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If $X = C\times C' \subset \mathbb{P}^3$ for $C$ of genus $g$ and $C'$ of genus $g'$ (both smooth), then we know from Hartshorne exercise V.1.5 that $$ 8(g-1)(g'-1) = d(d-4)^2 $$ where $d = \text{deg}(X)$.

Is there a similar formula for $X = C\times C' \subset \mathbb{P}^N$ for $N$ arbitrary?

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  • $\begingroup$ What kind of formula are you expecting? In 3-space, $X$ is a hypersurface and so it is very restricted. $\endgroup$ – Mohan Feb 19 '16 at 1:43
  • $\begingroup$ @Mohan I'm not sure; but the fact that the computation for $X \subset \mathbb{P}^3$ is so straight-forward doesn't mean there should't at least exist some results for $X \subset \mathbb{P}^N$. $\endgroup$ – Derek Allums Feb 19 '16 at 16:32
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    $\begingroup$ No, it has too many moving parts. It all depends on what information you want to put in. In general, it will depend on more numbers than just the two genus and the degree. I am unaware of such a simple formula and I do not even know of any worthwhile guess. $\endgroup$ – Mohan Feb 19 '16 at 17:38
  • $\begingroup$ @Mohan Could anything be said if we simplified to letting $X$ be ruled by a plane curve: $C \subset \mathbb{P}^2$ and $C' \simeq \mathbb{P}^1$? $\endgroup$ – Derek Allums Feb 19 '16 at 17:58
  • $\begingroup$ I am unsure what you are expecting. Given $C,C'$, there are infinitely many possible degrees attained by suitable embeddings of $C\times C'$ in 5-space and so, what would you like to say? $\endgroup$ – Mohan Feb 19 '16 at 20:28

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