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 $c_1(L)$.

$\endgroup$
  • 1
    $\begingroup$ 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 Feb 26 '15 at 17:16
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 $$ \deg(L^\vee) = -2 + 1 - 3(d-1)(k-1) + 9(d-1)(k-1) - 1 = 6(d-1)(k-1)-2. $$

$\endgroup$
  • $\begingroup$ 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 Feb 26 '15 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 Feb 26 '15 at 19:17
  • $\begingroup$ @Thanks for the clarification. $deg(L^{\vee})$ is indeed $-deg(L)$; sorry for asking that! $\endgroup$ – Ritwik Feb 26 '15 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 Feb 27 '15 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 Feb 27 '15 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.