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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 $G$.

Is $V_X$ necessarily irreducible? Is it indecomposable?

When $G$ is finite, then there is an easy character-theoretic argument. When $\mathbb F$ is any field and $G=\mathrm{GL}_2(\mathbb F)$ acts on $X=\mathbb P_{\mathbb F}^1$, this is answered here. What can be said in general?

EDIT (6/15): Although Noam gave a great example of a reducible $V_X$, the question still remains of whether $V_X$ can be decomposable. I suspect this can be proven by proving the only $G$-homomorphisms $V_X\to V_X$ are scalars, and such homomorphisms are uniquely determined by where $[y]-[x]$ is mapped to, where $x\ne y\in X$ and $[x]$ denotes the characteristic function of $\{x\}$. Then, the condition for such a map to be a homomorphism reduces to a complicated cocycle condition.

Further update: Per YCor's suggestion, I have posted a follow-up here regarding indecomposability.

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    $\begingroup$ I think you should accept Noam Elkies' answer and ask the follow-up question (on decomposability) in a new question, instead of here. $\endgroup$ Commented Jun 15, 2023 at 16:03

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$V_X$ can fail to be irreducible. If $X \subseteq {\bf C}$ is a field and $G$ is the group of affine linear transformations $x \mapsto ax+b$ (any $a,b \in X$ with $a \neq 0$) then $V_X$ has the proper subrepresentation consisting of functions $f : X \to {\bf C}$ of finite support such that $\sum_x f(x) = \sum_x x f(x) = 0$.

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    $\begingroup$ Thank you for the counter-example! This representation even has infinite length, since it has a filtration $V_n\supset V_{n+1}\supset\cdots$, where $V_n$ consists of functions $f$ such that $\sum_xx^if(x)=0$ for all $i\le n$. $\endgroup$ Commented Jun 14, 2023 at 20:16

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