Background: Let $X$ be a smooth complex projective algebraic variety, and let $V$ and $W$ be closed subvarieties. For simplicity, let's assume that $\dim V+\dim W=\dim X$.

Now Serre's famous Tor formula says that if $V\cap W$ has dimension zero, we have:

$$V\cdot W=\sum_{Z\subset V\cap W}\sum_{i=0}^\infty(-1)^i\operatorname{length}_{\mathcal O_{X,z}}\operatorname{Tor}_{\mathcal O_{X,z}}^i(\mathcal O_{X,z}/I_V,\mathcal O_{X,z}/I_W)$$

where the sum is over the irreducible components of $V\cap W$ (in this case, a finite number of points).

However (according to Wikipedia here), if $Z$ is an irreducible component of $V\cap W$ with positive dimension (this is called a nonproper intersection), then the alternating sum of $\operatorname{Tor}$'s, which I'll call $\mu(Z;V,W)$, is zero. Unfortunately, this means it cannot be used in exactly the same way as before to calculate $V\cdot W$. For example, if $V=W$, then the answer would always be zero, though certainly there exist half-dimensional varieties $V$ with $V\cdot V\ne 0$ (in fact it can even be negative).

Is it possible to remedy this situation? How does one count the intersection multiplicity if the intersection is not proper?

Comment: I know how to compute the self-intersection $V\cdot V$ as the top chern class of the normal bundle evaluated on $[V]$. I'm looking here for an answer that's the most general, i.e. applies to all $V$ and $W$ of complementary dimension, regardless of the dimension of their intersection.

  • 1
    $\begingroup$ One uses Chow's moving lemma to define the intersection pairing on the Chow ring. More precisely, let V and W be subvarieties of a smooth quasi-projective variety. Then the cycle defined by V is "rationally equivalent" to a cycle C which meets the cycle defined by W "properly". Then, we define the intersection product of V and W as the intersection product of W and C (via Serre's Tor formula). One can show that this is well-defined up to rational equivalence. That is, it gives an intersection pairing on the Chow ring. I'll post some references in the next comment. $\endgroup$ Jul 11, 2011 at 5:41
  • $\begingroup$ The following are notes by H. Gillet on K-theory and Intersection theory: math.uic.edu/~henri/preprints/K-Theory_Chow_Groups-6.pdf The relevant chapters are Chapter 2 and 3. More precisely, Section 2.6, 2.7 and Section 3.1. Note that Gillet also explains a different way to define the intersection pairing. Namely, via a technique called deformation to the normal cone. If your varieties are smooth and quasi-projective, Serre's Tor formula will suffice and gives the same intersection pairing. You could look at Fulton's book Intersection Theory for more on this. $\endgroup$ Jul 11, 2011 at 5:48
  • $\begingroup$ As has been mentioned in many other MathOverflow answers, you could also look at Hartshorne's appendix (Appendix C) in his book Algebraic Geometry. $\endgroup$ Jul 11, 2011 at 5:51

2 Answers 2


There is no formula which looks only at the generic point(s) of $V \cap W$; you need to understand the entire sheaf $\mathcal{T}or_j^{\mathcal{O}_X}(\mathcal{O}_V, \mathcal{O}_W)$. It might be worth explaining the $K$-theory perspective on this.

Let $K_0(X)$ be the Grothendieck group of coherent sheaves on $X$. There is a ring structure on $K_0(X)$, where $$[\mathcal{E}] [\mathcal{F}] = \sum (-1)^j [\mathcal{T}or_j^{\mathcal{O}_X}(\mathcal{E}, \mathcal{F}) ]$$ for any coherent sheaves $\mathcal{E}$ and $\mathcal{F}$. Here I am using $[\mathcal{A}]$ to mean "class of $\mathcal{A}$ in $K_0(X)$", and I am using that $X$ is smooth to guarantee that the sum is finite.

$K_0(X)$ has a descending filtration, $K_0(X) \supseteq K_0(X)_{1} \supseteq K_0(X)_{2} \supseteq \cdots \supseteq (0)$ where $K_0(X)_i$ is spanned by classes of sheaves with support in codimension $i$. This makes $K_0(X)$ into a filtered ring, meaning that $$K_0(X)_i K_0(X)_j \subseteq K_0(X)_{i+j} \quad (\ast)$$ Containment $(\ast)$ is NOT obvious, and we will return to this point.

Let $gr \ K_0(X)$ be the associated graded ring $\bigoplus_{i \geq 0} K_0(X)_j/K_0(X)_{j+1}$. Then there is a map of graded rings from $gr \ K_0(X)$ to the Chow ring $A^{\bullet}(X)$. This map sends $[\mathcal{O}_V]$ to $[V]$.

So, let $V$ and $W$ live in codimensions $i$ and $j$. We want to compute $[V] [W]$ in $A^{i+j}(X)$. From the above, we see that it would be enough to compute $$\sum (-1)^j [\mathcal{T}or_j^{\mathcal{O}_X}(\mathcal{O}_V, \mathcal{O}_W) ] \quad (\ast \ast)$$ as an element of $K_0(X)_{i+j}/K_0(X)_{i+j+1}$.

Every summand in $(\ast \ast)$ is supported on $V \cap W$. So, if $V \cap W$ lives in codimension $i+j$, then we can just compute the image of each summand separately in the quotient $K_0(X)_{i+j}/K_0(X)_{i+j+1}$. Working this out gives Serre's formula.

Suppose now that $V \cap W$ has codimension $k$, which is less than $i+j$. Then the individual Tor terms live in $K_0(X)_k$ and plugging into Serre's formula gives the image of $(\ast \ast)$ in $K_0(X)_k/K_0(X)_{k+1}$. But, by containment $(\ast)$, the sum $(\ast \ast)$ actually lives farther down the filtration, in $K_0(X)_{i+j}$. This is why simply plugging into the formula you quote gives $0$.

An example might be useful. Take $X = \mathbb{P}^2$. Then $K_0(X)$ is isomorphic as an additive group to $\mathbb{Z}^3$, and we'll take as a basis the structure sheaf of $X$, the structure sheaf of a line, and the structure sheaf of a point. The filtration is given by $$(\ast, \ast, \ast) \supseteq (0, \ast, \ast) \supseteq (0,0,\ast) \supseteq (0,0,0)$$

Consider intersecting a line $V$ with itself. $\mathcal{T}or_0$ is the tensor product $\mathcal{O}_V \otimes \mathcal{O}_V$, whose class is $(0,1,0)$. $\mathcal{T}or_1$, is the restriction, to $V$, of the ideal sheaf of $V$. This is $\mathcal{O}_V(-1)$ and, as you can work out, it is $(0,1,-1)$ in the basis I chose. The other Tor terms are all zero.

So the individual Tor terms are $(0,1,0)$ and $(0,1,-1)$, which each live in $K_0(X)_1$ Those leading $1$ terms correspond to the lengths of the Tor modules at the generic point of $V$. In order to compute the intersection multiplicity, you have to see farther down in the filtration, to the element $(0,1,0) - (0,1,-1)$ in $K_0(X)_2$. Indeed, $(0,1,0) - (0,1,-1) = (0,0,1)$, showing that a line in the projective plane intersects itself in the class of a point.

  • $\begingroup$ @David, do you have a reference on this? $\endgroup$
    – JME
    Jun 12, 2012 at 15:07
  • $\begingroup$ Brion, Positivity in the Grothendieck group of complex flag varieties, first paragraph on page 5. arxiv.org/abs/math.AG/0105254 I'm sure other people knew it earlier. $\endgroup$ Jun 12, 2012 at 15:12
  • $\begingroup$ Can you give a reference for the map $gr K_0(X) \to A^\bullet(X)$? I do not think it is well-defined. In fact, there is a well-defined map in the other direction, and if your map existed, they would clearly be mutually inverse. But $gr K_0(X) \cong A^\bullet(X)$ only with rational coefficients; the kernel of $A^\bullet(X) \to gr K_0(X)$ is torsion but nonzero in general. $\endgroup$
    – user147129
    Mar 18, 2020 at 16:59
  • $\begingroup$ By the way, did you mean to refer to a different paper of Brion in the comment above? The first paragraph on page 5 in that paper is just the definition of $K_0(X)$... $\endgroup$
    – user147129
    Mar 18, 2020 at 17:00

Let $ch_i(F)$ denote the $i$-th coefficient of the Chern character of a sheaf $F$. Then $$ V\cdot W = \sum_{i=0}^n (-1)^i ch_n(Tor_i(O_V,O_W)), $$ where $n = \dim X$, $O_V = O_X/I_V$, $O_W = O_X/I_W$, and one uses a natural identification of $H^{2n}(X,{\mathbb Z})$ with ${\mathbb Z}$.

  • $\begingroup$ Completely true, but is there any good way to compute $ch_n(Tor_i)$ without computing the entire class of $Tor_i$ in $K_0$? $\endgroup$ Jul 11, 2011 at 22:22
  • 2
    $\begingroup$ First, $ch_n$ depends only on the class of the object in the numerical $K_0$ which may be much smaller than the whole $K_0$. Second, of course, if you know Chern characters of $O_V$ and $O_G$ then $$ V\cdot W = ch_n(O_V\otimes^L O_W) = \sum_{i=0}^n ch_i(O_V)ch_{n-i}(O_W) $$, and formally speaking you don't need to compute other Chern coefficients, so it is a bit simpler. $\endgroup$
    – Sasha
    Jul 12, 2011 at 4:56
  • $\begingroup$ Thanks for your help; that's exactly the type of formula I was looking for. $\endgroup$ Jul 13, 2011 at 3:21

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