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Let $X$ be a degree $d$ hypersurface in $\mathbb{P}^n$. For $(d,n)=(3,4),(4,5)$, or $(5,6)$, Coskun and Starr proved in Rational curves on smooth cubic hypersurfaces that the Kontsevich space $\overline{\mathcal{M}}_{0,0}(X,e)$ of rational curves has two irreducible components ($e>1$):

1) $e$ to 1 covers of a line

2) the closure of the smooth rational curves

I'm confused about a detail in the proof that these are the only two components. The proposed proof is to specialize to a chain of lines, and then argue that the chain of lines is a smooth point of the Kontsevich space. (And they take care that if you start with a curve that's not a cover of a line, you don't degenerate it into a cover of a line.)

Why is it clear that the space of chains of lines (even restricted to those that are generically injective) is irreducible without some monodromy argument? Otherwise, even for the first case of conics on a cubic threefold, I worry that there might be that there are multiple components of conics, where for each component, we can specialize a general element to a chain of lines corresponding to a smooth point of the Kontsevich space, but elements of different components specialize to different chains of lines.

(Edit: I should mention that when $d=n-1$ as above, we would expect there to be finitely many lines through each point of the hypersurface)

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    $\begingroup$ For $(d,n)=(3,4)$ this is classical: the set of lines in a cubic threefold that intersect a fixed line is a degree $10$ cover of that line that is connected. I believe the geometry of this curve is discussed by Clemens and Griffiths in their study of the Picard group of the surface $S$ of lines. So the projection from the space of line pairs to the space of (unpointed) lines has connected fibers. We also know the dimension of the singular locus of the parameter space is at most $n-3$ (one of the lines must have special normal bundle), which is codimension $2$ in the full parameter space. $\endgroup$ Aug 12, 2017 at 21:45
  • $\begingroup$ Thanks for the reference! I'm interested in what happens more generally in the case $d=n-1$. Do you know what to do in higher degrees (4,5) and (5,6)? (Is it necessary to argue the space of chains of lines is irreducible to complete the proof that there are only two components of the Kontsevich space?) $\endgroup$
    – DCT
    Aug 12, 2017 at 22:03

1 Answer 1

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Answer under revision. The OP has now asked for the proof of irreducibility for all $e\geq 1$, not just for $e= 2.$ The argument for that uses the same basic ideas, but the edits are getting very long. As I have time, I will write this up as a PDF file and add a link to this post. What is written below is provisional, until I can finish the PDF document.

The arguments below are roughly the same as in the second chapter of my thesis. My thesis included some ad hoc arguments that apply for cubic threefolds. However, I think it is better to base the argument on a slight variant of Cristian Minoccheri's version of the Bertini Connectedness Theorem, Proposition 3.1 of the following.

Cristian Minoccheri
On the Arithmetic of Weighted Complete Intersections of Low Degree
https://arxiv.org/pdf/1608.01703.pdf

The following variant combines Minoccheri's result with Lemma 3.1 of the following.

A. J. de Jong and Jason Starr
Low degree complete intersections are rationally simply connected
https://www.math.stonybrook.edu/~jstarr/papers/nk1006g.pdf

Variant Statement. Let $X$ be a smooth $k$-scheme that is integral and étale simply connected. Let $M$ and $N$ be normal integral schemes. Let $i:N\to M$ and $h:M\to X$ be proper, finitely presented morphisms such that both $h$ and $h\circ i$ are surjective. For the smooth locus $M^o$ of $h$, if the complement of $i^{-1}(M^o)$ has codimension $\geq 2$ in $N$, then all fibers of $h$ are connected.

