Assume that for some $n \in \mathbb{N}$ I have a (possibly singular) irreducible, non-degenerate complex surface $X_n \subset \mathbb{P}^{N}$ with the following properties:

  1. for all $p \in X_n$ there exists a $1$-dimensional family $\mathcal{C}_p$ of rational normal curves of degree $n$ containing $p$;
  2. any two points $p, \, q \in X_n$ are joined by at least one rational normal curve of degree $n$ $C_{pq} \in \mathcal{C}_p \cap \mathcal{C}_q$.

Q1. Is it possible to completely classify those surfaces $X_n$ with the above properties?

An example is the $n^{\mathrm{th}}$ Veronese surface, namely $\mathbb{P}^2$ embedded in $\mathbb{P}^{n(n+3)/2}$ by $|\mathcal{O}_{\mathbb{P}^2}(n)|$. Are there more? Maybe singular examples?

In case Q1 turns out to be hopeless, let me ask

Q2. Is it possible to explicitly bound $N$ from above in function of $n$, i.e. finding an explicit numerical function $\varphi$ such that $N \leq \varphi(n)?$ For instance, is it true that $N \leq n(n+3)/2$, i.e. that the Veronese surface provides the example with the highest codimension?


Remark. If one replace "any" in the second property by "sufficiently general" then there are other examples: for instance we can consider the quadric surface in $\mathbb{P}^3$, in which every two points, not on the same ruling, are joined by a smooth conic (see potentially dense's and J. Starr's comments below).

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    $\begingroup$ I think your first example for $n=2$ fails. Conics on the quadric surface are exactly the plane sections, but if $p$ and $q$ are on the same ruling of the quadric surface, then any plane containing both of them also contains the line of the ruling that joins them. So there is no smooth conic containing both. Am I making sense? $\endgroup$ Jun 29, 2017 at 14:06
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    $\begingroup$ Replacing "any" in property 2 by "sufficiently general" addresses the issue identified by @potentiallydense. My recollection is that there is a classification of these surfaces, and that they are roughly the rational surface scrolls together with the Veronese surfaces. I think that Proposition 2.3.3 of Mingmin Shen's thesis might be relevant (but I need to look it over more carefully): math.columbia.edu/~thaddeus/theses/2010/shen.pdf $\endgroup$ Jun 29, 2017 at 14:11
  • $\begingroup$ You are both right, thanks. I was thinking that "sufficiently general point" was enough for my pourposes, but I realized now that I really need any point. I edited the question accordingly. $\endgroup$ Jun 29, 2017 at 14:21
  • $\begingroup$ The quadric surface example is fine if you take $n$ equal to $3$ (twisted cubics rather than plane conics). $\endgroup$ Jun 29, 2017 at 14:26
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    $\begingroup$ Let $\Sigma$ be the image of the embedding of $\mathbb{P}^1\times \mathbb{P}^1$ into $\mathbb{P}^N$ by the complete linear system of $\mathcal{O}(a,b)$. Let $(a',b')$ be a pair of integers such that $1\leq a'\leq a$, $1\leq b'\leq b$ and $a'b$ equals $ab'$. Define $n$ to be $a+b+ab'=a+b+a'b$. Rational normal curves of degree $n$ in $\Sigma$ are images of curves of type $(1,1+b')$ or $(1+a',1)$. Any two points are connected by one of these. $\endgroup$ Jun 29, 2017 at 14:34

1 Answer 1


I am writing up my comments as an answer. Let $f:\widetilde{X}\to X$ be a minimal desingularization of $X=X_n$. For a general member $\mathcal{C}$ of the family of curves $\mathcal{C}_{p,q}$, let $\widetilde{\mathcal{C}}$ be the strict transform of $\mathcal{C}$ in $\widetilde{X}$.

Lemma. The surface $\widetilde{X}$ is a smooth rational surface. The linear system of $\widetilde{\mathcal{C}}$ is basepoint free and big.

