I'm trying to understand the group of cycles (modulo numerical equivalence) contracted by a flopping contraction $f$.

More precisely, I'm in the setup of Definition 2.12 of this paper by Yukinobu Toda.

Let $f: X \to Y$ be a flopping contraction: $X$ is a smooth and projective CY3, f is birational, $Y$ is Gorenstein, $f$ is isomorphic in codimension one, $dim_\mathbb{R} N^1(X/Y)_\mathbb{R}=1$.

Where $N^1(X/Y)$ is the group of divisors of $X$ modulo numerical equivalence over $Y$ (viz. $D_1 \equiv D_2$ iff $D_1.C=D_2.C$ for all curves $C$ contracted by $f$).

(a side question is: what's the correct way to define "isomorphic in codimension d"?)

Denote, $N_1(X/Y)$ the group of 1-cycles contracted by $f$, modulo numerical equivalence.

What is $N_1(X/Y)$? (without tonsuring with Q or R)

In the paper cited above, it is written that the exceptional locus of $f$ is a tree of projective lines $C_1 \cup \ldots \cup C_m$

Is $C_i \equiv C_j$?

In the end I'm really hoping that $N_1(X/Y) = \mathbb{Z}$. If this is not the case, then I'm also interested in what happens after tensoring with $\mathbb{Q}$.



1 Answer 1


1) $f$ is isomorphic in codimension $d$ if it is an isomorphism near any codimension $d$ point in either $X$ or $Y$. Equivalently, there exists closed subsets $Z\subseteq X$ and $W\subseteq Y$ such that ${\rm codim}_XZ\geq d+1$, ${\rm codim}_YW\geq d+1$, and $f:X\setminus Z\overset{\simeq}{\longrightarrow} Y\setminus W$ is an isomorphism.

2) By the Theorem of the Base of Néron–Severi, if $f$ is proper of finite type, then $N_1(X/Y)_{\mathbb Q}$ and $N^1(X/Y)_{\mathbb Q}$ are finite-dimensional vector spaces of the same dimension. This is actually more than you need, because even without the finite type assumption it is true that the intersection pairing $N_1(X/Y)_{\mathbb Q}\times N^1(X/Y)_{\mathbb Q}\to {\mathbb Q}$ is non-degenerate.

  • $\begingroup$ Ah, thanks for 1). Regarding 2), I don't see how this tells me whether $C_i \equiv C_j$, even after tensoring with $\mathbb{Q}$. $\endgroup$
    – babubba
    Commented Oct 29, 2010 at 18:55
  • 4
    $\begingroup$ One of your assumptions is that $\dim_{\mathbb R}N^1(X/Y)=1$. Then 2) implies that $\dim_{\mathbb R}N_1(X/Y)=1$ or equivalently $\dim_{\mathbb Q}N_1(X/Y)=1$. In addition, the cycle of an algebraic subvariety can never be $0$, so it follows that $aC_i≡bC_j$ for some $0\neq a,b\in \mathbb Z$. Now $C_i$ and $C_j$ are part of a tree, so taking the reduced cycle that connects them, say $C$, one gets that $C_i\cdot C=C_j\cdot C =1$, so $a=b$ and you get that $C_i\equiv C_j$ over $\mathbb Q$. $\endgroup$ Commented Oct 29, 2010 at 20:10
  • $\begingroup$ great, thanks for that! Can I ask why at the end you write $C_1 \equiv C_2$ over Q? Doesn't your argument imply that the two curves are numerically equivalent even over Z? $\endgroup$
    – babubba
    Commented Oct 29, 2010 at 21:26
  • $\begingroup$ Actually, you are right. I was worried about torsion, but since you only care about numerical equivalence, you get it over $\mathbb Z$. $\endgroup$ Commented Oct 29, 2010 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.