# Topology of the projective symplectic group

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?

• Irreducible in the Zariski topology, I guess? – LSpice Mar 6 at 17:29
• What is the point of using $n+1$ instead of $n$ (or $2n$, if you want to emphasize parity). – Sasha Mar 6 at 17:37
• Like any semisimple group over $\mathbb C$, your group (which I, concerned with rationality issues, would prefer to call PGSp) is generated by irreducible unipotent subgroups, hence is irreducible. – LSpice Mar 6 at 17:56
• @LSpice: Yes, any connected semisimple group is generated by unipotent subgroups. Maybe we should understand the question as follows: Is the group ${\rm PSp}(2n,{\Bbb C})$ connected? – Mikhail Borovoi Mar 6 at 18:08
• The symplectic group is connected. Otherwise there would be a special name for its unit component... – YCor Mar 7 at 8:40

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 $$\mathrm{SL}_2$$.
• PS: connectedness still holds for the stabilizer of an arbitrary alternating form: indeed it has a matrix block-triangular decomposition with diagonal blocks $\mathrm{Sp}(2n)$ and $\mathrm{GL}_m$, and upper block abelian unipotent of dimension $2mn$ (the unipotent radical). – YCor Mar 7 at 10:37