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In his "Murphy's Law" paper, Vakil showed that every "singularity type" (with a precise meaning) occurs on certain Hilbert schemes; for instance, the Hilbert scheme of nonsingular curves in projective space. He also gives a method for constructing such singularities; however, the process to construct even, say, a singularity of nodal type would be extremely involved.

[As I understand it, one would have to blow up a plane at something like twenty points (conservative estimate), then take a certain eight-fold cover, then find an appropriate line bundle and take six "sufficiently general" sections,...]

For a smooth variety $X$, let $H_X$ denote its Hilbert scheme. A point of $H_X$ corresponds to a subscheme $V$ of $X$. I am interested in cases in which $V$ is also smooth. [Edit: I am also requiring that $H^1(T_X) = 0$, i.e., that the ambient variety $X$ admit no infinitesimal deformations. To put it another way, the complex structure on the smooth manifold $X$ cannot be deformed. This holds in particular if $X=\mathbb P^n$.] Certainly, explicit examples of such pairs $(V,X)$ corresponding to singular points of $H_X$ have been described; however, in the very few examples I have seen, the technique is to show that $V$ is contained in an irreducible component of $H_X$ that is generically non-reduced.

Can anyone give explicit examples of a smooth projective variety $X$ [such that $H^1(X,T_X) = 0$], together with a smooth subvariety $V$, such that the point $[V]\in H_X$ is both singular and reduced? [A method of constructing explicit examples will not suffice unless you can show, by example, that this method is actually practical to carry out.]

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Are you joking? Take a line connecting two points with conical hyperplane sections on a hypersurface of degree $d\geq n$ in $\mathbb{P}^n$. –  Jason Starr Jun 28 '12 at 21:08
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... with $n\geq 4$. And "conical hyperplane section" means that the tangent hyperplane section should be a cone. I guess I should also be more careful what I post! –  Jason Starr Jun 28 '12 at 21:12
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I should not have written that first comment the way that I did -- sorry! This is the sort of thing that is known, e.g., to "Joe Harris students", but it is not "common knowledge" (as suggested by my comment). –  Jason Starr Jun 30 '12 at 16:11
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Note: I am now including an additional condition that $H^1(T_X) = 0$, which in particular disqualifies the examples suggested above by Jason Starr. I would like to make it clear that these examples did answer the question as originally asked, and that I found them both helpful and insightful. However, since the question had no answers and no upvotes, I decided to edit it rather than ask a second, almost identical question. –  Charles Staats Jul 15 '12 at 18:44
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1 Answer

up vote 8 down vote accepted

Once again, I apologize for my previous comment! The new formulation of the problem is definitely more fun. Here is one solution which only uses smooth, rational curves.

Let $n$ be an integer, $n\geq 4$ (in fact, $n=2$ and $n=3$ make sense if you use the Kontsevich spaces instead of Hilbert schemes). Start with $\mathbb{P}^{n+1}\times \mathbb{P}^n$ with homogeneous coordinates $([x_0,\dots,x_n,x_{n+1}],[y_0,\dots,y_n])$. Consider the hypersurface $X$ of bidegree $(1,1)$ which is the zero scheme of $x_0y_0+\dots+x_ny_n$. In fact every smooth $(1,1)$-hypersurface in $\mathbb{P}^{n+1}\times \mathbb{P}^n$ is "conjugate" to this one under an automorphism of $\mathbb{P}^{n+1}\times \mathbb{P}^n$. Consider the projection $\pi_2:X\to \mathbb{P}^n$; this is a $\mathbb{P}^n$-bundle $\textbf{Proj}_{\mathbb{P}^n}\text{Sym}^\bullet(\mathcal{E})$ where $\mathcal{E}$ is the locally free, rank $n+1$, $\mathcal{O}_{\mathbb{P}^n}$-module $\mathcal{O}_{\mathbb{P}^n} \oplus Q$ and $Q$ is the cokernel of the tautological morphism $\mathcal{O}_{\mathbb{P}^n}(-1)\to \mathcal{O}_{\mathbb{P}^n}^{\oplus (n+1)}$. From this you can compute that $H^1(X,T_X)$ is zero; it is the same as proving that $H^1(\mathbb{P}^n,\textit{Hom}_{\mathcal{O}_{\mathbb{P}^n}}(\mathcal{E},\mathcal{E}))$ is zero.

