Ideas for calculating $K_0(l_{\infty})$ and $K_1(l_{\infty})$.

Question

Thank you for answering my question. I'm a bit new to K-theory.

So I was wondering how can I calculate $K_0(l_{\infty})$ and $K_1(l_{\infty})$.

I think if we have one, then by using bott periodicity we can have the other one.

Can anyone help me?

Answer 1

It can be shown that $K_0(\ell^\infty)$ is isomorphic to the collection of all bounded functions $\mathbb N\to\mathbb Z$, and $K_1(\ell^\infty)=0$.

To see the result about $K_0$, first observe that if $p\in M_n(\ell^\infty)$ is a projection, then $p$ is unitarily equivalent to $p_1\oplus\cdots\oplus p_n$ for some projections $p_k\in\ell^\infty$. Thus $K_0(\ell^\infty)$ is the $\mathbb Z$-linear span of the $K_0$-classes of projections in $\ell^\infty$. Now define a map $\varphi: K_0(\ell^\infty)\to\{f:\mathbb N\to\mathbb Z\text{ bounded}\}$ by linear extension of $$\varphi([\chi_E])=\chi_E$$ for any $E\subset\mathbb N$, where $\chi_E:\mathbb N\to\{0,1\}$ is the characteristic function of $E$. This map gives the desired isomorphism.

For $K_1$, we can cheat by observing that $\ell^\infty$ is a von Neumann algebra, so it has a Borel functional calculus (as do all matrix algebras over $\ell^\infty$). This implies that their unitary groups are path connected (if $u$ is a unitary, take a Borel measurable logarithm on its spectrum to find a self-adjoint element $a$ such that $u=e^{ia}$, and thus $u$ is homotopic to $1$). This then implies that $K_1(\ell^\infty)=0$. (This proof holds more generally for any von Neumann algebra.)