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In algebraic topology, we define a degree for a map $f: S^n \to S^n$ as where the induced map $f_*$ on the $n$-th homology group of $S^n$ sends $1$.

In differential topology, we have a different (same?) notion of degree for $f$. You take a regular value $b \in S^n$, consider $f^{-1} (b)$ (which is finite by the inverse function theorem and some compactness argument), and take the difference between the number of points in the preimage where the Jacobian of $f$ is positive and the number of points in the preimage where the Jacobian of $f$ is negative.

Geometrically, I can see that they are the same, but I couldn't convince myself rigorously. In Prop 2.30 of Hatcher, he mentions that the degree of $f$ is the sum of the local degrees of $f$ at each preimage point, and local degrees are either $\pm 1$. (Local degree is defined in the middle of page 136 in Hatcher.)

So, the final question is, must the sign of the local degree of $x \in f^{-1}(b)$ the same as the sign of the Jacobian of $f$ at $x$?

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There's technical details -- the jacobian needs to be non-degenerate for the formula to hold. But yes, they're the same. This is covered in Milnor's "topology from a differentiable viewpoint", Guillemin and Pollack's "differential topology", and it's also in Bredon's "geometry and topology". This construction can be viewed as something of a first step in the Pontriagin construction, covered in more detail in the Milnor text. –  Ryan Budney Nov 11 '09 at 9:17
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@Ryan: why didn't you make this an answer? –  Loop Space Nov 11 '09 at 12:39
    
The comment feature somehow seems more humble. I guess I'm getting unnaturally attached to it. –  Ryan Budney Nov 11 '09 at 19:13
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3 Answers

up vote 5 down vote accepted

I think what you need is the following lemma (usually called the "Stack of records" lemma):

Consider a smooth proper map of manifolds of the same dimension $f \colon M \to N$ and let $y \in N$ be a regular value of $f$.

Then there exists a neighbourhood $V \subset N$ of $y$ such that $f^{-1}(V) = \cup\_{i=1}^n U\_i$ with $U\_i \cap U\_j = \emptyset$ for $i \neq j$ and $f|\_{U_i} \colon U\_i \to V$ is a diffeomorphism for all $i$.



Now from this you can just sum up $\pm 1$ according to orientation on each $U_i$ to get the local degree of $f$ at $y$, and this works for both definitions of degree.

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I think so: it looks like the local degree according to Hatcher's definition measures whether $f$ preserves orientation or reverses it on the neighborhood of $x$. On page 233 he begins discussion of orientation using excision classes: an orientation for an neighborhood in an $n$-manifold at a point $x$ is just a choice of generator of $H_n(\mathbb R^n, \mathbb R^n-x)$, and a small neighborhood $U$ about $x$ is homeomorphic to $\mathbb R^n$. In his degree-counting, he takes a neighborhood $U$ of $x$ which is disjoint from other preimages $f^{-1}(f(U))$ and looks at the sign of the map $H_n(U, U-\{x\})\rightarrow H_n(f(U), f(U)-\{y\})$.

The sign of the Jacobian of $f$ should also tell you whether $f$ is locally orientation-preserving or reversing at at $x$.

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I think you can find a proof that the differentiable topological degree is the (co)homological degree in the book by Bott and Tu (Diffrential forms in algebraic topology). But there instead of homology they describe cohomology first. Then you need to translate everything to the homological setting (by de Rham isomorphism).

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