relationship between freeness and nefness of a rational curve

Question

I'm not able to figure out the precise relation between nefness and freeness of a rational curve $C$ on a projective variety $X$.

A curve is said to be nef if it intersects non-negatively every effective Cartier divisor on $X$.

A rational curve is said to be free if $f^{*}T_{X}$ is semipositive, where $f:\mathbb{P}^1 \rightarrow C$ is the normalization.

In general freeness implies nefness, since for any divisor $D$ there exists a deformation $C'$ of $C$ which doesn't lie on $D$ ($C$ can be "moved out" from any divisor).

Equivalence holds in special cases, such as the case of a rational curve on a smooth surface (Debarre proves in his book that in this case the freeness of $C$ is equivalent to $C^2 \geq 0$), but in general I think it shouldn't hold, since even if $C$ can't be moved out of a divisor $D$ it can however intersect non-negatively a divisor $D'$ numerically equivalent to $D$.

Do you know if equivalence holds in more general cases? Are there criteria for the equivalence?

Thank you

Answer 1

Let me give a series of examples with curves that are nef but not free according to your definitions (I never saw these definitions apart from the case of surfaces but this is possibly just my ignorance).

Consider any hypersurface $X_d$ in $\mathbb CP^n$ of degree $n\le d\le 2n-3$, $n\ge 4$. Such a hypersuface contains a line $L$. Now, by adjunction the anti-canonical bundle of $X_d$ is $O(n+1-d)$. Since $L$ is a line $c_1(T_{X_d}|L)=n+1-d$, at the same time $T_{X_d}|L$ contains a sub-bundle $O(2)$, so, since $n+1-d-2<0$, the bundle $T_{X_d}|L$ is not positive. Finally, since $Pic(X_d)=\mathbb Z$, the line $L$ intersects all divisors positively.