# Are any of these complex surfaces ever projective?

Let $$C$$ and $$T$$ be compact connected Riemann surfaces (or: smooth projective connected curves over $$\mathbb{C}$$) of genus at least two and let $$X:=C\times T$$. Let $$(c,t)$$ be a point of $$X$$, and let $$X'\to X$$ be the blow-up of $$X$$ in $$(c,t)$$. By Grauert's contraction theorem, we may contract the strict transform of $$\{c\}\times T$$ on $$X'$$ and obtain a normal complex-analytic surface $$X'\to S$$.

Under what conditions (if any) is $$S$$ projective?

Note that $$S$$ contains a unique rational curve (given by the image of the exceptional curve $$E$$ of $$X'\to X$$), and that $$S$$ has a unique singular point $$\sigma$$ in $$S$$.

My interest in this surface is related to Lang's conjectures, and I first learned about this surface from Frederic Campana. Indeed, the surface $$S$$ has the peculiar property that, for any point $$s$$ which does not lie on the rational curve and any pointed curve $$(D,d)$$, the set of pointed maps $$(D,d)\to (S,s)$$ is finite. However, for the pointed curve $$(C,c)$$ and the singular point $$\sigma$$, the space of pointed maps $$(C,c)\to (S,\sigma)$$ covers $$S$$.

I was not able to prove projectivity of $$S$$, not even whilst assuming it is proper (so that one might appeal to https://arxiv.org/abs/1112.0975 )

## 1 Answer

Here is a simple method for constructing projective examples:

Assume there exist maps $$f:C \to \mathbb{P}^1$$ and $$g:T \to \mathbb{P}^1$$ of the same degree which are totally ramified at $$c$$ and $$t$$. Let $$X = C \times T$$, $$Y = \mathbb{P}^1 \times \mathbb{P}^1$$, and consider the map $$p:=(f,g): X \to Y$$. Let $$X'$$ be the blowup of $$X$$ at $$(c,t)$$ and $$Y'$$ the blowup of $$Y$$ at $$(f(c), g(t))$$. An easy local computation shows that $$p$$ induces a morphism $$p': X' \to Y'$$.

Now let $$Y_1$$ be the blow down of the strict transform of $$\{f(c)\} \times \mathbb{P}^1$$ in $$Y'$$. The surface $$Y_1$$ is projective, so its normalisation $$X_1$$ in the function field of $$X'$$ is also projective. By using that the map $$X' \to Y_1$$ factors through $$X_1$$, it follows easily that $$X_1$$ is equal to the surface $$S$$, so $$S$$ is projective.

• Thank you for this! A small follow-up question (to double-check I understood correctly): Is the surface $Y_1$ actually smooth? – Ariyan Javanpeykar Oct 14 '20 at 10:48
• Yes, $Y_1$ is smooth since it is the blow down of a smooth rational curve with self-intersection $-1$. – naf Oct 14 '20 at 14:38