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$\renewcommand\Im{\operatorname{\mathcal{Im}}}\newcommand\Ker{\operatorname{\mathcal{Ker}}}$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.

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)?

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    $\begingroup$ As you suspect, I believe the statement should be about a continuous intertwiner, whose image and kernel are both closed. Could you point to where you have seen the theorem stated without this condition? $\endgroup$
    – LSpice
    Commented May 14, 2021 at 20:05
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    $\begingroup$ If you assume $M$ to be continuous, then $\mathcal{Ker}M$ is closed, but $\mathcal{Im}M$ need not be closed, and it could be a nontrivial proper non-closed $G$-invariant subspace. For representations of $C^*$-algebras, topological irreducibility implies algebraic irreducibility, so the two notions are the same. But in the group representation case, they rarely coincide. $\endgroup$
    – Masayoshi Kaneda
    Commented May 15, 2021 at 11:55
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    $\begingroup$ I've presented a more verbose version of Johannes Ebert's answer in math.SE (math.stackexchange.com/a/4140600/52543) for the purpose of that audience, while also trying to address the multiple points discussed in the comments here and there. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented May 16, 2021 at 6:46
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    $\begingroup$ @Michael_1812: Actually, "continuous=bounded" for linear operators, and I was talking about a general situation. Here, however, $M$ is an "intertwiner," so with this extra intertwining condition, one can prove that $\mathcal{Im}M$ is closed as Johannes Ebert and Pedro Lauridsen Ribeiro have already provided excellent full answers. $\endgroup$
    – Masayoshi Kaneda
    Commented May 27, 2021 at 10:12
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    $\begingroup$ @MasayoshiKaneda, thank you for this comment. Will think it over. $\endgroup$
    – Michael_1812
    Commented May 27, 2021 at 18:37

1 Answer 1

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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.

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  • $\begingroup$ Thank you for your answer. This classical result that the intertwiner $T$ is either zero or a multiple of identity is actually valid only for dimensions less than continuum. In mathoverflow.net/questions/336559/… , this is proven by a counterexample. (Please scroll to the bottom of that discussion.) $\endgroup$
    – Michael_1812
    Commented May 15, 2021 at 4:27
  • $\begingroup$ Also, your answer does not address the question asked. To prove Schur's lemma, is it sufficient to demonstrate the invariance of $\,{\cal{Im}} \,M\,$ and $\,{\cal{Ker}} \,M\,$, or is it necessary to prove the invariance of their closures? (And if the closures, then what additional assumptions should be made about the representations and/or the intertwiner?) $\endgroup$
    – Michael_1812
    Commented May 15, 2021 at 4:31
  • $\begingroup$ Finally, can we avoid requiring the representations to be unitary? $\endgroup$
    – Michael_1812
    Commented May 15, 2021 at 5:51
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    $\begingroup$ The argument assumed that T is bounded. In that case, it follows that im(T) is closed. $\endgroup$
    – Johannes Ebert
    Commented May 15, 2021 at 8:44
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    $\begingroup$ Johannes, your comment on the boundedness of T as a warrant of Im T being closed is exactly the answer I was looking for. Thank you for this. Can you kindly explain how the boundedness entails the closedness? (A physicist, and need help with things that may look simple to a mathematician.) $\endgroup$
    – Michael_1812
    Commented May 15, 2021 at 18:49

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