1
$\begingroup$

I have been told that degree of map can be defined by the combinatorial method instead of the differential one. Assume $M$, $N$ is orientiable pseduomanifold, and $M$ is compact, $N$ is connected. Let $f$ be a continuous map between them. $g$ is simplicial approximation from $\text{sd}^mM$ to $N$. Then we define $$\text{deg}f:=\text{number of }\{\sigma|g(\sigma)=\tau\}-\text{number of }\{\sigma|g(\sigma)=-\tau\}$$

Above is what our teacher tell us briefly. But I want to make up the definition of the combinatorial one. So I have some questions to ask.

  • The definition above is not very clear.

    • Are $\sigma$ and $\tau$ the arbitrarily orientable $n$-dimensional simplex?

    • Is the definition well-defined?

  • I search for the Internet that we can define it by homology group such

    http://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping

    However, there are actually two generated variables in $H_n(M)$ or $H_n(N)$. Then the degree induced by different variables will differ by $-1$.

  • I also find a book Mapping Degree Theory. It provides for axioms about degree. Does there any paper or book introducing the theory.

Any advice is helpful. Thank you.

$\endgroup$
  • $\begingroup$ There are lecture notes by Louis Nirenberg: "Topics in Nonlinear Functionalanalysis". I think that the theorey is introduced aciomaticly. $\endgroup$ – Quickbeam2k1 Nov 9 '14 at 14:33
1
$\begingroup$

In your definition, $\sigma$ is meant to be an oriented simplex of the domain. If you reverse the domain's orientation, then you'll replace $\sigma$ with $-\sigma$, and the degree will negate. And I believe that the same is true for $\tau$. The proof that this leads to a well-defined function is not simple.

Your observation about homology groups amounts to the same thing as the previous paragraph. For connected $M$, an "orientation" of $M$ is a choice of a generator of $H_n(M)$; without an orientation, degree isn't well-defined (except mod 2).

Probably much of the remainder of your course will involve proving the well-definedness and topological invariance of things like this, and the relationship to the smooth theory. If you have to get there sooner, you could look at Vick's *Homology Theory. *

$\endgroup$
  • $\begingroup$ $H_n(M)=\mathbb Z$. So there are two generated variables $1$ or $-1$. $\endgroup$ – gaoxinge Nov 9 '14 at 14:43
  • 1
    $\begingroup$ It's perhaps better to say that (for $M$ connected) $H_n(M)$ is isomorphic to $\mathbb Z$. For any generator $z$ of $H_n(M)$, there are two choices of isomorphism, one which sends $z$ to $+1$, another that sends it to $-1$. An "orientation" is a choice of a generator, together with the agreement that we'll map that generator to $+1 \in \mathbb Z$. $\endgroup$ – John Hughes Nov 9 '14 at 15:00
  • 1
    $\begingroup$ Suppose that $z$ generates $H_n(M)$. Then the elements of $H_n$ are $\ldots, -2z, -z, 0, z, 2z, 3z, \ldots$. The map you're looking for sends $kz$ to $k$ for $k \in \mathbb Z$. In a combinatorial manifold, you can find $z$ by taking a sum $\sum c_i \sigma_+$ of ALL the top-dimensional simplices, with each $c_i = \pm 1$. By picking the $c_i$ so that boundaries of adjacent simplices cancel, you get a cycle that generates $H_n(M)$. (Of course, negating all the $c_i$s given another, which is "the other orientation". $\endgroup$ – John Hughes Nov 9 '14 at 16:24
  • 1
    $\begingroup$ That's correct. IF you change the orientation on either $M$ or $N$, the degree negates. $\endgroup$ – John Hughes Nov 10 '14 at 1:34
  • 1
    $\begingroup$ @gaoxinge: That is not correct. In the differentiable case an orientable connected manifold also has two orientations. So the latest comment of John holds exactly as stated in the differentiable case as well. $\endgroup$ – Lee Mosher Nov 10 '14 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.