In the introduction of their 1993 paper (see reference below), Popa and Takesaki write
As it turns out, in these topologies [the weak and strong topology] $U(\mathscr{H})$ is again contractible (cf. [8]) by an argument that goes back to Kakutani's proof in 1943 of the contractibility of the unit sphere of the infinite dimensional Hilbert space [10]. Moreover, the same proof shows that $U(\mathscr{M})$ is contractible for any properly infinite yon Neumann algebra $\mathscr{M}$.
Here $U(\mathscr{H})$ denotes the unitary group of a separable Hilbert space, and the citation [8] is the book of Dixmier, where the contractibility of $U(\mathscr{H})$ is shown by taking $\mathscr{H} = L^2([0, 1])$ using the homotopy $$ H_t(u) = 1-S_t S_t^* + S_t u S_t^*, ~~~~~~u \in U(\mathscr{H}),$$ with $S_t$ the partial isometry given by $$S_t(f)(x) = \begin{cases} t^{-1/2}f(s/t) & s \in [0, t] \\ 0 & s \in [t, 1]. \end{cases}$$ I am trying to understand the last comment of the cited part above. To make this proof work for the unitary group of a von Neumann algebra on a Hilbert space $\mathscr{H}$, we would need a strongly continuous family $S_t \in \mathscr{M}$, $t \in [0, 1]$ of partial isometries that converge to strongly zero as $t \rightarrow 0$ and to the identity as $t \rightarrow 1$.
If $\mathscr{M}$ is properly infinite, then by definition, there exists a partial isometry $S \in \mathscr{M}$ such that $S^*S = 1$, but $SS^* < 1$. We can now look at the $n$-th powers $S^n$ of $S$. But these need not converge to zero strongly in general (the intersection of the images of the $S^n$ is a subspace on which acts as a unitary in general), and I don't see how to obtain a continuous family from this discrete one.
How does one fix the argument?
Popa, Sorin; Takesaki, Masamichi, The topological structure of the unitary and automorphism groups of a factor, Commun. Math. Phys. 155, No. 1, 93-101 (1993). ZBL0799.46074.
