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Let $X$ be a smooth projective variety of dimension $3$, and $L$ an ample line bundle with $h^0(X,L)\geq 2$. Let $s$ and $t$ be two generic linearly independent sections of $L$, and $C$ the curve that they define.

Assume we know all intersection numbers $L^3$, $L^2K_X$, $LK_X^2$ and $K_X^3$.

Of course we can compute the genus of $C$ by adjunction. Assume now that $C$ is irreducible but non-reduced, and of multiplicity $2$. In this case, we can also compute the genus of $C_{red}$ by adjunction: restricting $L$ to $\{s=0\}$ we get $2H$ for some line bundle $H$, and $C_{red}$ is defined by a section of $H$.

Now, take different sections and assume that the curve $C'$ you get has two irreducible components, say $A$ and $B$, with $A$ reduced and $B$ non-redued of multiplicity two.

Can I say something about the genus of $B_{red}$? For instance, is it smaller than the genus of $C_{red}$ computed in the previous paragraph?

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    $\begingroup$ Maybe this formula from Semple and Roth may help. Genus formula for two surfaces of degrees d,e in projective 3 space meeting in two curves C,C’, of degrees n,n’, and genera p,p’. (Use adjunction formulas for both curves on one surface.) 2(p-p’) = (n-n’)(d+e-4). $\endgroup$
    – roy smith
    Commented Apr 21, 2019 at 0:56
  • $\begingroup$ @RoySmith could you give me the precise reference please $\endgroup$
    – Giulio
    Commented Apr 22, 2019 at 13:43
  • $\begingroup$ Introduction to Algebraic Geometry, by J.G. Semple and L. Roth, Clarendon Press, Oxford, 1949, reprinted 1986, chapter IV, section 8.2, Intersection of two surfaces in two curves, pages 90-91. I found the old fashioned derivation mysterious, but was able to derive it myself from the adjunction formula, as I noted in my cryptic parenthetical comment above, which I found in my own notes and copied for what it is worth. Unfortunately I did not seem to save the derivation. $\endgroup$
    – roy smith
    Commented Apr 22, 2019 at 19:49

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This does not answer the question, since the question concerns both genera of residual and of non reduced curves, but addresses only the issue of residual genus formulas, and in the case of $\mathbb{P}^3$. The formula is taken from Semple and Roth, page 91. But it may help.

Ok, as in Semple and Roth, pages 90-91, consider a space curve $C$ of degree $n$ and genus $p$, lying on a surface $S$ of degree $d$. Then pass another surface $S’$ of degree $e$ through $C$, so that it cuts further on $S$, another residual curve $C’$ of degree $n’$ and genus $p'$. ($C’$ is not necessarily connected, so $p’$ could be negative.)

Then we claim: $2(p-p’) = (n-n’)(d+e-4)$.

This follows from the two adjunction formulas for $C$ and for $C’$, both taken say on $S$, and subtracted.

By the adjunction formula for the curve $C$ on the surface $S$, the degree of the canonical bundle $K$ of $C$, equals $C\cdot C + (d-4)n$, where $C\cdot C$ is the intersection number on the surface $S$. Similarly the degree of the canonical bundle of $C’$ computed as a curve on the same surface $S$, equals $C’\cdot C’ + (d-4)n’$.

Subtracting, the difference of the degrees of the two canonical bundles equals $\deg(K) - \deg(K’) = 2(p-p’) = (d-4)(n-n’) + C\cdot C - C’\cdot C’$.

Now since $S$ intersects $S’$ in the union $C+C’$, I claim $(C+C’)\cdot C = en$, and $(C+C’)\cdot C’ = en’$. Subtracting, and canceling, gives $C\cdot C - C’\cdot C’ = e(n-n’)$. Substituting above, gives $2(p-p’) = (d-4)(n-n’) + e(n-n’) = (d+e-4)(n-n’)$. Of course also $n+n’ = de$.

For example, if $C$ is two disjoint lines lying on a cubic surface $S$, and $S’$ is a quadric surface passing through both lines, then $C’$ is a quartic curve. Then $n=2$, $n’=4$, $p = -1$, $d=3$, $e=2$, so $2(-1 -p’) = -2$, so $p’ = 0$, and $C’$ is rational. This is example (2) at the top of p. 93, of Semple and Roth.

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  • $\begingroup$ The same argument seems to give in general 2(p-p') = (C-C').(D+E+K), where D,E are divisors on the 3 fold X, K is the canonical divisor on X, and these intersection numbers are taken on X. Thus even in the projective space case, it seems that to get a good hold on the genus of C', one needs not only the genus of C, but also its degree, i.e. its intersection numbers with K,S,S'. Still it is tempting to think the answer to your question is yes, even with no formulas. Oops, poor notation; my C,C' are your curves A,B. $\endgroup$
    – roy smith
    Commented Apr 23, 2019 at 2:21
  • $\begingroup$ Thank you very much!! My main problem is actually how to relate the genera of $B$ and $B_{red}$ when $B$ is not a complete intersection in $X$. (this seems easy to me just when $B$ itself is a complete intersection). $\endgroup$
    – Giulio
    Commented Apr 23, 2019 at 14:39
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    $\begingroup$ It seems the reduced curves can both have genus zero, but maybe you meant "less than or equal"? I.e. a rank 1 quadric cuts a quadric cone in a double conic, reduced genus zero. Then a rank 2 quadric containing a ruling line, cuts a double line and a residual conic. The double line again has reduced genus zero. These sections of O(2) are not generic, but it seems unlikely that generic sections would meet in a non reduced curve. Is it possible that B generally double covers its reduced form? I.e. choose the second section to contain the reduced component and let it move. $\endgroup$
    – roy smith
    Commented Apr 23, 2019 at 15:44
  • $\begingroup$ thanks! so here the upshot is that the complete intersection has genus 1 and $B_{red}$ has genus 0, which is exactly what I am looking for, i.e. the genus of $B_{red}$ is less than say something like $1/2$ or $1/4$ the genus of the complete intersection (this is exactly what happens when $B$ is a complete intersection!) $\endgroup$
    – Giulio
    Commented Apr 23, 2019 at 16:21

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