I'm reading Murphy's book on C*-algebras, and I'm a tad confused about the following statement in the book (rephrased):
Since $B(H)$ is the dual of $L^1(H)$ (the space of trace-class operators), then the ultraweak topology on $B(H)$ is just the weak$^*$ topology on $B(H)$
I'm a little bit confused here because the weak$^*$ topology is a topology on the dual of $B(H)$, not on $B(H)$. Can someone help me understand this? Thanks!
"I'm a little bit confused here because the weak$^∗$ topology is a topology on the dual of $B(H)$, not on $B(H)$"
You are misinterpreting the sentence that you quoted from Murphy. When one talks of the weak$^*$ topology on some Banach space, it is implied that this Banach space is actually the dual of some other space, and the mentioned weak$^*$ topology is the one our space inherits as a dual space.
In this particular setting: consider the space $L^1(H)$; as explained in the book, $B(H)\cong L^1(H)^*$, so $B(H)$ itself has a weak$^*$ topology, as the dual of $L^1(H)$: namely, for a net $(T_i)\subset B(H)$ and an operator $T\in B(H)$, we have that $T_i\to T$ in the weak$^*$ topology if and only if, for any trace class operator $S\in L^1(H)$, we have $\text{tr}(T_iS)\to\text{tr}(TS)$ (as numbers), where $\mathrm{tr}$ denotes the trace map.