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A bit of context for this question: as a project for my master's degree my supervisor asked me to understand the construction of Milnor's exotic spheres. After learning the heavy material (I knew very little algebraic topology so learning about characteristic classes counted as "heavy" for me) the proof of existence of these spheres is surprisingly short and easy.

I have finished early and I would like to go on and explore other ideas, ideally results which are not unreasonably difficult and preferably are "surprising" i.e. the conclusion is something other than the classification of vector bundles over a thing. Any pointers? :)

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    $\begingroup$ Milnor-Stasheff has lots of nice applications of characteristic classes, such as the fact that $\mathbb{R}P^2$ does not smoothly embed in $\mathbb{R}^3$ and that every compact orientable $3$-manifold is parallelizable. Though most such applications come down to arguing that "if you could do $X$ then there would be a vector bundle over a thing with property $Y$, and that can't happen because the characteristic classes of the thing are zero". $\endgroup$
    – Paul Siegel
    Commented May 19, 2016 at 20:29
  • $\begingroup$ There is the work of Thom about unoriented bordism being determined by SW numbers. $\endgroup$
    – Sean Tilson
    Commented May 23, 2016 at 2:10
  • $\begingroup$ Thank you all for your very interesting input! This has helped a lot. $\endgroup$
    – R Mary
    Commented May 24, 2016 at 11:44

3 Answers 3

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In this blog post you'll find a computation of the cohomology ring of a hypersurface of degree $d$ in $\mathbb{CP}^3$ using characteristic classes. This turns out to be a weirdly good exercise in using characteristic classes: the computation invokes, in order, Euler classes, Chern classes, Pontryagin classes, Wu classes, and Stiefel-Whitney classes, and doing it is what made me comfortable with characteristic class computations.

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    $\begingroup$ One of the easiest examples of cohomology computations using characteristic classes is by the Gysin sequence. If you know the Euler class of a sphere bundle, this sequence relates the cohomology of base and total space (in fact, it is a clever shortcut through calculations with the Leray-Serre spectral sequence). This way, you can determine the cohomology of complex and quaternionic projective spaces, for example. $\endgroup$
    – Sebastian Goette
    Commented May 20, 2016 at 8:57
  • $\begingroup$ Is your $\mathbb{CP}^3$ actually the 4-real dimension manifold? (Which people usually call it $\mathbb{CP}^2$ instead of $\mathbb{CP}^3$?) While the $\mathbb{CP}^1 \simeq S^2$ homeomorphic and diffeomorphic to a 2-sphere. $\endgroup$
    – wonderich
    Commented Jul 10, 2018 at 20:54
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I like this example. The Spheres $S^{2n}$ cannot be complex manifolds unless $n=0,1,3$. One proves that $TS^{2n}$ does not have the structure of a complex vector bundle in these cases. If $TS^{2n}$ were a complex vector bundle, then $c_{n}(TS^{2n}) = e (TS^{2n})$, so $c_n (TS^{2n})$ is twice a generator of $H^{2n} (S^{2n}; \mathbb{Z})$.

Case 1: $n=2m$ even. Then $p_m (TS^{2n}) = \pm c_{2m} (TS^{2n} \otimes \mathbb{C}) = c_{2m} (TS^{2n} \oplus \overline{TS^{2n}})$. By the product formula for Chern classes, this is $= \pm c_{2m}(TS^{2n}) + c_{2m} \overline{TS^{2n}}) =\pm 2 c_{2m }(TS^{2n}) \neq 0$. This is a contradiction since $TS^{2n} \oplus \mathbb{R}$ is trivial.

Case 2: $n \geq 4$. This is more difficult and relies on the Bott periodicity theorem, one of whose corollaries states that a the top Chern class $c_n (V)$ of a complex vector bundle $V\to S^{2n}$ is divisible by $(n-1)!$

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Secondary characteristic classes allow us to see the large scale behaviour of leaves of foliations of 3-manifolds by surfaces; see

MR1040572 (91h:57015) Reviewed
Ghys, Étienne(F-ENSLY)
L'invariant de Godbillon-Vey. (French) [The Godbillon-Vey invariant]
Séminaire Bourbaki, Vol. 1988/89.
Astérisque No. 177-178 (1989), Exp. No. 706, 155–181.
57R30 (58F18)
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  • $\begingroup$ Ben, I wanted to thank you for introducing this to me. I have accepted Qiaochu's answer as it probably fit best to the question but I have to say your answer was a lot closer to my personal interests :) $\endgroup$
    – R Mary
    Commented May 24, 2016 at 11:46

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