topologies defined on $B(H)$

Question

If $H$ is a Hilbert sapce,there are many topologies can be defined on $B(H)$:norm topology, strong operator topology,weak operator topology, σ-strong topology,σ-weak topology....

If $H$ is finite dimensional ,Can we deduce that all these topologies are the same?

Answer 1

Yes, they are all the same. The weak operator topology is the weakest of them all. One can show that in $M_n(\mathbb C)$ convergence in the wot is entrywise convergence. The latter is given by the norm $\|A\|_1=\sum_{k,j}|A_{kj}|$; and on a finite-dimensional vector space, all norms are equivalent.

Even more generally, it is not hard to show that on a finite-dimensional vector space there is a single topological-vector-space Hausdorff topology.

Answer 2

Yes: SOT, WOT and the norm topology are all equal when $H$ is finite-dimensional.

Clearly both the SOT and WOT are contained in the norm topology.

The basis elements for the SOT are given by $$U_{A, x_1, \ldots, x_n, \varepsilon} = \{B \in B(H) : \|(A - B)x_i\| < \varepsilon, \forall i = 1, \ldots, n\}$$ with $A \in B(H)$, $x_1, \ldots, x_n \in H$ and $\varepsilon > 0$.

The basis elements for the WOT are given by $$V_{A, x_1, \ldots, x_n, y_1, \ldots, y_n, \varepsilon} = \{B \in B(H) : \left|\langle (A - B)x_i, y\rangle\right| < \varepsilon, \forall i = 1, \ldots, n\}$$

with $A \in B(H)$, $x_1, \ldots, x_n, y_1, \ldots, y_n \in H$ and $\varepsilon > 0$.

Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis for $H$. See here that for $T \in B(H)$ we have

$$\|T\| \le \max\limits_{1 \le i \le j \le n}|\langle Te_i, e_j\rangle| \cdot \sqrt{n}$$

Let $A \in B(H)$ and $r > 0$ be arbitrary. We need to show that the open ball $B(A, r)$ w.r.t to the norm topology is contained in SOT and WOT.

The claim follows from

$$B(A, r) = U_{A, e_1, \ldots, e_n, \frac{r}{\sqrt{n}}} = \bigcap_{1 \le i \le j \le n} V_{A, e_i, \ldots, e_i, e_j, \ldots, e_j \frac{r}{\sqrt{n}}}$$

Let $B \in U_{A, e_1, \ldots, e_n, \frac{r}n}$. Then $$\|B - A\| \le \max_{1 \le i \le j \le n}|\langle (B-A)e_i, e_j\rangle| \cdot \sqrt{n} \le \max_{1 \le i\le n}\overbrace{\|(B-A)e_i\|}^{< \frac{r}{\sqrt{n}}} \cdot \sqrt{n} < r \implies B \in B(A, r)$$

Conversely, if $\|B-A\| < r$ then clearly $\|(B-A)e_i\| < r$ for all $1 \le i \le n$.

Let $B \in \bigcap_{1 \le i \le j \le n} V_{A, e_i, \ldots, e_i, e_j, \ldots, e_j \frac{r}{\sqrt{n}}}$. Then $$\|B - A\| \le \max_{1 \le i \le j \le n}\overbrace{|\langle (B-A)e_i, e_j\rangle|}^{< \frac{r}{\sqrt{n}}} \cdot \sqrt{n} < r \implies B \in B(A, r)$$

Conversely, if $\|B-A\| < r$ then clearly by CSB we have $|\langle (B-A)e_i, e_j\rangle| < r$ for all $1 \le i \le j \le n$.