Dependence of operator topologies in a $C^*$ algebra on the representation Let $A$ be a $C^*$ algebra. Given a faithful representation $\pi:A\to \mathcal{B}(H)$, we can define the weak operator topology with respect to $\pi$ as initial with respect to the maps $a\mapsto \langle \pi(a) x,y\rangle$ for each $x,y\in\mathcal{H}$. However, this definition depends on our representation $\pi$. 
Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?
 A: No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation. 
For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(\mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $\pi_1:A\to B(H_1)$ be the representation induced by the left-regular representation. Take $\pi_2:A\to B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple). 
Then we have 
$$
\overline{\pi_1(A)}^{\rm sot, wot, etc.}=L(\mathbb F_2),
$$
a II$_1$-factor. While 
$$
\overline{\pi_2(A)}^{\rm sot, wot, etc.}=B(H_2), 
$$
a type I factor. 
The example is a bit more dramatic if you take $A\subset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $\pi_1$ to be the identity, and $\pi_2$ to be an irreducible representation. Then $\pi_1(A)$ is sot/wot closed, while $\overline{\pi_2(A)}^{\rm sot, wot}=B(H_2)$. 
