THREE-DIMENSIONAL SCROLLS IN $\mathbb{P}^6$

MATEI TOMA

ABSTRACT. In this note it is proved that three-dimensional scrolls in $\mathbb{P}_C^6$ over curves are rational or elliptic.

By an $n$-dimensional *scroll* $X$ in $\mathbb{P}^N = \mathbb{P}_C^N$ we mean a $\mathbb{P}^{n-1}$-bundle over a complete nonsingular curve $C$ such that the fibres are linearly embedded. Thus $N \ge 2n - 1$.

If $N = 2n - 1$, then by the Barth-Lefschetz theorem the irregularity of $X$ is zero and $C$ must be rational. In fact, in this case $X$ is the image of a Segre morphism $\mathbb{P}^1 \times \mathbb{P}^{n-1} \to \mathbb{P}^{2n-1}$.

For $N = 2n$, there is one more scroll over $\mathbb{P}^1$ namely $\mathbb{P}^1 \times \mathbb{P}^n \cap \mathbb{P}^{2n}$, and Ionescu showed the existence of $n$-dimensional scrolls over elliptic curves too, [Ionescu(1990)]. It is unknown whether there are other families of such scrolls beside the three above mentioned, or equivalently (cf. **1** below) whether in this situation scrolls over higher genus curves exist. By filling a gap in a former proof of Lanteri ([Lanteri(1980)]), Aure settled the case $n = 2, N = 4$, showing that $C$ can only be rational or elliptic, [Aure(1987), Appendix B]. Following the idea of Aure we show in this note the same thing for $n = 3$:

**Theorem.** If $X$ is a 3-dimensional scroll in $\mathbb{P}^6$ over a smooth curve $C$, then $C$ is either rational or elliptic.

*Proof.** **1.** Let $\pi : X \to C$ denote the projection, $d$ the degree of $X$, $g$ the genus of $C$, $K$ the canonical class of $X$ and $N$ the normal bundle of $X$ in $\mathbb{P}^6$.

There exists a very ample divisor $L$ on $C$ such that $\pi^*L = K + 3H$, where $H$ denotes the hyperplane class on $X$. Using this in the self intersection formula

$$c_3(N) = d^2,$$

one gets

$$(1) \qquad 12g = (d-3)(d-4).$$

<footnote>Date: 26.05.1995.</footnote>

<footnote>Key words and phrases. Scrolls in projective space.</footnote>

<footnote>Acknowledgements: I wish to thank the Graduiertenkolleg "Komplexe Mannigfaltigkeiten" at Bayreuth University for its hospitality and financial support during the preparation of this paper. I also thank I. Coandă for useful discussions.</footnote>
2. Let $\alpha(C)$ denote the image of $C$ under the composition of the natural morphisms

$$C \to \mathbb{G}(3, 7) \xrightarrow{\text{Plücker}} \mathbb{P}^{34},$$

where $\mathbb{G}(3, 7)$ is the Grassmannian of planes in $\mathbb{P}^6$. Then $\deg(\alpha(C)) = d$.

To see this, choose a hyperplane in $\mathbb{P}^{34}$ cutting $\mathbb{G}(3, 7)$ along a Schubert variety $\overline{W}_1$ of codimension one. Its intersection points with $\alpha(C)$ correspond to the fibres of $X \to C$ cutting the 3-plane which defines $\overline{W}_1$. (Recall that $\overline{W}_1$ is the set of planes intersecting some fixed 3-plane in $\mathbb{P}^6$.) Choosing a general such 3-ptane we get $d = \deg(\alpha(C))$.

3. Using equation (1) and Castenuovo's inequality (cf. [Harris(1981)]) one finds that $\alpha(C)$ spans at most a 6-plane in $\mathbb{P}^{34}$.

Indeed, if $\alpha(C)$ spanned a 7-plane we should have for $M = [(d-1)/6]$ and $\epsilon = d - 1 - 6M$:

$$\frac{(d-3)(d-4)}{12} = g \le \binom{M}{2} 6 + M\epsilon,$$

hence $g = 0$ and $d \in \{3, 4\}$, a contradiction!

4. The 1-codimensional Schubert varieties of $\mathbb{G}(3, 7)$ correspond to 3-planes in $\mathbb{P}^6$ thus forming a 12-dimensional family. Since these Schubert varieties are hyperplane sections of $\mathbb{G}(3, 7)$ in $\mathbb{P}^{34}$, we find that there is at least a 5-dimensional subfamily of such varieties containing a 6-plane which contains $\alpha(C)$ or, equivalently, a 5-dimensional family of 3-planes in $\mathbb{P}^6$, $(P_i)$, intersecting every fibre of $X \to C$. It can be assumed that for each $i$, $P_i \cap X$ gives a section of $X \to C$ which we denote by $C_i$, otherwise we get a 2-dimensional scroll over $C$ embedded in $\mathbb{P}^3$ and hence $g = 0$.

5. Suppose all the $C_i$-s pass through the same point of the generic fibre of $X \to C$. This implies that they all coincide. If they are lines, then $g = 0$ and we are done. If not, they must be plane curves hence the $P_i$-s have a common plane. It would follow that the family $(P_i)$ is at most 3-dimensional contradicting 4.

