Timeline for For which finite projective planes can the incidence structure be written as a circulant matrix?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2018 at 20:51 | vote | accept | Wolfgang | ||
| Jan 29, 2018 at 20:29 | answer | added | Padraig Ó Catháin | timeline score: 4 | |
| Jan 27, 2018 at 13:29 | comment | added | Padraig Ó Catháin | 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. | |
| Jan 27, 2018 at 12:40 | comment | added | Wolfgang | @PadraigÓCatháin Thank you, I expected something quite easy about Desarguesian planes (projective geometry is not my specialty). What about uniqueness of the generating vector? | |
| Jan 27, 2018 at 11:39 | comment | added | Padraig Ó Catháin | 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. | |
| Jan 26, 2018 at 17:24 | history | asked | Wolfgang | CC BY-SA 3.0 |