$\renewcommand\Im{\operatorname{\mathcal{Im}}}\newcommand\Ker{\operatorname{\mathcal{Ker}}}$I was sure that this is a trivial question and placed it on Math Stackexchange
<https://math.stackexchange.com/questions/4136830/a-detail-in-the-proof-of-schurs-lemma-the-closures-of-the-cal-ker-and-cal>


Surprisingly, no one answered it there. So I am elevating it to MathOverflow.




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Consider two irreducibles of a topological group $G$, acting in respective Hilbert spaces $\mathbb V$ and $\mathbb V'$.


Schur's lemma says:


An intertwiner $M : \mathbb V\longrightarrow\mathbb V'$ of two irreducibles of a group is either zero or isomorphism.


To prove it, we first show that both $\Im M$ and $\Ker M$ are invariant subspaces. Then two observations are made.


**Observation 1.** For the invariance of $\Im M$ to agree with the irreducibility, $\Im M$ must coincide either with the space $\mathbb V'$ or with its zero subspace $\{\smash{\vec0}'\}$.


**Observation 2.** For the invariance of $\Ker M$ to agree with the irreducibility, $\Ker M$ must coincide either with the space $\mathbb V$ or with its zero subspace $\{\vec0\}$.


Summing up these observations, we conclude that $M$ is either zero or bijective and therefore invertible.


Now, my **question**. Is it really true that $\Im M$ itself must coincide either with $\mathbb V'$ or $\{\smash{\vec0}'\}$? Or is it rather the *closure* of $\Im M$ that must satisfy this requirement? I am enquiring, because a representation is always defined in a Hilbert space or in a *closed* thereof.


The same question about $\Ker M$. Should we prove that it is actually its *closure* that is invariant?


If it is the closures of $\Im M$ and $\Ker M$ whose invariance needs to be proven, do we have to impose additional requirements on the representations and/or on the intertwiner $M$ (say, boundedness)?



A detail in the proof of Schur's lemma: the closures of the $\mathcal{Ker}$ and $\mathcal{Im}$ of the intertwiner
Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be a bounded intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$. Claim: These two are multiples of the identity. It follows that $T$ is zero or a multiple of an isometric isomorphism.


To prove the claim, it is enough to show that a selfadjoint intertwiner $A$ of an irreducible represenation is a multiple of the identity.


But the spectral projections of $A$ are intertwiners. By irreducibility, they are either 0 or 1. The spectrsl theorem then proves that $A$ is a multiple of the identity.


