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

At least 3 years old · documented by 2022

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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The four-dimensional case is known, but an unverified explicit counterexample claims the general classification fails in six dimensions.

The conjecture, posed by Luchen Shi, Sunay Joshi, and Ritwick Bhargava in 2022, says that the invariants $s_k$ classify invertible skew-symmetric forms under symplectic congruence, apart from scalar multiples of $J$ . The primary source corrects the supplied statement’s use of “symmetric” matrices.

Known results

In dimension four, the Pfaffian and $-\frac{1}{2}\operatorname{tr}(JA)$ completely classify the orbits.
In dimension $2n$ , the paper proves that the $n$ invariants $s_k$ , defined from the coefficients of $\operatorname{Pf}(tJ+A)$ , are orbit invariants.
The general completeness assertion remains Conjecture 1 in the primary source.

Posted attempt

An explicit dimension-six pair is claimed to have identical Pfaffian polynomials, hence identical $s_0,s_1,s_2$ , but different eigenspace dimensions for $J^{-1}A$ and therefore distinct symplectic-congruence orbits. This claims a complete counterexample, but it has not been independently verified.

Current status (as of August 2026): The dimension-four classification and invariance of the general $s_k$ are established, while the general conjecture has an unverified dimension-six counterexample claim and no verified resolution.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.