# Joint Invariants of Linear Symplectic Actions
## 3 Invariants on Symplectic Vector Spaces
### The General Case: Algebra of Polynomial Invariants  
Minimal free resolutions can be computed in many examples for \(n\geq 1\). However, in what follows we restrict our attention to describing generators/relations of \(\mathcal{A}_{G}^{m}\). Let us count the number of local smooth invariants. The action of \(G\) on \(V\) is almost transitive, so the stabilizer of a nonzero point \(A_{1}\) has \(\dim G_{A_{1}}=\binom{2n+1}{2}-2n=\binom{2n}{2}\). For a generic \(A_{2}\) there is only one invariant \(a_{12}\) (the orbit has codimension 1) and the stabilizer of \(A_{2}\) in \(G_{A_{1}}\) has \(\dim G_{A_{1},A_{2}}=\binom{2n}{2}-(2n-1)=\binom{2n-1}{2}\). For a generic \(A_{3}\) there are two more new invariants \(a_{13},a_{23}\) (the orbit has codimension \(2+1=3\)) and the stabilizer of \(A_{3}\) in \(G_{A_{1},A_{2}}\) has \(\dim G_{A_{1},A_{2},A_{3}}=\binom{2n-1}{2}-(2n-2)=\binom{2n-2}{2}\). By the same reason for \(k\leq 2n\) the stabilizer of a generic \(k\)-tuple of points \(A_{1},\dots,A_{k}\) has \(\dim G_{A_{1},\dots,A_{k}}=\binom{2n-k+1}{2}\). Finally, for \(k=2n\) the stabilizer of generic \(A_{1},\dots,A_{2n}\) is trivial. Thus, we get the expected number of invariants \(a_{ij}\). For \(m\leq 2n+1\) there are no relations between them, and the first comes at \(m=2n+2\). These can be obtained by successively studying cases of increasing \(n\) resulting in the _Pfaffian relation_:  
\[b_{i_{1}i_{2}\ldots i_{2n+1}i_{2n+2}}:=\mathrm{Pf}(a_{i_{p}i_{q}})_{1\leq p \
eq 2n+2}=0.\]  
Recall that the Pfaffian of a skew-symmetric operator \(S\) on \(V\) with respect to \(\omega\) is \(\mathrm{Pf}(S)=\mathrm{vol}_{\omega}(Se_{1},\ldots,Se_{2n})\) for any symplectic basis \(e_{i}\) of \(V\). The properties of the Pfaffian are: \(\mathrm{Pf}(S)^{2}=\det(S)\), \(\mathrm{Pf}(TST^{t})=\det(T)\)\(\mathrm{Pf}(S)\). For \(n=1\) we get  
\[b_{1234}=\mathrm{Pf} \begin{pmatrix}0&a_{12}&a_{13}&a_{14}\\ -a_{12}&0&a_{23}&a_{24}\\ -a_{13}&-a_{23}&0&a_{34}\\ -a_{14}&-a_{24}&-a_{34}&0\end{pmatrix} =a_{12}a_{34}-a_{13}a_{24}+a_{14}a_{23}.\]  
Similarly, for \(n=2\) we get  
\[b_{123456}= a_{12}a_{34}a_{56}-a_{12}a_{35}a_{46}+a_{12}a_{36}a_{45}-a_{13}a_{24 }a_{56}+a_{13}a_{25}a_{46}-a_{13}a_{26}a_{45}+\] \[a_{14}a_{23}a_{56}-a_{14}a_{25}a_{36}+a_{14}a_{26}a_{35}-a_{15}a_ {23}a_{46}+a_{15}a_{24}a_{36}-a_{15}a_{26}a_{34}+\] \[a_{16}a_{23}a_{45}-a_{16}a_{24}a_{35}+a_{16}a_{25}a_{34}=0.\]  
Denote \(\boldsymbol{b}=\{b_{i_{1}i_{2}\ldots i_{2n+1}i_{2n+2}}:1\leq i_{1}<i_{2}< \cdots<i_{2n+1}<i_{2n+2}\leq m\}\). Theorem 3.1: _The algebra of \(G\)-invariants is generated by \(\boldsymbol{a}\) with syzygies \(\boldsymbol{b}\):_  
\[\mathcal{A}_{G}^{m}=\langle\boldsymbol{a}\,|\,\boldsymbol{b}\rangle.\]  
Proof. : Let us first prove that the invariants \(a_{ij}\) generate the field \(\mathcal{F}_{G}^{m}\) of rational invariants for \(m=2n\). We use the symplectic analog of Gram-Schmidt normalization: given points \(A_{1},\ldots,A_{2n}\) in general position, we normalize them using \(G=\mathrm{Sp}(2n,\mathbb{R})\) as follows. Let \(e_{1},\ldots,e_{2n}\) be a symplectic basis of \(V\), i.e., \(\omega(e_{2k-1},e_{2k})=1\) and \(\omega(e_{i},e_{j})=0\) else. At first \(A_{1}\) can be mapped to the vector \(e_{1}\). The point \(A_{2}\) can be mapped to the line \(\mathbb{R}e_{2}\), and because of \(\omega(OA_{1},OA_{2})=a_{12}\) it is mapped to the vector \(a_{12}e_{2}\). Next in mapping \(A_{3}\) we have two constraints \(\omega(OA_{1},OA_{3})=a_{13}\), \(\omega(OA_{2},OA_{3})=a_{23}\), and the point can be mapped to the space spanned by \(e_{1},e_{2},e_{3}\) satisfying those constraints. Continuing like this, we arrive to the following matrix with columns \(OA_{i}\):  
\[ \begin{pmatrix}1&0&-\frac{a_{23}}{a_{12}}&-\frac{a_{24}}{a_{12}}&\ldots&-\frac {a_{23}a_{12}}{a_{12}}&-\frac{a_{23}a_{12}}{a_{12}}\\ 0&a_{12}&a_{13}&a_{14}&\ldots&a_{1,2n-1}&a_{1,2n}\\ 0&0&1&0&\ldots&*&*\\ 0&0&0&\frac{b_{1234}}{a_{12}}&\vdots&*&*\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\ldots&1&0\\ 0&0&0&\ldots&0&a_{2n-1,2n}\end{pmatrix}