Proof. This is essentially the same proof as in Minoccheri's theorem. For the finite parts, $u:\widetilde{M}\to X$, resp. $v:\widetilde{N}\to X$, of $h$, resp. of $h\circ i$, there is an induced $X$-morphism $\widetilde{i}:\widetilde{N}\to \widetilde{M}$. For the smooth locus $\widetilde{M}^o$ of $u$, the inverse image in $N$ of $\widetilde{M}^o$ contains the inverse image of $M^o$. By hypothesis, the complement has codimension $\geq 2$. Thus, by the usual Seidenberg argument (e.g., Section 15 of Matsumura's "Commutative ring theory"), also the complement of $\widetilde{i}^{-1}(\widetilde{M}^o)$ in $\widetilde{N}$ has codimension $\geq 2$. Since $\widetilde{i}$ is a finite, surjective morphism of normal, integral schemes, the Going-Up and Going-Down Theorem hold. Therefore, the singular locus of $u$ has codimension $\geq 2$. By the Purity Theorem, Section X.3 of SGA 2, $u$ is finite and étale. By hypothesis, $X$ is étale simply connected. Thus $u$ is an isomorphism. So every (geometric) fiber of $h$ is connected. QED

Definition. For a closed subscheme $Y$ of $\mathbb{P}^n$, the nonlinear locus is the maximal open substack $\mathcal{M}_{0,r}(Y,e)^o\subset \mathcal{M}_{0,r}(Y,e)$ parameterizing stable maps with irreducible domain whose image is contained in no line.

Application 1. Irreducibility of the geometric generic fiber of $\text{ev}:\mathcal{M}_{0,1}(Y,2)^o \to Y$ for $Y\subset \mathbb{P}^n$ a general hypersurface of degree $n-1$.

Proposition [Harris, Roth, S]. For $Y\subset \mathbb{P}^n$ a sufficiently general hypersurface of degree $d\leq n-1$, $\mathcal{M}_{0,1}(Y,1)$ is irreducible and smooth of the expected dimension $2n-d-2$ and the evaluation morphism $\text{ev}:\mathcal{M}_{0,1}(Y,2)\to Y$ is flat of relative dimension $n-d-1$.

Sketch of proof. Inside the parameter space of Taylor expansions of a degree $d$ polynomial about a point of $\mathbb{P}^n$ that is a zero of the polynomial, for the closed locus where the common zero locus of the $d$ homogeneous parts has codimension strictly less than $d$, the codimension of this locus is $\geq n$ by a specialization argument. Since $Y$ has dimension $n-1$, this locus is disjoint from $Y$ for a general choice of $Y$. QED.

Let $n\geq 4$ be an integer. Let $d$ equal $n-1$.

Notation. Denote by $\mathcal{Y}$, resp. $\mathcal{Y}'$, the projective space parameterizing degree $d$ hypersurfaces $Y$ in $\mathbb{P}^n$, resp. pairs $(q,[Y])$ of $[Y]\in \mathcal{Y}$ and a point $q\in Y$. Denote by $\mathcal{W}$, resp. $\mathcal{W}'$, the parameter space of pairs $([C],[P])$ of a $2$-plane $P$ in $\mathbb{P}^n$ and a plane conic $C$ in $P$, resp. of triples $(q,[C],[P]).$ Denote by $\mathcal{V}\subset \mathcal{W}\times \mathcal{Y}$, resp. by $\mathcal{V}'\subset \mathcal{W}'\times \mathcal{Y}$, the parameter space of triples $(([C],[P]),[Y])$ such that $C$ is contained in $Y$, resp. of $4$-tuples $((q,[C],[P]),[Y])$ such that $C$ is contained in $Y$. Denote by $\Theta:\mathcal{V}\to \mathcal{Y}$, resp. $\Theta':\mathcal{V}'\to \mathcal{Y}',$ the natural projection.

