Proof or counterexample for isomorphism of group representations

Question

Let $(\pi , V_\pi)$ be an irreducible unitary representation of the (locally compact) group $G$. Let $V_\pi^n = V_\pi \oplus \ldots \oplus V_\pi$ be the $n$-fold direct sum of $V_\pi$ on which we have the (unitary) representation $\pi^n$ where $$\pi^n(g) \big(v_1 , \ldots , v_n \big) := \big(\pi(g) v_1 , \ldots , \pi(g) v_n \big) .$$ I strongly suspect, that for fixed $v \in V_\pi^n$ with $v_1 \neq 0$ the map $$\operatorname{Pr} \colon V_\pi^n \to V_\pi \colon (v_1 , \ldots , v_n ) \longmapsto v_1$$is a $G$-intertwining isomorphism bewtween$$\left( \pi^n , \overline{\langle \pi(g) v ~ : g \in G \rangle} \right) \longleftrightarrow \big( \pi , V_\pi \big) .$$ The $G$-intertwining property as well as surjectivity are clear but I'm having trouble showing injectivity. I'm grateful for any hints as well as for a counterexample, if there is one. If this makes it any easier, the solution to that problem would be sufficient for the case $G = \operatorname{SL}_2 \big(\mathbb{R}\big)$.

Answer 1

It's not true in general, even for finite dimensional representations of finite groups.

Take for instance $\mathfrak{S}_3$ acting on $\mathbb{R}^3$, take the irreducible subrepresentation $V=\{\mathbf{x}\in \mathbb{R}^3 \mid \sum x_i = 0\}$.

Then take $v =(v_1,v_2) \in V\oplus V$ with $v_1 = (1,1,-2)$, $v_2 = (1,2,-3)$.

Then with $\sigma = (1 2)$ you have $\sigma v - v$ which is sent to $\sigma v_1 - v_1=0$ but $\sigma v - v = (\sigma v_1 - v_1, \sigma v_2 - v_2) = (0, (1,-1,0))\neq 0$

(To construct the counterexample I thought of the easiest way for $\displaystyle\sum_g \lambda_g gv_1 = 0$ but $\displaystyle\sum_g \lambda_g g$ not $0$ on $V$, then it suffices to find $v_2$ not in the kernel; for this I looked for examples of $g-1$ not being $0$ (that is, $g$ is not in the kernel of the representation), but having a zero (that is, $g$ having a fixed point) - this led me to my example)