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If $M$ and $N$ are oriented $n$-manifolds and $f: M \to N$ then the degree of $f$ is given by $$ \deg f = \sum_{p \in f^{-1}(q)} sign_p f $$ where $q$ is a regular value and the sign is $+1$ if $f$ is orientation preserving at $p$ and $-1$ if is orientation reversing at $p$. If we identify $\mathbb S^2$ with the one point compactification of the complex plane, then any polynomial in $\mathbb C[X]$ is a smooth map from $\mathbb S^2$ to itself. Does the degree of a polynomial as a map between manifolds coincide the algebraic degree? Equal up to a sign? Easily related?

It's easy to see that they coincide for $f(z)=z^n$. The fundamental theorem of algebra assures that a regular point has exactly the algebraic degree number of preimages, but can the signs of the preimages lower the mapping degree?

spitespike
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    Related question that might interest you: http://math.stackexchange.com/questions/140123/how-do-different-definitions-of-degree-coincide – Jesse Madnick Mar 31 '13 at 02:49

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Yes, they coincide. Any polynomial (or indeed any holomorphic function) on the complex plane is orientation-preserving, so all of the "signs" of the preimage are positive.

Note, however, that the maps $f(n)=z^{-n}$ do not have negative degree. These maps are also holomorphic on the sphere, so the degrees are positive (with $z^{-n}$ having degree $n$). To get negative degree, you have to compose with the complex conjugation map, which has degree $-1$.

Jim Belk
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  • In Guillemin Pollack's book it says that $\operatorname{deg}(z^{m})=m$, where $m<0.$ Is correct that $\operatorname{deg}(z^{-n})=n?$ – Kalashinikov Nov 05 '21 at 20:54
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    @Bungmap Guillemin and Pollack consider the map $z\mapsto z^m$ on the unit cirlce, which indeed has degree $m$ for all integers $m$. But on the Riemann sphere the map $z\mapsto z^m$ has degree $|m|$. – Jim Belk Nov 06 '21 at 12:25
  • Why do we have the difference between Riemann's sphere and unit circle? – Kalashinikov Nov 06 '21 at 14:05
  • The difference is in how $S^1$ gets its orientation. If you want to use local degrees on $S^2$, you think of $S^1$ as getting a positive orientation as the boundary of a neighborhood of a point. The map $z \mapsto 1/z$ maps the positively oriented $S^1$ around $z=0$ to the positively oriented $S^1$ around $z=\infty$. So degree $+1$. But if you consider this is a map from that circle to itself, then it's a reflection with degree $-1$. – Ethan Dlugie Nov 09 '21 at 23:49