# The Euler class on a vector bundle and local degree

Bott and Tu introduce without giving much details a theorem relating the Euler class of a vector bundle to the singularities of a section as follows (pp 125):

This theorem can also be phrased in terms of vector bundles. $\pi: E \rightarrow M$ be an oriented rank k vector bundle over a manifold of dimension $k$ and $s$ a section of $E$ with a finite number of zeros. The multiplicity of a zero x of $s$ is defined to be the local degree of $x$ as a singularity of the section $s$ of the unit sphere bundle of E relative to some Riemannian structure on $E$. (This definition of the index is independent of the Riemannian structure because the local degree is a homotopy invariant.) Since the Euler class $e(E)$ of $E$ is a $k$-form on $M$, it is Poincare dual to $nP$, where $n = \int_M e(E)$ and $P$ is a point on $M$. Thus, we have the following:

Theorem 11.17. Let $\pi: E \rightarrow M$ be an oriented rank $k$ vector bundle over $q$ compact oriented manifold of dimension $k$. Let $s$ be a section of E with a finite number of zeros. The Euler class of $E$ is Poincare dual to zeros of $s$ counted with the appropriate multiplicities.

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Is it not true that an integral on a point is zero, and the $k$-form should produce a zero integral when wedged product with forms? What is the meaning of "$n$" in $np$ then? I generally honestly do not understand what the paragraph is trying to convey. The "zeros" of $s$ are not even in the domain, what is the meaning of their local degree? I appreciate any clarification on the paragraph and theorem. (I perfectly understand the same theorem for sphere bundles)

• Why do you say the zeroes of $s$ are not in the domain? We have a section $s : M \to E$, so its zeroes are elements of $M$. – Michael Albanese Sep 22 '17 at 11:59

By Poincaré duality, a closed $k$-form on a connected, oriented $k$-manifold $M$ corresponds to an element of $H_0(M)\cong\Bbb Z$. You can represent elements of this $\Bbb Z$ as integral multiples of any given point on $M$. In this case, the integer is the Euler number, which is $\int_M e(E)$. On the other hand, they're saying something more interesting: If you choose any smooth section $s$ of $E$ that is transverse to the zero-section, then its zero set (which will consist of points), counted with multiplicities, is homologous to $nP$ and is also Poincaré-dual to the Euler class. Indeed, this is how topologists and geometers think of the Euler class of a bundle (and analogously for Chern classes of complex vector bundles, etc.). (Moreover, the multiplicities are the local intersection numbers of $s$ with the zero-section of $E$.)