Lemma. The closed subscheme $\mathcal{V}$ with its projection to $\mathcal{W}$ is a projective subbundle of codimension $2d+1$ of the product projective bundle $\mathcal{W}\times \mathcal{Y}$, and similarly for the base change $\mathcal{V}'\to \mathcal{W}'$. In particular, $\mathcal{V}$, resp. $\mathcal{V}'$, is irreducible and it is smooth over $\mathcal{W}$, resp. over $\mathcal{W}'.$

Proof. This is equivalent to the statement that every $C$ parameterized by $\mathcal{W}$ "imposes independent conditions" on degree $d$ polynomials, i.e., the following restriction homomorphism is surjective, $$ H^0(\mathbb{P}^n,\mathcal{O}(d))\to H^0(C,\mathcal{O}(d)|_C).$$ This is equivalent to the statement that for the ideal sheaf $\mathcal{I}$ of $C$ in $\mathbb{P}^n$, the cohomology $h^1(\mathbb{P}^n,\mathcal{I}(d))$ is zero. Since $C$ is a complete intersection curve of type $(1,\dots,1,2)$, by the Koszul resolution, $h^q(\mathbb{P}^n,\mathcal{I}(e))$ equals zero for all $q>0$ and all $e\geq 0$. QED

Notation. For every locally closed subscheme $D$ of $\mathcal{W}$, resp. of $\mathcal{W}'$, denote by $\mathcal{V}_D$ the inverse image of $D$ in $\mathcal{V}$, resp. denote by $\mathcal{V}'_D$ the inverse image of $D$ in $\mathcal{V}'.$

Lemma. For every irreducible, locally closed subscheme $D$ of $\mathcal{W}$, resp. of $\mathcal{W}'$, that is smooth or even just regularly embedded, for the restriction $\Theta_D:\mathcal{V}_D\to \mathcal{Y},$ resp. $\Theta'_D:\mathcal{V}'_D\to \mathcal{Y}'$, the morphism $\Theta,$ resp. $\Theta',$ is smooth at every smooth point of $\Theta_D,$ resp. at every smooth point of $\Theta'_D.$

Proof. Since $\mathcal{W}$, resp. $\mathcal{W}'$ is smooth, the locally closed subscheme is a regular immersion of some codimension $c$. Since $\mathcal{V}\to \mathcal{W}$, resp. $\mathcal{V}'\to \mathcal{W}',$ is flat (even smooth), also $\mathcal{V}_D$ is a regular immersion of codimension $c$ in $\mathcal{V}$, resp. $\mathcal{V}'_D$ is a regular immersion of codimension $c$ in $\mathcal{V}'.$ QED

Lemma. For the locus $D$ in $\mathcal{W}$ parameterizing nonreduced conics ("double lines"), the morphism $\mathcal{V}_D\to \mathcal{Y}$ is generically smooth. Thus $\Theta$ is smooth at the generic point of $\mathcal{V}_D$.

Proof. One proof of this uses that for every $b$ with $2<b\leq n$, every smooth Fano hypersurface $Y$ of degree $b$ in $\mathbb{P}^n$ contains lines $L$ with $h^1(N_{L/Y}(-1)) >0.$ This follows by a parameter count (also Pandharipande has a proof of this using enumerative techniques). Thus $N_{L/Y}$ has a subbundle $\mathcal{O}(1)$. This subbundle of $N_{L/\mathbb{P}^n}$ equals $N_{L/P}$ for a unique $2$-plane $P$ that contains $L$ and is contained in $\mathbb{P}^n$. The intersection of $P$ with $Y$ contains a conic $C$ supported on $L$. Therefore $\mathcal{V}_D\to \mathcal{Y}$ is surjective. Now apply the previous lemma. QED

Lemma 2. For the universal family of pointed lines over $\mathcal{Y}'$, $\text{ev}_1:\overline{\mathcal{M}}_{0,1}(\mathcal{Y}'/\mathcal{Y},1) \to \mathcal{Y}'$, the unique irreducible component of the singular locus of $\ev_1$ that has codimension $1$ is irreducible and a general point of the singular locus parameterizes a triple with $Y$ smooth along $L$ and with $N_{L/Y}$ containing a unique subbundle $\mathcal{O}(1)$. The morphism $\text{ev}_1$ is generically finite.

