Let $K$ be a rank-2 symmetric complex matrix, such that the transpose $K^T=K$ is itself.

Let $V$ be a rank-2 unitary matrix, $V \in U(2)$.

Consider the identification between any K and K' of any rank-2 symmetric complex matrix, $$ K\sim K', $$ if it satisfies $$ V^T K V =K', $$ for any $V \in U(2).$

Originally, we can parametrize $K$ as $$ K=\begin{pmatrix}k1 & k\\ k& k2\end{pmatrix}, $$ with $$k1,k, k2 \in \mathbb{C}.$$ There are 6 real degrees of freedom in total.

After the identification of the $K\sim K'$, constrained by the $V\in U(2)$ which has 4 real degrees of freedom.

So totally there should be at least 2 real degrees of freedom left for the $d \geq 2$-dimensional space of the $K\sim K'$. (But $d \geq 2$ could be more due to the consideration of stabilizer.)

question:

  1. What is the real dimension of the new space of $K$ (under the $K\sim K'$ and $V^T K V =K'$, for any $V \in U(2)$ condition)?

  2. How do we parametrize this new space of $K$ in terms of a rank-2 matrix (mod out the redundancy under the $K\sim K'$ and $V^T K V =K'$, for any $V \in U(2)$ condition)?

(p.s. This space may be a called an orbifold space(?). i.e. The (orbifold) space of symmetric complex matrix after mod out a relation identifying a unitary matrix.)


According to the Autonne-Takagi factorization, for any symmetric complex matrix $K$ there exists a unitary matrix $V$ with $V^T K V = D$, where $D$ is a diagonal matrix with the non-negative real singular values of $K$ (i.e., the square roots of the eigenvalues of $K^*K$ on the diagonal). This means that your space is indeed real 2-dimensional and can be parametrized as the space of all matrices of the form $\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ with $a,b > 0$, factored by the $\mathbb{Z}/2\mathbb{Z}$ action which permutes $a$ and $b$. Geometrically, this is the first quadrant in the plane, factored by reflection in the line $a=b$.