2
$\begingroup$

This question is related to the matrices described in Deyi Chen's recent MO post (look at some examples there). The main difference: we are asking for a determinant evaluation instead of a permanent, plus we have added variables. Define the sequence $$f_n=\mathrm{det}\left[x_{i-j}\cdot\operatorname{sgn} \left(\tan\frac{(i+j)\pi}{2n+1} \right)\right]_{1\le i, \,j\le 2n},$$ where $n\geq1$ and $\operatorname{sgn}$ is the sign-function; i.e. $\operatorname{sgn}(y)=1$ if $y>0$; $\operatorname{sgn}(y)=-1$ if $y<0$; $\operatorname{sgn}(0)=0$.

If we set all $x_k=1$, user44191 offers an alternative description of the matrix: $n−1$ antidiagonals of $1$'s, $n$ antidiagonals of $-1$'s, an antidiagonal of $0$'s, $n$ antidiagonals of $1$'s, and $n-1$ antidiagonals of $-1$'s, in the given order.

QUESTION 1. Is this true? The determinant equals $f_n=(-1)^nP(\dots,x_{-2},x_{-1},x_0,x_1,x_2,\dots)^2$ for some squared polynomial $P$ of several variables (coefficients in $\mathbb{Z}$) so that each of its monomials is of the form $x_{i_1}x_{i_2}\cdots x_{i_m}$ with $i_1+i_2+\cdots+i_m=0$.

QUESTION 2 (specialization). Is this true? If we set $x_k=1$, for all $k$, then $f_n=(-1)^n$.

Example. For instance, $$f_2= \det\left[ \begin {array}{cccc} x_0&-x_{-1}&-x_{-2}&0\\ -x_1&-x_0&0&x_{-2} \\ -x_2&0&x_0&x_{-1}\\ 0&x_2&x_1&-x_0 \end {array} \right] =(x_{-2}x_2-x_{-1}x_1-x_0^2)^2.$$

Example. The matrix for $n=3$ and its determinant: $$ \left[ \begin {array}{cccccc} x_0&x_{-1}&-x_{-2}&-x_{-3}&-x_{-4}&0\\ x_1&-x_0&-x_{-1}&-x_{-2}&0&x_{-4}\\ -x_2&-x_1&-x_0&0&x_{-2}&x_{-3}\\ -x_3&-x_2&0&x_0&x_{-1}&x_{-2}\\ -x_4&0&x_2&x_1&x_0&-x_{-1}\\ 0&x_4&x_3&x_2&-x_1&-x_0 \end {array} \right], $$ \begin{align*} f_3&=-(x_{-4}x_0x_4 - x_{-4}x_{1}x_3 + x_{-4}x_2^2 - x_{-3}x_{-1}x_4 + x_{-3}x_0x_3 \\ & \qquad + x_{-3}x_1x_2 + x_{-2}^2x_4 + x_{-2}x_{-1}x_3 - 2x_{-2}x_0x_2 \\ & \qquad - x_{-2}x_1^2 - x_{-1}^2x_2 - x_0^3)^2. \end{align*}

$\endgroup$
4
  • $\begingroup$ In my preprint On some determinants involving the tangent function posted to arXiv two years ago (see arxiv.org/abs/1901.04837) , I proved that $$\det\left[\tan\pi\frac{j+k}{2n+1}\right]_{1\le j,k\le 2n}=(-1)^n(2n+1)^{2n-1}.$$ The method there might be helpful to your questions. $\endgroup$ Aug 28, 2021 at 0:32
  • $\begingroup$ Determinants are relatively easier to handle than permanents since there are many techniques for evaluating determinants. $\endgroup$ Aug 28, 2021 at 0:43
  • $\begingroup$ Note: You can replace the $m$ in Question 1 by an $n$. $\endgroup$ Aug 28, 2021 at 0:53
  • $\begingroup$ Question 2 seems easy: For $\sigma\in S_{2n}$ let $\bar{\sigma}(j)=2n+1-\sigma(j)$ for all $j=1,\ldots,2n$. Then $$\mathrm{sign}(\sigma)\times\mathrm{sign}(\bar{\sigma})=(-1)^{2n(2n+1)/2}=(-1)^n.$$ So it suffices to show that $$\det\left[\mathrm{sgn}\left(\tan\pi\frac{j-k}{2n+1}\right)\right]_{1\le j,k\le 2n}=1.$$ This should be easy as the determiant is close to a circular determinant. $\endgroup$ Aug 28, 2021 at 1:56

1 Answer 1

4
$\begingroup$

to Question 1: Yes.

