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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?

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    $\begingroup$ Irreducible in the Zariski topology, I guess? $\endgroup$ – LSpice Mar 6 at 17:29
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    $\begingroup$ What is the point of using $n+1$ instead of $n$ (or $2n$, if you want to emphasize parity). $\endgroup$ – Sasha Mar 6 at 17:37
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    $\begingroup$ 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. $\endgroup$ – LSpice Mar 6 at 17:56
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    $\begingroup$ @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? $\endgroup$ – Mikhail Borovoi Mar 6 at 18:08
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    $\begingroup$ The symplectic group is connected. Otherwise there would be a special name for its unit component... $\endgroup$ – YCor Mar 7 at 8:40
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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 $\mathrm{SL}_2$.

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    $\begingroup$ 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). $\endgroup$ – YCor Mar 7 at 10:37
  • $\begingroup$ Thank you very much to you all for the answer and the comments. I modified my question to better explain what is confusing me about the irreducibility of the symplectic group. $\endgroup$ – F_L Mar 8 at 10:21

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