Let $X$ be a smooth projective $\mathbb C$-variety and let $X^{(n)}$ denote the symmetric product $X^n/S_n$, parametrizing effective $0$-cycles of degree $n$ on $X$.

Question. Let $S$ be a noetherian $\mathbb C$-scheme, $\mathcal F$ a coherent sheaf on $X\times S$, flat and finite (of degree $n$) over $S$. Does its support define a family of $0$-cycles on $X$, i.e. a morphism $S\to X^{(n)}$?

I came up with this question studying the construction of the Hilbert-Chow morphism $X^{[n]}\to X^{(n)}$ in FGA explained, where a fundamental tool is the result that any flat sheaf $\mathcal F$ gives a relative effective Cartier divisor on $X\times S$ with good properties (essentially commutation with base change).

I am aware that the subscheme $\textrm{Supp }\mathcal F\subset X\times S$ may not be flat over $S$ (it is flat if $S$ is reduced). I am asking if $\mathcal F$ induces an $S$-valued point $S\to X^{(n)}$, possibly not factoring through a point $S\to X^{[n]}$.

Here the problem is that I do not know what the functor of points of $X^{(n)}$ looks like, in other words I do not know how to describe $$\textrm{Hom}(S,X^{(n)})$$ for an arbitrary scheme $S$.


  • $\begingroup$ Yes, there is an associated morphism to the symmetric product. You can read something about this in the first chapter of Koll'ar's book, "Rational Curves on Algebraic Varieties". Some other good references are Angeniol, Barlet (in the complex analytic setting), and, especially, David Rydh's papers. $\endgroup$ Commented Jul 11, 2016 at 12:06
  • $\begingroup$ I was looking in Kollar's book, but I did not find a reference involving sheaves directly. What he proves is Hilb $\to$ Chow, but I was hoping to "bypass" Hilb... Can I just use the result I quoted from FGA and apply the equivalence between cycles and Cartier divisors (on a projective space where $X$ is embedded)? Unfortunately I do not seem to have access to Angéniol's book. $\endgroup$
    – Brenin
    Commented Jul 11, 2016 at 13:22
  • $\begingroup$ If $X$ has dimension $1$, then you can certainly apply the construction from FGA (I believe this is the same as the "Div" construction of Fogarty and Knudsen-Mumford). In the case that $X$ has dimension $>1$, I do not see how to make that work. Rydh's papers are available online, and I believe that Rydh does give a functor of points for the Chow schemes. $\endgroup$ Commented Jul 11, 2016 at 13:35

1 Answer 1


I quickly reviewed the following article of David Rydh.

David Rydh.
Families of Cycles.

Rydh extends to positive characteristic the definition of Angeniol in characteristic $0$. I will explain the construction in your special case. The key algebra lemma has to do with the $n\times n$ determinant. For a unital, commutative ring $R$, for a free $R$-module $F$ of rank $n$, for the associated free $R$-module $E=\text{Hom}_R(F,F)$ of rank $n^2$ with its structure of associative, unital $R$-algebra, the usual determinant, $$ \text{det}_{F}:E \to R, $$ comes from an element $$ \text{det}'_F\in \text{Sym}_R^n(E^\vee), $$ where for every $R$-module $M$, $M^\vee$ is shorthand for $\text{Hom}_R(M,R)$. Assuming that $n!$ is invertible in $R$, this element is equivalent to a homomorphism of $R$-modules, $$ \widetilde{\text{det}}_F:(E^{\otimes n})^{\mathfrak{S}_n} \to R, $$ where for an $R$-module $M$, $M^{\otimes n}$ denotes the $n$-fold tensor product $M\otimes_R \dots \otimes_R M$, and where for an $R[\mathfrak{S}_n]$-module $N$, $N^{\mathfrak{S}_n}$ denotes the $R$-submodule of $\mathfrak{S}_n$-invariants. The determinant homomorphism is normalized so that for an ordered basis $(\mathbf{e}_1,\dots,\mathbf{e}_n)$ of $F$ and for the associated basis $(\mathbf{e}_{i,j})_{1\leq i,j\leq n}$ of $E$, this $R$-module homomorphism sends the $\mathfrak{S}_n$-invariant, idempotent element $$\epsilon_{\text{id}} := \sum_{\sigma \in \mathfrak{S}_n} \mathbf{e}_{\sigma(1),\sigma(1)}\otimes \dots \otimes \mathbf{e}_{\sigma(n),\sigma(n)}$$ to $1$.

