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The map $S^2 \to S^2$ is given in spherical coordinates as follows:

$$x = \cos(\phi)\cos(\psi),\ y = \cos(\phi)\sin(\psi),\ z = \sin(\phi).$$

$(\phi,\ \psi) \mapsto (n\phi,\ m\psi)$.

What is the degree of this mapping? Is it true that this mapping is indeterminate at the poles for even $m$? For odd $m$ from the geometric definition, i get the answer $m \cdot n$, so in every regular point displays exactly as many points and the function preserves the orientation in each of them.

But I have absolutely no idea how to use the definition through fundamental classes and even what is the fundamental class of a sphere.

I would be happy with any help or explanations.

Michael Hardy
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Akari Gale
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1 Answers1

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The definition through fundamental classes is hardly ever useful when you have actual equations at hand. It is rather abstract.

On the other hand, when you have equations like these, what you can do is pick an arbitrary point $P$ on the sphere and ask yourself:

  1. How many preimages does $P$ have? Make a list.
  2. Call these preimages $P_1, \ldots P_k$. Is $f$ a local homeomorphism near each of the $P_i$? If your initial point is arbitrary enough the answer should always be yes.
  3. If the answer is yes, then the final blow: does $f$ respect or on the contrary does it reverse the orientation near $P_i$? The answer may be different for each $P_i$.
  4. Count $+1$ for each $P_i$ where the orientation is preserved, and $-1$ where it is reversed. The sum of all these signs is your degree!

Intuitively, the degree is how many times you wrap your domain sphere around your codomain sphere. When the orientation is reversed you wrap in the wrong way, hence the $-1$.

Arnaud Mortier
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  • Thank you, this is how I acted in this case. – Akari Gale Jan 21 '18 at 23:19
  • Geometrically the fundamental class of a manifold is the manifold itself (seen as a patchwork of simplices glued to each other along their boundaries, so that it is really a homology class). Hence when you wrap the sphere around the other sphere, every simplex from the decomposition on the right gets covered $d$ times. Hence the induced map at the homological level is $\mathbb{Z}\simeq H_2\rightarrow H_2\simeq\mathbb{Z}$ defined by the multiplication by $d$. The fundamental class corresponds to the generator $1\in \mathbb{Z}$. This is useful to think but not to compute. – Arnaud Mortier Jan 21 '18 at 23:30
  • So I think, but I do not understand why wrapping $d$ times is equivalent to multiplying by $d$ as a coefficient – Akari Gale Jan 21 '18 at 23:37
  • That's because every time you wrap, it's like adding a new sphere. If $ f (1)=1+...+1 $ $ d $ times then $ f (1)=d\times 1 $. – Arnaud Mortier Jan 22 '18 at 00:28