Let $\mathbb F_q$ be a finite field with $q$ elements where $q$ is a power of a prime $p$. Consider the polynomial ring

$$\mathbb F[x_{ij}],$$ for $i,j=1,\ldots ,n$. Let $f$ and $g$ be polynomils in $\mathbb F[x_{ij}]$ defined by $$f=\sum_{\sigma\in S_n}sgn(\sigma)x_{1\sigma(1)}\ldots x_{n\sigma(n)}$$ and $$g=\sum_{\sigma\in S_n}x_{1\sigma(1)}\ldots x_{n\sigma(n)},$$ where $S_n$ denotes the set of all permutations of the set {1, 2, . . . , n} and $sgn$ is the signum of the permutation $\sigma$.

Let $X$ and $Y$ be the projective homogeneous varieties defined by $f$ and $g$, respectively.

how can I prove (I feel that it's true) that the varieties $X$ and $Y$ have not the same number of points?

  • $\begingroup$ A couple of comments: 1. Should the symbols a_{ij} in the definitions of f and g be x_{ij}? 2. You probably want to exclude the case p=2. – Artie Prendergast-Smith 0 secs ago $\endgroup$ – user5117 Apr 2 '12 at 20:44
  • $\begingroup$ 1. Yes, there was an error. Thanks for the correction. 2. Yes, p is a odd prime. If p=2 the two varieties have the same number of points. $\endgroup$ – miguel Apr 2 '12 at 20:53
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    $\begingroup$ If $n=2$, then the automorphism of ${\bf P}^3$ given by $x_{ij}\mapsto x_{ij}$ is $(ij)\not=(12)$ and $x_{ij}\mapsto -x_{ij}$ is $(ij)=(12)$ sends the hypersurface defined by $f$ onto the hypersurface defined by $g$. In particular, they will have the same number of points (tell if if I misunderstood something basic). $\endgroup$ – Damian Rössler Apr 2 '12 at 21:25
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    $\begingroup$ I claim that we can explicitly compute the number of points in $f$, since it's just the complement of $GL_n$. $GL_n$ is well-known to have $\prod_{j=1}^n (x^n-x^j)$ points. Your variety has $x^{n^2}$ minus that product number of points. $\endgroup$ – Will Sawin Apr 2 '12 at 21:43
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    $\begingroup$ It would be more appropriate to denote f by "det" and g, the permanent, by "per" or something similar. $\endgroup$ – Ralph Apr 2 '12 at 22:05

In a recent work http://arxiv.org/abs/1003.1984 the authors considered the sets $$P_n(\mathbb{F}) = \lbrace A \in \mathbb{F}^{n \times n} \mid g(A) = 0 \rbrace$$ $$D_n(\mathbb{F}) = \lbrace A \in \mathbb{F}^{n \times n} \mid f(A) = 0 \rbrace$$

They showed $|P_3(\mathbb{F})| = |D_3(\mathbb{F})| - q^2(q-1)^5 < |D_3(\mathbb{F})|$.

In case $n> 3$ they showed that there is a constant $k=k(n)$ such that for all $\mathbb F$ with $q > k$, $|P_n(\mathbb{F})| < |D_n(\mathbb{F})|$ holds.

Consequently, the same is true for $|V(g)|= |P_n(\mathbb{F})|/(q-1)$ and $|V(f)|= |D_n(\mathbb{F})|/(q-1)$.

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  • $\begingroup$ Thank you. Now I'm trying to determine the exact number of points of the varietie defined by g , that is, permanent. $\endgroup$ – miguel Apr 3 '12 at 9:03

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