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? 



$\textrm{GSp}_{4}^{\wedge} \cong \textrm{GSp}_4$
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 
$$(\phi C,{\ }^{T}\phi^{-1} D)=(A,B).$$


