix}1 & 1 &0\\1&1&-1\\0&-1&2\end{pmatrix}$.

Now $\phi\in{GL}\left(3,\mathbb{Z}\right)$ is an isomorphism of the root data, since $$(\phi C,{\ }^{T}\phi^{-1} D)=(A,B).$$


I still haven't figured out to do this using the root datum, but I just realized you can prove this in a much simpler way, if you take for granted the classification theorem for semisimple groups and that $\textrm{GSpin}_{2n+1} \cong \textrm{GSp}_{2n}^{\wedge}$.

The root systems for $\textrm{SO}_5$ and $\textrm{Sp}_4$ are the same (looking at the Dynkin diagrams). Then $\textrm{Sp}_4$ and $\textrm{Spin}_5$ are simply connected semisimple groups with the same root system, so they must be isomorphic. Then

$$\textrm{GSp}_4^{\wedge} \cong \textrm{GSpin}_5 \cong \textrm{GSp}_4$$


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