2
$\begingroup$

It is an easily proven theorem that if $f,g:M\to N$ are smooth maps that are homotopic maps between compact, connected, oriented, smooth manifolds of dimension $n$, then $\deg f=\deg g$. I was wondering about the converse statement to this: namely, if $\deg f=\deg g$, then $f,g:M\to N$ are homotopic. From Milnor's Topology from a Differentiable Viewpoint, we know (Hopf's Thereom), that if $f,g:M\to S^n$ have the same degree, then they are homotopic. However, what if $N\neq S^n$? Are there any obvious counterexamples?

$\endgroup$

1 Answer 1

4
$\begingroup$

There are lots of counterexamples: in general homotopy classes of maps between manifolds are much more complicated than this, and what makes $S^n$ special here is that it's highly connected.

For example, any $M \in M_n(\mathbb{Z})$ defines a map $T^n \to T^n$ from the $n$-torus to itself which induces the map $M$ on $H_1$ and hence whose degree is $\det M \in \mathbb{Z}$. And an integer matrix is far from being determined by its determinant.

More generally, if $M$ and $N$ are aspherical, e.g. tori or hyperbolic manifolds, then every conjugacy class of maps between their fundamental groups can be realized by a smooth map $f : M \to N$, and these will in general be far from determined by their degree. The simplest examples that aren't tori are surfaces of genus $\ge 2$.

$\endgroup$
2
  • $\begingroup$ @Moya: I'm not sure what you mean by "acting on the indices," but the point is that elements of $\text{GL}_n(\mathbb{Z})$ act on $\mathbb{R}^n$ and preserve $\mathbb{Z}^n$, so they act on the quotient. $\endgroup$ Commented May 12, 2015 at 6:16
  • $\begingroup$ Oh right, of course. Yes my comment about acting on the indices didn't make sense. $\endgroup$
    – Moya
    Commented May 12, 2015 at 6:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .