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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 $\mathbb{Q}$.


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up vote 5 down vote accepted

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.

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Ah, thanks for 1). Regarding 2), I don't see how this tells me whether $C_i \equiv C_j$, even after tensoring with $\mathbb{Q}$. – babubba Oct 29 '10 at 18:55
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 $\mathbb Q$. – Sándor Kovács Oct 29 '10 at 20:10
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? – babubba Oct 29 '10 at 21:26
Actually, you are right. I was worried about torsion, but since you only care about numerical equivalence, you get it over $\mathbb Z$. – Sándor Kovács Oct 29 '10 at 21:42
excellent, thank you very much. – babubba Oct 29 '10 at 22:06

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