Proof. First, it is $n-1$ conditions on $Y$ to be singular at some point of $L$. Since $n-1>1$, it suffices to consider triples such that $Y$ is smooth along $L$. Since $N_{L/\mathbb{P}^n}$ is $\mathcal{O}(1)^{\oplus(n-1)}$, and since $N_{Y/\mathbb{P}^n}|_L \cong \mathcal{O}(n-1),$ the bundle $N_{L/\mathbb{P}^n}(-1)$ has nonzero $h^1$ if and only if there is a subbundle isomorphic to $\mathcal{O}(1)$. This subbundle is $N_{L/P}$ for a unique $2$-plane $P$ in $\mathbb{P}^n$ that contains $L$. The parameter space of such $2$-plane is $\mathbb{P}^{n-2}$. For fixed $L$ and $P$, it is $d$ linear conditions on $Y$ for $N_{L/P}$ to be a subbundle of $N_{L/Y}$. Thus the singular locus is the image of a projective bundle over a $\mathbb{P}^{n-2}$-bundle over the flag variety $\text{Flag}(1,2;n+1)$. Counting parameters, this iterated projective bundle has dimension $1$ less than the dimension of $\overline{\mathcal{M}}_{0,1}(\mathcal{Y}'/\mathcal{Y},1).$ For fixed $(p,[L],[Y])$, if there are two or more choices of $P$, then there is a positive dimensional family of such $P$. Thus, the image of this locus in $\overline{\mathcal{M}}_{0,1}(\mathcal{Y}'/\mathcal{Y},1)$ has codimension $\geq 2$. QED.

Inside $\mathcal{X}\setminus \mathcal{X}_D$, denote by $\Delta$ the closed subset where $C$ is a union of two lines $L\cup L'$ with $(q,[L],[Y])$ and $(q,[L'],[Y])$ being singular points of $\Psi$.

Lemma 3. The codimension $1$ part of $\Delta$ is irreducible. The intersection of $\mathcal{X}\setminus \mathcal{X}_D$ with $\text{Sing}(\Phi)$ is a proper closed subset of this irreducible, codimension $1$ subset of $\mathcal{X}\setminus \mathcal{X}_D$.

Proof. The open subset $\mathcal{Z}\setminus \mathcal{Z}_D$ is a single orbit under $\text{Aut}(\mathbb{P}^n)$ parameterizing conics that are unions $L\cup L'$ of a pair of distinct lines in $P$. The locus where $Y$ is singular at a point of $L$, resp. $L'$, has codimension $n-1$. Since $n\geq 4$, this is $\geq 2$. Thus, it suffices to consider the case where $Y$ is smooth along $C$. By Lemma 2, it is a codimension $2$ condition for $N_{L/Y}$ to have worse than a single $\mathcal{O}(1)$, and similarly for $L'$. So it suffices to consider the locus where both $N_{L/Y}$ and $N_{L'/Y}$ have a subbundle $\mathcal{O}(1)$ and no worse.

Denote by $P$, resp. $P'$, the unique $2$-plane that contains $L$, resp. $L'$, with $N_{L/P}$ contained in $N_{L/Y}$, resp. with $N_{L'/P'}$ contained in $N_{L'/Y}$. There is a $\mathbb{P}^{d-1}$ parameterizing $P$, and there is a $\mathbb{P}^{d-1}$ parameterizing $P'$. Except for the degenerate case when $P$ equals $P'$, which has higher codimension, for fixed $P$ and $P'$, it is $2d-1$ linear conditions on $Y$ for $N_{L/P}$ to be contained in $N_{L/Y}$ and for $N_{L'/P'}$ to be contained in $N_{L'/Y}$. The parameter space for such data is the generically finite image of the total space of a projective bundle over the $\mathbb{P}^{d-1}\times \mathbb{P}^{d-1}$-bundle over $\mathcal{Z}\setminus \mathcal{Z}_D$. This image is irreducible.

