# Is rational equivalence of algebraic cycles “transitive”?

Let $X$ be a $k$-scheme, and let $A$ and $B$ be algebraic cycles on $X$ (i.e., irreducible subvarieties). Recall that $A$ is rationally equivalent to $B$ if there is a finite sequence of cycles $A_1, \dots, A_n$ on $X$ with $A_1 = A$ and $A_n = B$ such that for each $i$, there exists a flat family over $\mathbb{P}_k^1$ of cycles on $X$ interpolating between $A_i$ and $A_{i+1}$.

It follows immediately from the above definition that rational equivalence is transitive. However, I'm looking for something a bit stronger: if $A$ and $B$ are rationally equivalent cycles on $X$, does there exist a flat family over $\mathbb{P}_k^1$ of cycles on $X$ interpolating between $A$ and $B$? This is equivalent to asking whether the existence of such an interpolating family is a transitive condition.

• If $X$ is merely a $k$-scheme, then there are easy counterexamples (take $X$ to be non-integral). But I think the statement is false even if $X$ is integral, smooth and projective, but I don't know a counterexample offhand. Is it even true for $\mathbb{P}^n$? – Jake Levinson Dec 27 '16 at 2:44
• +1 for this naïve, well-formulated and difficult question: the very best combination in my eyes :-) – Georges Elencwajg Dec 27 '16 at 16:53

It's false. In $\mathbb{P}^3$, a twisted cubic $C$ is rationally equivalent to an elliptic curve $E$ contained in a plane. But there is no flat family interpolating between $C$ and $E$.

Reason: the corresponding Hilbert scheme $H$ has two irreducible components, and $C$, $E$ lie in their interiors, so no single $\mathbb{P}^1$ links them. To be precise, the first component $H_1$ parametrizes twisted cubics, and the second $H_2$ parametrizes "plane cubics union a point". The intersection $H^1 \cap H^2$ corresponds to singular plane cubics with an embedded point at the node (the nonreduced structure pokes outside of the plane into the ambient $\mathbb{P}^3$).

• Thanks for the great answer and the reference, Jake. – Georges Elencwajg Dec 27 '16 at 16:51

To add to Jake Levinson's wonderful answer, it seems that Piene's and Schlessinger's original motivation was to find a suitable compactification of the space of twisted cubic curves which would permit them to make enumerative computations (à la Schubert) in a modern setting. The main motivation for compactifying any kind of moduli space of algebraic-geometric object is that in order to have a well-functioning intersection theory, one needs a compact space. (Sometimes a partial compactification works.) So for example, to answer questions in enumerative geometry of twisted cubics, one needs a compact parameter space. The particular interest of twisted cubics is maybe that they are the simplest space curves that are not planar. Schubert studied their enumerative properties, and Hilbert asked for a rigorous foundation of Schubert calculus.

At least one of Piene or Schlessinger must have been an expert in deformation theory, as the paper looks at this question in terms of flat deformations (Hilbert schemes).

I suggest that you study not only Hilbert schemes, but also Chow schemes, in order to understand the difference between these two parameter spaces. Here are some suggested readings.

David Rydh's master's thesis: https://people.kth.se/~dary/Chow.pdf.

Chapter 4 in Gelfand, Kapranov, and Zelevinsky's book "Determinants, resultants, and multidimensional determinants".

A construction for the case of curves in $\mathbb{P}^3$ is found here: http://www.mat.uniroma3.it/users/sernesi/hilbdott07.pdf.

I think there is also a construction or definition in János Kollár's book "Rational curves on algebraic varieties".