Henceforth we shall exclude this situation.

6. Assume now that all the $C_i$-s are plane curves.

If two such planes intersect each other, then they span at most a 4-plane which will contain the 2-dimensional scroll over $C$ generated by the two corresponding sections $C_1$ and $C_2$, say. In this case $C$ is rational or elliptic by [Aure(1987)].

Since the family of 3-planes containing a given 2-plane is 3-dimensional, we must have at least a 2-dimensional family of planes, $(P'_j)$, cutting sections $(C_j)$
of $X \to C$. Let $T$ be the (reduced, irreducible) base of this family. We consider now the case when all its members are disjoint.

Then any $P'_j$ intersects any fibre, excepting possibly one, in a point. Their intersections with the general fibre cover this fibre and we get a morphism $\mathbb{P}^2 \to T$. Since it cannot contract a line, it must be even one-to-one and thus coincide with the normalization of $T$. Considering now the commutative diagram of natural morphisms

<figure><img src="image_1.png" /></figure>

in which $I$ is the incidence variety of our family, we get a projection, $p: X \to \mathbb{P}^2$, whose fibres are exactly the curves $C_j$. Then $X \cong C \times \mathbb{P}^2$ and the embedding in $\mathbb{P}^6$ is given by a line bundle $\pi^*\mathcal{L} \otimes p^*\mathcal{O}_{\mathbb{P}^2}(1)$, where $\mathcal{L}$ is a very ample line bundle on $C$ of degree $d'$. It follows that $d = 3d'$ and by (1) and Castelnuovo's inequality we have

$$ \frac{(3d' - 3)(3d' - 4)}{12} = g \leq \frac{(d' - 1)(d' - 2)}{2}, $$

hence $d' = 1, g = 0$ and $X$ is Segre embedded.

7. We can assume now there exists a section $C_1$ in our family which is a space curve and let $P_1$ denote the 3-plane in which it lies.

$C_1$ admits a secant which is not a multisecant. Denote by $A$ and $B$ its intersection points with $C_1$. There is a 1-dimensional subfamily of $(P_i)$ of 3-planes which pass through $A$ and $B$. We may take it irreducible.

There exists another 3-plane, $P_2$, through $A$ and $B$ giving a space curve $C_2$ as a section of $X \to C$. This holds because in **6** we have in fact proven that the subfamily of $P_j$-s whose corresponding sections, $C_j$, are plane curves, is at most 4-dimensional (hence it forms a thin subset in any 5-dimensional component). We may also assume that $P_1$ and $P_2$ don't contain a common plane since in this case they would span a 4-plane, etc.

Now $C_1$ and $C_2$ generate a 2-dimensional scroll $S$ over $C$ in the hyperplane $P$ spanned by $P_1$ and $P_2$. If $H$ and $F$ denote the hyperplane class on $S$ and the class of a fiber of $S \to C$ respectively, we have numerically

$$ H = C_1 + r_1 F = C_2 + r_2 F, $$

for some integers $r_1, r_2$. Then

$$ d \geq H^2 = C_1 C_2 + r_1 + r_2, $$
$$ \deg C_1 + \deg C_2 = HC_1 + HC_2 = 2C_1C_2 + r_1 + r_2 \le d + C_1C_2 $$

and we have for instance

$$ \deg C_1 \le \frac{1}{2}(d + C_1C_2). $$

But $C_1C_2 \le 2$ since these two curves intersect at most in A and B and then transversally. By the Castelnuovo inequality applied to $C_1 \subset \mathbb{P}^3$ we now get

$$ \frac{1}{12}(d-3)(d-4) = g \le \frac{1}{4}(\frac{d}{2}+1)^2 - (\frac{d}{2}+1) + 1 $$

hence $d \in \{3, 4, 7, 12\}$ and $g \in \{0, 1, 6\}$.

8. Assume finally that $d = 12$, $g = 6$.

In this case one sees that all bounds in the inequalities from **7** are attained, more precisely:

$$ \deg X = \deg S = 12, \deg C_1 = \deg C_2 = 7, r_1 = r_2 = 5. $$

In particular $C_1$ lies on a quadric in $\mathbb{P}_1$. As such it may be a divisor of type (3, 4) on a non-singular quadric or it may lie on a quadric cone. In both cases it admits a $g_3^1$, i.e. a linear system of dimension 1 and degree 3. To see this in the second case, consider the projection of $C_1$ from the vertex of the cone onto its base which is a conic. The linear system giving this projection has degree six and is the pullback of $|\mathcal{O}_{\mathbb{P}_1}(2)|$. The wanted system will be the pullback of $|\mathcal{O}_{\mathbb{P}_1}(1)|$.

