In a (right) finite dimensional quaternionic Hilbert space there is an analogue of the spectral theorem (see theorem 4.6 in Farenick and Pidkowich) for normal matrices in $\mathbb{H}^{m\times m}$, allowing for a convenient notion of rank of a normal matrix over quaternions (the number of positive singular values). Further one can also map matrices over quaternions to matrices over complex numbers of twice the dimension by use of symplectic representations. Vectors have images under the $*$-homomorphism, $\Psi:\mathbb{H}^m\rightarrow \mathbb{C}^{2m}$ given by $$\xi=\xi_{1}+\xi_{2}j \mapsto \begin{pmatrix} \ \ \xi_{1} \\ -\overline{\xi_{2}}\end{pmatrix},$$ where $\xi_{1},\xi_{2}$ are complex quaternions (that is, with zero $j$ and $k$ parts), and the image of (right) linearly independent vectors over $\mathbb{H}$ remain linearly independent over $\mathbb{C}$ under this mapping. Further for matrices over quaternions the mapping $\Phi:\mathbb{H}^{m\times m}\rightarrow\mathbb{C}^{2m\times 2m}$ $$\Gamma_{1}+\Gamma_{2}j\mapsto \begin{pmatrix} \Gamma_{1} & \Gamma_{2} \\ -\overline{\Gamma_{2}} & \overline{\Gamma_{1}} \end{pmatrix}$$ the rank of the image under $\Phi$ of a matrix is doubled.

Without taking consideration the form of matrices in the image of $\Phi$, the set of $2m\times 2m$ matrices of rank at most $2k$ over $\mathbb{C}$ is an irreducible projective algebraic variety of co-dimension $4(m−k)^2$. Is there a way to meaningfully say something about the dimension of the space of hermitian $m\times m$ matrices over quaternions with rank less than or equal to $k$?

Quaternions being non-commutative means that the condition of all size $k+1$ minors vanishing which can define a rank at most $k$ matrix over $\mathbb{C}$ is not handled similarly for $\mathbb{H}$, with the closest thing resembling being 'quasideterminants' (see the paper by Gelfand et al from 2002 ). Does this question make sense in non-commutative algebraic geometry, or is it more natural to come from another perspective?