Proof. This argument is essentially the same as in Mingmin Shen's thesis. By varying $p$ and $q$, it follows that $\widetilde{\mathcal{C}}$ deforms in a family of curves on $\widetilde{X}$ that connect two general points. By the classification of surfaces in characteristic $0$, $\widetilde{X}$ is a rational surface.

Moreover, the normal bundle of $\widetilde{\mathcal{C}}$ in $\widetilde{X}$ is "generically globally generated". By the classification of line bundles on $\mathbb{P}^1$, the normal bundle is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(m)$ for some integer $m>0$. Thus, the normal bundle is globally generated, not just generically globally generated. Therefore, for every point $p\in \widetilde{\mathcal{C}}$, there exists a first-order deformation of $\widetilde{\mathcal{C}}$ that deforms away from $p$. Since the normal bundle has vanishing $h^1$ (by the computation of cohomology of line bundles on $\mathbb{P}^1$), this first-order deformation extends to an honest deformation. Since rational surfaces have $h^1(\widetilde{X},\mathcal{O}_{\widetilde{X}})$ equal to $0$, all of these deformations are in the complete linear system of $\widetilde{\mathcal{C}}$. Thus, the complete linear system is basepoint free and big. QED.

Proposition. The integer $N$ is bounded above by $n(n+3)/2$.

Proof. Consider the restriction of $f^*\mathcal{O}(1)(-r\widetilde{\mathcal{C}})$ to $\widetilde{\mathcal{C}}$. This is an invertible sheaf on a smooth genus $0$ curve that has degree $n-rm$, where $m$ is the self-intersection number of $\widetilde{\mathcal{C}}$ on $\widetilde{X}$. Notice, $m$ is positive since the linear system is basepoint free and big. Thus, the total degree is negative for $r>n/m$. Since $\widetilde{\mathcal{C}}$ gives a basepoint free linear system, it has nonnegative intersection number with every effective divisor. Thus, $f^*\mathcal{O}(1)(-r\widetilde{\mathcal{C}})$ has only the zero section when $r>n/m$. Therefore we can bound the dimension of the vector space of global sections of $f^*\mathcal{O}(1)$ on $\widetilde{X}$ by the sum over $r=0,\dots,\lfloor n/m \rfloor$ of the dimension of global sections of $f^*\mathcal{O}(1)(-r\widetilde{\mathcal{C}})|_{\widetilde{\mathcal{C}}}$ on $\widetilde{\mathcal{C}}$.

For an invertible sheaf of nonnegative degree $d$ on $\mathbb{P}^1$, the dimension $h^0$ equals $d+1$. Thus, the sum above evaluates to, $$N+1 \leq (\lfloor n/m \rfloor + 1)(n+1 - (m/2) \lfloor n/m \rfloor).$$ This is maximized when $m$ equals $1$. This gives the bound, $$N+1\leq (n+1)(n+2)/2, \ \text{ i.e., } N\leq n(n+3)/2.$$ QED.

It is possible to get a classification as in Question 1, but it might not be very explicit. By the computation above, $m\leq n$. Thus, by Mingmin Shen's theorem, the surface $X'$ with hyperplane section $\widetilde{\mathcal{C}}$ is a surface of minimal degree equal to $m$. For each of the finitely many such minimal degree surfaces of degrees $m=1,\dots,n$ in the del Pezzo-Bertini classification, we can list all of the linear systems whose intersection number with $\widetilde{\mathcal{C}}$ has degree $n$. Then we consider sublinear systems of that complete linear system with specified basepoints (basepoints of the linear system on $X'$ give negative self-intersection curves on $X$).

Edit. As a corollary of the proof of the proposition, $N$ equals $n(n+3)/2$ only if $X$ is a Veronese $n$-uple surface. Indeed, we need $m$ to equal $1$, so that the surface $X'$ of minimal degree is just $\mathbb{P}^2$ with hyperplane section a line. Moreover, we need the linear system to be the complete linear system of $f^*\mathcal{O}(1)$, or else $N$ would be strictly smaller than $n(n+3)/2$. Thus, there are no base points, and $X$ equals $X'$ embedded by the $n$-uple Veronese map.


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