Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I read everywhere :"By definition, the characteristic classes of smooth manifolds are invariant under diffeomorphisms." Does it follow from de Rham cohomology? If this is so, then what about topological manifolds? In this case we cannot use diffeomorphism invariance of de Rham cohomology. Besides even if we know that cohomology classes are preserved under diffeomorphisms (is this correct for the cohomology with all coefficients (not only Z, but also Q and R coefficients)?), how can we be sure that these classes are also preserved? I cannot see it from the definitions.

share|cite|improve this question
1. What is your definition of characteristic class? 2. When you say "the characteristic classes of a smooth manifold" you mean "the characteristic classes associated to the tangent bundle." What plays the role of "tangent bundle" in the topological manifolds setting? – Jason DeVito Feb 22 '12 at 21:56
What characteristic classes do you have in mind? There are many, with different invariances... – Mariano Suárez-Alvarez Feb 22 '12 at 21:56
up vote 0 down vote accepted

To say it crisply, characteristic classes are invariant under diffeomorphisms because they are invariant under homeomorphisms!

A little more in detail, characteristic classes for real vector bundles (for example) are defined for paracompact topological spaces , which needn't even be topological manifolds .
Given a real vector bundle $E$ of rank $r$ on a paracompact space $X$, there exists a continuous map $ f:X\to G_r(\mathbb R^\infty)$ such that $f^*(\gamma)\cong E$ and that map is unique up to homotopy.
Here $G_r(\mathbb R^\infty)$ denotes the Grassmannian of real $r-$planes in infinite-dimensional affine space $\mathbb R^\infty$ and $\gamma$ is the tautological $r$- bundle on that Grassmannian.
The isomorphism above is between the pull-back $f^*(\gamma)$ of this bundle and the given bundle $E$ .

Now if you have chosen a ring of coefficients $\Lambda$ and an element $c\in H^*(G_r(\mathbb R^\infty),\Lambda$), the pull-back $f^*(c)\in H^*(X,\Lambda)$ will get you the corresponding characteristic class of $E$.

There is no need for differentiability in this theory, but of course if $X$ happens to be a manifold you'll get a whole bunch of new bundles derived from the tangent bundles and characteristic classes will tell you a lot about them.
A favourite reference on the subject is Milnor-Stasheff's Characteristic Classes

share|cite|improve this answer
thanks. (as topological invariance of them was proved (by Novikov) Later than the fact that they are smooth invariants, I just wanted to see it without considering homeomorphisms.) – wqr Feb 23 '12 at 8:40

You can only talk meaningfully about diffeomorphisms on smooth manifolds, since only they guarantee smooth transition functions between patches. In that case, de Rham's Theorem guarantees an isomorphism between de Rham cohomology and singular cohomology.

share|cite|improve this answer

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.