# What is the definition of quasi-equivalent representations by subrepresentations?

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 $$x\in X$$.

• 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. Jun 15, 2023 at 11:14
• @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. Dec 11, 2023 at 17:16

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 $$\pi_2$$.