K-theory of classifying spaces

Question

Can someone help me calculate the following groups in $ K $-theory

1) $ KU^0 (B\mathbb{S}^1) $

2) $ KU^0 (\mathbb{RP}^\infty) $

where $ B \mathbb{S}^1$ is the classifying space of $ \mathbb{S}^1 $

Thank you

Answer 1

This is an application of the Atiyah-Segal completion theorem.

Let $G$ be a compact Lie Group and let $R_{\mathbf{C}}(G)$ be its representation ring of finite dimensional continuous complex representations and let $I$ be the augmentation ideal i.e. the kernel of the canonical morphism $R_{\mathbf{C}}(G)\to \mathbf{Z}$ which sends every representation to its dimension. If we let $BG$ denote the classifying space of this group, then the content of the Atiyah-Segal Completion theorem is that $$K^0(BG)\cong R_{\mathbf{C}}(G)_{\hat{I}}$$ and that $K^{1}(BG)=0$, where $R_{\mathbf{C}}(G)_{\hat{I}}$ is the completion of the representation ring at $I$. We note that the Completion theorem says more than just this, but this will suffice for our purposes.

As a computational tool note that if $R$ is a noetherian ring and $I=(a_1,\ldots, a_n)$, then $$R_{\hat{I}}=R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots, x_n-a_n).$$

Now in the harder second case we have that $\mathbf{RP}^{\infty}=B\mathbf{Z}/2$. Notice that the representation ring of $\mathbf{Z}/2$ can canonically be identified with its character ring, since each irreducible representation is one dimensional and there are two of them. This shows that we must have $R_{\mathbf{C}}(\mathbf{Z}/2)=\mathbf{Z}[x]/(x^{2}-1)$. The augmentation ideal is generated by $(x-1)$. Set $x-1=u$, and we see that we are completing the ring $\mathbf{Z}[u]/((u+1)^{2}-1)$ at the ideal $u$ but this is just isomorphic to $\mathbf{Z}[[u]]/((u+1)^{2}-1)$ . Thus we have $$K^{0}(\mathbf{RP}^{\infty})=\mathbf{Z}[[u]]/((u+1)^{2}-1)$$ and $K^1(\mathbf{RP}^{\infty})=0.$

In the other case we have $G=S^{1}$. One can show that $R_{\mathbf{C}}(S^{1})=\mathbf{Z}[x,x^{-1}]$. Running the above argument once again we see that the augmentation ideal is really just $(x-1)$ again. Changing coordinates $x-1=c$ we see that we are completing the ring $\mathbf{Z}[c+1,(c+1)^{-1}]$ at $(c)$. But clearly this is just $\mathbf{Z}[[c]]$.

Thus $$K^0(BS^{1})=\mathbf{Z}[[c]]$$ and $K^1(BS^{1})=0.$