Let $f:X \hookrightarrow \mathbb P^n$ be a $(d_1,\ldots ,d_r)$ smooth complete intersection over an algebraically closed field $k$. Let $\ell$ be a prime number different from the characteristic of $k$, and let $\Lambda = \mathbb Z/\ell^h \mathbb Z$ for some $h\geq 1$.

The normal bundle $\mathcal N_X(\mathbb P^n)$ is well known to be isomorphic to $\bigoplus_{i=1}^r \mathcal O_X(d_i)$. Let $c_r(\mathcal N_X(\mathbb P^n)) \in H^{2r}(X,\Lambda(r))$ be the $r$-th Chern class of this normal bundle. By the theory of Chern polynomials, if $\zeta = c_1(\mathcal O_X(1)) \in H^2(X,\Lambda(1))$, then we can directly compute $$c_r(\mathcal N_X(\mathbb P^n)) = d_1\ldots d_r \zeta^r,$$ where $\zeta^r = \zeta \cup \ldots \cup \zeta$ is the $r$-fold cup-product of $\zeta$ with itself.

Besides, let $[X] \in H^{2r}(\mathbb P^n,\Lambda(r))$ be the image of $X$ via the cycle class map. Equivalently, $[X] = f_*(1_X)$ is the image of $1_X \in H^0(X,\Lambda) \simeq \Lambda$ via the Gysin map. By general theory, we have $f^*[X] = c_r(\mathcal N_{X}(\mathbb P^n))$, where $f^*$ denotes the restriction map on etale cohomology.

In regards to the formula above, this seems to suggest that we have an equality $[X] = d_1\ldots d_r c_1(\mathcal O(1))^r$ in $H^{2r}(\mathbb P^n,\Lambda(r))$, where $\mathcal O(1)$ is the twisted sheaf on $\mathbb P^n$. Indeed, we do have $f^*c_1(\mathcal O(1)) = \zeta$ in $H^2(X,\Lambda(1))$. Is this equality true?

Edit: This seems to hold if $r \leq n-r = \dim(X)$. Indeed, in this case it is known that the restriction map $f^*: H^{2r}(\mathbb P^n,\Lambda) \to H^{2r}(X,\Lambda)$ is injective. However, if $r > n-r$ then $H^{2r}(X,\Lambda) = 0$ (and in particular $c_r(\mathcal N_X(\mathbb P^n)) = 0$ as well). In this case my starting observation becomes empty, however my question on the identity $[X] = d_1\ldots d_rc_1(\mathcal O(1))^r$ still makes sense. Is it true or not then?

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    $\begingroup$ I might be mistaken (don't know much about étale cohomology), but in singular cohomology, $[X]$ would be a class in $H^{2c}(\mathbb P^n)$, not $H^{2r}$. Similarly for Chow rings, $[X]\in A^c(\mathbb P^n)$. $\endgroup$ Commented Apr 11 at 7:38
  • $\begingroup$ @red_trumpet You are absolutely right, this is an awful typo. Let me edit this. In fact, $r$ is already the codimension of $X$ in $\mathbb P^n$. The integer $c$ I defined made no sense. Hence, I removed it $\endgroup$
    – Suzet
    Commented Apr 11 at 7:50

1 Answer 1


I claim that the equality you propose is true. I think it boils down to the fact that your equalities in the cohomology ring already hold at the level of Chow groups because the same formal properties hold there. Then you apply the cycle class map of whatever Weil cohomology theory you like and you win. More precisely: for any $E$ vector bundle of rank $r$ on $Y$ and any integer i, one has cycles $$c_i(E) \in CH^{2i}(Y)$$ which satisfy all the formal relations we know and love. In particular, just like what you wrote for cohomology classes, in the case $E=\bigoplus_{j=1}^r L_j$ we have $$c_r(E)=c_1(L_1)\dots c_1(L_r)$$ where the product between classes is the intersection product on $CH^*(X)$. Taking the cycle class of such chow-theoretic Chern classes recovers cohmological Chern classes.

I propose two arguments, both in the language and using the theory of Fulton's Intersection Theory book.

The first uses more theory but is slightly more general. If $i: X\hookrightarrow Y$ a regular embedding (for example, a complete intersection closed subscheme) then $$[X]=i^{*}[Y]$$ where $i^*$ is the Gysin pullback for regular embeddings.(c.f. end of example 6.2.1 [Fult]). The fact that your $X$ is in fact the zero section of some vector bundle $E$ allows you to express its class as the top chern class of this vector bundle, i.e. $$[X]=c_r(E)\cap [Y]$$ (You can find something much more general to this effect in Example 6.3.4 (a), plugging in $Y=Y'$ $\alpha=[Y]$ and $[X]=i^{*}[Y]=i^{!}[Y]$ as mentioned above). Then conclude by the decomposition of the top chern class in the first chern classes of line bundles.

The second approach is much simpler: you only need to know up to chapter 2 of Fulton, Just use the fact that for $D$ Cartier, $$c_1(O_Y(D))\cap [Y]=[D]$$ and a bunch of iterations of the projection formula to reduce $[X]$ to a product of first chern classes of line bundles $\mathcal{O}(d_1) , \dots , \mathcal{O}(d_r)$.

For the second approach you really are using complete intersection, the first on the other hand only needs your $X$ to be a regular zero section of some vector bundle.

I think one could probably carry out the first approach just with the theory up to chapter 5: if someone has a precise reference less general than my 6.3.4, i'd be interested to know!

  • $\begingroup$ Dear Teovvv, thank you so much for you answer. It convinces me that I need to start reading Fulton's Intersection Theory in order to progress. I have a little question to check my understanding. In your first approach, $[X]$ and $[Y]$ are understood as elements of $\mathrm{CH}^{*}(Y)$, right? In this case, isn't $[Y]$ the neutral element for the intersection product? So that when writing $[X] = c_r(E) \cap [Y]$, we really have $[X] = c_r(E)$? With $Y = \mathbb P^n$ and $E \simeq \bigoplus_{i} \mathcal O(d_i)$, I deduce $[X] = d_1\ldots d_r c_1(\mathcal O(1))^r$ as desired $\endgroup$
    – Suzet
    Commented Apr 12 at 4:18
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    $\begingroup$ Ah, I think I got it. $[Y]$ is the identity when it is irreducible. Otherwise it decomposes into the sum of the cycles given by its irreducible components, with multiplicities. If I specialize to $Y = \mathbb P^n$ (or anything irreducible), then sure $[X] = c_r(E)$. $\endgroup$
    – Suzet
    Commented Apr 12 at 5:08
  • $\begingroup$ @Suzet yes, but i would add that in some very concrete sense, once you believe that $[X]=c_i(E)\cap [Y]$, then $[X]=c_r(E)$ should be true by definition: this because original way that Fulton defines Chern classes is as things that act on cycles. In this sense, the symbol $\cap$ is just a way to write "the image of $[Y] $ through $c_i(E)$. At that point, the only reasonable way to define $c_i(E)$ as a cycle on $Y$ is to just take it to be the action of $c_i(E)$ onto $[Y]$. $\endgroup$
    – teovvv
    Commented Apr 12 at 13:47

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