Dimension (as an algebraic variety) of the symplectic group over a finite field The dimension of the symplectic group Sp(2n, F) is $2n^2+n$ if F is $\mathbb{R}$ or $\mathbb{C}$. I would like to know if the same is true if F is the algebraic closure of a finite field.
To elaborate: if I understood correctly, if F is a finite field, the dimension of the group Sp(2n, F) (as a algebraic variety) is the dimension of Sp(2n,$\tilde{F}$), where $\tilde{F}$ is the algebraic closure of F. Given the formula for the order of Sp(2n, F), (https://groupprops.subwiki.org/wiki/Order_formulas_for_symplectic_groups) and the result of Lachaud and Rolland ( On the number of points of algebraic sets over finite fields, Journal of Pure and Applied Algebra 219 (11) ,  5117-5136 (2015)) bounding the number of solutions of an algebraic equation on a finite field in terms of the dimension of the variety, the dimension of Sp(2n,F) is at least $2n^2+n$. Is it equal to $2n^2+n$?
This is not my field (I am a geometer/topologist) so I hope I am not talking nonsense! Thank you very much for your answers.
 A: Yes, it is also $2n^2+n$. Whatever $F$ is, $Sp_{2n}$ is a smooth linear algebraic group over $F$, so its dimension is also the dimension of the vector space $Lie(Sp_{2n})=\ker(Sp_{2n}(F[\varepsilon])\to Sp_{2n}(F))$, where $F[\varepsilon]=F[X]/(X^2)$ (and $\varepsilon$ is the class of $X$), and the map is the one induced by $a+b\varepsilon\in F[\varepsilon]\mapsto a\in F$.
Now, for any commutative $F$-algebra $R$, $Sp_{2n}(R)=\{ M\in M_{2n}(R)\mid M^t H_{2n}M=H_{2n}\}$, where $H_{2n}$ is the block diagonal matrix whose diagonal blocks are all equal to $\begin{pmatrix}\phantom{-}0 & 1 \cr -1 & 0\end{pmatrix}$.
Writing an element of $M_{2n}(F[\varepsilon])$ as $M_0+\varepsilon M_1,M_i\in M_{2n}(F)$, and using $\varepsilon^2=0$, we see easily that $Lie(Sp_{2n})=\{I_{2n}+\varepsilon M_1\mid H_{2n}M_1+M^t_1H_{2n}=0\}$.
Now, $H_{2n}=-H_{2n}^t$, and is invertible, so $M_1=H_{2n}^{-1}S,$ where $S$ is symmetric. All in all, $Lie(Sp_{2n})$ is isomorphic as a vector space to the subspace of symmetric matrices of $M_{2n}(F)$, whose dimension is $\frac{2n(2n+1)}{2}=2n^2+n$.
