# irreducibility punctual Hilbert scheme of relative subschemes of length $2$

Let $$X$$ be an irreducible projective variety over $$\mathbb{C}$$ (note that I do not assume $$X$$ smooth) and let $$p : X \longrightarrow S$$ be a projective surjective morphism. For any open $$U \subset S$$, I consider the natural map:

$$\pi : \mathrm{Hilb}^2_{U}(p^{-1}(U)) \longrightarrow S^2(p^{-1}(U)/U)),$$ where $$\mathrm{Hilb}^2_U(p^{-1}(U))$$ is the punctual Hilbert scheme parametrizing relative subschemes of $$p^{-1}(U)$$ of length $$2$$ and $$S^2(p^{-1}(U)/U))$$ is the relative symmetric square of $$p^{-1}(U)$$ over $$U$$. Let me finally denote by $$\mathcal{H}_U$$ the closure in $$\mathrm{Hilb}^2_U(p^{-1}(U))$$ of : $$\pi^{-1}\left(S^{2}(p^{-1}(U)/U) \backslash \Delta_{p^{-1}(U)/U} \right),$$ where $$\Delta_{p^{-1}(U)/U)}$$ is the relative diagonal.

Question : I would like to know if there exists a non-empty $$U$$ such that $$\mathcal{H}_U$$ is irreducible?

The comment below this question seems to suggest (by generic smoothness) that it could be true if $$X$$ is smooth. However I am interested in the general case. I would also be interested in a reference (or a short proof) if the answer to the question happens to be positive.

Edit : In a former (naive) version of the question, I asked if the whole relative Hilbert scheme could be irreducible over some non-empty open $$U$$. The answer is trivially "no", as observed by Jason Starr in the comment below.

• Without a smoothness hypothesis, that is usually not true. Consider the case that $X$ is a curve with a non-planar triple point. Jan 14, 2022 at 21:51
• @JasonStarr : Indeed : I will edit my question. Jan 14, 2022 at 22:02

If I understand the question correctly, the subset $$\mathcal{H}_U$$ is the closure of the locus parametrizing two distinct points. Over this locus, the Hilbert-Chow morphism $$\pi$$ is an isomorphism so irreducibility of $$\mathcal{H}_U$$ is equivalent to irreducibility of $$S^2(p^{-1}(U)/U \setminus \Delta_{p^{-1}(U)/U})$$ which it turn would follow from irreducibility of $$S^2(p^{-1}(U)/U)$$.

Lemma: Suppose $$p : X \to S$$ is a flat and proper surjective morphism of varieties such that $$S$$ is irreducible and the generic fiber of $$p$$ is geometrically irreducible. Then the symmetric powers $$S^n(X/S)$$ are irreducible.

Proof: Since flatness is stable under base change and composition, the $$n$$-fold fiber powers $$X^n_S := X \times_S \ldots \times_S X$$ are flat over $$S$$. Moreover, the generic fiber of $$X^n_S \to S$$ is the $$n$$-fold product of the generic fiber of $$p$$ and thus geometrically irreducible. By Tag 0559, there exists a non-empty open subvariety $$U \subset S$$ such that the pullback $$X^n_U \to U$$ has irreducible fibers. By flatness, $$X^n_S$$ is the closure of $$X^n_U$$ and by this answer, $$X^n_U$$ is irreducible. Thus $$X^n_S$$ is irreducible and we conclude that $$S^n(X/S) = X^n_S/\mathfrak{S}_n$$ is irreducible.

Corollary: Suppose $$p : X \to S$$ is a proper surjection of varieties with geometrically irreducible generic fiber. Then there exists a nonempty open subset $$U \subset S$$ such that $$S^n(p^{-1}(U)/U)$$ is irreducible.

Proof By generic flatness, there exists a nonempty open subset $$U \subset S$$ over which $$p$$ is flat and this open set does the job by the Lemma.

If we don't assume that the generic fiber of $$p$$ is geometrically irreducible, then the result is false. For example we can take $$p : X \to S$$ to be the map $$\mathbb{G}_m \to \mathbb{G}_m$$ given by $$z \mapsto z^5$$. Then I believe that $$S^2(X/S)$$ has 3 irreducible components which all map $$5$$-to-$$1$$ onto $$S$$. One of them is the diagonal which we remove but then whats left is still not irreducible and doesn't become irreducible if we shrink $$S$$.

• Thanks a lot for your answer. I am very interested by your example. Just to be sure, the irreducible components of $S^2(X/S)$ are also connected components, aren't they? Otherwise it would contradict the comment I linked to in my question, since your morphism $p$ is generically smooth. Or do I miss something? Jan 15, 2022 at 9:47
• @Libli yes that's right $S^2(X/S)$ is the disjoint union of 3 irreducible varieties which are each finite étale over $S$. One way to see this is that $X^2_S$ is the subvariety $x^5 = y^5$ inside $\mathbb{G}_m^2$ and the $\mathfrak{S}_2$ action swaps $x$ and $y$. Jan 15, 2022 at 18:13
• Maybe a better way to think of it is in terms of the monodromy action though. $p$ is a cyclic covering space so the branches are cyclically ordered. Then $S^2(X/S)$ consists of the diagonal, a component parametrizing two distinct points on adjacent branches of the covering space, and another component parametrizing two distinct points on non-adjacent branches. The non-diagonal components correspond to the two orbits of the $\mathbb{Z}/5\mathbb{Z}$ action on the set of $2$-element subsets of $\{1, \ldots, 5\}$ induced by the cyclic permutation. Jan 15, 2022 at 18:17