@MoisheKohan: Ah! I forgot to mention that I’m supposed to solve the problem without assuming that topological $ 2 $-manifolds (with or without boundary) can be triangulated. The reason for my post is that a similar result was used by P. H. Doyle and D. A. Moran to establish the main objective of their paper A Short Proof that Compact $ 2 $-Manifolds Can Be Triangulated, but they neglected to prove that result, merely saying that it was a standard result in geometric topology.

@MoisheKohan: Thanks! However, Epstein does assume that the topological $ 2 $-manifold comes with a triangulation, which I can’t.

@NikWeaver: This may sound like a silly question, Nik. Do you happen to know if every strongly continuous action of $ G $ on $ \mathbb{K}(\mathcal{H}) $ extends to a strongly continuous action of $ G $ on $ \mathbb{B}(\mathcal{H}) $? I’m assuming $ G $ to be an arbitrary locally compact Hausdorff group.

Yes, we always get a projective representation.

@NikWeaver: Shouldn’t it be $ \mathbb{U}(\mathcal{H}) $ modulo the circle group instead?

Thanks for your counterexample, Nik! Actually, I just found out that my question has a negative answer by way of what’s called the “Mackey obstruction”. It’s an element of $ {H^{2}}(G,\mathbf{T}) $ associated to every strongly continuous action $ \alpha $ of $ G $ on $ \mathbb{K}(\mathcal{H}) $ with the property that if it isn’t trivial, then $ \alpha $ can’t be implemented by even an algebraic homomorphism from $ G $ to $ \mathbb{U}(\mathcal{H}) $, much less a norm-continuous one.

@YCor: I apologize for any confusion that I may have caused you. I’ve clarified in my post what I mean by a group action on a $ C^{\ast} $-algebra.