The algebra fact is that for a commutative $R$-subalgebra $A$ of $E$, the restriction of $\widetilde{\text{det}}_F$ to $(A^{\otimes n})^{\mathfrak{S}_n}$ is a homomorphism of commutative $R$-algebras. For a choice of ordered basis as above, for the subalgebra $A$ spanned by $\mathbf{e}_{1,1},\dots,\mathbf{e}_{n,n}$, this can be checked directly. The $R$-algebra $(A^{\otimes n})^{\mathfrak{S}_n}$ is isomorphic to a direct product of factors of $R$ indexed by the nondecreasing functions $u:\{1,\dots,n\}\to \{1,\dots,n\}$ (with the standard total order). The corresponding idempotent is $$\epsilon_u := \frac{1}{\# \text{Aut}(u)}\sum_{\sigma\in \mathfrak{S}_n} \mathbf{e}_{u(\sigma(1)),u(\sigma(1))}\otimes \dots \otimes\mathbf{e}_{u(\sigma(n)),u(\sigma(n))}.$$ The idempotent $\epsilon_{\text{id}}$ corresponds to the identity function on $\{1,\dots,n\}$. The determinant $R$-module homomorphism sends the idempotent $\epsilon_{\text{id}}$ to $1$, and it sends every other idempotent to $0$.

Returning to the setup in the question, denote by $\text{pr}_S:X\times S \to S$ the projection morphism. By hypothesis, the $\mathcal{O}_S$-module $F:=\text{pr}_{S,*}\mathcal{F}$ is a locally free $\mathcal{O}_S$-module of rank $n$. For every open affine subscheme $U\subset X$, there exists a maximal open subscheme $V\subset S$ such that $\text{Supp}(\mathcal{F}) \cap (V\times X)$ is contained in $V\times U$. Since $X$ is projective, as $(U,V)$ varies, the open subsets $V$ form an open cover of $S$. Thus, it suffices to construct the "Hilbert-Chow morphisms" on the open subsets $V$, provided those local morphisms satisfy the glueing condition on overlaps.

For every element $a\in \Gamma(U,\mathcal{O}_U)$, left multiplication by $a$ on $\mathcal{F}|_{V\times U}$ defines an element $\widetilde{a}$ in the endomorphism algebra, $$E := \text{Hom}_{\mathcal{O}_V}(F|_V, F|_V).$$ This defines a homomorphism of $\mathcal{O}_V$-algebras of quasi-coherent sheaves, $$ \phi : \Gamma(U,\mathcal{O}_U)\otimes_{\mathbb{C}} \mathcal{O}_V \to E.$$ Taking tensor products, this defines a homomorphism of $n$-fold tensor product $\mathcal{O}_V$-algebras, $$ \phi^{\otimes n}: \Gamma(U,\mathcal{O}_U)\otimes_{\mathbb{C}} \dots \otimes_{\mathbb{C}}\Gamma(U,\mathcal{O}_U) \otimes_{\mathbb{C}} \mathcal{O}_V \to E\otimes_{\mathcal{O}_V} \dots \otimes_{\mathcal{O}_V} E.$$ There is a natural action of the finite symmetric group $\mathfrak{S}_n$ on the domain and the target, and $\phi^{\otimes n}$ is equivariant for these actions. In particular, there is a restriction homomorphism on the subrings of $\mathfrak{S}_n$-invariants, $$\phi^{\otimes n}_{\mathfrak{S}_n} : (\Gamma(U,\mathcal{O}_U)^{\otimes n})^{\mathfrak{S}_n}\otimes_{\mathbb{C}} \mathcal{O}_V \to (E^{\otimes n})^{\mathfrak{S}_n}.$$ Finally, the determinant above defines a homomorphism of $\mathcal{O}_V$-modules, $$\widetilde{\text{det}}_F: (E^{\otimes n})_{\mathfrak{S}_n} \to \mathcal{O}_V.$$ Composing with $\phi^{\otimes n}_{\mathfrak{S}_n}$ defines a homomorphism of $\mathcal{O}_V$-modules, $$\text{det}(\phi):(\Gamma(U,\mathcal{O}_U))^{\otimes n})^{\mathfrak{S}_n}\otimes_{\mathbb{C}} \mathcal{O}_V \to \mathcal{O}_V.$$ By the algebra fact above, this is a homomorphism of commutative $\mathcal{O}_V$-algebras. This algebra homomorphism defines a morphism $V\to U^{(n)} \subset X^{(n)}$. These morphisms glue to a morphism $S\to X^{(n)}.$

Edit. Proof of the Algebra Fact. For certain choices of $R$, there are some funny commutative subalgebras $A$ of $E$, cf. my answer to the following question: Simultaneous triangularizability over a commutative ring. Thus, it makes sense to give a proof of the algebra fact. Actually it follows from a more general fact about $\widetilde{\text{det}}_F$ on the associative $F$-algebra $(E^{\otimes n})^{\mathfrak{S}_n}$.

