TITLE: Sharply $k$-transitive actions on spheres
QUESTION [5 upvotes]: A nice fact from complex analysis is that the mobius group acts sharply 3-transitively on the Riemann sphere. I am wondering if other sharply k-transitive (continuous) actions are known on any $S^n$, and if it's possible to classify them (perhaps up to conjugation by a homeomorphism of the sphere).
If $k=1$, I believe this is the same as asking which spheres have a lie group structure, which just gives possibilities with $S^1$ and $S^3$, but besides actions these and the mobius group example, I know no others fulfilling my criteria.
edit: I believe real mobius transformations act sharply 3-transitively on the circle (real line with point at infinity) just like in the complex case, and I read here that there are not infinite sharply k-transitive groups for $k \ge 4$

REPLY [4 votes]: The purpose of this post is to classify manifolds (not just spheres) which are sharply $3$-transitive.  The main result is that the examples in my other answer are the only ones.

Suppose $M$ is a closed manifold which is sharply $3$-transitive.  Then $M$ is diffeomorphic to $S^1$ or $S^2$.

Proof:  Suppose $G$ acts on $M$ sharply $3$-transitively.  From the classifcation of closed $1$-manifolds, we may assume $\dim M\geq 2$.   Pick $x\in M$.  Then the isotropy group $G_x$ must act transitively on $M\setminus \{x\}$.  Pick $y\in M\setminus\{y\}$.  Then the isotropy group $(G_x)_y$ must act simply transitively on $N:=M\setminus \{x,y\}$.
This implies that $N$ is diffeomorphic to the Lie group $(G_x)_y$.  Note also that $N$ is non-compact and it is connected since $\dim N =\dim M \geq 2$.  A connected Lie group is always diffeomorphic to a product $\mathbb{R}^k\times K$ where $K$ is a compact Lie group, so $N\cong \mathbb{R}^k\times K$.
Note that $N$ has a two-point compactification $(M)$, so by this MO answer, $N$ must have at least two ends.  This implies $k=1$ because for $k\geq 2$, $N$ has just one end.
Now, the two point compactification of $N$ obtained by adding a point at each end (that is, $M$), is nothing but the suspension of $K$, $\Sigma K$.  However, according to this MSE question if $\Sigma K$ is a manifold, then $K$ must have been a sphere.  (Technically, the linked answer assumed $\dim K \geq 3$, but if $\dim = 1,2$, then a closed homology sphere must be homeomorphic to a sphere by the low dimensional classification of manifolds.)
Thus, $K$ is a compact Lie group and a sphere.  It follows that $K = S^3$ or $K= S^1$.  If $K= S^1$, then $M$ is the suspension of $S^1$, so is $S^2$.  It $K = S^3$, then $M\cong S^4$ is the suspension of $S^3$.  But this was ruled out in the previous answer.  $\square$
The result for $k=2$ is the following.

Suppose $M$ is a closed manifold which is sharply $2$-transitive.  Then $M$ is diffeomorphic to $S^1$ or $S^3$.

Proof:  Suppose $G$ acts on $M$ in a sharply $2$-transitive manner.  Picking $x\in M$, we deduce that the isotropy group $G_x$ acts simply transitively on $N:=M\setminus\{x\}$.  In particular, $N$ is a non-compact connected manifold, which is diffeomorphic to $G$.  Just as above, it follows that $N\cong \mathbb{R}^\ell\times K$ for some compact Lie group $K$.
Now, $M$ is the one point compactification of $N$.  So, we may express $M = S^k\times K/(\{\infty\}\times K$.  Using the long exact sequence of the pair $(\{\infty\}\times K, M)$, it follows easily that $H^{\ell + s}(K)\cong \overline{H}^s(M)$ for all $s > 0$.  (This is just a simple manifestation of the Thom isomorphism applied to the bundle $\mathbb{R}^\ell\rightarrow N\rightarrow K$.)
On the other hand, computing the local homology of the pair $(M,N)$ yields $H^\ast(M,N)\cong \mathbb{Z}$ when $\ast = \dim M$, and $0$ otherwise (since $M$ is a manifold).  From the long exact sequence of cohomology of a pair, together with the fact that $N$ deformation retracts to $K$, we see that $H^s(M)\rightarrow H^s(K)$ is an isomorphism for any $0 < s < \dim M$.  Combining this with the result in the previous paragraph, we see $H^s(K) \cong H^s(M)\cong H^{s+\ell}(K)$ is valid for any $0<s< \dim M$.
If $\dim K\neq 0$, set $s = \dim K$.  Then $H^{\dim K}(K)\neq 0$ because $K$ is orientable (since it's a Lie group), while $H^{s + \ell}(K) = 0$ because $\ell > 1$ (since $\mathbb{R}^\ell\times K$) is non-compact.
Thus, we have a contradiction unless $\dim K = 0$.  Now it follows that $N$ is diffeomorphic to $\mathbb{R}^{\ell}$, so $M$, being the one point compactification of $N$, is $S^{\ell}$.  Having reduced to the case where $M$ is a sphere, the previous answer now gives the result. $\square$