Be warned, this may be a ridiculous question.

I understand characteristic classes of principal $G$-bundles (and associated vector bundles) over a space $X$ arise from the classifying maps $f\colon X \to BG$ of those bundles by pulling back classes in $H^k(BG;R)$ along $f$, for some $k \in \mathbb{N}$ and coefficient ring $R$. For example, the Chern classes $c_k$ pull back from $H^{2k}(BU(n);\mathbb{Z})$ for big enough $n$, or from the colimit $H^*(BU;\mathbb{Z})$.

If $X$ is an $n$-manifold, then taking a cup product, with total degree $n$, of characteristic classes of the tangent bundle $TX$, and then evaluating it against the fundamental class, one gets a characteristic number of the manifold.

I know Pontrjagin numbers, Chern numbers, Stiefel–Whitney numbers, and the Euler characteristic arise in this way. But there are more characteristic classes I've seen named (although I couldn't tell you off the top of my head what $H^*(BG;R)$ they come from, and some I've never seen described that way; that would be nice to see too).

But it seems like there should be more characteristic numbers—there are after all more $G$ than $O(n)$ and $U(n)$ and more $R$ than $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$. Or maybe I'm just embarrassing myself on the internet.

What are some other characteristic numbers?


When you say "characteristic number of a manifold" instead of "characteristic class of a principal $G$-bundle" you're gliding over some important points. First, your manifold needs to be closed and oriented for it to have a fundamental class. Second, a priori the tangent bundle of an $n$-dimensional manifold only gives you a principal $O(n)$-bundle (up to homotopy), and in order to get a principal $G$-bundle out of this for $G$ some other interesting group you need to supply some extra data, usually a choice of reduction of the structure group from $O(n)$ to $G$, which also requires specifying a morphism $G \to O(n)$. In particular, you can't talk about the Chern numbers of a $2n$-manifold until you pick an almost complex structure on it, or equivalently a reduction of the structure group from $O(2n)$ to $U(n)$.

It's also worth pointing out that the classical cases of characteristic numbers you describe happen to be particularly important: they are complete cobordism invariants (for oriented and complex cobordism respectively depending on the characteristic number).

So, having said that, here's an example that depends on a choice of principal $G$-bundle. Let $G$ be a finite group and let $\alpha$ be a class in $H^n(BG, U(1))$ where $U(1)$ has the discrete topology. If $M$ is a closed oriented $n$-manifold, then associated to the classifying map $f : M \to BG$ of any principal $G$-bundle on $M$ there is a characteristic number $f^{\ast}(\alpha) [M]$ called the Dijkgraaf-Witten invariant of $f$. This construction is important in producing a topological quantum field theory called Dijkgraaf-Witten theory; the next step is to integrate the characteristic numbers above over the space of all principal $G$-bundles on $M$ in a suitable sense.

  • $\begingroup$ Thanks! It's really interesting that these characteristic numbers exist. $\endgroup$ – jdc Jul 15 '14 at 17:42
  • 1
    $\begingroup$ What role does the topology on $U(1)$ play here? I don't normally use topological structure on the coefficient group, so I'm not sure what role this assumption plays. $\endgroup$ – jdc Jul 15 '14 at 17:48
  • $\begingroup$ What sort of things does one learn from these invariants? Does one have to go to something as complicated as a space of all principal $G$-bundles to define other characteristic numbers? $\endgroup$ – jdc Jul 15 '14 at 17:50
  • 1
    $\begingroup$ This is all interesting. Thank you! For the $U(1)$-topology, I think I'm unfortunately asking a simpler question than the one you've answered. In the isomorphism $H^n(M;U(1)) \cong H^{n+1}(M;\mathbb Z)$, how is the cohomology group on the left defined to make this true? Are the cochains required to be continuous functions from the free abelian group on the singular simplicial complex to $U(1)$? Is the isomorphism somehow tied to $U(1)$ being $B\mathbb Z$? $\endgroup$ – jdc Jul 15 '14 at 20:53
  • 1
    $\begingroup$ @jdc: it's convenient, but yes, it's unnecessary. The point is that the natural map $BO(n) \to BGL_n(\mathbb{R})$ is a homotopy equivalence, so up to homotopy specifying a classifying map into the latter is equivalent to specifying a classifying map into the former. More concretely this reflects the fact that the space of Riemannian metrics on a vector bundle is contractible. $\endgroup$ – Qiaochu Yuan Jul 15 '14 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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