# Rational curves in projective spaces

Let $X\subset(\mathbb{P}^{N})^n$ be the variety defined as follows: $(p_1,...,p_n)\in (\mathbb{P}^{N})^n$ such that there exists a rational curve $C$ of degree $d$ with $p_1,...,p_n\in C$.

Is there a bound $n\geq f(d,N)$ ensuring that $X\neq (\mathbb{P}^{N})^n$? For instance, if $d = N$ we have $n\geq N+4$.

In this case is it true that $X$ is normal and that given a general point $(p_1,...,p_n)\in X$ there exists a unique rational curve of degree $d$ in $\mathbb{P}^N$ passing through $p_1,...,p_n$?

In will rephrase my question in terms of moduli spaces of stable maps. Let us consider the moduli space of stable maps $\overline{M}_{0,n}(\mathbb{P}^N,d)$, with the evaluation map $$ev:\overline{M}_{0,n}(\mathbb{P}^N,d)\rightarrow (\mathbb{P}^{N})^n$$ Does there exists a number $f(d,N)$ such that $ev$ is birational onto its image for any $n\geq f(d,N)$? In this case is its image $X:=ev(\overline{M}_{0,n}(\mathbb{P}^N,d))$ normal?

• I am not sure I understand your question, but: through $N+3$ general points in $\mathbb{P}^N$ there passes a unique rational curve of degree $N$, necessarily normal; see for instance Theorem 1.18 of Harris' "Algebraic Geometry: a first course". – abx May 28 '15 at 15:46
• I rephrased my question in terms of moduli spaces of stable maps. Perhaps now it is more clear. – japin May 28 '15 at 17:28
• That morphism is obviously never birational onto its image if $N$ equals $1$ and $d\geq 2$. For $N>1$, the morphism is birational to its image for $n\gg 0$. This follows, for instance, from birationality of the Hilbert-Chow morphism. Why do you ask about this? – Jason Starr May 29 '15 at 8:52
• Yes, if $N=1$ the morphism is birational if and only if $d = 1$. For $n>>0$ I would like to conclude that $ev_{*}\mathcal{O}_{\overline{M}_{0,n}(\mathbb{P}^N,d)} = \mathcal{O}_{X}$ where $X$ is the image of $ev$. That's why I would like to have $$ev:\overline{M}_{0,n}(\mathbb{P}^N,d)\rightarrow X\subset(\mathbb{P}^{N})^n$$ birational and $X$ normal. – japin May 29 '15 at 16:50
• @japin: Birationality should be fine for $N$ equals $2$, and this should imply birationality for all $N \geq 2$ via linear projection. However, for $N$ equals $2$ and $d\geq 3$, the image should not be normal. The problem, of course, is that the image will fail to be unibranch along the images of the boundary divisors when any of the $n$ points crosses a node. For $N$ at least $3$ and $d$ at least $3$, the image is probably normal. – Jason Starr May 30 '15 at 10:32

This answer is basically a summary of the comments above. First, for $N$ equals $1$ and for $d\geq 2$, for every integer $n$, the morphism is not birational. One can analyze the pushforward of the structure sheaf; there is something about this in "The Kodaira dimension of spaces of rational curves on low degree hypersurfaces".
For $N$ equals $2$, denote by $\mathbb{P}H^0(\mathbb{P}^2,\mathcal{O}(d))$ the projective space parameterizing degree $d$, effective Cartier divisors in $\mathbb{P}^2$. Denote by $\Sigma^o(d)$ the locally closed subset of $\mathbb{P}H^0(\mathbb{P}^2,\mathcal{O}(d))$ parameterizing Cartier divisors that are reduced, irreducible, with precisely $\delta = (d-1)(d-2)/2$ ordinary double points and no other singularities, i.e., the Severi variety of degree $d$, genus $0$ curves in $\mathbb{P}^2$. Denote by $\Sigma(d)$ the Zariski closure of $\Sigma^o(d)$ in $\mathbb{P}H^0(\mathbb{P}^2,\mathcal{O}(d))$. Then there is a "Hilbert-Chow morphism", $$FC:\overline{\mathcal{M}}_{0,0}(\mathbb{P}^2,d) \to \Sigma(d).$$ This morphism "factors" the evaluation morphisms above. This morphism is birational, and thus also the evaluation morphisms are birational for $n\gg 0$. Via linear projections, it follows that for every $N\geq 2$, for every $d\geq 1$, for every $n\gg 0$, the evaluation morphism is birational.
However, $\Sigma(d)$ is not normal, and thus also the images of the evaluation morphisms will not be normal when $N$ equals $2$, when $d\geq 3$, and when $n\gg 0$. Consider, for instance, the irreducible closed subset $D_{1,d-1}$ of $\Sigma(d)$ whose general points parameterize reducible Cartier divisors, one component of which is a line $L$, and the other component is a Cartier divisor $D$ of degree $d-1$ parameterized by $\Sigma(d-1)$. The issue is that $L\cap D$ consists of $d-1$ points. For each point, we can deform just that node of $L\cup D$ to obtain a general point of $\Sigma(d)$. Thus, $\Sigma(d)$ has $d-1$ branches along $D_{1,d-1}$. Therefore $\Sigma(d)$ is not normal.
Actually, now I see that this problem persists for $N\geq 3$. Consider curves in $\mathbb{P}^N$ of the form $L\cup D$, where $D$ is a general curve of genus $0$ and degree $d-1$ in $\mathbb{P}^N$, and where $L$ is a line that intersects $D$ in $2$ points. Locally near such a point, the image $\Sigma(d)$ (not a closed subset of the Chow variety) has two branches. Since $\Sigma(d)$ is not normal, also the images of the evaluation morphisms are not normal.