I've been having problem actually restricting a Line bundle $L$ defined on some projective space $\mathbb C \mathbb P^{N-1}$ to a subvariety $X$.

I know how to do this on an abstract level, but actually computing what's going on, seems quite mysterious.

From Fulton's "Intersection Theory" I have

$c_1(L) \cap [X] = [C]$ where $C$ is the divisor corresponding to $\mathcal O_X(C) \simeq L\vert_X$

Now, I have $c_1(L)$ given by $-N[H]$ where $[H]$ denotes the hyperplane class in $\mathbb C \mathbb P^{N-1}$. I also have some polynomial $P$ whose zero locus defines $X$. I even know $c_1(TX) = 0$ and have computed that if $X$ is taken to be a divisor in $\mathbb C \mathbb P^{N-1}$, the corresponding line bundle would satisfy $c_1(\mathcal{O}(X)) = N[H]$ (Not quite sure yet if this helps).

But, what I really would like to know is $c_1(L\vert_X)$?

My attempts so far have been to find an actual section $s$ of $L$, use the equation for the zero locus and actually intersect that with $X$. However, finding such a section has proven difficult.

How would one normally go about this? Am I on the right track? Any help is highly appreciated!


1 Answer 1


In general (i.e. for restricting line bundles from a variety to a subvariety) one has $c_1(L_{\vert X}) = c_1(L) \cdot X$ (the intersection, thought of as a divisor on $X$).

In your case one can be more explicit, since $L = \mathcal O(d) = \mathcal O(1)^{\otimes d}$ for some $d$ (all line bundles on projective space are of this form). Thus, if we let $H_X$ denote a generic hyperplane section of $X$, then $c_1(L_{\vert X}) = d H_X$.

  • $\begingroup$ Thanks very much! I was kind of hoping that the first chern class vanishes. I guess the only case where this happens is if $d=0$, or are there other possibilities? $\endgroup$
    – Mike
    Jun 29, 2012 at 5:20
  • 1
    $\begingroup$ @Mike: Dear Mike, No, there are no possibilities besides the trivial line bundle ($d = 0$). A basic fact is that $H_X$ gives a non-trivial class in $H^2$ of $X$. (The so-called Kahler class.) Regards, $\endgroup$
    – Matt E
    Jun 29, 2012 at 12:35

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