# For which finite projective planes can the incidence structure be written as a circulant matrix?

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.

• 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. – Padraig Ó Catháin Jan 27 '18 at 11:39
• @PadraigÓCatháin Thank you, I expected something quite easy about Desarguesian planes (projective geometry is not my specialty). What about uniqueness of the generating vector? – Wolfgang Jan 27 '18 at 12:40
• 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. – Padraig Ó Catháin Jan 27 '18 at 13:29

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.