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$PSL(2,7)$ acts on the projective plane over $\mathbb{F}_2$ (the Fano plane) through its identification with $GL(3,2)$. It also acts on the projective plane over $\mathbb{C}$ through either of its pair of 3-dimensional complex representations. Does the Fano plane embed in $\mathbb{P}_\mathbb{C}^2$ so that the action restricts? To make the question precise:

Does there exist a 7-point orbit in $\mathbb{P}_\mathbb{C}^2$ so that the permutation representation of $PSL(2,7)$ obtained from the action on this orbit is conjugate to the permutation representation of $PSL(2,7)$ on the 7 points of the Fano plane?

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    $\begingroup$ What have you tried? The permutation action on the Fano plane is transitive, so is determined by its stabilizer. So it's necessary and sufficient to find an element of the complex projective plane with the same stabilizer. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 12, 2016 at 1:46
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    $\begingroup$ Restricting the action to the Klein quartic in ${\mathbb C}{\mathbf P}^2$ might help to invoke some geometric intuition. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 12, 2016 at 6:18
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    $\begingroup$ I think the answer is "no". ${\rm PSL}(2,7)$ has two conjugacy classes of subgroups of index $7$, and all subgroups of index $7$ are isomorphic to the symmetric group $S_{4}$. Each of its non-trivial $3$-dimensional complex representations restrict irreducibly to all subgroups isomorphic to $S_{4}$,, so no such subgroup stabilizes a one-dimensional subspace ( to see irreducibility of $S_{4}$ note that the normal Klein $4$-subgroup acts trivially in the irreducible representations of $S_{4}$ of degree less than $3$- or look at the character). $\endgroup$
    – Geoff Robinson
    Commented Mar 12, 2016 at 9:20
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    $\begingroup$ I confirmed Geoff's comment with a character table computation in Magma. The degree $3$ characters of $L_2(7)$ both restrict to irreducible characters of $S_4$. But of course this is an easy calculation! $\endgroup$
    – Derek Holt
    Commented Mar 12, 2016 at 9:31
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    $\begingroup$ @GeoffRobinson I agree entirely. I am afraid that my instinct these days is to try the computer first and think later if necessary. As you observed, it is also very easy to see that $S_4$ has no faithful complex representation of degree $2$. You should probablymake your comment into an answer. $\endgroup$
    – Derek Holt
    Commented Mar 12, 2016 at 16:54

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(As suggested, comment turned into answer) : The answer is "no". ${\rm PSL}(2,7)$ has two non-conjugate subgroups of index $7$, each isomorphic to the symmetric group $S_{4}$. Either of the three dimensional irreducible representations of ${\rm PSL}(2,7)$ restrict irreducibly to each subgroup of ${\rm PSL}(2,7)$ isomorphic to ${\rm S}_{4}$ (a direct and easy calculation), so no such subgroup stabilizes any one-dimensional subspace.

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