2
$\begingroup$

Let $E\to X$ be an oriented real $n$-plane bundle. Then the Euler class $e(E)\in H^n(X;\Bbb Z)$ can be defined, using Thom isomorphism (https://en.wikipedia.org/wiki/Euler_class#Formal_definition). I am curious about its properties, given here: https://en.wikipedia.org/wiki/Euler_class#Properties.

  1. As Chern classes or Stiefel-Whitney classes, does the four properties (functoriality, Whitney sum formula, normalization, and orientation) uniquely characterize the Euler class?

  2. It is also written that if $X$ is an oriented smooth $d$-manifold and $\sigma:X\to E$ is a smooth section that intersects the zero section transversally, then $e(E)$ is the Poincare dual of the class in $H_{d-r}(X;\Bbb Z)$ represented by the zero locus of $\sigma$. How can this be proved? Is there a reference of a proof for this statement?

$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

The standard reference for characteristic classes is the book by Milnor and Stasheff "Characteristic Classes", and I recommend reading it.

Your second question is also answerd in these seminar notes ( Theorem 3.2) by Matthias Görg.

The answer to your first question is no. They don't make sure we just have $e(E)=0$ for all bundles.

Add a fifth axiom, saying something like $e(\gamma)[\mathbb{CP}^1] = -1 $ , where $\gamma \to \mathbb{CP}^1$ is the tautological bundle ( as an oriented $2$-plane bundle) and you can proove uniqueness, using a real splitting principle.

Edit: As pointed out in the comments, the uniqueness part is not working the way I thought. I posted this as a new question here.

$\endgroup$
3
  • $\begingroup$ Another reference for question 2 is Proposition 12.8 of Bott and Tu's Differential Forms in Algebraic Topology. $\endgroup$
    – Michael Albanese
    Oct 10, 2021 at 13:33
  • $\begingroup$ Can you explain what you mean by proving uniqueness using a real splitting principle? The usual splitting principle for real vector bundles involves pulling back to a Whitney sum of line bundles, but in general they won't be orientable, so they don't have Euler classes. $\endgroup$
    – Jack Lee
    Oct 11, 2021 at 15:20
  • $\begingroup$ I thought of the "oriented splitting principle" (Theorem 6.1 in The transfer map and fibre bundles by Becker & Gottlieb). But checking this again, I realise its not enough for the general case. We can always produce a map $f:Y \to X$ which pulls back $E$ to a sum of oriented 2-plane bundles and possibly a trivial line bundle (if $n$ is odd), but the pullback on cohomology does not need to be injective for integral coefficients... $\endgroup$
    – Jonas
    Oct 11, 2021 at 20:43
4
$\begingroup$

(Edited after some discussion in the comments.)

As pointed out by Jonas, the four properties listed on Wikipedia do not characterise the Euler class as setting $e(E) = 0$ for all $E$ also satisfies those properties.

I just want to point out that the property which Wikipedia calls normalization is redundant: it follows from the Whitney sum formula and the orientation property. To see this, note that if $E \to X$ has a nowhere-zero section, then $E \cong E_0\oplus\varepsilon^1$ where $\operatorname{rank} E_0 = \operatorname{rank} E - 1$ and $\varepsilon^1$ is a trivial line bundle. If $E$ is orientable, then so is $E_0$, so by the Whitney sum formula $e(E) = e(E_0\oplus\varepsilon^1) = e(E_0)\cup e(\varepsilon^1)$. As $\varepsilon^1$ admits an orientation-reversing isomorphism, we have $e(\varepsilon^1) = e\left(\overline{\varepsilon^1}\right) = -e(\varepsilon^1)$, so $e(\varepsilon^1)$ is $2$-torsion, but $e(\varepsilon^1) \in H^1(X; \mathbb{Z})$ which is torsion-free, so $e(\varepsilon^1) = 0$ and hence $e(E) = e(E_0)\cup e(\varepsilon^1) = 0$.

$\endgroup$
7
  • $\begingroup$ Jack Lee made me realise that my argument for uniqueness via splitting principle does not work. It seems picking the "usual" euler class in $H^{2n}(BSO(2n),\mathbb{Z})$ and zero in $H^{2n+1}(BSO(2n+1),\mathbb{Z})$ satisfies the 4 axioms. Is it possible that one needs a further normalisation ? $\endgroup$
    – Jonas
    Oct 11, 2021 at 20:51
  • $\begingroup$ @Jonas: You're right. This works for bundles of even rank by a version of the splitting principle (see Proposition III.11.2 of Spin Geometry by Lawson & Michelsohn). I'm not sure how to nail down the Euler class for odd-rank bundles. It's an interesting problem. Maybe you should ask it as a new question (possibly on MathOverflow). $\endgroup$
    – Michael Albanese
    Oct 12, 2021 at 15:03
  • 1
    $\begingroup$ @Jonas: I don't think setting odd classes to zero actually satisfies these axioms. If $E\to X$ is an odd-rank bundle such that $E\oplus E$ has nonzero Euler class, then $e(E)\cup e(E)$ must be nonzero. I think you can construct such an example by adapting this answer of Qiaochu Yuan, replacing $\mathbb R\mathbb P^2 \times \mathbb R\mathbb P^2$ with $\mathbb R\mathbb P^\infty \times \mathbb R\mathbb P^\infty$. It's still not clear to me, though, whether the axioms uniquely determine the Euler class. $\endgroup$
    – Jack Lee
    Oct 12, 2021 at 22:04
  • $\begingroup$ @Jonas: BTW, I looked at the proof of the even-rank splitting principle in Lawson & Michelsohn, and there's something about the proof that I don't understand. I've posted this as a separate question here. If you have any insight about that proof, I'd appreciate your thoughts. $\endgroup$
    – Jack Lee
    Oct 13, 2021 at 0:08
  • $\begingroup$ @Jonas: Another property of the Euler class listed in Milnor & Stasheff is that the Euler class reduces mod 2 to the corresponding Stiefel-Whitney class. Maybe including this as an axiom would be sufficient. $\endgroup$
    – Michael Albanese
    Oct 13, 2021 at 11:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .