Let a matrix be $(m+n)$ by $(m+n)$ be all reals with determinant $1.$ Call the matrix $S.$ Name the top $m$ rows $A$ and the next $n$ rows $V.$ Now, $S$ has an inverse $T,$ $ST = I.$ Name the left $m$ columns $W^T$ and the right $n$ columns $B^T.$
We get two Gram matrices, first $G = A A^T$ is $m$ by $m.$ Next, $H = B B^T$ is $n$ by $n.$
All the examples I have done so far have $$\det(G) = \det(H)$$
Theorems:
(I) true when integer entries
(II) true when $S \in SO_n$
(III) true for 2 by 2, that would be $m=n=1$
The question is, is this thing always true with real entries? i have a limited ability to look for counterexamples in low dimension, none so far. I have tried a few examples up to $m+n \leq 4,$ however rational entries with small denominators. I also cannot prove it, it is in the area of Schur complement but I do not see a proof either. I should add that, should the thing be true over the reals, i would expect a proof to have appeared in the (few) places where there is a proof with integer entries.
I did one by hand last night, irrational entries. Let $w = 2 \cos \frac{2 \pi}{9},$ so that $w > 1$ and $w^3 - 3w + 1 = 0.$ The initial matrix with determinant $1$ is
$$ S = \left( \begin{array}{ccc} w & 1 & 1 \\ 1 & w & 1 \\ 1 & 1 & w \\ \end{array} \right) $$ with $$ T = \left( \begin{array}{ccc} w^2-1 & 1-w & 1-w \\ 1 -w & w^2-1 & 1-w \\ 1-w & 1-w & w^2-1 \\ \end{array} \right) $$ From $$ A = ( w,1,1)$$ we get $G = \left( w^2 + 2 \right)$ and $$ \det G = w^2 + 2. $$ Then $$ B = \left( \begin{array}{ccc} 1 -w & w^2-1 & 1-w \\ 1-w & 1-w & w^2-1 \\ \end{array} \right) $$ gives, using relations $w^3 = 3 w - 1$ and $w^4 = 3 w^2 - w,$ $$ H = \left( \begin{array}{cc} 3 w^2 - 5 w + 3 & 3w^2-6w+1 \\ 3w^2 - 6 w + 1 & 3 w^2 - 5 w + 3 \\ \end{array} \right) $$ and, after using the same relations, $$ \det H = w^2 + 2 = \det G $$
Example with integer coefficients I had partly typed up. Detail that could be important, could be irrelevant: here, i did not work out a complete matrix $S$ that contained $A$ as the top two rows. i went directly to putting $AR$ in Hermite form by column operations. Hmmm: with the square matrix $R$ far below, the top two rows of $R^{-1}$ really do turn out to be $A.$ maybe this is not a problem.
The whole point of the exercise is that the determinant of the Gram matrix for $M$ is the same as the determinant of the Gram matrix for $M^\perp \; .$
We begin with a lattice $M$ with basis given by the two rows of $$ A = \left( \begin{array}{ccccc} 1&1&0&0&2 \\ 1&1&1&0&3 \end{array} \right) $$ The inner products give the Gram matrix $G = A A^T$ as $$ G = \left( \begin{array}{cc} 6&8 \\ 8&12 \end{array} \right) $$ This gives $$ G^{-1} = \left( \begin{array}{cc} \frac{3}{2}&-1 \\ -1&\frac{3}{2} \end{array} \right) $$ This allows us to define a basis with rational elements, for the dual lattice $M^\ast,$ given by $A^\ast = G^{-1}A.$ The dual lattice is integer valued linear functionals on $M,$ but we can expres it using rational vectors. $$ A^\ast = \left( \begin{array}{ccccc} \frac{1}{2}&\frac{1}{2}&-1&0&0 \\ -\frac{1}{4}&-\frac{1}{4}&\frac{3}{4}&0&\frac{1}{4} \end{array} \right) $$ We have, automatically, $A^\ast A^T = I$
We also need the Smith Normal Form of $G,$ in symbols $USV = G.$ this one came out with $U = I$ and $$ S = \left( \begin{array}{cc} 2&0 \\ 0&4 \end{array} \right) , $$ $$ V = \left( \begin{array}{cc} 3&4 \\ 2&3 \end{array} \right) . $$
This allows us to give an improved basis for $A^\ast,$ while keeping $A$ because $U^{-1} = I$ and $U^{-1}A=A.$ Let vectors $x_1, x_2$ be the rows of $$ VA^\ast = \left( \begin{array}{ccccc} \frac{1}{2}&\frac{1}{2}&0&0&1 \\ \frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&\frac{3}{4} \end{array} \right) $$ as in the proof of Lemma (2.3) in\cite{looijenga}, while vectors $a_1, a_2$ are the rows of $$ U^{-1}A = \left( \begin{array}{ccccc} 1&1&0&0&2 \\ 1&1&1&0&3 \end{array} \right) . $$ Note that the basis vectors are aligned. We have $a_1 = 2 x_1$ and $a_2 = 4 x_2.$ The quotient group $M^\ast / M $ is therefore $Z/2Z \oplus Z/4Z.$ This follows the method on page 36 of Newman~\cite{newman}. We also need vectors $z_1, z_2$ in the ambient lattice such that $z_i \cdot a_j = x_i \cdot a_j. $ This was easy enough, I took $z_1, z_2$ as the rows of $$ Z = \left( \begin{array}{ccccc} 0&1&0&0&1 \\ 1&1&1&0&0 \end{array} \right) $$
We are finally ready to solve the linear Diophantine system $A B^T = 0,$ where $B$ will be a basis matrix for the full rank lattice $M^\perp.$
Alright, our lattice $M$ is primitively embedded. We take the definition to be that the matrix $A$ can be completed to become a square matrix of integers with determinant $1.$ What we want is to follow\cite{gilbert} and use column operations to force $A$ into Hermite Normal Form. We find integer matrix $R$ such that $\det R = 1,$
$$ R = \left( \begin{array}{ccccc} 1&0&1&0&-2 \\ 0&0&-1&0&0 \\ -1&1&0&0&-1 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ \end{array} \right). $$ we get $$ AR = \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&1&0&0&0 \end{array} \right) $$
So, what is a basis for $M^\perp \; ?$ We know that an integer column vector $X$ that solves $ARX =0$ can be any $$ X = \left( \begin{array}{c} 0 \\ 0 \\ p \\ q \\ r \\ \end{array} \right) $$ Thus $RX$ can be any integer linear combination of the three right columns of $R.$ Writing as rows, we find a basis of $M^\perp \;$ given by $$ B = \left( \begin{array}{ccccc} 1&-1&0&0&0 \\ 0&0&0&1&0 \\ -2&0&-1&0&1 \end{array} \right) $$
The Gram matrix for $B$ is $$ H = \left( \begin{array}{ccc} 2&0&-2 \\ 0&1&0 \\ -2&0&6 \end{array} \right) $$ which has determinant $8,$ same as $\det G.$ Who Knew?