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Let $X\subset\mathbb{P}^2$ be a complex manifold defined by a homogeneous polynomial of degree $d>3$. Let $$\phi:\mathbb{P}^1\rightarrow X$$ be a holomorphic map. Show that $\phi$ is constant.

Let $\mathcal{O}_{\mathbb{P}^n}(-1)$ be the Tautological line bundle on $\mathbb{P}^n$. We know the following facts:

  • The Canonical line bundle of X is the restriction of $\mathcal{O}_{\mathbb{P}^2}(d-2-1)=\mathcal{O}_{\mathbb{P}^2}(d-3)$ to $X$.
  • The Canonical line bundle of $\mathbb{P}^1$ is $\mathcal{O}_{\mathbb{P}^1}(-2)$.

From this we can conclude that $X$ and $\mathbb{P}^1$ are certainly not isomorphic.

Now the problem is solved straightforwardly by means of the Riemann-Hurwitz formula:
if $\phi$ is non constant then the genus of $X$ is less or equal the genus of $\mathbb{P}^1$, which is zero. Thus $g_X=0$, and we know this implies $X$ to be isomorphic to $\mathbb{P}^1$. This is a contradiction.

My goal is instead to solve the problem without using the theory of Riemann Surfaces but by means of other methods, e.g. the properties of holomorphic mappings.

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1 Answer 1

Claim: If a non-constant holomorphic map $$\phi:\mathbb{P}^1\rightarrow X$$ exists with $X$ a hypersurface in $\mathbb{P}^2$ of degree $d$, then $d \le 2$ and $\phi$ is biholomorphic.

First, I prove that the existence of $\phi$ implies that $X$ is biholomorphic to $\mathbb{P}^1$: The holomorphic map $$\phi:\mathbb{P}^1\rightarrow X$$ is surjective: The image is compact, hence closed. In addition, the image of a non-constant holomorphic map of one variable is open. Being surjective, $\phi$ induces an inclusion

$$\phi^*: \mathscr{M}(X) \longrightarrow \mathscr{M}(\mathbb{P}^1) $$

of the corresponding function fields. The field extension $\mathscr{M}(\mathbb{P}^1)$ of $\mathbb{C}$ is pure transcendental with transcendence degree $1$. By Lueroth’s theorem every subfield, which properly contains $\mathbb{C}$, is pure trancendental, too. Hence

$$\mathscr M (\mathbb{P}^1) \cong \mathscr{M}(X) $$

from an abstract point of view. As a consequence, a bimeromorphic map

$$X \longrightarrow \mathbb{P}^1$$

exists. In general, a meromorphic map is defined outside an analytic subset, which has codimension at least 2. Hence the bimeromorphic map is already biholomorphic.

Secondly, I prove that a hypersurface in $\mathbb{P}^2$ of degree $d > 2$ is not biholomorphic to $\mathbb{P}^1$: The hypersurface $X$ is the zero-variety of a section $s \in H^0(\mathbb {P}^2, \mathscr{O}_{\mathbb {P}^2}(d))$ according to the following exact sequence of sheaves

$$ 0 \longrightarrow \mathscr O_{\mathbb{P}^2}(-d) \mathop{\longrightarrow} \limits^{s} \mathscr O_{\mathbb{P}^2} \longrightarrow \mathscr O_X \longrightarrow 0$$

The long exact cohomology sequence contains the following segment

$$ H^1(\mathbb {P}^2, \mathscr{O}_{\mathbb {P}^2}) \longrightarrow H^1(X, \mathscr{O}_X) \longrightarrow H^2(\mathbb {P}^2, \mathscr{O}_{\mathbb {P}^2}(-d)) \longrightarrow H^2(\mathbb {P}^2, \mathscr{O}_{\mathbb {P}^2})$$

The cohomology of projective spaces is well-known. Hence

$$H^1(X, \mathscr{O}_X) \cong H^2(\mathbb {P}^2, \mathscr{O}_{\mathbb {P}^2}(-d))$$

and $H^1(X, \mathscr{O}_X) \neq 0$ for $d > 2$, but $H^1(\mathbb {P}^1, \mathscr{O}_{\mathbb {P}^1}) = 0$, q.e.d.

One question: In my opinion the proof using Riemann-Hurwitz formula is fine. Why do you want to avoid it?

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