Let $H$ be a degree $d$ hypersurface of $\mathbb P^2$. I want to calculate the integral $\int_H c_1 (\mathcal O(k)|_H)$.

I think I got a proof for this, but im am not sure if I got all the details right. I would appreciate if you could take a look on it:

  1. First we note that as $c_1(\mathcal O(k))= k \cdot c_1(\mathcal O(1))$, it suffices to consider the case $k=1$.

  2. If we denote by $i$ the inclusion map, then $\mathcal O(1)|_C$ The integral above is the pairing $$\langle c_1(i^*\mathcal O(1)),[H]\rangle = \langle i^*c_1(\mathcal O(1)),[H]\rangle= \langle c_1(\mathcal O(1)),i_*[H]\rangle.$$

  3. $i_*[H]$ is Poincaré dual to $c_1(\mathcal O[H])$. As $H$ is of degree $d$, we have $\mathcal O[H])=\mathcal O(d)$. Thus $$\langle c_1(i^*\mathcal O(1)),[H]\rangle = \langle c_1(\mathcal O(1))\cup c_1(\mathcal O(d)),[\mathbb P^2]\rangle $$
  4. As further $c_1(\mathcal O(1))$ is Poincaré dual to a linear hyperplane which intersects the degree $d$ hypersurface $H$ in exactly $d$ points (counting multiplicities), the Poincaré dual $P$ of $c_1(\mathcal O(1))\cup c_1(\mathcal O(d))$ is the linear combination of these $d$ points.
  5. Applying the definition of Poincaré duality on the constant $1$-function, we get $$ \langle c_1(\mathcal O(1))\cup c_1(\mathcal O(d)),[\mathbb P^2]\rangle = \langle P, 1|_P\rangle =d. $$
  6. Summarizing we get $\int_H c_1 (\mathcal O(k)|_H)=d\cdot k.$

Is what I wrote right? Is there a shorter proof for this?

  • 1
    $\begingroup$ Look fine to me. $\endgroup$
    – Kenny Wong
    Commented May 20, 2018 at 16:54

1 Answer 1


It looks correct. I'm not sure if there is a quicker way or not, especially with characteristic classes, when they are many definitions (I hope mine match with yours). Here is a try :

1) Let $\alpha \in H^*(Y)$, $\alpha = \alpha^0 + \dots + \alpha^n$ where $\alpha^i \in H^i(X)$. We define $\int_Y \alpha := \deg(\text{PD}(\alpha^n))$ where $\text{PD}$ is the Poincaré dual.

2) We define $c_1(L) = [D] \in H^2(X)$ if $L = O_X(D)$. Then, we have $$\int_Y c_1(L_{|Y}) = \int_Y D_{|Y} = \int_Y D \cdot Y$$

Now you are done since you're looking at $\deg(\text{PD}(D \cdot Y)) = dk$.

Remark : If you are working in the Chow ring, it's even more easier since by definition $CH^k(Y) := CH_{n-k}(Y)$, in this case you define it by $\deg(\alpha^n)$. Also intersection theory is way more easy in the Chow ring, see the discussion in the first chapter of the book by Eisenbud and Harris.

  • $\begingroup$ I am not familiar with some of your definitions. Probably this is because I come from the complex geometrical viewpoint and I am not that familiar with the algebraic one. What is $A^n$? How do you define the degree of the Poincare dual? By looking at is as a Divisor? But how does this work for general values of $i$?. What does $D\cdot Y$ denote, seems to be their intersection to me? And why is the last equality evident? $\endgroup$
    – klirk
    Commented May 20, 2018 at 17:29
  • $\begingroup$ @klirk $A^n$ was a typo I meant $\alpha^n$, this is corrected. The degree map is going from the 0-cycles to $\Bbb Z$ simply counting them. More formally, define $\deg([p]) = 1$ for any point $p \in Y$ and extend by linearity. For general value of $i$, the integral is zero, so $\int_Y \alpha = \int_Y \alpha^n$. Yes this is the intersection product. In topology the definition is technical but in algebraic geometry, there is often possibility to "move" $Y$ so that $D,Y$ are transverse (let's assume everything is smooth) and then we take the set-theoretic intersection. For the last equality ... $\endgroup$ Commented May 20, 2018 at 17:33
  • $\begingroup$ ... for the last equality it's because restricting a divisor to a subvariety is exactly taking the intersection of the divisor with this subvariety (again not the set-theoretic intersection ) $\endgroup$ Commented May 20, 2018 at 17:35
  • $\begingroup$ @klirk : so I guess if you're not interested by the AG perspective you can ignore my answer. If you are, reading the first pages of "Eisenbud, Harris : 3264 and all that" can be very enlightning. $\endgroup$ Commented May 20, 2018 at 17:38
  • $\begingroup$ Thank you for clarifying. With "last equality" i meant $\deg(\text{PD}(D \cdot Y)) = dk$. This is still not clear to me. $\endgroup$
    – klirk
    Commented May 20, 2018 at 17:49

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