details in the proof: G a commutative Lie group and $(\pi, V)$ a finite-dim unitary representation of G. Then $\pi$ is irreducible iff dim $V = 1$

Question

I am getting lost at filling in the details in the proof that

Let G a commutative Lie group and $(\pi, V)$ a finite-dim unitary representation of G. Then $\pi$ is irreducible iff dim $V = 1$

I having trouble with $\implies$

Assume $\pi$ is irreducible. Fix a $g\in G$, I claim that the map $T:V\to V : v \mapsto \pi(g)v$ is an intertwiner from $(V,\pi(g))$ to itself:

• T is linear because $\pi(g)$ is linear.
• The diagram in the definition commutes.
• T is continuous (Question1 how do I justify this?)

(This part in the solution that I have is actually phrased in an odd way: The operator $\pi(g)$ commutes with $\pi(G)\subset Aut(V)$.How can a map commute with a set? A map should commute with another map, otherwise it makes no sense Question2 Can you clear this up?)

So by Schur's lemma, $\exists \lambda_g \in \Bbb C$ s.t $T=\lambda_gId_V$ Hence span $\pi(G)=\Bbb C Id_V$

Every linear subspace of V is a subrepresentation. By the irreducibility of V, there are no subrepresentations apart from V and 0. There for V must be one-dimensional.

Question3 I can't understand this last paragraph, this is my main problem

Why is every linear subspace of V is a subrepresentation (I assume this follow from the previous paragraphs but I just don't see how)?

Why does the fact that there are no subrepresentations imply that $V must be one-dimensional ? No subrepresentations would imply that the are no non trivial subspaces, but then I am just getting a circular argument

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These are the definitions and theorem that I have:

Schur's lemma: If G is a loc compact group; $(\pi, V)$ a cont fin-dim irreducible representation. Then Hom$_G(V,V)=\{$ intertwiners from $(\pi,V)$ to itself $\}=\Bbb C Id_V$

Intertwiner: An intertwiner between 2 representations $\pi_1$ and $\pi_2$ is a continuous linear map $T:V_1 \to V_2$ s.t. the diagram commutes for every $h \in G$:

$$\begin{array} .V_1 & \stackrel{T}{\longrightarrow} & V_2 \\ \downarrow{\pi_1(h)} & & \downarrow{\pi_2(h)} \\ V_1 & \stackrel{T}{\longrightarrow} & V_2 \end{array} $$

Subrepresentation or invariant subspace of $V$ is a linear subspace $W \subset V$ s.t $\pi(g)W \subset W, \forall g \in G$

Answer 1

1. A representation assumes that the map $\pi:G\times V\to V$ is continuous. So the composition $V\to \{g\}\times V\subseteq G\times V \xrightarrow{\pi} V$ is continuous. The topologies taken are the one on $G$ as a Lie group, and on $V$ as a vector space over $\Bbb{C}$ there is a unique vector-topology compatible with the field-topology of $\Bbb{C}$. The topology on $G\times V$ is the product topology.
2. It's shorthand for saying that $\pi(g)$ commutes with $\pi(h)$ for all $h\in G$.
3. Because each $\pi(g)$ acts as scalar-multiplication by $\lambda_g$ (dependent on $g)$. If you restrict $\pi(g)$ to a subspace, each $\pi(g)$ is still scalar multiplication by $\lambda_g$ within the subspace. Since every subspace is a subrepresentation, and the full representation is assumed irreducible, the only subspaces are $0$ and $V$ so $V$ must be one-dimensional (the only non-zero vector space with such property).