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I've recently read about a number of different notions of "degree." Reading over Javier Álvarez' excellent answer for the thousandth time finally prompted me to ask this question:

How exactly do the following three notions of "degree" coincide?

(1) Algebraic Topology. Let $f\colon X \to Y$ be a continuous map between compact connected oriented $n$-manifolds.

Wikipedia tells me that $H_n(X) \cong H_n(Y) \cong \mathbb{Z}$, and that a choice of orientations for $X$ and $Y$ amount to choices of generators $[X], [Y]$ for $H_n(X), H_n(Y)$, respectively. We then define $\deg f$ via $$f_*([X]) = (\deg f)[Y].$$

(2) Differential Topology. Let $f\colon X \to Y$ be a smooth map between oriented $n$-manifolds, where $X$ is compact and $Y$ is connected.

Let $y \in Y$ be a regular value of $f$ (which exists by Sard's Theorem), let $D_xf\colon T_xX \to T_yY$ denote the derivative (a.k.a. pushforward), and define $$(\deg f)_y = \sum_{x \in f^{-1}(y)}\text{sgn}(\det D_xf).$$ It can be shown that $(\deg f)_y$ is independent of the choice of $y \in Y$, so we can talk meaningfully about a single quantity $\deg f = (\deg f)_y$.

(3) Riemann Surfaces. Let $f\colon X \to Y$ be a holomorphic map between compact connected Riemann surfaces.

For $x \in X$, we let $\text{mult}_x(f)$ denote the multiplicity of $f$ at $x \in X$. For $y \in Y$, we define $$(\deg f)_y = \sum_{x \in f^{-1}(y)} \text{mult}_x(f).$$ As in (2), it can be shown that $(\deg f)_y$ is independent of the choice of $y \in Y$. (Does this generalize to arbitrary complex manifolds?)


Thoughts: As was mentioned in my topology class last semester (and also on Wikipedia), there is this concept of "local homology" which lets us compute (1) as a sum of "local degrees."

I imagine that in the case of (2), each of these local degrees is, in fact, equal to $\text{sgn}(\det D_xf)$ because $f$ is a local diffeomorphism at each regular point $x$. I also imagine that in the case of (3), each of these local degrees is, in fact, equal to $\text{mult}_x(f)$ because the degree of $\mathbb{S}^n \to \mathbb{S}^n$, $z \mapsto z^k$ is $k$. (Does this also mean that $f$ is not regular at any point where $\text{mult}(f) \geq 2$? This would make sense, but what is the proof?)

This all seems correct in my head, but I would really like more details if possible.

  • 1
    That different versions of the same concept coincide is usually the stuff of major theorems. See for instance [Atiyah–Singer index theorem](http://en.wikipedia.org/wiki/Atiyah–Singer_index_theorem).2012-05-02
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    I think comparing this to the ASIT is a little over the top. (1) and (2) are related by the de Rham theorem (cf. Bott & Tu, as always), and (3) is really (2) in disguise once you acknowledge that you're using the "local model" theorem for holomorphisms. The Weierstrauss preparation theorem is the higher-dimensional analog; I haven't thought through it, but I'd imagine a similar argument applies.2012-05-04
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    What about, for example, local and global degrees of a vector field?2012-05-12
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    @Neal: What is your question exactly? For any vector bundle on a smooth manifold (not just the tangent bundle), the algebraic sum of the local degrees of a transverse section is an invariant, namely the "global degree". If you think through it, all we're actually doing is taking the intersection number of the manifold with itself inside of the total space! Moreover, this recovers Euler characteristic of the manifold when we specialize to the tangent bundle. If you like this sort of thing, I highly recommend Thurston's "Geometry and Topology" (or something like that).2012-05-14
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    I'm just throwing out a couple more definitions of "degree" for the author to consider. Thanks for the recommendation!2012-05-15

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