The Degree of a Map

This will be a nice, simple definition. As we saw last time, the top-degree compactly-supported de Rham cohomology is easy to calculate for a connected, oriented manifold $M$ . Specifically, we know that $H_c^n(M)\cong\mathbb{R}$ .

So, let’s say that $f:M\to N$ is a smooth map from one connected, oriented $n$ -manifold to another. The pullback $f^*$ induces a map on cohomology: $f^*:H_c^n(N)\to H_c^n(M)$ . But since each of these cohomology spaces is (isomorphic to) $\mathbb{R}$ , $f^*$ must act on them by multiplication by some number. That is, if $\omega\in\Omega_c^n(N)$ is a closed $n$ -form, and if the integral of $\omega$ over $N$ is $I$ , then $f^*\omega\in\Omega_c^n(M)$ is also a closed $n$ -form, and the integral of $f^*\omega$ over $M$ is $d_fI$ , where $d_f$ is some real number that depends only on the function $f$ and not on the form $\omega$ . That is, the same $d_f$ works for all $\omega$ .

We call this number the “degree” of $f$ , and write $\deg f$ . It turns out that when $f$ is a proper map — one for which the preimage of any compact set is compact, which is always the case when $M$ is itself compact — that we have a good way of calculating it. If $q\in N$ is a regular value of $f$ , some of which which will always exist, then the preimage of $q$ consists of a finite collection of points $p\in M$ . For each one we calculate the “signum” of $f$ at $p$ — written $\mathrm{sgn}_pf$ — to be $+1$ if $f_{*p}$ is orientation-preserving and $-1$ if $f_{*p}$ is orientation-reversing. I say that

$\displaystyle\deg f=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf$

While we will not prove this until next time, there is one immediate consequence: if $f$ is not onto, then any point in $N\setminus f(M)$ has an empty preimage, and so we find that $\deg f=0$ . In a sense, the degree of $f$ is counting how many times the image of $M$ covers that of $N$ , considering orientation.

December 9, 2011 - Posted by John Armstrong | Differential Topology, Topology

3 Comments »

Are you sure “f” doesn’t just need more cowbell? 🙂

Comment by Rhonda ROO | December 9, 2011 | Reply
[…] I asserted yesterday, there is a simple formula for the degree of a proper map . Pick any regular value of , some of which must exist, since the critical values […]

Pingback by Calculating the Degree of a Proper Map « The Unapologetic Mathematician | December 10, 2011 | Reply
[…] can use the concept of degree to prove the (in)famous “hairy ball theorem”. That is, any smooth vector field defined […]

Pingback by The “Hairy Ball Theorem” « The Unapologetic Mathematician | December 13, 2011 | Reply

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