Weak operator topology and matrix representation

Question

For simplicity let $H$ be the separable Hilbert space and $(e_i)_i$ be a fixed orthonormal basis. Weak operator topology on $B(H)$ is usually defined using seminorms $|(Tu,v)|$ where $u,v$ are vectors in $H$. Let us denote $\mathcal{T}$ the topology induced by the seminorms $|(Te_i,e_j)|$. I believe this is also the topology induced by all $|(Tu,v)|$ where $u,v$ are finite linear combination of $e_i$'s. What is the relation between $\mathcal{T}$ and WOT? Do they agree on norm bounded sets? I guess yes because

$||(Tu,v)|-|(Tu_0,v_0)||\leq|(Tu,v)-(Tu_0,v_0)|=|(Tu-Tu_0,v)+(Tu_0,v-v_0)|$

$\leq||T||\cdot||u-u_0||\cdot||v||+||T||\cdot||u_0||\cdot||v-v_0||$

So for any fixed $u,v$ and $\epsilon>0$, if $u_0,v_0\in \text{span}\{e_i\mid i\in\mathbb{N}\}$ are close enough to $u,v$ and there is a uniform bound on $||T||$, then $|(Tu_0,v_0)|\leq \epsilon$ implies $|(Tu,v)|\leq 2\epsilon$.

Now since we fix a basis $(e_i)_i$ we can regard operators as infinite matrices and $(Te_i,e_j)$ is the $j,i$-th entry. Therefore if the above calculation is correct then on bounded sets WOT can be viewed as “entrywise convergence topology”. This looks more comfortable and intuitive for me, for example it becomes even clearer that unit ball is compact because we can view it as a subspace of $[-1,1]^{\mathbb{N}^2}$ (I found that I don't know how to argue it is a closed subspace without using the fact that it is same as WOT). Similarly strong operator topology may be viewed as “columnwise convergence topology”.

Answer 1

The answer is no. Although $\mathcal T$ coincides with the WOT on bounded sets, as observed by the OP, these topologies do not coincide over the whole of $B(H)$.

One very nice point of view from which to understand this question is the theory of duality. Suppose $B$ is a complex vector space (I am using the letter $B$ here because the application will be for $B=B(H)$). Suppose also we are given a set $\Phi $ of linear functionals on $B$.

One may then define a topology $\mathcal T$ on $B$ using the seminorms $$ T\in B\mapsto |\varphi (T)|\in \mathbb R, $$ for every $\varphi $ in $\Phi $.

It is evident that every $\varphi $ in $\Phi $ is $\mathcal T$-continuous and hence also every linear combination of functionals in $\Phi $.

The key fact is that nothing else is continuous! In other words, the topological dual of $(B,\mathcal T)$ is precisely the linear span of $\Phi $ within the algebraic dual of $B$.

Taking $B=B(H)$ and $\Phi $ the set of linear functionals of the form $$ T\in B(H)\mapsto \langle Te_i, e_j\rangle , \tag{1} $$ we may ask whether or not a linear functional of the form $$ T\in B(H)\mapsto \langle Tu, v\rangle , \tag {2} $$ is $\mathcal T$-continuous, for given vectors $u$ and $v$ in $H$.

The answer will be negative if $u$ and $v$ are infinite linear combinations of basis elements since one may easily prove that (2) is not a finite linear combination of functionals as in (1).