To prove this, let me fix a positive integer $n$ and denote your matrix (whose determinant $f_{n}$ is) by $A$. The notation $\left[ k\right] $ shall be used for the set $\left\{ 1,2,\ldots,k\right\} $ whenever $k$ is an integer. The notation $M_{i,j}$ will be used for the $\left( i,j\right) $-th entry of any matrix $M$. Thus, \begin{equation} A_{i,j}=x_{i-j}\cdot\operatorname*{sgn}\left( \tan\dfrac{\left( i+j\right) \pi}{2n+1}\right) \label{eq.darij1.1} \tag{1} \end{equation} for any $i,j\in\left[ 2n\right] $.

Let $B$ be the $2n\times2n$-matrix obtained by "turning $A$ upside down", i.e., reversing the order of the rows of $A$. Explicitly, this means that \begin{equation} B_{i,j}=A_{2n+1-i,j}\qquad\text{for all }i,j\in\left[ 2n\right] . \label{eq.darij1.2} \tag{2} \end{equation} We note that $B$ can be obtained from $A$ by $n$ row-swaps (i.e., by $n$ steps, where each step swaps a pair of rows). Indeed, all we need to do is to swap the $1$-st and the last row, then to swap the $2$-nd and the $2$-nd-to-last row, etc., until we reach the middle of the matrix. Since each of these swaps multiplies the determinant by $-1$, this entails that \begin{equation} \det B=\left( -1\right) ^{n}\det A. \label{eq.darij1.3} \tag{3} \end{equation}

Now, I claim that the matrix $B$ is alternating -- i.e., that \begin{equation} B_{i,i}=0\qquad\text{for all }i\in\left[ 2n\right] \label{eq.darij1.4} \tag{4} \end{equation} and \begin{equation} B_{i,j}=-B_{j,i}\qquad\text{for all }i,j\in\left[ 2n\right] . \label{eq.darij1.5} \tag{5} \end{equation}

Indeed, in order to prove \eqref{eq.darij1.4}, it suffices to observe that \begin{align*} B_{i,i} & =A_{2n+1-i,i}=x_{\left( 2n+1-i\right) -i}\cdot\operatorname*{sgn} \left( \tan\dfrac{\left( \left( 2n+1-i\right) +i\right) \pi} {2n+1}\right) \\ & =x_{2n+1-2i}\cdot\underbrace{\operatorname*{sgn}\left( \tan\dfrac{\left( 2n+1\right) \pi}{2n+1}\right) }_{=\operatorname*{sgn}\left( \tan\pi\right) =\operatorname*{sgn}0=0}=0. \end{align*} The proof of \eqref{eq.darij1.5} is not much harder (using the fact that $\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\pi-\dfrac{\left( i-j\right) \pi}{2n+1}$ and therefore \begin{align} \tan\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\tan\left( \pi-\dfrac{\left( i-j\right) \pi}{2n+1}\right) =-\tan\dfrac{\left( i-j\right) \pi}{2n+1}, \end{align} and furthermore $\tan$ is an odd function).

