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Let $p$ be an odd prime, and let $(\frac{\cdot}p)$ denote the Legendre symbol. Motivated by my question http://mathoverflow.net/questions/310301, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^-]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$?

Based on my computation, I conjecture that the question has a positive answer.

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The conjectures are true. If you negate the first row of $A_p^{-}$ you get a cofactor of the matrix in "Chapman's evil determinant". In particular you can get the answer from the same matrix decomposition used by Vsemirnov in his paper: On the evaluation of R. Chapman’s “evil determinant”.

I couldn't deduce the result for $A_p^{+}$ from Vsemirnov's paper so I'm sketching a different approach here. The proof essentially consist of (1) using quadratic Gauss sums to write the matrix as a (rank one perturbation of) a product of shifted Vandermonde matrices, and (2) repeated use of the Lagrange interpolation formula to evaluate various sums of rational functions evaluated at roots of unity. Here are the details:

Throughout this post $p$ is a prime $3\pmod{4}$ and $\zeta=e^{\frac{2\pi i}{p}}$.

Lemma 1: Given $a_1,a_2,\dots,a_{\frac{p-1}{2}},b_1,b_2,\dots,b_{\frac{p-1}{2}}\in \mathbb Z/p\mathbb Z$ we have $$\left[\left(\frac{a_i +b_j}p\right)\right]_{1\le i,j\le \frac{p-1}{2}}=\frac{1}{i\sqrt{p}}\left(J+2M_aM_b^{\top}\right)$$ where $M_a=[\zeta^{a_i k^2}]_{1\le i,k\le \frac{p-1}{2}}$ and $M_b=[\zeta^{b_jk^2}]_{1\le j,k\le \frac{p-1}{2}}$ and $J$ is the matrix with all entries $1$ which can also be written as $\textbf{1}^{\top}\textbf{1}$, with $\textbf{1}$ being the row vector of all $1$'s.

Proof: This follows entry-wise from the quadratic Gauss sum $$\left(\frac{a_i +b_j}p\right)=\frac{1}{i\sqrt p}\left(1+2\sum_{k=1}^{\frac{p-1}{2}}\zeta^{(a_i+b_j)k^2}\right)=\frac{1}{i\sqrt p}\left(1+2\sum_{k=1}^{\frac{p-1}{2}}\zeta^{a_ik^2}\cdot \zeta^{k^2 b_j}\right).$$

Corollary 1: By the matrix determinant lemma, we have $$\det \left[\left(\frac{a_i +b_j}p\right)\right]_{1\le i,j\le \frac{p-1}{2}}=\frac{1}{\left(i\sqrt p\right)^{\frac{p-1}{2}}}\left(2^{\frac{p-1}{2}}\det M_a \det M_b+\textbf{1}^{\top}\textbf{Adj} (2M_aM_b^{\top}) \textbf{1}\right)$$ $$=\frac{1}{\left(i\sqrt p\right)^{\frac{p-1}{2}}}\left(2^{\frac{p-1}{2}}\det M_a \det M_b+2^{\frac{p-3}{2}}\sum_{r=1}^{\frac{p-1}{2}}\det M_a^{(r)}\det M_b^{(r)}\right)$$ Where we use the notation $M^{(r)}$ to denote the matrix obtained from $M$ by replacing the $r$-th row by $\textbf{1}$.


The matrix $A_p^{+}$ corresponds to the choice $(a_1,\dots,a_{\frac{p-1}{2}})=(0,2,3,\dots,\frac{p-1}{2})$ and $(b_1,\dots,b_{\frac{p-1}{2}})=(1,2,\dots,\frac{p-1}{2})$. Computing $\det M_a,\det M_b$ or $\det M_a^{(r)},\det M_b^{(r)}$ for these vectors is an evaluation of Vandermonde determinants or simple instances of Schur polynomials, so I am leaving it as an exercise, for the sake of brevity. What you will end up getting from corollary 1 on these specific vectors, is the following:

$$\det A_p^{+}=$$ $$\frac{\left(\prod_{k_1<k_2} (\zeta^{k_1^2}-\zeta^{k_2^2})\right)^2}{\left(i\sqrt p\right)^{\frac{p-1}{2}}}\left(-2^{\frac{p-3}{2}} \left(i\sqrt{p}+1\right)+2^{\frac{p-3}{2}}\sum_{r=1}^{\frac{p-1}{2}}F_{+}(\zeta^{r^2})\left(\prod_{k\neq r} \frac{1-\zeta^{k^2}}{\zeta^{r^2}-\zeta^{k^2}}\right)^2\right)$$

where $F_{+}(\zeta^{r^2})=\frac{1-i\sqrt p}{2\zeta^{2r^2}}-\frac{1}{\zeta^{3r^2}}$. What follows are some facts that help us simplify this expression:


Lemma 2: We have $\left(\prod_{k_1<k_2}(\zeta^{k_1^2}-\zeta^{k_2^2})\right)^2=(i\sqrt p)^{\frac{p-3}{2}}$.