Now consider the Hilbert scheme of (flat families of) curves $C$ in $X$ which have constant Hilbert polynomial $1$ with respect to the invertible sheaf $\pi_1^*\mathcal{O}_{\mathbb{P}^{n+1}}(1)$ and which have Hilbert polynomial $t\mapsto nt+1$ with respect to the invertible sheaf $\pi_2^*\mathcal{O}_{\mathbb{P}^n}(1)$. In other words, $C$ is contained in a fiber of $\pi_1$, and under $\pi_2$ the curve $C$ projects isomorphically to a curve of degree $n$ and arithmetic genus $0$. Consider the smooth parameterized curve $V$ which is the image of the following closed immersion. $$v:\mathbb{P}^1 \to \mathbb{P}^{n+1}\times \mathbb{P}^n, \ \ [t_0,t_1] \mapsto ([0,0,\dots,0,1],[0,t_0^n,t_0^{n-1}t_1,\dots,\widehat{t_0^kt_1^l},\dots,t_0t_1^{n-1},t_1^n])$$ where $(k,l)$ is any pair of integers with $2\leq k,l\leq n-2$ and $k+l=n$. So the image of $V$ in $\mathbb{P}^n$ is a smooth, rational, degree $n$ curve spanning a $\mathbb{P}^{n-1}$ (in particular, it is not a rational normal curve since it is "linearly degenerate"). Just to remark, since the morphism above is $\mathbb{G}_m$-equivariant for the standard action of $\mathbb{G}_m$ on $\mathbb{P}^1$, it suffices to check that $v$ is a closed immersion at $[1,0]$ and $[0,1]$, and this follows since we included the coordinates $t_0^n$, $t_0^{n-1}t_1$, $t_0t_1^{n-1}$ and $t_1^n$.

The deformation theory of smooth integral curves in smooth varieties is described, e.g., in Theorem II.1.7, pp. 95-96, of Rational Curves on Algebraic Varieties by János Kollár. In particular, the Hilbert scheme is a local complete intersection at $[C]$ if its (local) dimension equals the "expected dimension", i.e., $\text{deg}_C(c_1(T_X))+(\text{dim}(X)-3)(1-g(C))$. In this case, that works out to be $(n+1)^2-4$. Inside the Hilbert scheme there is an open subscheme $U$ which parameterizes smooth, integral curves; in particular $[V]$ is contained in $U$.

The claim is that $U$ has precisely two irreducible components each having dimension $n^2+2n-3$, and the Hilbert scheme is reduced at the generic point of each irreducible component. Moreover, $[V]$ is contained in the intersection of the two irreducible components. Assuming the claim, it follows that $U$ is a local complete intersection. Since $U$ is a local complete intersection that is generically reduced, $U$ is everywhere reduced. Since $[V]$ is contained in the intersection of the two irreducible components, $[V]$ is a singular point of $U$.

The first irreducible component is the closure ${U}_1$ in $U$ of the open subset $W_1$ parameterizing smooth curves $C$ such that $\pi_1([C])$ is not $[0,\dots,0,1]$. The point is that $\pi_1:X\to \mathbb{P}^{n+1}$ is a $\mathbb{P}^{n-1}$-bundle away from $[0,\dots,0,1]$. Thus $W_1$ is Zariski locally over $\mathbb{P}^{n+1}\setminus \{[0,\dots,0,1]\}$ equal to a product with fiber isomorphic to the Hilbert scheme of smooth, rational, degree $n$ curves in $\mathbb{P}^{n-1}$. That fiber space is well-described in many places: it is a smooth, rational variety of dimension $n^2+n-4$. Hence $W_1$ is a smooth, rational variety of dimension $(n^2+n-4)+(n+1)=n^2+2n-3$.

