Let $X$ be a smooth, integral and projective $d$-dimensional variety over a field $k$ of characteristic 0. Let $H$ be a very ample line bundle over $X$. Assume that there exists a smooth and irreducible curve $C$ rationally equivalent to the cycle $c_1(H)^{d-1}\cdot X$. Can we compute the genus of $C$ explicitly? If $X$ is a surface, then the question is quite easy by adjunction formula. For the general case, as @abx mentioned in comment, in $\mathbb{P}^3_k$, the curves of degree $n^2$ will deny the existence of such computation. But is it possible to give an upper estimate of the genus just like the space curves case?
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2$\begingroup$ Use the adjunction formula to compute the degree of the canonical class of the curve. $\endgroup$– SashaCommented Jun 20, 2021 at 8:51
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$\begingroup$ @Sasha You are right. Assume a flag of subvarieties, I can roughly say the degree of the canonical divisor of the curve could be KH^{d-1}+(d-1)H^d. But I don’t know how to deal with the conormal bundle in general. $\endgroup$– Pickle LiobeCommented Jun 20, 2021 at 9:30
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2$\begingroup$ It seems very unlikely to me that you can compute the genus from just knowing the rational equivalence class, even for projective space (if $H$ is not $\mathcal{O}(1)$). Is this really trivial? $\endgroup$– nafCommented Jun 20, 2021 at 10:16
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2$\begingroup$ This is already false for projective space: it says for instance that all (smooth) curves of degree $n^2$ in $\mathbb{P}^3$ have the same genus (taking $H=\mathscr{O}(n)$). All you can get is a bound on the genus. $\endgroup$– abxCommented Jun 20, 2021 at 10:23
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$\begingroup$ @naf sorry, I actually mean $H=O(1)$. $\endgroup$– Pickle LiobeCommented Jun 20, 2021 at 11:12
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