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This is a very classical flavoured question, and probabaly it is not difficult. I would like to know the shape of the space of rational degree 7 curves in $P^4$ that pass through 10 fixed points. By "shape" I mean the dimension and possibly also more structure (maybe it is a projective space, maybe a simple variety like a quadric, etc.)

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The expected dimension of the space of stable maps $M_0(P^4,7)$ is 5*7+1 = 36, and each point imposes a codimension 3 condition so I would expect the dimension of this space to be 6. – Jim Bryan Mar 21 '14 at 17:07
Just out of curiosity, why do you ask about this? – Jason Starr Mar 21 '14 at 18:40
@IMeasy: I think it will also be difficult to rationalize the variety of degree 7 curves containing 10 nodes. Maybe the Gale transform could help . . . – Jason Starr Mar 21 '14 at 19:40
@IMeasy: "it is the cubic that I want to rationalize ..." How are you going to do that? There is certainly a dominant morphism from the corresponding fiber of $\text{ev}$ to the Segre cubic by taking the "residual 21st point". But now you have to produce a rational subvariety of the fiber of $\text{ev}$. Since this is a less concrete variety than the Segre cubic, to me this seems to go in the wrong direction. – Jason Starr Mar 23 '14 at 0:59
up vote 6 down vote accepted

This is an expansion of Jim Bryan's answer. There is a smooth stack, $\overline{\mathcal{M}}_{0,10}(\mathbb{P}^4,7)$, parameterizing $10$-pointed stable maps from genus $0$ curves to $\mathbb{P}^4$ with curve class $7[\text{line}]$. If the characteristic is $>7$, this stack is a Deligne-Mumford stack (otherwise it could be an Artin stack with finite diagonal). The stack has a dense open substack that is a scheme of dimension $46$, and the complement has codimension $4$. The stack is irreducible, and the coarse moduli space is rational (so the stack is rational). There is an evaluation morphism, $$\text{ev}:\overline{\mathcal{M}}_{0,10}(\mathbb{P}^4,7)\to (\mathbb{P}^4)^{10}.$$ This morphism is dominant and separable. This morphism is dominant, and it is separable away from small characteristics in all characteristics. Then, by the generic smoothness theorem, a general fiber is a smooth stack of dimension $6$. I expect one can prove irreducibility of the geometric generic fiber (i.e., a general fiber) by the same combinatorial strategy as in my paper with Harris and Roth, but let me come back to that. Using either the early papers of Pandharipande, or using my computation with de Jong, the divisor class of the anti-canonical bundle of $\overline{\mathcal{M}}_{0,10}(\mathbb{P}^4,7)$, and hence of a general fiber of $\text{ev}$, equals $$\frac{50}{7}\mathcal{H} + \sum_{e'+e''=7}(\frac{25}{14}e'e''-2)\Delta_{e',e''} + \frac{1}{9}\sum_{A\sqcup B =\{1,\dots,9\}}|A|(10-|A|)\Delta_{A,B\sqcup\{10\}}, $$ where $\mathcal{H}$ is the "Hilbert-Chow divisor class", and where the $\Delta$s are the boundary divisor classes for decompositions of the curve class $7[\text{line}] = e'[\text{line}] + e''[\text{line}]$, respectively, for partitions of $\{1,\dots,9\}$ into $A\sqcup B$. In particular, it appears to me that the anticanonical divisor class is big, so that a general fiber should be rationally connected by the work of Qi Zhang and of Hacon-McKernan (there are other heuristics for this as well).

Regarding irreducibility, since the fiber dimension is $6$, using the "rigidity theorem" or "bend and break", every irreducible component of a general fiber should parameterize some stable maps with 6 nodes, i.e., with 7 irreducible components. Since all 10 points have to be contained in these 7 components, this limits the combinatorial possibilities. That bounds the number of irreducible components: at most one for each combinatorial type of a tree of 7 lines containing the 10 points. Moreover, by considering what happens by deforming one node of a line pair to get a conic, one should be able to show that many of the combinatorial possibilities lie on common irreducible components. My guess is that this will prove that a general fiber is irreducible.

Edit. Actually, I am having trouble finding any tree of 7 lines containing these 10 points. That may be an indication that $\text{ev}$ is not dominant. I will look at this more closely and try to answer this one way or another.

Second edit. There are, in fact, trees of 7 lines. Choose an ordering of the points as $(p_1,q_1,r_2,r_3,p_4,q_4,r_5,r_6,p_7,q_7)$. Define $L_1=\text{span}(p_1,q_1)$, $L_4=\text{span}(p_4,q_4)$, and $L_7=\text{span}(p_7,q_7)$. Define $\Pi_{1,2} =\text{span}(p_1,q_1,r_2)$, $\Pi_{3,4} = \text{span}(r_3,p_4,q_4)$, $\Pi_{4,5}=\text{span}(p_4,q_4,r_5)$, and $\Pi_{6,7}=\text{span}(r_6,p_7,q_7)$. Let $s_{2,3}$, resp. $s_{5,6}$, be the unique point of intersection of $\Pi_{1,2}\cap \Pi_{3,4}$, resp. of $\Pi_{4,5}\cap \Pi_{6,7}$. Finally, define $L_2=\text{span}(r_2,s_{2,3})$, $L_3=\text{span}(s_{2,3},r_3)$, $L_5=\text{span}(r_5,s_{5,6})$, and $L_6 = \text{span}(s_{5,6},r_6)$. Then $(L_1,L_2,L_3,L_4,L_5,L_6,L_7)$ is a chain of 7 lines that contains the 10 points. By deforming a single node to a conic, and also by deforming the "central" chain of 3 lines to a twisted cubic, via the strategy from Harris-Roth-Starr, one can prove that there is a unique irreducible component of the fiber containing all chains of 7 lines as above. Also, since this construction depends only on "linear algebra" of lines and 2-planes in $\mathbb{P}^4$, it is valid in all characteristics. Thus $\text{ev}$ is, in fact, dominant and separable in all characteristics.

Third edit. I realize now that the paragraph above proves that the geometric generic fiber is irreducible. Since $\overline{\mathcal{M}}_{0,10}(\mathbb{P}^4,7)$ is irreducible, the natural action on the set of irreducible components of the geometric generic fiber of $\text{ev}$ by the Galois group of the function field of $(\mathbb{P}^4)^{10}$ is transitive. Hence, since one irreducible component of the geometric generic fiber contains a chain of $7$ lines as above, every irreducible component of the geometric generic fiber contains such a chain of $7$ lines. However, by the "smooth nodes" technique, there is a unique irreducible component of the geometric generic fiber that contains all such chains of $7$ lines. Hence the geometric generic fiber is irreducible.

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