---
tag: BW678BX9ZX
kind: definition
name: Ultraweak topology
msc: [46L10, 47L05]
defines: CA38T5ZH14
uses: [HM2HHE3J51, BMDG4JQPEH, ZVFXQ7S6FY, Z2CP530WDE, 4XM1VN1C1N]
sources:
  - url: https://en.wikipedia.org/wiki/Ultraweak_topology
---

## Definition

Let $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.
For sequences $x = (x_n)$ and $y = (y_n)$ in $H$ with $\sum_n \|x_n\|^2 < \infty$ and $\sum_n \|y_n\|^2 < \infty$, let
$$\omega_{x,y}(T) = \sum_{n} \langle T x_n, y_n \rangle, \quad T \in B(H).$$
The *ultraweak topology*, or *$\sigma$-weak topology*, on $B(H)$ is the [[topology generated by the seminorms|ZVFXQ7S6FY]] $T \mapsto \abs{\omega_{x,y}(T)}$ for all such pairs of sequences.

## Remarks

The series defining $\omega_{x,y}(T)$ converges absolutely, by the Cauchy–Schwarz inequality.
Equivalently, the ultraweak topology is the weak-$*$ topology of the duality $B(H) \cong \mathcal{L}^1(H)^*$, where $\mathcal{L}^1(H)$ is the space of [[trace class operators|Z2CP530WDE]] and $T$ acts by $\rho \mapsto \operatorname{tr}(T \rho)$.

Example: for a trace class operator $\rho$, the functional $T \mapsto \operatorname{tr}(T\rho)$ is ultraweakly continuous.
Indeed, writing $\rho = \sum_n s_n \langle \cdot, e_n \rangle f_n$ in Schmidt form with $\sum_n s_n < \infty$ gives $\operatorname{tr}(T\rho) = \omega_{x,y}(T)$ with $x_n = \sqrt{s_n} f_n$ and $y_n = \sqrt{s_n} e_n$.

Non-example: let $H$ be infinite-dimensional and let $\varphi$ be a state on $B(H)$ that vanishes on all compact operators.
Such states exist because the quotient algebra by the compact operators has states.
Then $\varphi$ is norm-continuous but not ultraweakly continuous: every ultraweakly continuous functional has the form $T \mapsto \operatorname{tr}(T\rho)$, vanishing on all rank-one operators forces $\rho = 0$, and yet $\varphi(I) = 1$.

Every [[weak operator topology|4XM1VN1C1N]] open set is ultraweakly open, and the two topologies agree on every norm-bounded subset of $B(H)$.
They differ on all of $B(H)$ when $H$ is infinite-dimensional.