Since $n!$ is invertible in $R$, there is a basis of $(E^{\otimes n})^{\mathfrak{S}_n}$ as follows. For every ordered $n$-tuple $(I,J) = ((i_1,j_1),\dots,(i_n,j_n))$ of pairs $(i,j)\in \{1,\dots,n\}\times \{1,\dots,n\}$, define $$\epsilon_{I,J} = \sum_{\sigma\in \mathfrak{S}_n} \mathbf{e}_{i_{\sigma(1)},j_{\sigma(1)}}\otimes \dots \otimes \mathbf{e}_{i_{\sigma(n)},j_{\sigma(n)}}.$$ The element only depends on $(I,J)$ up to permutation by an element in $\mathfrak{S}_n$. By direct computation, $\widetilde{det}_F$ is nonzero on this element if and only if $(I,J)$ is the graph of a bijection, i.e., $((\tau(1),1),\dots,(\tau(n),n))$ for some $\tau\in \mathfrak{S}_n$. In this case, denote the element by $\epsilon_\tau$. By direct computation, $\widetilde{\text{det}}_F(\epsilon_\tau)$ equals $\text{sgn}(\tau)$. Also $\epsilon_\tau \cdot \epsilon_\rho$ equals $\epsilon_{\tau\cdot \rho}$ for $\rho,\tau\in \mathfrak{S}_n$. Finally, if $\epsilon_\tau\cdot \epsilon_{I,J}$, resp. $\epsilon_{I,J}\cdot \epsilon_\tau$, equals $\epsilon_\rho$, then $\epsilon_{I,J}$ equals $\epsilon_{\tau^{-1}\cdot \rho}$, resp. $\epsilon_{\rho\cdot \tau^{-1}}$. Altogether, this implies that $\widetilde{\text{det}}_F(\epsilon_{I,J}\cdot \epsilon_{I',J'})$ is zero unless both $\epsilon_{I,J} = \epsilon_\tau$ and $\epsilon_{I',J'}=\epsilon_\rho$ for some $\rho,\tau\in \mathfrak{S}_n$. Moreover, in this case, $\widetilde{\text{det}}_F(\epsilon_\tau\cdot \epsilon_\rho)$ equals $\text{sgn}(\tau\cdot \rho)$. Since the sign of a permutation is a group homomorphism, it follows that in all cases, $$\widetilde{\text{det}}_F(\epsilon_{I,J}\cdot \epsilon_{I',J'}) = \widetilde{\text{det}}_F(\epsilon_{I,J})\cdot \widetilde{\text{det}}_F(\epsilon_{I',J'}).$$ Therefore, already the $R$-module homomorphism $$\widetilde{\text{det}}_F:(E^{\otimes n})^{\mathfrak{S}_n} \to R,$$ is a homomorphism of associative, unital $R$-algebras. Thus, the restriction to any commutative $R$-subalgebra of $(E^{\otimes n})^{\mathfrak{S}_n}$ is a homomorphism of commutative $R$-algebras.

  • $\begingroup$ Thank you so much for providing so many details. So, over $\mathbb C$, did the map $S\to X^{(n)}$ already come from Angéniol's theory? As for the algebra fact, it seems to be nontrivial in char $0$, too. I still have a hard time seeing the relation between supports and determinants (in other words why the map $\det(\phi)$ should correspond to $s\mapsto Supp(\mathcal F_s)$), is that clear from the construction somehow? $\endgroup$
    – Brenin
    Commented Jul 12, 2016 at 20:49
  • $\begingroup$ Unfortunately I also do not have Angeniol's book. However, based on what Rydh writes, I believe that Angeniol is defining the functor $\text{Chow}_{0,n}(X)$ as above. For checking that the map does what you expect on points, you may as well assume that $S$ is $\text{Spec}(\mathbb{C})$. Then you can filter $\mathcal{F}$ by coherent subsheaves whose associated subquotients are skyscraper sheaves of points. Now you can use that the determinant is multiplicative for short exact sequences. That should prove that the cycle of $\mathcal{F}$ is the sum of the length over points in the support. $\endgroup$ Commented Jul 12, 2016 at 21:20

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