Since in von Neumann algebra having seperable predual any projection is countable union cyclic projections, why don't we closely look at cyclic projections and study their properties to get the general information of projections?? Because analogue to spectral theorem we study the cyclic representation it is more simpler in that set up. Are cyclic pretty much complicated to study?


1 Answer 1


There is nothing left to understand about a projection: the equality $P^*P=P$, and the fact that on a Hilbert space it corresponds to the orthogonal projection onto a subspace, say all that can be said about it.

Projections are interesting because they are the building blocks of any operator, but more importantly because of the relations between them (Murray-von Neumann equivalence, dimension functions, traces, etc.) It is not that much about the projection itself. Examples

  • Consider $M=M_2(\mathbb C)$, $P=E_{11}$, $Q=E_{22}$. Then $P\sim Q$. Now consider $N=\mathbb C\oplus\mathbb C\subset M$; in the subalgebra $N$, $P$ and $Q$ are not equivalent.

  • In an infinite-dimensional von Neumann algebra $M$, let $P$ be an infinite projection. In the subalgebra $N=\mathbb CP$, the projection $P$ is finite.

  • Let $M=\bigoplus_{n-1}^\infty M_2(\mathbb C)$. Then the identity $I$ is an infinite sum of pairwise orthogonal cyclic projections. Still, $I$ is a finite projection.

Note that the result is that an arbitrary projection is a union of cyclic projections. Unions of projections are not pretty, very little can be said about them in general.

Finally, an imperfect analogy: we know all there is to know about the number one (or even about the whole set of natural numbers). It doesn't help much in understanding the set of prime numbers.

  • $\begingroup$ Thanks for clarifying the ideas. Properties of projections depend on the structure of algebra when we are going to subalgebra properties get deformed. $\endgroup$
    – user548061
    Apr 12, 2018 at 15:23
  • 1
    $\begingroup$ Sorry Martin, I did not read carefully enough. I'll just remove the previous comment. $\endgroup$ Apr 13, 2018 at 22:30

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