# 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.)

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Is it obvious that you can't draw the Fano plane with lines in $\mathbb{R}^2$? – Dustin G. Mixon May 15 '13 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. – Noam D. Elkies May 15 '13 at 2:59
...and conversely, if $k$ does have characteristic $2$ then $A',B',C'$ are always collinear... – Noam D. Elkies May 15 '13 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 May 16 '13 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. – Noam D. Elkies May 17 '13 at 16:16

## 1 Answer

Does this one work?

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Seems it does - very nice! – Seva May 15 '13 at 6:06