Out of this we infer that $C$ cannot be realized as a plane curve. Indeed, with respect to an embedding in $\mathbb{P}^2$ we should get for the canonical bundle on $C$

$$ \mathcal{K}_C = \mathcal{O}_C(2) $$

and letting $D$ be a member of a $g_3^1$ we have by Riemann-Roch and [Hartshorne (1977), IV Lemma 5.5]

$$ \dim |(\mathcal{O}_C(1)(-D)| = \dim |(\mathcal{O}_C(1)(D)| - 3 \ge \dim |(\mathcal{O}_C(1)| + \dim |D| - 3 \ge 0. $$

Hence $\mathcal{O}_C(1) = \mathcal{O}_C(D + D')$ for some effective divisor $D'$ of degree 2. It would follow that $\dim |\mathcal{O}_C(1)(-D')| > \dim |\mathcal{O}_C(1)| - 2$, contradicting the very ampleness of $\mathcal{O}_C(1)$.

Consider now a hyperplane $P$ in $\mathbb{P}^6$ spanned by two fibers $\pi^{-1}(Q_1), \pi^{-1}(Q_2)$ of $X \to C$, for $Q_1, Q_2$ in $C$. Its intersection with $X$ contains a 2-dimensional scroll $S'$ of degree $d' \le 10$. Applying to $S' \subset P$ the procedure from **2** and **3**, we get that $\alpha'(C) \subset \mathbb{P}^{14}$ spans at most a 5-plane.

Assume first that $\alpha'(C)$ is even contained in a 4-plane. Then, as in **4**, we find a 3-dimensional family, $(P'_j)$, of 3-planes in $P$ cutting every fiber of $S' \to C$, hence a corresponding 3-dimensional family of sections $(C'_j)$. Since through two
points A, B, of $S'$ we still get a 1-dimensional subfamily of sections, working on $S'$ as in 7 we obtain

$$ \deg C'_1 \leq \frac{1}{2}(d' + C'_1 C'_2) \leq 6, $$

and $C'_1$ must be a plane curve, a contradiction!

It follows that $\alpha'(C)$ spans a 5-plane in $\mathbb{P}^{14}$, which we denote by $\mathbb{P}_{\alpha'}^5$. Castel-
nuovo's inequality tells us then that $d' = \deg(\alpha'(C)) = 10$. By Riemann-Roch

$$ h^0(\mathcal{K}_{\alpha'(C)}(-1)) = h^0(\mathcal{O}_{\alpha'(C)}(1)) + 1 - 6 \geq 1, $$

hence $\mathcal{K}_{\alpha'(C)} \cong \mathcal{O}_{\alpha'(C)}(1)$. Realizing $C$ as the intersection $C'$ of $S' \subset P^5$ with a generic hyperplane $P'$ and using the definition of $\alpha'$ we get

$$ \mathcal{K}_{C'} \cong \mathcal{O}_{P'}(1)|_C. $$

More precisely, the hyperplane section divisors of $C' \subset P'$ correspond through $\alpha'$ to hyperplane section divisors of $\alpha'(C) \subset \mathbb{P}_{\alpha'}^5$.

Applying now $\pi_*$ to

$$ 0 \to \pi^*(\mathcal{O}_C(Q_1 + Q_2)) \to \mathcal{O}_X(1) \to \mathcal{O}_{S'}(1) \to 0 $$

we get the following exact sequence on $C$

$$ 0 \to \mathcal{O}_C(Q_1 + Q_2) \to \pi_*(\mathcal{O}_X(1)) \to \pi_*(\mathcal{O}_{S'}(1)) \to 0 $$

and similarly

$$ 0 \to \mathcal{O}_C \to \pi_*(\mathcal{O}_{S'}(1)) \to \mathcal{K}_C \to 0. $$

Thus $\det \pi_*(\mathcal{O}_X(1)) = \mathcal{O}_C(Q_1 + Q_2 + K)$. Since this happens for any choice of distinct points $Q_1, Q_2$ on $C$, it follows that $\mathcal{O}_C(Q_1 + Q_2 - Q_2 - Q_3) = \mathcal{O}_C(Q_1 - Q_3)$ for any $Q_1, Q_3$ on $C$, hence $C$ is rational, a contradiction!

This ends the proof.

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## REFERENCES

[Aure(1987)] Aure, A., On surfaces in projective 4-space, Ph.D. thesis, Oslo (1987).

[Harris(1981)] Harris, J., Ann. Sc. Norm. Super. Pisa, Cl. Sci., IV. Ser. 8, 35 (1981).

[Hartshorne(1977)] Hartshorne, R., Algebraic geometry, Graduate Texts in Mathematics, Vol. 52 (Springer-Verlag, New York, 1977) pp. xvi+496.

[Ionescu(1990)] Ionescu, P., "Embedded projective varieties of small invariants. iii," Algebraic geometry, Proc. Int. Conf., L'Aquila/Italy 1988, Lect. Notes Math. 1417, 138-154 (1990). (1990).

[Lanteri(1980)] Lanteri, A., Atti Accad. Naz. Lincei, VIII. Ser., Rend., Cl. Sci. Fis. Mat. Nat. 69, 223 (1980).

MATEI TOMA, UNIVERSITÉ DE LORRAINE, CNRS, IECL, F-54000 NANCY, FRANCE

Email address: Matei.Toma@univ-lorraine.fr

URL: https://iecl.univ-lorraine.fr/membre-iecl/toma-matei/