Quillen's excess-intersection formula

The formula in the topological setting is due to Quillen (1971).

Now, suppose we are given Y″ → Y' and suppose i': X' = X ×_Y Y' → Y' is regular of codimension d' so that one can define i'^! as before. Let F be the excess bundle of i and i^'; that is, it is the pullback to X″ of the quotient of N by the normal bundle of i'. Let e(F) be the Euler class (top Chern class) of F, which we view as a homomorphism from A_k−d' (X″) to A_k−d(X″). Then

where i^! is determined by the morphism Y″ → Y' → Y.

Finally, it is possible to generalize the above construction and formula to complete intersection morphisms; this extension is discussed in § 6.6. as well as Ch. 17 of loc. cit.

Proof: One can deduce the intersection formula from the rather explicit form of a Gysin homomorphism. Let E be a vector bundle on X of rank r and q: P(E ⊕ 1) → X the projective bundle (here 1 means the trivial line bundle). As usual, we identity P(E ⊕ 1) as a disjoint union of P(E) and E. Then there is the tautological exact sequence

${\displaystyle 0\to {\mathcal {O}}(-1)\to q^{*}E\oplus 1\to \xi \to 0}$

on P(E ⊕ 1). We claim the Gysin homomorphism is given as

${\displaystyle A_{k}(E)\to A_{k-r}(X),\,x\mapsto q_{*}(e(\xi ){\overline {x}})}$

where e(ξ) = c_r(ξ) is the Euler class of ξ and ${\displaystyle {\overline {x}}}$ is an element of A_k(P(E ⊕ 1)) that restricts to x. Since the injection q^*: A_k−r(X) → A_k(P(E ⊕ 1)) splits, we can write

${\displaystyle {\overline {x}}=q^{*}y+z}$

where z is a class of a cycle supported on P(E). By the Whitney sum formula, we have: c(q^*E) = (1 − c₁(O(1)))c(ξ) and so

${\displaystyle e(\xi )=\sum _{0}^{r}c_{1}({\mathcal {O}}(1))^{i}c_{r-i}(q^{*}E).}$

Then we get:

${\displaystyle q_{*}(e(\xi )q^{*}y)=\sum _{i=0}^{r}s_{i-r}(E\oplus 1)c_{r-i}(E)y}$

where s_I(E ⊕ 1) is the i-th Segre class. Since the zeroth term of a Segre class is the identity and its negative terms are zero, the above expression equals y. Next, since the restriction of ξ to P(E) has a nowhere-vanishing section and z is a class of a cycle supported on P(E), it follows that e(ξ)z = 0. Hence, writing π for the projection map of E and j for the inclusion E to P(E⊕1), we get:

${\displaystyle \pi ^{*}q_{*}(e(\xi ){\overline {x}})=\pi ^{*}(y)=j^{*}q^{*}y=j^{*}({\overline {x}}-z)=j^{*}({\overline {x}})=x}$

where the second-to-last equality is because of the support reason as before. This completes the proof of the explicit form of the Gysin homomorphism.

The rest is formal and straightforward. We use the exact sequence

${\displaystyle 0\to \xi '\to \xi \to r^{*}F\to 0}$

where r is the projection map for . Writing P for the closure of the specialization of V, by the Whitney sum formula and the projection formula, we have:

${\displaystyle i^{!}(V)=r_{*}(e(\xi )P)=r_{*}(e(r^{*}F)e(\xi ')P)=e(F)r_{*}(e(\xi ')P)=e(F){i'}^{!}(V).}$

${\displaystyle \square }$

One special case of the formula is the self-intersection formula, which says: given a regular embedding i: X → Y with normal bundle N,

${\displaystyle i^{*}i_{*}=e(N).}$

(To get this, take Y' = Y″ = X.) For example, from this and the projection formula, when X, Y are smooth, one can deduce the formula:

${\displaystyle i_{*}(x)i_{*}(y)=i_{*}(e(N)xy).}$

in the Chow ring of Y.

Let ${\displaystyle f:{\widetilde {Y}}\to Y}$ be the blow-up along a closed subscheme X, ${\displaystyle {\widetilde {X}}}$ the exceptional divisor and ${\displaystyle g:{\widetilde {g}}:{\widetilde {X}}\to X}$ the restriction of f. Assume f can be written as a closed immersion followed by a smooth morphism (for example, Y is quasi-projective). Then, from ${\displaystyle f^{*}i_{*}=i^{!}g^{*}}$, one gets:

Example: intersection of two plane curves containing the same component

Let ${\displaystyle C_{1}=Z(x_{0}x_{1})}$ and ${\displaystyle C_{2}=Z(x_{0}x_{2})}$ be two plane curves in ${\displaystyle \mathbb {P} ^{2}}$. Set theoretically, their intersection