Finally, for a pair as above, since $N_{C/Y}|_L$ is the elementary transform up of $N_{L/Y}$ at the point $q$ in the normal direction $T_q L'$ of $L'$, $N_{C/Y}(-1)$ has vanishing $h^1$ unless $T_q L'$ is in the subbundle of $N_{L/Y}$ generated by global sections, cf. Graber-Harris-Starr. This is one additional linear condition on $L'$. Thus, the locus is a proper closed subset of $\Delta$. QED

Theorem. The morphism $\Phi:\mathcal{X}\to \mathcal{Y}$ is a generically smooth morphism between smooth, projective varieties. The target $\mathcal{Y}$ is a projective space, thus algebraically simply connected. The singular locus of $\Phi$ has codimension $\geq 2$ everywhere in $\mathcal{X}$. The geometric generic fiber of $\Phi$ is irreducible.

Proof. The intersection of the singular locus with $\mathcal{X}_D$ has codimension $\geq 2$ in $\mathcal{X}$ by Lemma 1. The intersection of the singular locus with $\mathcal{X}\setminus \mathcal{X}_D$ is a proper closed subset of $\Delta$. The codimension $1$ part of $\Delta$ is irreducible. Thus, $\text{Sing}(\Phi)$ has codimension $2$ in $\mathcal{X}\setminus \mathcal{X}_D$. By Cristian Minoccheri's version of the Bertini Connectedness Theorem, the geometric generic fiber of $\Phi$ is connected. It is also LCI, since $\Phi$ is a surjective morphism between smooth schemes. Finally, by the computation above, the singular locus of the fiber has codimension $\geq 2$. Thus, the geometric generic fiber of $\Phi$ is normal by Serre's criterion. Since it is also connected, it is irreducible. QED

Original post.I cannot promise that the following is the argument that we had at that time, but this is how we used to prove similar results. Let $Y$ be a general hypersurface of degree $d$ in $\mathbb{P}^n$. Consider the evaluation morphism from the parameter space of pointed lines to $Y$, $$f: X \to Y.$$ This is a finite, surjective morphism from a smooth variety $X$ to a smooth variety $Y$ that is branched over a divisor $B$. By parameter counts (similar to the ones in Chapter V of Kollár's book), the morphism has simple branching at general points of $B$ if $Y$ is general, i.e., the locus where $h^1$ of the normal bundle is $\geq 2$ has codimension $2$ in the parameter space $X$ of pointed lines in $Y$, and the locus of line pairs both having positive $h^1$ of the normal bundle is also codimension $2$ in the parameter space $\text{Sym}^2(X/Y)$ of line pairs in $Y$: for a fixed line, resp. line pair, in $\mathbb{P}^n$, consider the linear conditions on the partial derivatives of a defining equation of $Y$ in order for $h^1$ to be $\geq 2$, resp. for $h^1$ to be $\geq 1$ for both lines.

Thus, by the same answer as in the following MathOverflow question, the restriction of this finite morphism over a general $2$-plane section of $Y$ has full monodromy.

Monodromy representation of elementary simple covers

To say one more word about this: the monodromy group $H$ of the cover is a transitive subgroup of $\mathfrak{S}_n$ that is generated by transpositions. So all elements of $\{1,\dots,n\}$ are equivalent, the partition is trivial, and $H$ is the full symmetric group.

Since the parameter space of line pairs is the (closure of the) complement of the diagonal in the relative $\text{Sym}^2$ of the morphism $f$ with its projection to $Y$, this monodromy result implies that the inverse image of a general $2$-plane section of $Y$ in the space of line pairs is irreducible. Thus, the parameter space of line pairs is irreducible.

