Let S(2n,R) denote the space of real symmetric 2n× 2n matrices, let J be the standard symplectic matrix, and let rho be the symplectic-group action on S(2n,R). For 0≤ k≤ n-1, let s_k denote the n invariants described in Conjecture 1. Orbit-space classification conjecture. If A,B∈ S(2n,R) satisfy A≠ cJ quadfor all c∈R,qquad B≠ cJ quadfor all c∈R, and s_k(A)=s_k(B)quadfor all 0≤ k≤ n-1, then A and B are equivalent under the group action rho; equivalently, these n invariants determine the orbit space S(2n,R)/operatornameSp(2n). This conjecture proposes that the stated invariants completely classify the relevant symplectic-group orbits, extending the observed classification in dimension four; its resolution would clarify the general linear orbit problem.

Orbit-space classification conjecture for symmetric matrices under the symplectic group

From papers

Let $S(2n,\mathbb{R})$ denote the space of real symmetric $2n\times 2n$ matrices, let $J$ be the standard symplectic matrix, and let $\rho$ be the symplectic-group action on $S(2n,\mathbb{R})$ . For $0\leq k\leq n-1$ , let $s_k$ denote the $n$ invariants described in Conjecture 1. Orbit-space classification conjecture. If $A,B\in S(2n,\mathbb{R})$ satisfy

A\neq cJ\quad\text{for all }c\in\mathbb{R},\qquad B\neq cJ\quad\text{for all }c\in\mathbb{R},

and

s_k(A)=s_k(B)\quad\text{for all }0\leq k\leq n-1,

then $A$ and $B$ are equivalent under the group action $\rho$ ; equivalently, these $n$ invariants determine the orbit space $S(2n,\mathbb{R})/\operatorname{Sp}(2n)$ . This conjecture proposes that the stated invariants completely classify the relevant symplectic-group orbits, extending the observed classification in dimension four; its resolution would clarify the general linear orbit problem.

Progress summary

Open

The conjecture that a short list of quantities completely determines these symplectic orbits remains unproved, and the source may concern skew-symmetric rather than symmetric matrices.

The relevant paper formulates Conjecture 1 for invertible real skew-symmetric matrices under symplectic congruence, whereas the supplied statement says symmetric matrices. It asserts that the invariants $s_k$ completely determine the orbit whenever neither matrix is a scalar multiple of $J$ , but presents this only as a conjecture.

Known results

For $n=2$ , the paper states that the Pfaffian and the sum function completely classify the orbits.
For general $n$ , the invariants $s_k$ , $0\leq k\leq n-1$ , are defined and proposed as complete, but no proof is given.

Current status (as of August 2026): The two-dimensional classification is reported as known, while the general conjecture has no verified proof or resolution in the retrieved sources.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Luchen Shi, Sunay Joshi and Ritwick Bhargava, “Congruence of Linear Symplectic Forms by the Symplectic Group”, arXiv:2212.08360 (2022).

1 week ago

In the primary source, $S(2n,\mathbb R)$ denotes invertible skew-symmetric matrices representing symplectic forms; the word “symmetric” in the problem description is a transcription error.

The original conjecture is false already in dimension six. Put

J_2= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}, \qquad J=\operatorname{diag}(J_2,J_2,J_2).

Consider

A_0= \begin{pmatrix} 0&1&0&0&0&0\\ -1&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&-1&0&0&0\\ 0&0&0&0&0&2\\ 0&0&0&0&-2&0 \end{pmatrix},

and

A_1= \begin{pmatrix} 0&1&0&0&0&0\\ -1&0&-1&0&0&0\\ 0&1&0&1&0&0\\ 0&0&-1&0&0&0\\ 0&0&0&0&0&2\\ 0&0&0&0&-2&0 \end{pmatrix}.

Both matrices are skew-symmetric,

\det A_0=\det A_1=4,

and neither is a scalar multiple of $J$ . Their complete Pfaffian polynomials agree:

\operatorname{Pf}(tJ+A_0) = \operatorname{Pf}(tJ+A_1) = (t+1)^2(t+2) = t^3+4t^2+5t+2.

Thus all three proposed invariants $s_0,s_1,s_2$ coincide.

Nevertheless, define

K_i=J^{-1}A_i.

Direct calculation gives

\dim\ker(K_0-I_6)=4, \qquad \dim\ker(K_1-I_6)=2.

Indeed,

K_0=\operatorname{diag}(1,1,1,1,2,2),

whereas the eigenvalue-one part of $K_1$ has nontrivial Jordan blocks.

If the forms were symplectically equivalent, some $P$ would satisfy

P^{\mathsf T}JP=J, \qquad A_1=P^{\mathsf T}A_0P.

The first identity implies

J^{-1}P^{\mathsf T}=P^{-1}J^{-1}.

Consequently,

K_1 = J^{-1}P^{\mathsf T}A_0P = P^{-1}J^{-1}A_0P = P^{-1}K_0P,

contradicting the unequal eigenspace dimensions. Hence the two forms have identical proposed invariants but lie in different symplectic-congruence orbits.

The proposed invariants cannot classify orbits because they do not detect the Jordan structure.

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Shivam Patel · 1 day ago