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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

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 $\pi(g)W \subset W, \forall g \in G$

some_math_guy
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1 Answers1

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  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).
Chad K
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  • A linear space that has no subspaces different from 0 and iteself has to have dim=1 necesarily right? Then it looks like that can be concluded independently of the rest of the proof – some_math_guy Nov 30 '23 at 19:35
  • @some_math_guy: A non-zero linear space that has no proper subspaces has dimension one. That's not 'concluded'. It's a result from linear-algebra relied upon. – Chad K Nov 30 '23 at 19:36
  • Ok, so why do I need the rest of the proof to say this? – some_math_guy Nov 30 '23 at 19:37
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    @some_math_guy: The rest of the proof shows each linear subspace is a subrepresentation. That's not trivial. subspace $\ne$ subrepresentation (usually.) – Chad K Nov 30 '23 at 19:39
  • I forgot this little part in the proof :" Hence span $\pi(G)=\Bbb C Id_V$ " that I added after Schur's lemma application. If you don't mind, why is this true and relevant?, the proof seemed to flow without writting that. – some_math_guy Nov 30 '23 at 20:04
  • @some_math_guy: It's true because the proof shows $\pi(G)\subseteq \Bbb{C}^* 1_V$ (where $\Bbb{C}^* = \Bbb{C}\setminus{0}$), and therefore span $\pi(G)\subseteq \Bbb{C} 1_V$, but $\Bbb{C}1_V$ is one-dimensional so they're equal. Then every subspace $W\subseteq V$ such that $\pi(G)W\subseteq W$ must satisfy $\Bbb{C}1_V W = \Bbb{C} W=W$ so every subrepresentation is a subspace. It's not necessary, just another way of saying the same thing and I guess the author finds it more elegant. – Chad K Nov 30 '23 at 20:18
  • Regarding the first point my definition of representation does not include continuity, is it possible to show that T is continuous without that?. On the other hand if we assume a continuous representation, my definition of irreducibility says that a representation is irreducible if the only closed invariant subspaces are V and 0, I am not sure if this changes anything – some_math_guy Nov 30 '23 at 21:26
  • @some_math_guy: Since you assume the representation to be a unitary representation, each $\pi(g)$ acts as a unitary operator on the underlying subspace $V$ which must have some inner product for 'unitary' to be defined. A unitary operator is always continuous. – Chad K Dec 01 '23 at 09:19
  • At the very end, the only subspaces are 0, and V, how do we rule out that the subspace is 0? – some_math_guy Dec 02 '23 at 23:27