5
$\begingroup$

Let $M$ and $N$ be compact complex manifolds of the same dimension ($m$) and $\mu: M \rightarrow N$ a holomorphic map. Let $D \subset M$ be the subset of points of $M$, where $d\mu|_p$ fails to be injective. Assume that $D$ is a smooth complex submanifold of $M$ of the expected dimension $m-1$ (more precisely, $d\mu|_p$ is not injective means a certain determinant is zero, assume that determinant vanishes transversally). Furthermore assume that on all points of $D$, the Kernel of $d\mu|_p$ is $\textit{exactly}$ one dimensional.

$\textbf{Question:}$ Define the line bundle over $D$, given by $L:= Ker(d\mu) \rightarrow D$. How does one compute $c_1(L)$?

The specific example where I need to compute $c_1(L)$ is as follows: $M:= \mathbb{P}^1 \times \mathbb{P}^1$, $N:= \mathbb{P}^2$ and $\mu:M \rightarrow N$ is a map of type $(d,k)$, i.e. $\mu^*\mathcal{O}(1) = \mathcal{O}(d,k)$.

$\textbf{Added Later:}$ My main interest is in the specific example I asked. Its being pointed out that in general there may not be any explicit/reasonable formula for $c_1(L)$.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Off the top of my head, I doubt that there is a completely general formula. The divisor class $c_1(L)$ on $D$ may not be the pullback to $D$ of any divisor class on $M$, in which case, in what terms are you expecting to describe $c_1(L)$? In the specific example, I vaguely remember that Steven Kleiman's article on "The enumerative theory of singularities" has something about this. $\endgroup$
    – Jason Starr
    Commented Feb 26, 2015 at 17:16

1 Answer 1

Reset to default
3
$\begingroup$

On $M$ there is an exact sequence $$ 0 \to \mu^*\Omega_N \to \Omega_M \to i_*L^\vee \to 0, $$ where $i:D \to M$ is the embedding. This allows to understand the class of $D$ since $D = c_1(i_*L) = c_1(\Omega_M) - \mu^*c_1(N)$. In your case it is equal to $$ (-2,-2) - (-3d,-3k) = (3d-2,3k-2). $$ Thus $D$ is a curve of bidegree $(3d-2,3k-2)$. In particular, its genus is equal to $g = 9(d-1)(k-1)$. To understand the degree of $L$ you can use Riemann--Roch: $$ \deg(L^\vee) + 1 - g = \chi(i_*L^\vee) = \chi(\Omega_M) - \chi(\mu^*\Omega_N). $$ In your case $\chi(\Omega_M) = -2$, while the pullback of the Euler sequence $$ 0 \to \mu^*\Omega_N \to O(-d,-k)^{\oplus 3} \to O \to 0 $$ allows to compute $\chi(\Omega_N) = 3(d-1)(k-1) - 1$. In the end $$ \deg(L^\vee) = -2 + 1 - 3(d-1)(k-1) + 9(d-1)(k-1) - 1 = 6(d-1)(k-1)-2. $$

Improve this answer
$\endgroup$
6
  • $\begingroup$ A few questions about the notations: What is $i_*L^{\vee}$? Is it the cokernel sheaf (which happens to be a line bundle restricted to $D$)? Is the degree of $L^{\vee}$ the same as degree of $L$? $\endgroup$
    – Ritwik
    Commented Feb 26, 2015 at 18:54
  • $\begingroup$ $L^\vee$ is the cokernel of the map on $\Omega$'s, which is dual to the kernel of the map on $T$'s, which as far as I understnad you denote by $L$. Of course $\deg(L^\vee) = - \deg(L)$. $\endgroup$
    – Sasha
    Commented Feb 26, 2015 at 19:17
  • $\begingroup$ @Thanks for the clarification. $deg(L^{\vee})$ is indeed $-deg(L)$; sorry for asking that! $\endgroup$
    – Ritwik
    Commented Feb 26, 2015 at 19:22
  • $\begingroup$ I am really sorry to ask a very basic question; can you explain how you got $\chi(\Omega_M) = -2$? I am assuming everything is restricted to $D$. I am aware that $c_1(T^*M) = -2(a_1+a_2)$ where $a_1$ and $a_2$ are the generators of $H^*(M)$. But how does that give $\chi(\Omega_M)$? $\endgroup$
    – Ritwik
    Commented Feb 27, 2015 at 10:19
  • $\begingroup$ I am also confused about one further thing; isn't $i_*L^{\vee}$ the normal bundle of $D$ inside $M$? In which case isn't the degree of the bundle same as $D.D$? Said differently, the degree is $<c_1(L^{\vee}), [D]>$. Since the poincare dual of $[D]$ is $c_1(L^{\vee})$, the number should be $[D].[D]$. Isn't it? $\endgroup$
    – Ritwik
    Commented Feb 27, 2015 at 12:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .