## Action of PGL(2) on Projective Space

Let $k$ be a field, let $G = PGL_2(k)$ be the projective general linear group of $k$, and let $X = k \cup { \infty }$ be one-dimensional projective space over $k$. Then $G$ acts on $X$ (via fractional linear transformations). This action has the following properties:

1) The action of $G$ on $X$ is simply 3-transitive. That is, it acts simply transitively on the set of 3-tuples of distinct elements of $X$. (Edited as indicated in the comments.)

2) Suppose that $x,y \in X$ are distinct elements and that $g \in G$ satisfies $gx = y$, $gy = x$. Then $g$ has order $2$.

Is the converse true? (That is, if we are given an action of a group $G$ on a set $X$ satisfying 1) and 2), does it follow that $G = PGL_2(k)$ for some field $k$, with its natural action on $k \cup { \infty }$?

(This is true at least when $G$ and $X$ are finite: it can be deduced from the theorem of Frobenius on Frobenius groups.)

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Perhaps I am missing something silly here--isn't the one-element set acted on by the trivial group a counterexample? (Both conditions are vacuous here). Only slightly more seriously, what about the two-element set acted on faithfully by $\mathbb{Z}/2\mathbb{Z}$? (1) is vacuous again, and (2) is satisfied... But clearly $X$ is too small in both cases to be a projective space over any field. – Daniel Litt Jun 4 2011 at 2:34
Apart from projectve lines over fields there are projective lines over (not necessarily commutative) division rings, such as the quaternions. These examples are not there if $\#X<\infty$ since all finite division rings are fields, by Wedderburn's theorem. – algori Jun 4 2011 at 2:50
Daniel, your silliness can be generalized: $X$ could be empty; and if it is empty or a singleton then $G$ can be arbitrary. – Tom Goodwillie Jun 4 2011 at 3:19
And the remedy is to redefine $3$-transitive as meaning that the action of $G$ on the set of ordered triples of distinct elements is transitive, i.e. has exactly one orbit. – Tom Goodwillie Jun 4 2011 at 3:19
Yes, that is what I should have said. – Jacob Lurie Jun 4 2011 at 5:03

A KT-field $(F,+,\times,\sigma)$ consists of a neardomain $(F,+,\times)$ together with an involutionary automorphism $\sigma$ satisfying $$\sigma(1 + \sigma(x)) = 1 - \sigma(1 + x)$$ for all $x \in F \setminus {0,1}$. (My impression is that neardomains are quite weak entities, e.g. $F^{\times}$ is required to be a group but it may not be commutative, $(F,+)$ is not even necessarily a group. Industrious MO reader adds the definition of a neardomain to this answer if they wish.) Sharply $3$-transitive groups are determined up to isomorphism as permutation groups on $\mathbf{P}^1(F) = F \cup { \infty }$ consisting of maps of the form:

(i): $x \mapsto a + m x, \quad \infty \mapsto \infty$

(ii): $x \mapsto a + \sigma(b + m x), \quad \infty \mapsto a, \quad - m^{-1} b \mapsto \infty$,

where $a,b \in F$ and $m \in F^{\times}$.

Consider the set of elements $\gamma \in G$ such that $\gamma(0) = \infty$ and $\gamma(\infty) = 0$. They are given exactly by mappings of the form: $$\gamma: x \mapsto \sigma( \lambda x)$$ for any $\lambda \in F^{\times}$. If all such $\gamma$ have order two, then $$\sigma(\lambda \sigma(\lambda x)) = x$$ for all $x, \lambda \in F^{\times}$. Setting $x = \lambda^{-1}$, it follows that $\sigma(\lambda) = \lambda^{-1}$ for all $\lambda \in F^{\times}$. Since $\sigma$ is an automorphism, it follows that $F^{\times}$ is commutative. From a Theorem of Kirby (see below), it follows that $(F,+,\times)$ is actually a commutative field, and $G = \mathrm{PGL}_2(F)$.

All the results and definitions of this answer can be gleamed from the math review: MR0997066 (91b:20004a) of a paper by William Kerby. The paper is only $3$-pages long, so I assume that is is relatively elementary - although I can't access it myself, and it may refer to previous results. (Full disclosure, all I did was type "sharply 3-transitive" into mathscinet, I don't actually know what a neardomain actually is.) In case your actual purpose is to generalize this result to $(\infty,\pi)$-whatzit categories with creamy rice pudding centres, you might want to take a glance at the actual paper.

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 Sounds like exactly what I'm looking for. Thanks. – Jacob Lurie Jun 5 2011 at 11:09