# Fano plane drawings: embedding PG(2,2) into the real plane

By a drawing of the Fano plane I mean a system of seven simple curves and seven points in the real plane such that

• every point lies on exactly three curves, and every curve contains exactly three points;
• there is a unique curve through every pair of points, and every two curves intersect in exactly one point;
• the curves do not intersect except in the seven points under consideration.

The familiar picture

does not count as a drawing, since the last requirement is not satisfied: there are two "illegal" intersections. In fact, this is easy to fix:

However, this drawing is degenerate in the sense that two of the curves just "touch" each other, without crossing, at some point. And here, eventually, my question goes:

Is every drawing of the Fano plane degenerate?

(Although I can give a topological definition of degeneracy, it is a little technical and, may be, not the smartest possible one, so I prefer to suppress it here.)

• Is it obvious that you can't draw the Fano plane with lines in $\mathbb{R}^2$? Commented May 15, 2013 at 1:32
• Yes, a line drawing is impossible, over ${\bf R}$ or any field $k$ not of characteristic $2$. Let $A,B,C,O$ be non-collinear points of the Fano plane, and $A',B',C'$ the intersections of $AO,BO,CO$ with $BC,CA,AB$ respectively. By Ceva's theorem (actually proved by Al-Mutaman centuries earlier, and extended algebraically to the case where $O$ is outside the triangle, and indeed to arbitrary $k$), points $A',B',C'$ divide segments $BC,CA,AB$ in signed ratios whose product is $1$. But by Menealus' theorem, $A',B',C'$ are collinear iff that product is $-1$. Since $1 \neq -1$ we're done. Commented May 15, 2013 at 2:59
• ...and conversely, if $k$ does have characteristic $2$ then $A',B',C'$ are always collinear... Commented May 15, 2013 at 3:00
• @Noam: I see, the basic idea is that (1) any line not passing thorough a vertex of a triangle intersects an even number of its edges, while (2) for any triangle $ABC$, and any point $O$ not on its boundary, the three lines $OA$, $OB$, and $OC$ intersect an odd number of the edges.
– Seva
Commented May 16, 2013 at 8:57
• Actually I'm not using anything lke that (certainly not for an arbitrary field). Another way to say this is to choose projective coordinates so $A$, $B$, and $C$ are at the unit vectors $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$, and then scale those coordinates so that $O$, which must have all three coordinates nonzero (else it's on one of the lines $AB$, $AC$, $BC$) is on $(1:1:1)$; then $OA$ is the line $y=z$, so $A'=OA \cap BC$ is $(0:1:1)$, and likewise $B = (1:0:1)$ and $C = (0:1:1)$. Now calculate that the determinant of $A,B,C$ is $2$, so $ABC$ are collinear iff we're in characteristic 2. Commented May 17, 2013 at 16:16

Does this one work?

• Seems it does - very nice!
– Seva
Commented May 15, 2013 at 6:06

Here is a bit more symmetric version of the picture in the accepted answer:

I always remember this diagram that seems impossible to find online. I have no source for it (except that I remember that it is/was engraved in the sidewalk near the Weber Building on the Colorado State University campus in Fort Collins, CO).

Does this one count as nondegenerate or is it still bad because of the outer circle only touching the red, blue and magenta curves?

• Pretty but degenerate. Commented Jun 14, 2018 at 9:13
• Thanks! That's what I thought because of the curves touching, but I wasn't sure if closing the loop made it okay.
– Tom
Commented Jun 15, 2018 at 16:31
• Degenerate but beautiful! Commented Nov 26, 2021 at 22:25