MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I don't know the answer to the question, but I wonder if the rational case is due to a spectrum level effect. Does anyone know if $K\wedge Q$ splits as a product of Eilenberg-MacLane spectra? – Greg Friedman Jul 2 '10 at 20:38
$E \wedge HQ$ always splits (up to equivalence) as a product of Eilenberg-Maclane spectra. This is true for any spectrum E. – Chris Schommer-Pries Jul 2 '10 at 21:07
Is it ok to replace "Cech cohomology" with just "cohomology" here? – Kevin H. Lin Jul 3 '10 at 3:35
Yes. Perhaps it's the more natural recipient of the Chern character, but for manifolds and many other spaces it's equivalent to singular cohomology. – Tom Goodwillie Jul 3 '10 at 3:38
One way to think about this is that the Chern character isomorphism implies that the Atiyah-Hirzebruch spectral sequence collapses rationally. So you're just looking for an example in which something interesting happens to the torsion in this spectral sequence. Tom's answer shows that there can be non-trivial extensions. I suppose it's then natural to ask for an example with non-zero differentials. – Dan Ramras Jul 3 '10 at 4:39
up vote 18 down vote accepted

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.

share|cite|improve this answer
Another reference for the calculation of the complex K-theory of real projective spaces is Atiyah's K-theory book, Proposition 2.7.7. – Allen Hatcher Jul 3 '10 at 5:14
Are there any easy orientable examples? – Greg Friedman Jul 5 '10 at 19:44
$\mathbb RP^5$ is just as good an example as $\mathbb RP^4$. The inclusion of the latter in the former induces isomorphisms in $H^{even}$ and in $K^0$. – Tom Goodwillie Jul 5 '10 at 23:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.