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N.P. Landsman (2017) defines quasi-equivalent representations as:

Two representations $\pi_1,\,\pi_2$ are quasi-equivalent if every subrepresentation of $\pi_1$ has a subrepresentation that is (unitarily) equivalent to some subrepresentation of $\pi_2$, and vice versa;

Is it necessary to talk about a subrepresentation of a subrepresentation of $\pi_1$? Or was it some blunder of the writer?

I understand that this is not equivalent to the definition given upwards, but shouldn't the definition of quasi-equivalent representations be:

Two representations $\pi_1,\,\pi_2$ are quasi-equivalent if every subrepresentation of $\pi_1$ is (unitarily) equivalent to some subrepresentation of $\pi_2$, and vice versa;

Landsman's definition seems to imply that there are subrepresentations of $\pi_1$ that are not unitarily equivalent to any subrepresentation of $\pi_2$, but there are always subrepresentations of these subrepresentations that are unitarily equivalent to some subrepresentation of $\pi_2$. Is this the sense in which they are quasi-equivalent? That is, if what I wrote was true would they be equivalent? That is, one could then be able to create a unitary transformation composed by each of these unitary transformations of each subrepresentations therefore implying full on equivalence, or just being able to talk about subrepresentations makes it weaker in a way that you cannot talk about it globally?

I looked at other places the definition of quasi-equivalent representations, but it is always given in a somewhat convoluted way that makes it obscure to understand what that would mean in terms of subrepresentations, like:

  1. there exist unitarily-equivalent representations $\rho_1$ and $\rho_2$ such that $\rho_1$ is a multiple of $\pi_1$ and $\rho_2$ is a multiple of $\pi_2$.

  2. the non-zero subrepresentations of $\pi_1$ are not disjoint from $\pi_2$, and the non-zero subrepresentations of $\pi_2$ are not disjoint from $\pi_1$.

  3. $\pi_2$ is unitarily equivalent to a subrepresentation of some multiple representation $\rho_1$ of $\pi_1$ that has unit central support.

  4. there exists an isomorphism Φ of the von Neumann algebra generated by the set $\pi_1(X)$ onto the von Neumann algebra generated by the set $\pi_2(X)$ such that $Φ(π_1(x))=π_2(x)$ for all $x\in X$.

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    $\begingroup$ According to Landsman's own website, he has no papers from 2017. Please provide a source for this, because this is an overly convulted way of describing what I know quasi-equivalent representations to be. $\endgroup$ Jun 15, 2023 at 11:14
  • $\begingroup$ @DavidA.Craven it is not in a paper, it is in his book "Foundations of Quantum Theory From Classical Concepts to Operator Algebras" of 2017. $\endgroup$ Dec 11, 2023 at 17:16

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I understand that this is not equivalent to the definition given upwards, but shouldn't the definition of quasi-equivalent representations be:

Two representations $\pi_1$,$\pi_2$ are quasi-equivalent if every subrepresentation of $\pi_1$ is (unitarily) equivalent to some subrepresentation of $\pi_2$, and vice versa;

No. Your definition isn't even an equivalence relation, for example.

To see what is going on, just stick to finite groups, which makes life a lot easier. Here, quasi-equivalent means that every irreducible subrepresentation that appears in one must appear in the other. So if we take the cyclic group of order $2$, which has two irreducible representations, $\pm 1$, then any two representations that contain both are quasi-equivalent. So for this group there are three quasi-equivalence classes:

  1. Multiples of the trivial representation;
  2. Multiples of the $-1$ representation;
  3. All other representations.

In general, for finite groups with $n$ irreducible representations, there are $2^n-1$ quasi-equivalence classes of representations, which is the power set of the set of irreducible representations minus the empty set.

Once one moves away from finite-dimensional representations of finite groups things get more convoluted, but the finite case should help explain where your reasoning went wrong.

Edit: To specifically address whether it is a blunder, no certainly not. The condition is that every subrepresentation has a property P. In particular, $\pi_1$ itself has property P. Thus removing the second 'subrepresentation' would have the effect of claiming that $\pi_1$ is a subreprsentation of $\pi_2$.

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