Proof: This calculation follows from the fact that for any $n\neq 0\pmod p$, the equation $n=x^2-y^2$ has exactly $\frac{p-3}{4}$ solutions with $x,y\in \{1,2,\dots,\frac{p-1}{2}\}$.


Lemma 3 We have the following equalities $$\sum_{r=1}^{\frac{p-1}{2}}\frac{1}{\zeta^{2r^2}}\left(\prod_{k\neq r} \frac{1-\zeta^{k^2}}{\zeta^{r^2}-\zeta^{k^2}}\right)^2=-2$$ $$\sum_{r=1}^{\frac{p-1}{2}}\frac{1}{\zeta^{3r^2}}\left(\prod_{k\neq r} \frac{1-\zeta^{k^2}}{\zeta^{r^2}-\zeta^{k^2}}\right)^2=-2+i\sqrt{p}$$

Proof: For the first equality we will us Lagrange interpolation on the polynomial $G(x)=\frac{1}{x}\left(\prod_{k=1}^{\frac{p-1}{2}} (x-\zeta^{-k^2})-\prod_{k=1}^{\frac{p-1}{2}}(x-\zeta^{k^2})\right)$ at the points $x=\zeta^{1^2},\dots,\zeta^{(\frac{p-1}{2})^2}$. On one hand we have $G(1)=2\prod_{k=1}^{\frac{p-1}{2}}(1-\zeta^{-k^2})$, and on the other $$G(1)=\sum_{r=1}^{\frac{p-1}{2}}\frac{1}{\zeta^{r^2}}\prod_{k=1}^{\frac{p-1}{2}}(\zeta^{r^2}-\zeta^{-k^2})\prod_{k\neq r} \frac{1-\zeta^{k^2}}{\zeta^{r^2}-\zeta^{k^2}}$$ $$=\sum_{r=1}^{\frac{p-1}{2}}\frac{1}{\zeta^{r^2}}\left(\frac{-p\zeta^{(p-1)r^2}}{\prod_{k=1}^{\frac{p-1}{2}}(1-\zeta^{k^2})}\right)\left(\prod_{k\neq r} \frac{1-\zeta^{k^2}}{\zeta^{r^2}-\zeta^{k^2}}\right)^2$$ So we have $$\sum_{r=1}^{\frac{p-1}{2}}\frac{1}{\zeta^{2r^2}}\left(\prod_{k\neq r} \frac{1-\zeta^{k^2}}{\zeta^{r^2}-\zeta^{k^2}}\right)^2=\frac{-2}{p}\prod_{k=1}^{\frac{p-1}{2}}(1-\zeta^{-k^2})(1-\zeta^{k^2})=-2.$$ The second identity can be similarly obtained by applying Lagrange interpolation to $G(x)$ at the points $x=0,\zeta^{1^2},\zeta^{2^2},\dots,\zeta^{(\frac{p-1}{2})^2}$. For this you will need that $$G(0)=\zeta^{-1^2-\cdots-(\frac{p-1}{2})^2}\left(\sum_{k=1}^{\frac{p-1}{2}}\zeta^{-k^2}\right)-\zeta^{1^2+\cdots+(\frac{p-1}{2})^2}\left(\sum_{k=1}^{\frac{p-1}{2}}\zeta^{k^2}\right)$$ $$=\frac{-1-i\sqrt{p}}{2}-\frac{-1+i\sqrt p}{2}=-i\sqrt{p}.$$


Applying the calculations from the previous two lemmas to our expression gives $$\det A_p^{+}=\frac{2^{\frac{p-3}{2}}}{i\sqrt{p}}\left(-i\sqrt{p}-1-2\left(\frac{1-i\sqrt{p}}{2}\right)+2-i\sqrt{p}\right)=-2^{\frac{p-3}{2}}$$ as desired.

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    $\begingroup$ Very good job here. $\endgroup$ – T. Amdeberhan Sep 18 '18 at 2:34

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