${\displaystyle {\begin{aligned}C_{1}\cap C_{2}&=Z(x_{1},x_{2})\cup Z(x_{0})\\&=[1:0:0]\cup \{[0:a:b]\in \mathbb {P} ^{2}\}\\&=Z_{1}\cup Z_{2}\end{aligned}}}$

is the union of a point and an embedded ${\displaystyle \mathbb {P} ^{1}}$. By Bézout's theorem, it is expected this intersection should contain ${\displaystyle 4}$ points since it is the intersection of two conics, so interpreting this intersection requires a residual intersection. Then

${\displaystyle (C_{1}\cap C_{2})^{Z_{1}}=\left\{{\frac {c(N_{C_{1}/\mathbb {P} ^{2}})c(N_{C_{2}/\mathbb {P} ^{2}})}{c(N_{Z_{1}/\mathbb {P} ^{2}})}}\right\}_{0}\in A_{0}(Z_{1})}$ ${\displaystyle (C_{1}\cap C_{2})^{Z_{2}}=\left\{{\frac {c(N_{C_{1}/\mathbb {P} ^{2}})c(N_{C_{2}/\mathbb {P} ^{2}})}{c(N_{Z_{2}/\mathbb {P} ^{2}})}}\right\}_{1}\in A_{1}(Z_{2})}$

Since ${\displaystyle C_{1},C_{2}}$ are both degree ${\displaystyle 2}$ hypersurfaces, their normal bundle is the pullback of ${\displaystyle {\mathcal {O}}(2)}$, hence the numerator of the two residual components is

${\displaystyle {\begin{aligned}c({\mathcal {O}}(2))c({\mathcal {O}}(2))&=(1+2[H])(1+2[H])\\&=1+4[H]+4[H]^{2}\end{aligned}}}$

Because ${\displaystyle Z_{1}}$ is given by the vanishing locus ${\displaystyle Z(x_{1},x_{2})}$ its normal bundle is ${\displaystyle {\mathcal {O}}(1)\oplus {\mathcal {O}}(1)}$, hence

${\displaystyle {\begin{aligned}c(N_{Z_{1}/\mathbb {P} ^{2}})&=c({\mathcal {O}}(1)\oplus {\mathcal {O}}(1))\\&=(1+[H])(1+[H])\\&=1+2[H]+[H]^{2}\\&=1\end{aligned}}}$

since ${\displaystyle Z_{1}}$ is dimension ${\displaystyle 0}$. Similarly, the numerator is also ${\displaystyle 1}$, hence the residual intersection is of degree ${\displaystyle 1}$, as expected since ${\displaystyle Z_{1}}$ is the complete intersection given by the vanishing locus ${\displaystyle Z(x_{1},x_{2})}$. Also, the normal bundle of ${\displaystyle Z_{2}}$ is ${\displaystyle {\mathcal {O}}(1)}$ since it's given by the vanishing locus ${\displaystyle Z(x_{0})}$, so

${\displaystyle c(N_{Z_{2}}/X)=1+[H]}$

Inverting ${\displaystyle c(N_{Z_{2}}/X)}$ gives the series

${\displaystyle {\frac {1}{1+[H]}}=1-[H]+[H]^{2}}$

hence

${\displaystyle {\begin{aligned}{\frac {c(N_{C_{1}/\mathbb {P} ^{2}})c(N_{C_{2}/\mathbb {P} ^{2}})}{c(N_{Z_{2}/\mathbb {P} ^{2}})}}=&(1+4[H]+4[H]^{2})(1-[H]+[H]^{2})\\=&(1-[H]+[H]^{2})+(4[H]-4[H]^{2})+4[H]^{2}\\=&1+3[H]+[H]^{2}\\=&1+3[H]\end{aligned}}}$

giving the residual intersection of ${\displaystyle 3[H]}$ for ${\displaystyle Z_{2}}$. Pushing forward these two classes gives ${\displaystyle 4[H]^{2}}$ in ${\displaystyle A^{*}(\mathbb {P} ^{2})}$, as desired.

Example: the degree of a curve in three surfaces

Let ${\displaystyle X_{1},X_{2},X_{3}\subset \mathbb {P} ^{3}}$ be three surfaces. Suppose the scheme-theoretic intersection ${\displaystyle \bigcap X_{i}}$ is the disjoint union of a smooth curve C and a zero-dimensional schem S. One can ask: what is the degree of S? This can be answered by #formula.

