Do doubly-transitive actions give rise to indecomposable representations for infinite groups?

Question

This is a follow-up to this question.

Let $G$ be a group acting doubly-transitively on a set $X$. Then the vector space $V_X$ of functions $f\colon X\to\mathbb C$ with finite support such that $\sum_{x\in X}f(x)=0$ carries an action of $G$.

There are examples where $V_X$ can be reducible. Is $V_X$ necessarily indecomposable?

One possible approach is to prove the only $G$-homomorphisms $\varphi\colon V_X\to V_X$ are scalars. Such a homomorphism is determined by the image $f\in V_X$ of $[x]-[y]$, where $x\ne y\in X$ and $[x]$ is the characteristic function of $\{x\}$. Now, $G$-equivariance in particular implies a strange cocyle condition:

Let $g\in G_x$. Then $f-\pi(g)f=\pi(h)f$ for any $h\in G_y$ such that $hx=gy$ (which exists by double-transitivity).

I wonder if this is enough to prove $\mathrm{supp}(f)=\{x,y\}$.

Answer 1

Yes.

Let $\varphi$ be a $G$-homomorphism $V_X \to V_X$. As mentioned in the question, it suffices to show $\varphi$ is a scalar.

For $x, y\in X$ with $x\neq y$, let $c_{x,y}$ be obtained by evaluating the function $\varphi([x]-[y])$ at $x$. Then $c_{g(x),g(y)}$ is the evaluation of $$\varphi([g(x)]-[g(y)]) = \varphi( g \cdot [x] - g\cdot[y])=\varphi( g\cdot ([x]-[y]) = g\cdot \varphi([x]-[y])$$ at $g(x)$ and hence equals $c_{x,y}$.

Since the action of $G$ on $X$ is doubly-transitive, this implies $c_{x,y}$ is a constant function of $x,y\in X$ with $x \neq y$, say $c_{x,y}=c$ for all $x,y$.

Then the value at $x$ of $\varphi([x]-[y])$ is $c$, the value at $y$ is $-c$ since $[x]-[y]= - ([y]-[x])$, and the value at $z$ with $z\neq x,z\neq y$ is $0$ since $[x]-[y]=([z]-[y])-([z]-[x])$.

So $\varphi( [x]-[y]) = c ([x]-[y])$ and since these elements span $V_X$, $\varphi$ is scalar multiplication by $c$, as desired.