# A graphic representation of classical unitals on 28 points

I would like to understand the geometry of the classical unitals. They are block designs containing $$q^3+1$$ points and whose blocks have cardinality $$q+1$$, where $$q$$ is a prime power. For $$q=2$$ (if I understand properly), the classical unital is isomorphic to the affine plane over the $$3$$-element field, so its geometric structure is clear. What is the geometric structure of the next unital, namely that of cardinality $$3^3+1=28$$ containing lines of cardinality $$3+1=4$$? Is there any (nice symmetric) graphical representation of this $$28$$-element unital that can be helpful for understanding the geometric structure of this unital, its symmetries, etc. It has only 28 points, 63 lines, so those can be drawn somehow?

• Perhaps the book Configurations from a Graphical Viewpoint by Pisanski and Servatius might contain something that interests you (it's about representing combinatorial configurations as configurations of points and lines, possibly by preserving symmetries, which is sort of what you're talking about). I've reads bits of it some time ago, my memory is hazy, but I think it's rather pleasant to read (and it could be of interest to you independently of this specific question). Commented Nov 19, 2023 at 20:00
• @Gro-Tsen's reference, clickably: Pisanski and Servatius - Configurations from a graphical viewpoint. Commented Nov 19, 2023 at 23:33
• @Gro-Tsen Thank you for the info about this book. Indeed, very interesting, but I have not seen unitals there. Commented Nov 19, 2023 at 23:41

Using the answer of Taras Banakh, I have implemented an interactive example using HTML and Javascript. It shows all lines of the unital with random colors. Every line is clickable. After clicking on a line it shows the line together with points that belong to this line. To reset — click outside any line. I don't guarantee that this example will last forever (or even for a long time), so anybody interested in it can save the source of this page and host it where he wants. If somebody wants to improve or change this example, PM me for the source and how-to-launch (I'm too lazy to write documentation, shortly it's compiled from Java source using TeaVM).

Now I have understood how to draw the classical unital, $$U_q$$ at least for $$q=3$$. It is known that this unital is isomorphic to the subset of the projective plane $$PG(2,9)$$, defined by the equation $$X^4+Y^4+Z^4=0$$ in homogeneous coordinates. The projective plane $$PG(2,9)$$ is just the affine plane $$\mathbb F_9\times\mathbb F_9$$ with the attached projective line $$\mathbb F_9\cup\{\infty\}$$. The affine plane and the projective lines are over the $$9$$-element field $$\mathbb F_9=\{-1-i,-1,-1+i,-i,0,i,1-i,1,1+i\}$$where $$i\in\mathbb F_9$$ is an element such that $$i^2=-1$$. The addition in $$\mathbb F_9$$ is by modulo $$3$$. The list of the elements of the field $$\mathbb F_9$$ determines the linear order on $$\mathbb F_9$$ that will be used in our drawing later.

In the field $$\mathbb F_9$$, the elements $$1$$ and $$-1$$ have four roots of 4-th order: $$\sqrt[4]{1}=\{-1,-i,i,1\}\quad\mbox{and}\quad\sqrt[4]{-1}=\{-1-i,-1+i,1-i,1+i\}.$$

Then the $$28$$-element classical unital $$U_3$$ can be identified with the subset $$U_3=(\{0\}\times\sqrt[4]{-1})\cup(\sqrt[4]{-1}\times\{0\})\cup(\sqrt[4]{1}\times\sqrt[4]{1})\cup \sqrt[4]{-1}$$of the projective plane $$PG(2,9)=(\mathbb F_9\times\mathbb F_9)\cup\mathbb F_9\cup\{\infty\}$$.

In this model, we can draw lines and see whatever we would like to see. For example, that the unital $$U_3$$ satisfies the O'Nan condition:

$$\forall o,x,y\in U_3\;\forall p\in\overline{xy}\setminus(\overline{ox}\cup\overline{oy})\;\forall u\in \overline{oy}\setminus\{o,y\}\;(\overline{up}\cap\overline{ox}=\varnothing).$$

At the following drawing, this is illustrated on the example of the points $$\color{blue}{o=(-1,-1)}$$, $$\color{blue}{x=(1,-1)}$$, $$y=(-1,1)$$ and $$p=(i,-i)\in\overline{xy}$$. For the points $$\color{magenta}{u=(-1,-i)}$$ and $$\color{red}{v=(-1,i)}$$ on the line $$\overline{oy}$$ we can see that the lines $$\color{magenta}{\overline{up}}$$ and $$\color{red}{\overline{vp}}$$ are indeed disjoint with the line $$\color{blue}{\overline{ox}}$$.