Thus, we know that the matrix $B$ is alternating. Hence, as for any alternating $2n\times2n$-matrix, its determinant is the square of its Pfaffian. In other words, \begin{equation} \det B=\left( \operatorname*{Pf}B\right) ^{2}, \label{eq.darij1.6} \tag{6} \end{equation} where $\operatorname*{Pf}B$ denotes the Pfaffian of $B$. The latter Pfaffian is a polynomial in the entries of the matrix with coefficients in $\mathbb{Z} $. Since the entries of the matrix belong to $\mathbb{Z}\left[ \ldots ,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $, we thus conclude that the Pfaffian belongs to $\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1} ,x_{2},\ldots\right] $ as well. In other words, \begin{equation} \operatorname*{Pf}B\in\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1} ,x_{2},\ldots\right] . \label{eq.darij1.7} \tag{7} \end{equation}

Now, \eqref{eq.darij1.3} yields \begin{align} \det A=\left( -1\right) ^{n}\det B=\left( -1\right) ^{n}\left( \operatorname*{Pf}B\right) ^{2} \end{align} (by \eqref{eq.darij1.6}). Because of \eqref{eq.darij1.7}, this shows that $\det A$ equals $\left( -1\right) ^{n}\cdot P^{2}$ for some polynomial $P\in\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $ (namely, for $P=\operatorname*{Pf}B$), exactly as claimed in Question 1.

In order to complete the answer to Question 1, we now need to show that each monomial in $P=\operatorname*{Pf}B$ is of the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ with $i_{1}+i_{2}+\cdots+i_{n}=0$. This can be done in various ways, but the easiest is probably the following: Let us equip the polynomial ring $\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $ with a $\mathbb{Z}$-grading in which each indeterminate $x_{i}$ is homogeneous of degree $i$. Now, recall the explicit formula for the Pfaffian as a sum over all perfect matchings on the set $\left[ 2n\right] $ (see Definition 3 in Michel Goemans, 18.438 in Spring 2014, Lectures 4 and 6, or any good textbook on Pfaffians). If \begin{equation} M=\left\{ \left\{ a_{1},b_{1}\right\} ,\left\{ a_{2},b_{2}\right\} ,\ldots,\left\{ a_{n},b_{n}\right\} \right\} \label{eq.darij1.9o} \tag{9} \end{equation} is such a perfect matching, then the corresponding addend in $\operatorname*{Pf}B$ is \begin{equation} \pm B_{a_{1},b_{1}}B_{a_{2},b_{2}}\cdots B_{a_{n},b_{n}}. \label{eq.darij1.9} \tag{10} \end{equation} Each of the $n$ factors $B_{a_{i},b_{i}}$ in this product can be rewritten as \begin{align} B_{a_{i},b_{i}}=A_{2n+1-a_{i},b_{i}}=x_{\left( 2n+1-a_{i}\right) -b_{i} }\cdot\left( 1\text{ or }-1\text{ or }0\right) , \end{align} and thus (using our weird grading) is homogeneous of degree $\left( 2n+1-a_{i}\right) -b_{i}=2n+1-a_{i}-b_{i}$. Hence, the entire product \eqref{eq.darij1.9} is homogeneous of degree \begin{align*} \sum_{i=1}^{n}\left( 2n+1-a_{i}-b_{i}\right) & =n\left( 2n+1\right) -\underbrace{\sum_{i=1}^{n}\left( a_{i}+b_{i}\right) } _{\substack{=1+2+\cdots+2n\\\text{(since \eqref{eq.darij1.9o} is a}\\\text{perfect matching of }\left[ 2n\right] \text{)}}}\\ & =n\left( 2n+1\right) -\left( 1+2+\cdots+2n\right) =0. \end{align*} This means that this product is a $\mathbb{Z}$-linear combination of monomials of the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ with $i_{1}+i_{2} +\cdots+i_{n}=0$. Clearly, the same must therefore holds for the polynomial $\operatorname*{Pf}B$ (since this polynomial is a sum of such products). This concludes the answer to Question 1.

Answering Question 2 requires proving that $\det B=1$ when all $x_{i}$ are set to $1$. This should be easy given that $\operatorname*{sgn}\left( \tan \dfrac{\left( i+j\right) \pi}{2n+1}\right) $ can be explicitly computed (and the matrix $B$ becomes a circulant when all $x_{i}$ are $1$); but it's late here and I have too many things on my list until the quarter begins. Sorry!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.