Help needed with evaluating a matrix delta-function. I am reading the following paper (https://arxiv.org/abs/1008.2075) and I am struggling to verify Equation 1. They have a matrix delta function and I believe they are claiming that
$$\delta\left(\sqrt{1 - A^T A} \,  \left(C^T C - 1\right) \sqrt{1 - A^T A}\right) = \left| \det(1- A^T A)\right|^{-\frac{M+1}{2}} \, \delta\left(C^T C - 1\right)$$
$A^T A$, $C^T C$ and $1$ (identity) are $M \times M$ matrices. $0 < A^T A < 1$ and so this expression is well-defined (by "$A > B$" I mean "$A - B$ is positive definite"). This is really confusing me because it seems to me the correct factor is $\left| \det(1- A^T A)\right|^{-M}$ and not $\left| \det(1- A^T A)\right|^{-\frac{M+1}{2}}$.
This is because the matrix delta-function is a delta-function on the space of $M \times M$ matrices, which can be identified with $\mathbb{R}^m \otimes (\mathbb{R}^m)^\ast$, and thus the map $E \mapsto RER^T$ can be canonically identified with $R \otimes R^T$. We have that $\det(R \otimes R^T) = \det(R)^{2M}$. In our case $R = \sqrt{1-A^T A}$. I cannot see where the exponent $\frac{M+1}{2}$ comes from.
 A: The mistake I am making is the following. With a matrix delta function, this should be interpreted as a product over independent matrix elements. $C^T C - 1$ is clearly symmetric and so this shouldn't be a product over all matrix elements, only the $i,j$ elements with $i \leq j$. We need the following lemma.
For $A$,$B$ symmetric $n \times n$ matrices, the product $ABA$ is also symmetric. Then we want to compute the Jacobian of the map $\mathcal{C}_A : B \mapsto ABA$. We claim that $\det(\mathcal{C}_A) = \det(A)^{n+1}$. If we could show this then the claim in the paper would follow.
Let $\omega$ be a top form on the linear space of symmetric matrices. If $e_i$ is a basis for $\mathbb{R}^n$ then symmetric matrices are spanned by $m_{ij} = e_i e_j^T + e_j e_i^T$, and we could take $\omega = \wedge_{i \leq j}(e_i e_j^T + e_j e_i^T)$.
Then there is natural map $\omega \mapsto \mathcal{C}_A \omega = \wedge_{i \leq j} \mathcal{C}_A m_{ij} = \det(\mathcal{C}_A) \omega$. Clearly $ \mathcal{C}_{A A^\prime} = \mathcal{C}_{A} \circ  \mathcal{C}_{A^\prime}$ and thus $\det(\mathcal{C}_{A A^\prime}) = \det(\mathcal{C}_{A }) \det(\mathcal{C}_{A^\prime})$. Thus we may write without loss of generality $A = U \Lambda U^T$ where $U$ is orthogonal and $\Lambda = \mathrm{diag}(\lambda_1, \dots , \lambda_n)$. Thus $\det(\mathcal{C}_{A}) = \det(\mathcal{C}_{\Lambda})$.
$\mathcal{C}_{\Lambda} m_{ij} = \Lambda m_{ij}\Lambda = \lambda_i \lambda_j m_{ij}$. From this one finds that $\mathcal{C}_{\Lambda} \omega = \left( \prod_{1 \leq i \leq j \leq n} \lambda_i \lambda_j\right) \omega$. Finally we see that $\left( \prod_{1 \leq i \leq j \leq n} \lambda_i \lambda_j\right) = \det(A)^{n+1}$.
