2
$\begingroup$

It is well known that the projective plane of order $2$ can be represented by the circulant matrix $M_2:=circ(x,x,1,x,1,1,1)= \begin{pmatrix} x&x&1&x&1&1&1\\ 1&x&x&1&x&1&1\\ 1&1&x&x&1&x&1\\ 1&1&1&x&x&1&x\\ x&1&1&1&x&x&1\\ 1&x&1&1&1&x&x\\ x&1&x&1&1&1&x\\ \end{pmatrix}.$

For the one of order $3$ we can take $M_3:=circ(x,x,1,1,x,1, x, 1, 1, 1,1,1,1)= \begin{pmatrix} x& x& 1& 1& x& 1& x& 1& 1& 1& 1& 1& 1\\ 1& x& x& 1& 1& x& 1& x& 1& 1& 1& 1& 1\\ 1& 1& x& x& 1& 1& x& 1& x& 1& 1& 1& 1\\ 1& 1& 1& x& x& 1& 1& x& 1& x& 1& 1& 1\\ 1& 1& 1& 1& x& x& 1& 1& x& 1& x& 1& 1\\ 1& 1& 1& 1& 1& x& x& 1& 1& x& 1& x& 1\\ 1& 1& 1& 1& 1& 1& x& x& 1& 1& x& 1& x\\ x& 1& 1& 1& 1& 1& 1& x& x& 1& 1& x& 1\\ 1& x& 1& 1& 1& 1& 1& 1& x& x& 1& 1& x\\ x& 1& x& 1& 1& 1& 1& 1& 1& x& x& 1& 1\\ 1& x& 1& x& 1& 1& 1& 1& 1& 1& x& x& 1\\ 1& 1& x& 1& x& 1& 1& 1& 1& 1& 1& x& x\\ x& 1& 1& x& 1& x& 1& 1& 1& 1& 1& 1& x \end{pmatrix}.$

  • Can the incidence structure of a finite projective plane always be written as a circulant matrix?
  • Is this known at least for the Desarguesian planes?
  • Is this circulant matrix essentially unique for a given plane? (i.e. up to cyclic permutation and reflection)

The determinant of this matrix for any projective plane of order $d$ is $$\det M_d=d^{d(d+1)/2}(x-1)^{d(d+1)}[(d+1)x+d^2],$$ but the structure of this expression does not seem to give more hints.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ A projective plane with circulant incidence matrix is exactly equivalent to a cyclic difference set with $\lambda = 1$. In a Desarguesian plane, a Singer cycle acts regularly on points and on lines, so these all have circulant incidence matrices. $\endgroup$
    – Padraig Ó Catháin
    Commented Jan 27, 2018 at 11:39
  • $\begingroup$ @PadraigÓCatháin Thank you, I expected something quite easy about Desarguesian planes (projective geometry is not my specialty). What about uniqueness of the generating vector? $\endgroup$
    – Wolfgang
    Commented Jan 27, 2018 at 12:40
  • 1
    $\begingroup$ The generating vector is not unique in general. Identify the 1s in the first row of the incidence matrix with a subset of a cyclic group (this is the difference set). If this set is $D$, then subsequent rows are the incidence vectors of the sets $g^{i}D$. You can apply any automorphism of the cyclic group to $D$, and the rows will still give the incidence matrix of a projective plane. You may wish to look up equivalence of difference sets - Baumert's "Cyclic difference sets" or Hall's "Combinatorial Theory" discuss this. $\endgroup$
    – Padraig Ó Catháin
    Commented Jan 27, 2018 at 13:29

1 Answer 1

Reset to default
4
$\begingroup$

To answer your questions:

1) A projective plane admits a circulant incidence matrix if and only if the automorphism group contains a cyclic group acting regularly on points and regularly on blocks. Equivalently, the projective plane comes from a difference set. The automorphism group of the projective plane coming from the Dickson near-field of order 9 is not divisible by 91, so it cannot have a circulant incidence matrix. Most likely it is the case that 'most' projective planes do not admit a circulant incidence matrix.

2) Yes, for the Desarguesian planes, a conjugacy class of such cyclic subgroups exist, they are called Singer cycles.

3) No, the circulant form for the incidence matrix is not unique. As mentioned in the comments, if one identifies the 1s in the first row of the incidence matrix with elements of the cyclic group of order $v$, then allowing the automorphism group of this cyclic group to act in the natural way gives different incidence matrices (but all come from equivalent difference sets). A single plane could have multiple inequivalent realisations as a difference set, these will give further realisations as circulant incidence matrices.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .