Intersection on Singular Varieties

Question

Let $X$ be a variety over some algebraically closed field $k$. In order to define the intersection product of the Chow ring one usually demands $X$ to be smooth. This is for example well explained by Fulton in his book Intersection Theory.

I was wondering whether it is possible to weaken this assumption. I would like it to work on normal varieties but I guess that is too much to hope for. Generalizations that only work on small dimensional varieties might also be helpful for me.

In my specific situation I also have a birational map $f:X \to Y$ where $Y$ is a smooth variety. However, I'm unsure whether this helps at all for defining the product.

Answer 1

One thing that works under pretty general conditions is intersecting curves with $\mathbb Q$-Cartier divisors. This covers for instance the example mentioned by David.

Just in case:
Definition Let $X$ be a normal variety and $D$ a Weil divisor on $X$. Then $D$ is called $\mathbb Q$-Cartier if there exists an $m\in\mathbb N$ such that $mD$ is Cartier. A normal variety $X$ is called $\mathbb Q$-factorial if all Weil divisors are $\mathbb Q$-Cartier.

Now if $X$ is a normal variety, $D$ is a $\mathbb Q$-Cartier divisor on $X$ and $C\subset X$ is a proper curve, then one can define $$ C\cdot D := \frac {C\cdot mD}m $$ for any $m$ such that $mD$ is a Cartier divisor and of course $C\cdot mD$ is defined the usual way: If $L$ is a Cartier divisor and $\gamma:\widetilde C\to C$ is the normalization of $C$, then $$ C\cdot L:= \deg_{\widetilde C} \ \gamma^*\left(\mathscr O_X(L)|_C\right). $$

In David's example a line through the singular point is not Cartier, but twice the line is, because it is linearly equivalent to the hyperplane section. The intersection of a hyperplane section and a line is clearly $1$, so the intersection of two lines through the singular point has to be $\dfrac 12$.

It turns out that $\mathbb Q$-factorial varieties exist in large numbers. For instance the minimal model program (this is currently in char $0$) produces $\mathbb Q$-factorial varieties and there is even a process of $\mathbb Q$-factorialization, which is a partial resolution and on not too bad singularities it is actually a small resolution, so the divisors are not effected.

In the particular case of a normal surface this means that if it is $\mathbb Q$-factorial, then you can intersect any two $1$-cycles, but of course the intersection numbers may be rational numbers and not integers even between integral curves. Although the intersection between a Cartier divisor and an integral curve will still always be an integer.

For a concrete example one can mention that rational surface singularities are always $\mathbb Q$-factorial. The example in David's answer falls in this category. (This is not true in higher dimensions).

As far as trying to use the resolution, the problem is that pulling back cycles does not respect linear equivalence. The only reasonable way to do it is that Cartier divisors can be pulled back as line bundles and then you can extend this numerically to $\mathbb Q$-Cartier divisors. However, then you're back to the same restriction as above.

Alternatively, one can try numerical pull-back as in Sakai's paper mentioned by Francesco. I am not sure how well that works in higher dimensions. I think Batyrev had some work on that and more recently Araujo. I don't remember off the top of my head.

For problems with trying to intersect with non-$\mathbb Q$-Cartier divisors see Tom G's answer.

Answer 2

Intersections of Weil divisors in normal surfaces wind up lying in $\mathbb{Q}$ rather than $\mathbb{Z}$. Here is the basic example of why:

Consider the projective closure of $\mathrm{Spec} \ k[x,y,z]/(xz-y^2)$. The group of Weil divisors is generated by the ideal class $D:=\langle x,y \rangle$. The group of Cartier divisors is the subgroup of index $2$, generated by $\langle x \rangle$.

Suppose that we want to intersect two generators of the group of Cartier divisors, say $\langle x \rangle$ and $\langle z \rangle$. The answer is $2$. Here are two ways to obtain this: (Method 1) The subvarieties are both locally principal (i.e. they are Cartier) and their supports are transverse, so there is no higher Tor between them. So the intersection is just the length of $k[x,y,z]/(xz-y^2, x,z) = k[y]/y^2$, which has length $2$. (Method 2) Perturb the subvarieties to $x-\epsilon$ and $z-\epsilon$. Then they meet transversely, away from the singularity, at the two points $(x,y,z)=(\epsilon, \pm \epsilon, \epsilon)$.

Similarly, one can check that $\langle x,y \rangle$ and $\langle z \rangle$ meet with multiplicity $1$.

So, in order to be linear, the intersection product of $\langle x,y \rangle$ and $\langle y,z \rangle$ must be $1/2$.

By the way, I'll mention a musing I've had and never pursued. Let $R = k[x,y,z]/(xz-y^2)$. One can show that $\mathrm{Tor}^j(R/\langle x,y \rangle, R/\langle y,z \rangle) \cong k$ for all $j$. So trying to use Serre's intersection formula produces $1-1+1-1+1-1+\cdots$. Any student of divergent series knows that this sum is $1/2$. Is there a theorem hiding here?

Answer 3

For normal surfaces there is a version of the intersection product on the group of (Weil) divisors, developed by Sakai using some ideas of Mumford.

See F. Sakai, Weil divisors on normal surfaces, Duke Math. J. Volume 51, Number 4 (1984), 877-887.

Answer 4

Normal is not enough to define an intersection product in higher dimensions. In Fulton, I think you can find the example of a cone over a quadric surface on which it is impossible to define a well behaved intersection theory. (Intersecting a plane over a line of one ruling with a line of the opposite ruling should give one point, but intersecting with a line of the same ruling should give zero. Unfortunately, a line in one ruling is rationally equivalent to a line in the other ruling, since you can deform either to be a line through the vertex.)

On the other hand, if you have a morphism to a smooth variety Y, you can at least intersect any cycle on X with the pullback of a cycle on Y. Perhaps this will be enough for you?