Consider the finite field ${\bf F}_p$ and its cubic extension ${\bf F}_{p^3}$. The multiplicative group ${\bf G}_m({\bf F}_{p^3})$ contains the multiplicative group ${\bf G}_m({\bf F}_p) \cong {\bf Z}/(p-1){\bf Z}$ as a subgroup. The quotient $A_p = {\bf G}_m({\bf F}_{p^3})/{\bf G}_m({\bf F}_p)$ is an abelian group. On the other hand, ${\bf F}_{p^3}$ can be considered simply as a $3$-dimensional vector space over ${\bf F}_p$, so this quotient is naturally a projective plane ${\bf P}^2({\bf F}_p)$. Its cardinality equals $p^2+p+1$, thus divisible by $3$ for $p=6k+1$. Therefore the group $A_p$ contains a subgroup of order $3$. Its cosets are represented by some triangles in ${\bf P}^2({\bf F}_p)$. Of course all the triangles in the projective plane go to each other under the projective transformations, but they should have some particular geometry with respect to all the markings on the plane coming from the identification of a $3$-dimensional vector space with the field ${\bf F}_{p^3}$. Say, can one speak of triangle centers for this case (at least if one fixes the line on the infinity)? It seems like the least possible case $p=7$, when ${\bf P}^2({\bf F}_p)$ contains $57$ points, is of interest.
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55$\begingroup$ Are you trying to convince Grothendieck? $\endgroup$– darij grinbergCommented Nov 15, 2017 at 4:36
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7$\begingroup$ In case there's anyone not in on darij's joke, see the bottom of the first page of ams.org/notices/200410/fea-grothendieck-part2.pdf $\endgroup$– Gerry MyersonCommented Nov 15, 2017 at 6:20
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18$\begingroup$ That reminds me of a professor of mine. In a lecture, he was once asked to give an example with actual numbers. Not really understanding how the students could possibly not see the beauty of the theorems he just wrote down, but of course willing to give such an example, he started to write on the blackboard: "Let a,b,c be actual numbers...". $\endgroup$– DirkCommented Nov 15, 2017 at 11:19
1 Answer
I'm not sure if this does what you want, but the subgroup of order $3$ in the additive group $(\mathbb{Z}_{57},+)$ is $\{0,19,38\}.$
To expand on that, one construction for the plane $\mathbb{P}_{7}$ of order $7$ is to take as points the elements of $\mathbb{Z}_{57}$ with lines $$\ell_k=[k,k+1,k+3,k+13,k+32,k+36,k+43,k+52]$$ for $k \in \mathbb{Z}_{57}.$ Of course the lines are unordered sets but keeping the order might make the cyclic collineation $\phi:x \mapsto x+1$ easier to follow.
The triangle $T=T_0=\{0,19,38\}$ (mentioned above) has $19$ cosets $T_i=\{i,i+19,i+38\}$.
The lines obtained by extending the sides are
$\ell_6=[6,\ \ 7\ ,9,\ \ {\large 19 },{\large 38},42,49,1]$
$\ell_{25}=[25,26,34,{\large 38 },{\large 0},4,11,20]$
$\ell_{44}=[44,45,47,{\large 0 },{\large 19},23,30,39]$
If "the" center of a triangle $\{P,Q,R\}$ in some plane (like $\mathbb{R}^2$ or $\mathbb{P}_{7}$) is "the" solution $C$ to $C+C+C=P+Q+R$ then we see that in $\mathbb{P}_{7}$ (as constructed here) a triangle has three centers. $C,C+19,C+38$ for a unique $0 \leq C \leq 18.$ These are just the vertices of $T_C.$
So the special triangles could be described those which are the triangle of centers for some triangle. OR as those triangles which coincide with their own triangle of centers.
