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A search of the literature reveals that for a curve $C$ of genus $\geq 2$, determining the effective cone of $C \times C$ is hard. My question is this: do we know a single example of a curve $C$ of genus $\geq 2$ for which we understand the effective cone of $C \times C$ and its image in $\text{Num}(C \times C)$? In particular, when can we say that the image in $\text{Num}(C \times C)$ is finitely generated?

Edit: Since there haven't been any answers to my question about finite generation, perhaps I can be more specific. Let $k = \mathbb{C}$. If $C = \mathbb{V}(x^n + y^n + z^n) \subset \mathbb{P}^2_k$, are there any techniques to calculate the effective cone of $C \times C$ when $n \geq 4$?

I ask this because if $k$ is a field of characteristic $p > 0$, it's well-known that this cone is not finitely generated since we have infinitely many extremal curves $\Gamma_q := \{(u,v) | \text{ Frob}_q(u) = v \}$ where $q = p^r$.

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Since I asked this question several months ago, I wrote a paper proving that the Mori cone of $C \times C$ for a smooth projective complex curve $C$ of genus $g \geq 2$ is not polyhedral. I thought I'd include a link to that for completeness.

http://arxiv.org/abs/1502.06061

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You should like at Section $5$ of this paper

http://www.ams.org/journals/tran/1993-337-01/S0002-9947-1993-1149124-5/S0002-9947-1993-1149124-5.pdf

for bounds on the effective an ample cones of $C\times C$.

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  • $\begingroup$ Question edited. Also, Kouvidakis' paper is about $\text{Sym}^2(C)$, not $C \times C$. Is there an easy result that one can deduce from $\text{Sym}^2(C)$? Since $N^1(C \times C)$ is generically three-dimensional, it might be harder. $\endgroup$
    – Ashwath Rabindranath
    Commented Aug 10, 2014 at 1:12

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