Why shift operator is not homotopic to 1 ($K_1$-approach)?

Question

Let us recall that via fourier transform it holds true that $C^*(S)\cong C(\mathbb{T})$, with map given by $S\mapsto e^{2\pi i x}$ (considering $\mathbb{T}=\mathbb{R}/\mathbb{Z}$). It is also true that $K_1(C(\mathbb{T}))=\mathbb{Z}[e^{2\pi i x}]$, so it follows that $$K_1(C^*(S))=\mathbb{Z}[S],$$ and of coure this implies that $S$ is not homotopic to the identity (if it were, $[S]=[1]$ and then $K_0(C^*(S))$ would be the trivial group).

On the other side, let us define $u(t)=e^{2\pi i x t}$ and observe that

• $u(0)=1$ and $u(1)=e^{2\pi i x}$
• $u(t)$ is norm continuous since \begin{equation} \begin{array}{rl} ||u(t)-u(s)||&=||e^{2\pi i x t}(1-e^{2\pi i x (s-t)})||\\ &\leq||1-\sum_{k\geq 0}\frac{(2\pi i x (s-t))^k}{k!}||\\ &=||\sum_{k\geq 1}\frac{(2\pi i x (s-t))^k}{k!}||\\ &\leq||(2\pi i x (s-t))\sum_{k\geq 0}\frac{(2\pi i x (s-t))^k}{(k+1)!}||\\ &\leq||(2\pi i x (s-t))||\sum_{k\geq 0}\frac{||(2\pi i x (s-t))^k||}{(k+1)!}\\ &\leq|s-t|\sum_{k\geq 0}\frac{(2\pi)^k}{k!} = e^{2\pi}|t-s| \end{array} \end{equation}
• $u(t)$ is unitary for every $t$, since $u(t)^*=e^{-2\pi i x t}$ and then $u(t)u(t)^*=1$.
• $u(t)\in C(\mathbb{T})$ by functional calculus, since $u(t)$ is a continuous function of $e^{2\pi i x}$ for every $t\in[0,1]$.

This proves that $e^{2\pi i x}$ is homotopic to $1$ and of course that the shift operator $S$ is homotopic to the identity, which can't be true, so... where is the issue?

Answer 1

What you've found is a path connecting $S$ and $1$ in the unitary group of $\mathbb B(\ell^2(\mathbb Z))$. What you've failed to show is that for all $t\in[0,1]$, $u(t)\in C^*(S)$, i.e., you haven't shown that $S$ and $1$ are homotopic in $C^*(S)$. But this is a good thing, because it's not true, and one way to see this is through $K$-theory: the class of $S$ in $K_1(C^*(S))\cong\mathbb Z$ is a generator, while the class of $1$ is the identity.