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I have asked the following question in math stackexchange(https://math.stackexchange.com/questions/4389626/fourier-transform-of-inverse-of-determinant-of-1-skew-symmetric-matrix), but did not receive a satisfactory answer. So I try to put the question here and in case it is not of research level, I will delete the question immediately.

This is a specific question concerning the Fourier transform of certain determinant function of a skew-symmetric matrix: Write $SSym(n)$ for the space of $n\times n$ skew-symmetric real matrices. I want to evaluate the Fourier transform of the function $F(x):=\mathrm{det}(1+x)^{-1}$ for $x\in SSym(n)$, in other words, for $y\in SSym(n)$, what is the following integral?

$$\widehat{F}(y):=\int_{SSym(n)}e^{-2i\mathrm{Tr}(x\cdot\,^{\mathrm{t}}y)}\frac{1}{\mathrm{det}(1+x)}dx$$

For the special case $n=2$, this is easy: write $y=\begin{pmatrix}0 & b \\-b & 0\end{pmatrix}$, then it is easy to see $\widehat{F}(y)=e^{-4|b|}$(perhaps up to some explicit non-zero constant multiple).

Note that $\mathrm{det}(1+x)^2=\mathrm{det}(1+x)\mathrm{det}(1+\,^{\mathrm{t}}x)=\mathrm{det}(1+x\cdot\,^{\mathrm{t}}x)>0$ because $1+x\cdot\,^{\mathrm{t}}x$ is always positive definite. So $F(x)$ is well-defined everywhere and a smooth function on $SSym(n)$. Concerning the convergence of the integral, one can argue as follows: each skew-symmetric matrix $x$ is of the form $x=\,^{\mathrm{t}}AXA$ for some orthogonal matrix $A\in\mathrm{O}(n)$ and $X$ is a block-diagonal matrix, whose diagonal blocks are of the form $\begin{pmatrix}0 & b \\-b & 0\end{pmatrix}$(the converse is clearly true). Moreover $\mathrm{det}(1+x)=\mathrm{det}(1+X)$. Write $S$ for the set of all such $X$. Then we can rewrite $\widehat{F}(y)$ as follows:

$$\widehat{F}(y)=\int_{S}\frac{1}{\mathrm{det}(1+X)}dX\int_{\mathrm{O}(n)}e^{-2i\mathrm{Tr}(\,^{\mathrm{t}}AXA\,^{\mathrm{t}}y)}dA$$

(Here the inner integral should be over $\mathrm{O}(n)/\mathrm{O}(2)^{\lfloor n/2\rfloor}$, but we ignore such issue)Now the inner integral is over a compact group and the integrand has bounded absolute value, thus the integral, as a function on $y$ and $X$, is uniformly bounded above. Then the convergence of the integral $\int_S\frac{1}{\mathrm{det}(1+X)}dX$ is reduced to the case $\int_{\mathbb{R}}\frac{1}{1+t^2}dt$ because $\mathrm{det}(1+\begin{pmatrix}0 & t \\-t & 0\end{pmatrix})=1+t^2$. This last integral is clearly convergent.

Of course, maybe one way is to compute first the inner integral $\int_{\mathrm{O}(n)}e^{-2i\mathrm{Tr}(\,^{\mathrm{t}}AXA\,^{\mathrm{t}}y)}dA$, but except for some obvious properties of this function on $X$ and $y$, I could not go further. Moreover it is easy to see $\widehat{F}(\,^{\mathrm{t}}AyA)=\widehat{F}(y)$ for any $A\in\mathrm{O}(n)$, thus we can just assume that $y$ is an element in $S$.

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  • $\begingroup$ You are implicitly using that the integration measure $dx$ is invariant under the action of $O(n)$. To me that is not so obvious if you just separately integrate over the matrix components of $x$. Have you worked out how $dx$ changes under the transformation ${}^{\mathrm{t}} A x A$ when $A$ is not necessarily orthogonal? If you've implicitly somehow defined $dx$ to be $O(n)$ invariant, please clarify that. $\endgroup$
    – Igor Khavkine
    Commented Mar 13, 2022 at 23:13
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    $\begingroup$ @IgorKhavkine good question. Here for the variable skew-symmetric matrix $x=(x_{i,j})_{i,j=1,\cdots,n}$, the differential form $dx$ is given as $dx_{1,2}\wedge dx_{1,3}\cdots dx_{1,n}\wedge dx_{2,3}\cdots$. For a diagonal invertible matrix $A=diag(a_1,\cdots,a_n)$ or a unipotent matrix $A=1_n+E_{I,j}(I\neq j)$, we make a change of variables $x'=\,^tAxA$. In both cases, one can check by direct computation that $dx'=|det(A)^{n-1}|dx$. So this last identity also holds for any invertible matrix $A$. $\endgroup$
    – Zhan
    Commented Mar 14, 2022 at 10:34

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