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Let $C_1, C_2$ be rationally equivalent curves in a smooth projective variety $P$. Let $$N_i: = \mathcal{H}om(I_{i}/I^2_{i}, \mathcal{O}_{C_i})$$ be the normal bundle of $C_i$, where $I_i$ is the ideal sheaf of $C_i$ in $P$.

Recall that a vector bundle $E$ over $P$ is called nef if $\mathcal{O}_{\mathbb{P}(E)}(1)$ is nef on the total space $\mathbb{P}(E)$.

My question is the following:

Suppose $N_1$ is nef, then does this imply that $N_2$ is also nef?

(I wish there are some Chern calculation which only depends on the rationally or even numerically equivalent class.)

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  • $\begingroup$ Take a bundle on $\mathbb{P}^1\times\mathbb{P}^1$ so that the restriction to ${x}\times\mathbb{P}^1$ depends on $x$. Now take the total space of this bundle and compactify it. Then taking $C_i=x_i\times\mathbb{P}^1,$ I believe this construction gives for any pair of vector bundles $M_1,M_2$ on $\mathbb{P}^1$ with the same degree two curves $C_1$,$C_2$ with $N_i=\mathcal{O}\oplus M_i$. Now I believe this immediately gives a counterexample to your statement by taking $M_1=\mathcal{O}(-1)\oplus\mathcal{O}(1)$ and $M_2=\mathcal{O}\oplus\mathcal{O}$, but my knowledge of nef vector bundles is bad... $\endgroup$
    – dhy
    Commented Feb 23, 2015 at 22:04
  • $\begingroup$ Sorry, but I did not follow your construction. Which is space $P$? $\endgroup$
    – Li Yutong
    Commented Feb 24, 2015 at 2:41
  • $\begingroup$ The compactified total space of the chosen bundle on $\mathbb{P}^1\times\mathbb{P}^1.$ $\endgroup$
    – dhy
    Commented Feb 24, 2015 at 5:28

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I just now noticed this question. Of course dhy is correct, as usual. I want to remark that there are also examples that arise "in nature". Let $X$ be any smooth, projective hypersurface in $\mathbb{P}^n$ of degree $d$. If $3\leq d \leq n$, then there exist lines $L\subset X$ such that $N_{L/X}$ is not nef. If the characteristic is $0$, or even just $>d$, and if $d\leq n-1$, then for a general line $L'$ in $X$, $N_{L'/X}$ is nef, i.e., $L'$ is "free". Thus some lines in $X$ have nef normal sheaf, while others do not.

If $X$ is sufficiently general, then the Fano subscheme $F\subset \text{Grass}(\mathbb{P}^1,\mathbb{P}^n)$ parameterizing lines contained in $X$ is smooth of dimension $2n-d-3$ with dualizing sheaf $\omega_{F/k} \cong \mathcal{O}_{G}(-n-1+d(d+1)/2)|_F$, where $\mathcal{O}_G(1)$ is the ample generator of the Picard group of $\text{Grass}(\mathbb{P}^1,\mathbb{P}^n)$. With the single exception of $(n,d) = (3,2)$, if $2n-d-3$ is positive then $F$ is connected. In particular, if $d(d+1)/2 \leq n$, with the two exceptions $(n,d) = (2,1)$ and $(3,2)$, then $F$ is a Fano manifold. Thus, by Campana and Kollár-Miyaoka-Mori (or just direct computation in this explicit case), $F$ is rationally chain connected. Therefore any two lines in $X$ are rationally equivalent, yet some have nef normal sheaf and others do not.

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  • $\begingroup$ Thank you so much for this more conceptual explanation!! This makes a lot sense! $\endgroup$
    – Li Yutong
    Commented Jun 18, 2015 at 13:48

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