Consider the projective symplectic group $\mathrm{PSp}(n+1)$ over $\mathbb{C}$. This parametrizes $(n+1)\times (n+1)$ symplectic matrices modulo scalar multiplication.


Is $\mathrm{PSp}(n+1)$ irreducible?


Consider $4\times 4$ symplectic matrices. A matrix $A$ has a symplectic representative (modulo scalar) if and only if $A^{t}\Omega A = \lambda\Omega$ for some $\lambda\in\mathbb{C}^{\*}$, where $\Omega$ is the standard symplectic form. Set $N = A^{t}\Omega A$. The condition $N = \lambda\Omega$ translates into the following equations in the entries of $A$:
$$N\_{12} = a\_{00}a\_{21}-a\_{01}a\_{20}+a\_{10}a\_{31}-a\_{11}a\_{30}=0$$
$$N\_{14} = a\_{00}a\_{23}-a\_{03}a\_{20}+a\_{10}a\_{33}-a\_{13}a\_{30}=0$$
$$N\_{23} = a\_{01}a\_{22}-a\_{02}a\_{21}+a\_{11}a\_{32}-a\_{12}a\_{31}=0$$
$$N\_{34} = a\_{02}a\_{23}-a\_{03}a\_{22}+a\_{12}a\_{33}-a\_{13}a\_{32}=0$$
$$N\_{13}-N\_{24} = a\_{00}a\_{22}-a\_{01}a\_{23}-a\_{02}a\_{20}+a\_{03}a\_{21}+a\_{10}a\_{32}-a\_{11}a\_{33}-a\_{12}a\_{30}+a\_{13}a\_{31}=0$$


Consider the variety $X$ defined by these equations in the $\mathbb{P}^{15}$ of $4\times 4$ matrices modulo scalar. MacAulay2 tells me that $X = X\_1\cup X\_2$ has two irreducible components both of dimension $10$ and of degree $12$ and $20$ respectively. 


This is where the confusion comes from. What am I missing here?



Topology of the projective symplectic group
The symplectic group $G=\mathrm{Sp}(2n)=\mathrm{Sp}(V)$ is connected (say, in characteristic zero, as algebraic group), and hence so is its quotient $\mathrm{PSp}(2n)$. Let $K$ be the ground algebraically closed field, and $(V,\langle\cdot,\cdot\rangle)$ the given symplectic space. 


Indeed, we have to check that every $g\in G$ belongs to the component of $1$. The Zariski closure of $\langle g\rangle$ is product of a unipotent (hence connected) group and a diagonalizable one. This reduces to the case when $g$ is diagonalizable, with eigenvalue decomposition $V=\bigoplus\_{\lambda\in K^\*}V\_\lambda$. One sees that $\langle V\_\lambda,V\_\mu\rangle=0$ for $\lambda\mu\neq 1$. Hence, $$V\_1\oplus V\_{-1}\oplus\bigoplus\_{\{\lambda,\lambda^{-1}\}\text{ of card 2}}(V\_\lambda\oplus V\_\lambda^{-1})$$
is an invariant orthogonal decomposition. This reduces to the case when this decomposition is trivial.


If $g=\pm 1$, there is nothing to do. Otherwise, say $g$ has eigenvalues $\lambda^{\pm 1}$ with $\lambda\neq\pm 1$; then one easily checks that there is a $g$-invariant orthogonal decomposition into 2-dimensional subspaces. This, in turn, reduces to the case $\dim(V)=2$. Then we're all set since then the symplectic group is just the connected group $\mathrm{SL}\_2$. 


