Generation of a non-accidental degenerate eigenspace carrying out the symmetry operations on one eigenstate

Question

Let us consider a system described by an Hamiltonian $H$ over an Hilbert space $M$, and the finite group $G$ of symmetry operations w.r.t $H$, i.e.

\begin{equation} R_g : M \rightarrow M \qquad g\in G \end{equation}

such that

\begin{equation} [R_g,H]=0\qquad \forall g\in G \end{equation}

Considered an eigenvalue $\epsilon$ of $H$, and the corresponding eigenspace $V$, the $R_g$ are internal to this subspace \begin{equation} R_g(V)\subseteq V\qquad\forall g \end{equation} Therefore, the $R_g$ restricted to $V$ are a representation of $G$. By definition, this representation is irreducible iff it does not exsist a proper $W\subset V$ such that \begin{equation} R_g(W)\subseteq W\qquad\forall g \end{equation}

My question, inspired by the Tinkham's book about group theory, is this: are the two statements

1. The $R_g$ restricted to $V$ are an irreducible representation of $G$

2. Given an element $e\in V-\{0\}$, the space vector generated by the vectors $\{R_g(e):\,\,g\in G\}$ is $V$

equivalent? If yes, how is it demonstrated?

p.s. Edited after comments

Answer 1

It's a general property of irreducible representations of a group (the context of degenerate eigenspaces is superfluous). The claims are equivalent if you assume $e\in V\setminus\{0\}$ as Sebastian Riese pointed out.

The proof only requires the definition and to notice that $V_e:= \text{Vec}(\{R_g(e):g\in G\})$ is smallest subspace (in the sense of inclusion) stable by the group action containing $e$. You can prove it by double inclusion. $V_e$ contains $e$ and is stable because its generating elements stay in $V_e$ by the action, and for any element, it is also true by linearity. Conversely, let $W$ be a stable subspace containing $e$, it must contain $\{R_g(e):g\in G\}$, by stability and therefore contain its span being a vector subspace.

You can now prove the equivalence by double implication:

• $1.\implies2.$ : $V_e$ is a stable subspace by the previous remark. It is not $\{0\}$ since it contains $e\neq 0$, so by irreducibility, $V_e = V$.
• $2.\implies1.$ : Let $W\neq \{0\}$ a stable subspace. Let $e\in W$ such that $e\neq0$. $V_e\subset W$ by minimality, and $V_e=V$, so $W=V$.

Hope this helps.