What is $K_1(B(H))$? (operator K-theory)

Question

Let $H$ be a (separable) infinite-dimensional Hilbert space. Can someone give me a reference or a proof why the identity $$K_1(B(H)) = 0$$ is true? The more elementary the proof, the happier I am. Thanks.

Answer 1

This is a standard proof, which should be available in any K-theory book (for operator algebras). Note that $M_n(B(H)) \simeq B(H^{\oplus n}) \simeq B(H)$ if $H$ is infinte-dimensional. So realistically, you can convince yourself that you can compute the $K_1$-group by computing $U(B(H))/U(B(H))_0$, the unitary gourp modulo the connected component.

So to see that the $K_1$ group is 0, you just need to show that every unitary in $B(H)$ is connected to the identity (via a path of unitaries). But since this is a von Neumann algebra, you've got the $L^{\infty}$-functional calculus for normal elements, which you can use to convince yourself that every unitary is of the form $e^{ia}$ for some self-adjoint $a \in B(H)$. Then the path $(e^{tia})_{t \in [0,1]}$ is a continuous path from 1 to $e^{ia}$.

Note that this actually gives that the $K_1$-group of any von Neumann algebra is trivial.