Example: conics tangent to given five lines

The plane conics are parametrized by ${\displaystyle \mathbb {P} ^{{\binom {2+2}{2}}-1}=\mathbb {P} ^{5}}$. Given five general lines ${\displaystyle \ell _{1},\ldots ,\ell _{5}\subset \mathbb {P} ^{2}}$, let ${\displaystyle H_{\ell _{i}}\subset \mathbb {P} ^{5}}$ be the hypersurfaces of conics tangent to ${\displaystyle \ell _{i}}$; it can be shown that these hypersurfaces have degree two.

The intersection ${\displaystyle \bigcap _{i}H_{\ell _{i}}}$ contains the Veronese surface ${\displaystyle Z\simeq \mathbb {P} ^{2}}$ consisting of double lines; it is a scheme-theoretic connected component of ${\displaystyle \bigcap _{i}H_{\ell _{i}}}$. Let ${\displaystyle h=c_{1}({\mathcal {O}}_{Z}(1))}$ be the hyperplane class = the first Chern class of O(1) in the Chow ring of Z. Now, ${\displaystyle Z=\mathbb {P} ^{2}\hookrightarrow \mathbb {P} ^{5}}$ such that ${\displaystyle {\mathcal {O}}_{\mathbb {P} ^{5}}(1)}$ pulls-back to ${\displaystyle {\mathcal {O}}_{\mathbb {P} ^{2}}(2)}$ and so the normal bundle to ${\displaystyle H_{\ell _{i}}}$ restricted to Z is

${\displaystyle N_{H_{\ell _{i}}/\mathbb {P} ^{5}}|_{Z}={\mathcal {O}}_{\mathbb {P} ^{5}}(H_{\ell _{i}})|_{Z}={\mathcal {O}}_{\mathbb {P} ^{5}}(2)|_{Z}={\mathcal {O}}_{Z}(4).}$

So, the total Chern class of it is

${\displaystyle c(N_{H_{\ell _{i}}/\mathbb {P} ^{5}}|_{Z})=1+4h.}$

Similarly, using that the normal bundle to a regular ${\displaystyle X\hookrightarrow Y}$ is ${\displaystyle T_{Y}|_{X}/T_{X}}$ as well as the Euler sequence, we get that the total Chern class of the normal bundle to ${\displaystyle Z\hookrightarrow \mathbb {P} ^{5}}$ is

${\displaystyle c(N_{Z/\mathbb {P} ^{5}})=c(T_{\mathbb {P} ^{5}}|_{Z})/c(T_{Z})=c({\mathcal {O}}_{\mathbb {P} ^{5}}(1)^{\oplus 6}|_{Z})/c({\mathcal {O}}_{\mathbb {P} ^{2}}(1)^{\oplus 3})=(1+2h)^{6}/(1+h)^{3}.}$

Thus, the Segre class of ${\displaystyle Z\hookrightarrow \mathbb {P} ^{5}}$ is

${\displaystyle s(Z,\mathbb {P} ^{5})=c(N_{Z/\mathbb {P} ^{5}})^{-1}=1-9h+51h^{2}.}$

Hence, the equivalence of Z is

${\displaystyle \deg((1+4h)^{5}(1-9h+51h^{2}))=160-180+51=31.}$

By Bézout's theorem, the degree of ${\displaystyle \bigcap _{i}H_{\ell _{i}}}$ is ${\displaystyle 2^{5}=32}$ and hence the residual set consists of a single point corresponding to a unique conic tangent to the given all five lines.

Alternatively, the equivalence of Z can be computed by #formula?; since ${\displaystyle \deg(c_{1}(T_{\mathbb {P} ^{2}}))=\deg(c_{2}(T_{\mathbb {P} ^{2}}))=3}$ and ${\displaystyle \deg(Z)=4}$, it is:

${\displaystyle 3+4(3)+(40-10(6)+21)\deg(Z)=31.}$

Example: conics tangent to given five conics

Suppose we are given five plane conics ${\displaystyle C_{1},\ldots ,C_{5}\subset \mathbb {P} ^{2}}$ in general positions. One can proceed exactly as in the previous example. Thus, let ${\displaystyle H_{C_{i}}\subset \mathbb {P} ^{5}}$ be the hypersurface of conics tangent to ${\displaystyle C_{i}}$; it can be shown that it has degree 6. The intersection ${\displaystyle \bigcap _{i}H_{C_{i}}}$ contains the Veronese surface Z of double lines.

Example: functoriality of construction of a refined Gysin homomorphism

The fuctoriality is the section title refers to: given two regular embedding ${\displaystyle X{\overset {i}{\hookrightarrow }}Y{\overset {j}{\hookrightarrow }}Z}$,

${\displaystyle (j\circ i)^{!}=j^{!}\circ i^{!}}$

where the equality has the following sense: