12 votes

Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

The answer is "yes" if $S$ is the Riemann sphere. This is because a map $f$ of degree $d$ from the sphere to itself is homotopic to $z \mapsto z^d$. The answer is "basically no" ...
Sam Nead's user avatar
  • 23.9k
11 votes
Accepted

Determine whether a (1,2) tensor is Nijenhuis tensor

Yes, there are pointwise algebraic conditions on a section $N$ of $T\otimes\Lambda^2(T^*)$ in order for $N$ to equal $N_J$ for some almost complex structure $J$, but there are differential conditions ...
Robert Bryant's user avatar
6 votes
Accepted

Holomorphic Gauss normal map

Gauss map is holomorphic (as a map to the Riemann sphere) if the surface is minimal. This is Lemma 8.3 in the book of Osserman, A Survey of Minimal Surfaces. In fact, if you replace "embedded&...
Alexandre Eremenko's user avatar
5 votes

Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

The answer is no. This is a corrected version of Nicolast's comment. Let $E$ be an elliptic curve, let $f: E \to E$ be an endomorphism and let $H_1(f) : H_1(E) \to H_1(E)$ be the induced map on $H_1$. ...
David E Speyer's user avatar
5 votes

Finding global sections of a sheaf of sets using (some kind of) sheaf cohomology?

It is indeed true that the sequence $$1 \to \mathscr F_H \to \mathscr F_G \to \mathscr F_M \to 1\tag{1}\label{1}$$ of sheaves of pointed sets is exact. The only nontrivial thing to check is ...
R. van Dobben de Bruyn's user avatar
1 vote

Holomorphic Gauss normal map

Let $S$ be the embedded surface (assumed compact and not homeomorphic to a sphere) and let $S_1$ and $S_2$ be the subsets of $S$ where the Gaussian curvature is positive and negative, respectively. By ...
Vik78's user avatar
  • 254
1 vote
Accepted

Common holomorphic forms for two distinct complex structures

Any $C^1$ function $g$ on a connected Riemann surface is holomorphic if and only if $dg\wedge \omega=0$, for $\omega$ holomorphic and not the zero form. Where $\omega\ne 0$ this is clear, expanding ...
Ben McKay's user avatar
  • 24.7k

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