-2
$\begingroup$

Let $G$ be an infinite discrete group and $\pi$ a representation of $G$ on the Hilbert space $H$.

Suppose that the group representation is transitive on an orthonormal basis $B = \{e_j\}_{j=1}^{\infty}$ of $H$, that is, for all $e_m$ and $e_n$ in $B$, there is a $g \in G$ such that $\pi(g)e_m = e_n$.

Irreducible representations are representations without non-trivial invariant subspaces. I would say that being transitive on $B$ is not enough to ensure irreducibility of $\pi$, since nothing ensures that the space spanned by every vector not in $B$, for instance, the vector $e_1 + e_2$, is $H$.

Is there any example of a non-irreducible representation of a group acting transitively on some basis?

Improve this question
$\endgroup$
2
  • 5
    $\begingroup$ $C_2$ acts on $\mathbb C^2$ by swapping the coordinates, and acts transitively on the standard basis, but is reducible. $\endgroup$
    – Wojowu
    Commented Jun 4 at 12:11
  • 1
    $\begingroup$ @Wojowu I think the OP assumes $H$ to be infinite-dimensional, because in finite dimensions the 1d subspace spanned by $\sum_i e_i$ is always invariant. $\endgroup$
    – gmvh
    Commented Jun 4 at 12:18

1 Answer 1

Reset to default
5
$\begingroup$

This is indeed not sufficient. Consider the group of permutations on the set $\mathbb{Z}$ generated by $\sigma,\tau$, where $\sigma(2n)=2n+1$, $\sigma(2n+1)=2n$ and $\tau(n)=n+2$. Then this is transitive on $\mathbb{Z}$ and hence its permutation representation on the Hilbert space with basis $e_n, n\in \mathbb{Z}$ is transitive on that basis. But the subspace spanned by the vectors $e_{2n}+e_{2n+1}$, $n\in\mathbb{Z}$, is a proper invariant subspace.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Just to add, the point is that this action is imprimitive, and so the blocks allow us to construct an invariant subspace. $\endgroup$
    – Max Horn
    Commented Jun 4 at 19:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .