4
$\begingroup$

From what I gather, the degree of a map originally arose in the context of studying maps $f:S^n\rightarrow S^n$. Since $H_n(S^n)\cong \mathbb{Z}$, the induced map $f_\star$ has the form $x\mapsto kx$ for some integer $k$, and this number ends up telling us a lot about the map, so we give it a name.

The most generalized form of this definition I've encountered is for maps $f:X\rightarrow Y$ between closed connected oriented manifolds of equal dimension $n$. Since $H_n(X)$ and $H_n(Y)$ are both isomorphic to $\mathbb{Z}$, we can define degree just like with a sphere.

The real meat of this generalization seems to be that endomorphisms of $\mathbb{Z}$ are completely determined by choice of an integer, and the induced map $f_\star$ gives us an obvious place to find an endomorphism of $\mathbb{Z}$. But I can think of some other situations where the same thing would happen.

  1. The top relative homology group $H_n(X,\partial X)$ of a compact orientable manifold with boundary is still isomorphic to $\mathbb{Z}$, and its generator is even still called a fundamental class. Is "relative degree" a thing?

  2. If $X$ and $Y$ are closed connected nonorientable manifolds of dimension $m_X$ and $m_Y$, then $H_{m_X}(X)$ and $H_{m_X}(Y)$ can't be $\mathbb{Z}$ (lest they be orientable). However, it could still be that their highest non-trivial homology groups have the same dimension and are both isomorphic to $\mathbb{Z}$, and in that case there would still be an obvious choice of integer to associate with a map $f:X\rightarrow Y$.

  3. if $X$ and $Y$ aren't both manifolds, but the dimension of their highest nontrivial homology groups match up, and both happen to be isomorphic to $\mathbb{Z}$, then again, we can get an integer in the same way as above. (Unlike the above, though, the number of the highest homology group could still correspond to its dimension, in whatever sense we wish to think of it.)

I've never seen these constructions discussed. Is that because they are not actually well-defined, or are otherwise problematic (e.g. not homotopy invariant)? If not, do these generalizations just lose all the important properties normally associated with degree (i.e. arbitrarily taking the $\mathbb{Z}\rightarrow\mathbb{Z}$ thing out of context misses the point of what degree is supposed to mean conceptually)?

$\endgroup$
1
$\begingroup$

For manifolds with boundary, I think you need that $\partial X$ is mapped to $\partial Y$. Then everything works.

For non orientable maps, one cheaply has the mod 2 degree : $H_n(X;\mathbb{Z}/2\mathbb{Z})\rightarrow H_n(Y;\mathbb{Z}/2\mathbb{Z})$. Another way of defining a integer valued degree is to use local coefficients. One should assume that the line bundle $(\Lambda^{\mathrm{n}}TX)^*\otimes f^{*}(\Lambda^nTY)$ is orientable if I remember correctly. This is an assumption on the map $f$, and basically tells you that it matches the non-orientability of the manifolds nicely. This is due to Paul Olum.

For connected open (orientable) manifolds you can study proper maps $f:X\rightarrow Y$. They induce maps in compactly supported cohomology. These groups will be isomorphic to $\mathbb{Z}$, and one can define a degree.

$\endgroup$
1
$\begingroup$

Of course the invariants you are asking for seem to be interesting and of course they are interesting in the sense that they contain information about the map $f$ but then you could just look at the element $f_*$ in the corresponding homomorphism spaces between the homology groups. This would be something like the finest invariant of this kind.

The main point of the degree is that it is easy to compute (This generalizes to the non-orientable case if you look at everything mod $2$). You just have to look at a point with a finite preimage and basically look at the local homology there and then sum it up. This is very easy, but also gives you a nice invariant of the map.

$\endgroup$

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.