Cyclic vector for an operator algebra ($\overline{A\Omega }=H$)

A vector is cyclic for an operator algebra when its orbit under that algebra has dense linear span. This is distinct from requiring the powers of one chosen operator to span the space. If the algebra is linear, its orbit itself is already a vector subspace.

A vector is separating for an operator algebra when evaluation at that vector is injective. A cyclic vector for the commutant of an operator algebra is separating for the original algebra: if $a\Omega =0$, then $a$ vanishes on its dense commutant orbit.

For a Von Neumann algebra with a cyclic separating vector, this is a densely defined antilinear operator. It is closable: the commutant orbit lies in the domain of its antilinear adjoint and is dense. Write $S$ for its closure and $S=J{\Delta }^{1/2}$ for its polar decomposition. In finite dimension it is everywhere defined and invertible. For a tracial vector, $S_0$ is isometric and extends to an antiunitary involution.

The positive operator in the polar decomposition $S=J{\Delta }^{1/2}$ of the Tomita operator. In a matrix-algebra standard representation with cyclic vector $D>0$, it is $\Delta (X)=D^{2}X{D}^{-2}$. For a tracial vector it is the identity.

The antiunitary factor in the polar decomposition of the Tomita operator. In the finite-dimensional and tracial settings it satisfies $J^2=I$ and $JMJ=M^{\prime }$. In a matrix-algebra standard representation by left multiplication, with positive invertible cyclic vector $D$, it acts by $J(X)=X^{\ast }$.