What is an example of a compact smooth manifold whose K-theory and Cech cohomology are not isomorphic?

Question

On page 283 of Max Karoubi’s book, “K-theory,” he states that for any compact Hausdorff space $X$, the Chern character determines an isomorphism from the group $K^0(X) \otimes Q$ to $H^{even}(X; Q)$, the direct sum of the even-dimensional Cech cohomology groups of X with rational coefficients. In particular, this theorem implies that $K^0(X)$ and $H^{even}(X; Z)$ are isomorphic up to torsion.

Does anyone know of an example (and also a reference, preferably) of a smooth compact manifold $X$ with the property that $K^0(X)$ and $H^{even}(X; Z)$ are not isomorphic?

Answer 1

For $X=\mathbb RP^4$ the groups $K^0(X)$ and $H^{even}(X)$ respectively are $\mathbb Z\oplus \mathbb Z/4$ and $\mathbb Z\oplus \mathbb Z/2\oplus \mathbb Z/2$. More generally, $K^0(\mathbb RP^{2k})\cong \mathbb Z\oplus \mathbb Z/2^{k}$. These computations of real and complex $K$-groups of real and complex projective spaces can be found in an early paper of J F Adams, I believe the one about vector fields on spheres. Let $k\to\infty$ so that $\mathbb RP^k$ becomes $BG$ for $G$ of order $2$. The Atiyah-Segal Completion Theorem says that for a finite or more generally compact Lie group $G$ the ring $K^0(BG)$ is the completion of the complex representation ring $RG$ with respect to the augmentation ideal (kernel of rank homomorphism $RG\to\mathbb Z$).

Even when the homology groups are torsion-free, so that $K^0$ is abstractly isomorphic to $H^{even}$, there is in some sense a difference between $K^0$ and $H^{even}$; they are then two slightly different lattices in the same rational vector space. For example, if $x$ generates $H^2(\mathbb CP^k,\mathbb Z)$ then a $\mathbb Z$-basis for $H^{even}(\mathbb CP^k)$ is $1,x,\dots,x^k$ while a $\mathbb Z$-basis for (the image under the Chern character of) $K^0(\mathbb CP^k)$ is $1,e^x,\dots,e^{kx}$ (or $1, e^x-1,\dots, (e^x-1)^k$). When $X$ is a sphere $S^{2k}$, they are the same lattice.