The second irreducible component $U_2$ is the Hilbert scheme of smooth, rational, degree $n$ curves in the fiber $D:=\pi_1^{-1}([0,\dots,0,1])\cap X$. Note that $\pi_2:D \to \mathbb{P}^n$ is an isomorphism. Again the Hilbert scheme of smooth, rational, degree $n$ curves in $\mathbb{P}^n$ is a smooth, rational variety of dimension $n^2+2n-3$. The scheme $U$ equals the union of its closed subsets $U_1\cup U_2$, and even $W_1\cup U_2$, from which it follows that $\text{dim}(U)$ equals $\text{dim}(U_1)=\text{dim}(U_2)=n^2+2n-3$. Thus $U$ is indeed a local complete intersection. Moreover, $U_1$ contains the dense open subscheme $W_1$, which is smooth, hence reduced. Thus $U$ is reduced at the generic point of $U_1$.

The last issue is whether or not $U$ is reduced at the generic point of $U_2$. This is the same as the question of whether or not $H^1(C,N_{D/X})$ is zero for a rational normal curve (of degree $n$) in $D \cong \mathbb{P}^n$. In fact $N_{D/X}\cong \pi_2^*Q^\vee$, with $Q$ as above. By an explicit computation, for a rational normal curve $f:\mathbb{P}^1\to \mathbb{P}^n$, the pullback $f^*Q$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(1)^{\oplus n}$, and hence $f^*N_{D/X}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus n}$. In particular, $H^1(\mathbb{P}^1,f^*N_{D/X})$ is zero. From this it follows that $U$ is smooth at the generic point of $U_2$.

Finally, why is $[V]$ in the intersection $U_1\cap U_2$? Clearly, $[V]$ is contained in $U_2$. To see that $[V]$ is contained in $U_1=\overline{W}_1$, consider the family, parameterized by the coordinate $s$, of image curves $V_s$ of the following closed immersions.$$v_s:\mathbb{P}^1 \to \mathbb{P}^{n+1}\times \mathbb{P}^n, \ \ [t_0,t_1] \mapsto ([s,0,\dots,0,1],[0,t_0^n,t_0^{n-1}t_1,\dots,\widehat{t_0^kt_1^l},\dots,t_0t_1^{n-1},t_1^n]).$$ For $s\neq 0$, the images are curves in $W_1$. For $s=0$, the image equals the curve $V$. Thus $[V]$ is the image of $0$ under the corresponding morphism $$\tilde{v}:\mathbb{A}^1\to U, \ s\mapsto [V_s],$$ and thus is in the closure of $\tilde{v}(\mathbb{G}_m) \subset W_1$.

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By the way, this variety $X$ and the example curve $V$ illustrate that there is a big difference between a (rationally connected) variety being "infinitesimally rigid" and being "convex" (in the sense of Kontsevich). Conjecturally convex varieties are all projective homogeneous varieties. As this example illustrates, there are very many non-convex varieties which are infinitesimally rigid. –  Jason Starr Aug 2 '12 at 18:49
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This answer is great (and very thorough)! –  Charles Staats Aug 2 '12 at 20:16
    
What just happened ?! I used community wiki to avoid reputation points, but somehow I still got them. –  Jason Starr Aug 3 '12 at 16:51
    
Community Wiki prevents you from receiving reputation for upvotes, but does not interfere with bonuses. If you are determined to get rid of the points you gained this way, you can ask a Community Wiki question, put a bonus on it, and accept your own answer. –  Charles Staats Aug 3 '12 at 22:24
    
("bonus" should read "bounty".) Next time you want to answer a question with a bounty, you might add a note that if the OP wants to accept your answer, he should instead write a "dummy answer" of his own, directing readers to your answer, and accept that. –  Charles Staats Aug 3 '12 at 22:42
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