Ultrastrong-star topology - Imported Knowlpedia Content

Definition

Let $M$ be a von Neumann algebra. The ultrastrong-star topology, or $\sigma$ -strong-star topology, is the locally convex topology generated by

x\longmapsto \bigl(\varphi(x^*x)+\varphi(xx^*)\bigr)^{1/2},

where $\varphi$ ranges over the positive normal functionals on $M$ . Equivalently, a net $x_i$ converges to $x$ ultrastrong-star exactly when both $x_i\to x$ and $x_i^*\to x^*$ in the ultrastrong topology. This definition makes involution continuous by construction and is intrinsic to the von Neumann algebra.

Concrete seminorms

For $M\subseteq B(H)$ , the inherited topology is generated by

q_{\boldsymbol{\xi}}(T)= \left(\sum_{n=1}^{\infty}\|T\xi_n\|^2+ \sum_{n=1}^{\infty}\|T^*\xi_n\|^2\right)^{1/2},

where $(\xi_n)$ ranges over square-summable sequences in $H$ . The two sums separately measure ultrastrong convergence of $T$ and $T^*$ . A faithful normal representation gives the same intrinsic topology Takesaki, Chapter III, §2.

Comparison with neighboring topologies

Ultrastrong-star convergence implies ultrastrong convergence and strong-operator convergence. On norm-bounded subsets it agrees with the strong-star operator topology, but the two topologies differ on the whole algebra in infinite dimension. The ultrastrong-star topology is therefore not merely another name for strong-star convergence.

Algebraic behavior

Involution is globally continuous, while multiplication is jointly continuous on norm-bounded subsets. These properties make ultrastrong-star convergence convenient when limits must respect both adjoints and products. Without boundedness, multiplication need not be jointly continuous, so convergence of $x_i$ and $y_i$ alone does not justify convergence of $x_iy_i$ .

References

Masamichi Takesaki, Theory of Operator Algebras I, Springer, 1979. Springer DOI record. Relevant: Chapter III, §2 on the $\sigma$ -strong and $\sigma$ -strong-star topologies.