Given a prime power $q$, I would like to enumerate (preferably up to isomorphism*) all the permutation polynomials $f(x)$ on $K = GF(q^3)$ satisfying the following conditions:

  1. $f(ax) = af(x)$ for all $a \in GF(q)$,
  2. $Tr_{K/F}(x f(x)) = 0$. ($F := GF(q)$)

Such polynomials correspond to perfect matchings in the incidence graph of the projective plane $PG(2,q)$ as follows: Look at $GF(q^3)$ as a $3$-dimensional vector space over $GF(q)$ equipped with the symmetric bilinear form $(x,y) \mapsto Tr_{K/F}(xy)$. Let the points of the projective plane be $1$-dimensional subspaces. And let the lines be given by the $2$-dimensional null spaces of the maps $x \mapsto Tr(ax)$, i.e. $a^\perp$, for a given $a \in GF(q^3)$. Therefore, a point $p$ is incident with a line $L$ if $Tr(x_p x_L) = 0$ where $x_p$ and $x_L$ are corresponding members of $GF(q^3)$. Now, every permutation polynomial on $GF(q^3)$ defined by the two conditions above maps $1$-dimensional subspaces to $1$-dimensional subspaces giving rise to a bijection between points and lines of $PG(2,q)$, i.e., a perfect matching in the incidence graph of $PG(2,q)$. We can see that this is an onto map where fiber of each perfect matching is a set of $(q-1)^{1+q+q^2}$ permutation polynomials.

By [1] a $k$-regular bipartite graph of size $2n$ has at least $$\left( \frac{(k-1)^{(k-1)}}{k^{(k-2)}} \right)^n$$ perfect matchings. Therefore there is a lower bound on the number of perfect matchings given by $\frac{q^{qn}}{(q+1)^{n(q-1)}}$ where $n = 1 + q + q^2$.

I also know that for for $q = 3$ there are exactly $3852$ perfect matchings giving rise to $31555584 = 3852 \times 2^{13}$ such permutations. This is from computer computations of the perfect matchings. For $q = 4$ there are $18534400$ of them.

To the finite field experts: Is there a way of theoretically classifying such polynomials? Can we give some bounds on the total number? Any suggestions on what approach might work?

*Look at the action of $\Gamma L(3,q)$ on $GF(q^3)$ viewed as a vector space over $GF(q)$. For a permutation polynomial $f$ and a group element $\sigma$ define $f^\sigma (x) = \hat{\sigma} (f(\sigma(x))$ where $\hat{\sigma}$ is the adjoint of $\sigma$ w.r.t. the symmetric bilinear form $(x, y) \mapsto Tr_{K/F}(xy)$.

[1] http://homepages.cwi.nl/~lex/files/countpms2.pdf

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  • $\begingroup$ Edited. Is it clear now? $\endgroup$ – Anurag May 26 '14 at 20:29
  • $\begingroup$ The number of perfect matchings in the incidence graph of $PG(2,3)$ is certainly $3852$. You can also check it here: oeis.org/A000794. So, there might be some error in the interpretation of a perfect matching as a permutation polynomial. I would check it again and see if I can find the error. $\endgroup$ – Anurag May 26 '14 at 21:24
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    $\begingroup$ I want to clarify that the current (7th) version of this question is internally inconsistent. For $q=3$ there are 31555584 permutations $f$ of $GF(q^3)$ which satisfy conditions (1) and (2). The OP's assertion that there are only 3852 such permutations is wrong. It appears that the OP did not compute these permutations, but got the number 3852 from the number of perfect matchings in the incidence graph of $PG(2,3)$. So the real question is to find the mistake in the OP's proof (given in version 7 of the question) that the number of such permutations equals the number of perfect matchings. $\endgroup$ – Michael Zieve May 26 '14 at 22:43
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    $\begingroup$ $31555584/2^{13}$ = 3852. I think for $q = 3$ each perfect matching corresponds $2^{(1+3+3^2)}$ such polynomials because for every pair $U$, $V$ of $1$-dimensional subspaces with $f(U) = f(V)$ we can take any permutation of the non-zero elements of $V$ to define another permutation polynomial corresponding to the same perfect matching and since there are $13$ $1$-dimensional subspaces we have that factor. In general I'll have to be careful about this and I think I should re-formulate the conditions. (Thank you for pointing it out and I apologise for so many errors) $\endgroup$ – Anurag May 26 '14 at 23:49
  • $\begingroup$ @MichaelZieve: I have edited the question details again. Hopefully this is the last edit. $\endgroup$ – Anurag May 27 '14 at 1:19

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