Statement

Let MB(H)M\subseteq B(H) be a and let TT be a closed on HH, with polar decomposition T=vTT=v|T|. Then TT is if and only if the vv belongs to MM and every

ET(Δ)=1Δ(T)E^{|T|}(\Delta)=\mathbf 1_\Delta(|T|)

belongs to MM for every Borel set Δ[0,)\Delta\subseteq[0,\infty). Equivalently, vMv\in M and (T+i)1M(|T|+i)^{-1}\in M. For self-adjoint TT, affiliation is equivalent to all spectral projections of TT lying in MM.

Why the criteria agree

The defining commutation relation for affiliation says that TT commutes, as an unbounded operator, with every unitary in MM'. Uniqueness of polar decomposition then forces both vv and the spectral measure of T|T| to commute with MM'; the bicommutant theorem places them in MM. Conversely, these bounded data reconstruct TT by spectral integration and make it commute with MM'. See Takesaki, Chapter V, §5.

Self-adjoint and positive cases

For positive self-adjoint TT, the polar partial isometry is the support projection of TT, so membership of the spectral projections alone is sufficient. For self-adjoint TT, the bounded resolvent (Ti)1(T-i)^{-1} generates the same von Neumann algebra as the spectral projections, giving a concise resolvent criterion for affiliation.

Scope and cautions
References
  1. Masamichi Takesaki, Theory of Operator Algebras I, Springer, 1979. DOI record. Relevant: Chapter V, §5 on closed operators affiliated with von Neumann algebras and their polar and spectral data.
  2. Edward Nelson, “Notes on Non-Commutative Integration,” Journal of Functional Analysis 15 (1974), 103–116. DOI record. Relevant: §§1–2 on affiliated operators and spectral truncations.