$K_0(C(\mathbb{T}^{n})) \cong \mathbb{Z}^{2^{n-1}}$

Question

I am new to this website and I have a question.

I want to show that $K_0(C(\mathbb{T}^{n})) \cong \mathbb{Z}^{2^{n-1}}$ but first I want to show that $C(\mathbb{T}^{n}) \cong C(\mathbb{T} \rightarrow C(\mathbb{T}^{n-1}))$, where $C(\mathbb{T}^{n})$ is the continuous functions on the $n$-torus.

Can anybody help me out?

Answer 1

The algebras $C(\mathbb{T}^{n})$ and $C(\mathbb{T}^{n} , C(\mathbb{T}^{n-1}))$ are not isomorphic, as their spectra ($\mathbb T^n$ and $\mathbb T^n\times\mathbb T^{n-1}$ respectively) are not homeomorphic.

For computing $K_0(C(\mathbb{T}^{n}))$, cutting a circle out of $\mathbb T^n$ in a nice way yields an exact sequence $$0\to C_0(\mathbb T^{n-1}\times\mathbb R)\to C(\mathbb T^n)\leftrightarrows C(\mathbb T^{n-1})\to 0$$ And thus \begin{align*} K_i(C(\mathbb T^n))&\cong K_i(C(\mathbb T^{n-1}))\oplus K_i(C_0(\mathbb T^{n-1}\times\mathbb R))\\ &\cong K_i(C(\mathbb T^{n-1}))\oplus K_{1-i}(C(\mathbb T^{n-1})). \end{align*} Now apply induction.

EDIT To see that $C(\mathbb T^n)$ and $C(\mathbb T,C(\mathbb T^{n-1}))$ are isomorphic, note that if $f\in C(\mathbb T^n)$ then for each $z\in\mathbb T$, the map $f_z:\mathbb T^{n-1}\to\mathbb C$ given by $f_z(z_1,\ldots,z_{n-1})=f(z,z_1,\ldots,z_{n-2})$ is in $\mathbb C(\mathbb T^{n-1})$, and the map $z\mapsto f_z$ is continuous from $\mathbb T$ to $C(\mathbb T^{n-1})$. Now define $*$-homomorphism $\varphi:C(\mathbb T,C(\mathbb T^{n-1}))\to C(\mathbb T^n)$ by $$\varphi(f)(z_0,z_1,\ldots,z_{n-1})=f_{z_0}(z_1,\ldots,z_{n-1}).$$ This map is the (or one of the possible) desired isomorphism.