Can homomorphisms be Borel in some weaker topology?

Question

Let $X$ be a Banach space with separable dual and let $A$ be a Banach algebra. Consider a norm continuous homomorphism $h$ from $L(X)$, the Banach algebra of bounded operators on $X$ onto $A$. In $L(X)$ we have all sorts of weaker topologies such as the SOT or WOT. Can it be the case that $h$ is Borel with respect to either topology?

Answer 1

It doesn't look so. There is a separable reflexive space $X$, so that $L(X)$ has $2^{2^{\aleph_0}}$ homomorphisms to $A = \mathbb C$ (see this thread).

As the space is reflexive, the unit ball of $L(X)$ is compact in WOT/SOT, and since $X$ is separable, it is metrisable. Each homomorphism is determined by its behaviour on the unit ball of $L(X)$, but you can't have more than continuum complex Borel functions on a compact metric space, so most of them are not Borel.