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Fix a $k\times k$ positive definite symmetric real matrix $Y=(Y_{pq})$. Denote by $I_p$ the $p\times p$ identity matrix, $0_{p\times q}$ the $p\times q$ zero matrix, and $\delta_{pq}$ the Kronecker delta. Now let $i,j=1,...,k-1$. Define $B_{ij}$ to be the $(k-i)\times (k-j)$ (block) matrix
$$ \begin{pmatrix} 0_{(k-i)\times(i-j-1)} & -(Y_{pj})_{p=i+1,...,k-1} & Y_{ij}I_{k-i} \end{pmatrix} $$ if $i>j$, and $$ Y_{ij}I_{k-i} +(Y_{pq})_{p,q=i+1,...,k}$$ if $i=j$, and define $B_{ij}:=B_{ji}^T$ if $i<j$. Note that $Y_{pq}$ are real numbers while $B_{ij}$ are matrices.

The problem is to find the determinant of the $k(k-1)/2\times k(k-1)/2$ (block) matrix $$(B_{ij})_{i,j=1,...,k-1}.$$

For example when $k=4$, $M_4$ is the $(3+2+1)\times (3+2+1)$ matrix $$\begin{pmatrix} Y_{11}+Y_{22} & Y_{23} & Y_{24}&-Y_{31}&-Y_{41}&0\\ Y_{32}&Y_{11}+Y_{33}&Y_{34}&Y_{12}&0&-Y_{14}\\ Y_{42}&Y_{43}&Y_{11}+Y_{44}&0&Y_{12}&Y_{13}\\ -Y_{31}&Y_{21}&0&Y_{22}+Y_{33}&Y_{34}&-Y_{24}\\ -Y_{41}&0&Y_{21}&Y_{43}&Y_{22}+Y_{44}&-Y_{23}\\ 0&-Y_{41}&Y_{31}&-Y_{42}&Y_{32}&Y_{33}+Y_{44}\\ \end{pmatrix}.$$

Motivation. This monster determinant arises from finding the $mk-k(k+1)/2$ dimensional volume of the manifold $\{X\textrm{ is an }m\times k\textrm{ real matrix}:X^TX=Y\}$. Namely, this volume divided by the volume of $\{X\textrm{ is an }m\times k\textrm{ real matrix}:X^TX=I_k\}$ (which is known), equals $$\sqrt{\det M_k}\sqrt{\det Y^{m-k}}/2^{k(k-1)/4}.$$ This has been checked to be consistent with the case $k=1$, that is just a sphere, and the case $k=m=2$.

Partial results. Let $M_k$ be the matrix concerned. $\det M_2=trace (Y)$ is easy, and after some computation we have $\det M_3=a_0-a_1a_2$ where $a_i$ is the coefficient of $x$ in the characteristic polynomial $\det(xI-Y)$. It is thus expected the answer can be written in terms of $a_i$, or even better, the zonal polynomial of $Y$.

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    $\begingroup$ I think I can see the construction here, but I feel like an example illustrating the $k=3$ or $k=4$ cases would help substantially. $\endgroup$ Aug 26, 2018 at 16:09
  • $\begingroup$ @Stadnicki: The case k=4 is now given. $\endgroup$
    – JSCB
    Aug 27, 2018 at 4:30

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