Edit to original answer. The OP asks the valid question: how do we know that the branching is simple over $B$? Consider the incidence correspondence $$\mathcal{X} = \{ (p,[L],[Y]) : p\in L\subset Y\},$$ where $L$ is a line and $Y$ is a hypersurface of degree $d$ in $\mathbb{P}^n$, $n=d+1$. For the incidence correspondence $\mathcal{Y}=\{(p,[Y]) : p\in Y\}$, there is a universal forgetful morphism, $$\Phi:\mathcal{X}\to \mathcal{Y}, \ \ (p,[L],[Y])\mapsto (p,[Y]).$$ There is an open dense subset $\mathcal{X}^0$ of $\mathcal{X}$ parameterizing triples such that $Y$ is smooth along $L$. Via projection to the flag variety of pointed lines, the locus in $\mathcal{X}^0$ where $N_{L/Y}$ has nonzero $h^1$ is an irreducible divisor $D$ in $\mathcal{X}^0$. The open subset $D^0$ where $h^1$ equals $1$ is open in $D$. It suffices to find an arc in $\mathcal{X}$ that intersects $D^0$ in one point and such that the branching of $\Phi$ along that arc is simple.

Choose homogeneous coordinates on $\mathbb{P}^n,$ $[x,y,z,w,u_4,\dots,u_n]$. Inside the projective $3$-space, $\mathbb{P}^3=\text{Zero}(u_4,\dots,u_n)$, consider the following singular cubic surface, $$\Sigma = \text{Zero}(H), \ \ H = xz^2-yw^2.$$ For every $[s,t]\in \mathbb{P}^1$, consider the following line in $\Sigma,$ $$L_{s,t} = \text{Zero}(tw-sz,t^2x-s^2y).$$ Consider this as a pointed line via the point $p_{s,t}=[s^2,t^2,0,0].$ The union of these lines is a line $M=\text{Zero}(z,w)$. For each $p\in M$ other than $[1,0,0,0]$ and $[0,1,0,0]$, there are precisely two values $[s,t]$ and $[s,-t]$ such that the corresponding point equals $p$. Thus, as $p$ in $M$ approaches $[1,0]$, this is a family of pointed lines with simple branching.

Now consider the following degree $d$ polynomial, $$F(x,y,z,w,u_4,\dots,u_{d+1}) = x^{d-3}H + u_4(x^{d-1}+w^{d-1}) + $$ $$\left[u_5x^{d-4}w^3 + \dots + u_{d+1}w^{d-1}\right].$$ The last terms in brackets are only included if $d\geq 4$. It is straightforward to compute that $Y=\text{Zero}(F)$ is smooth along the line $L_{1,0}$, and the normal bundle of the line in $Y$ is $\mathcal{O}(-1)\oplus \mathcal{O}^{\oplus(d-3)}\oplus \mathcal{O}(1)$. Thus the family of pointed lines gives an arc in $D^0$. So the fiber of $f$ is curvilinear at this point. We have the two branches $L_{s,t}$ and $L_{s,-t}$ permuted by the local monodromy. If there were any worse branching on the branch containing $(p_{1,0},[L_{1,0}])$, then the fiber would not be curvilinear.

To avoid any mention of branching (this is somewhat problematic in positive characteristic), it should be possible to compute the Hessian of $\Phi$ at the point $(p_{1,0},[L_{1,0}],Y)$ and see that this point is a simple ordinary double point of the fiber of $\Phi$.

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  • $\begingroup$ Thanks for the explanation! Can I ask why h^1(normal bundle)=1 implies simple branching? Do we have to worry that a fiber of $Y\to X$ over $B$ might contain a curvilinear scheme in general, so the tangent space is still only 1-dimensional? $\endgroup$
    – DCT
    Aug 13, 2017 at 2:07
  • $\begingroup$ @DCT. I added an explanation to the answer. $\endgroup$ Aug 13, 2017 at 12:00
  • $\begingroup$ Sorry for reviving this, but what did you have in mind in the case where $e$ is large, so a rational curve specializes to a union of many lines? $\endgroup$
    – DCT
    Aug 18, 2017 at 20:58

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