4
$\begingroup$

Suppose I have a singular projective variety $X\subset \mathbb{P}^n$ that is reducible with $X=\bigcup_i X_i$ smooth irreducible components. That is, the irreducible components are smooth but $X$ is singular along the intersections of the $X_i$.

I would like to compute the Segre classes $s_t(X,\mathbb{P}^n)$. In particular I (optimistically) wonder whether one can obtain these classes in terms of the Segre classes $s_j(X_i,\mathbb{P}^n)$ of the irreducible componentes. If not, what would I need more to compute $s_t(X,\mathbb{P}^n)$?

Improve this question
$\endgroup$
2
  • $\begingroup$ Segre classes of what vector bundle do you want to compute? $\endgroup$
    – Sasha
    Commented Sep 28, 2016 at 9:45
  • $\begingroup$ It is the Segre classes of $X$ inside $\mathbb{P}^n$, so I think they are the classes of the normal cone of $X$. $\endgroup$
    – IMeasy
    Commented Oct 1, 2016 at 17:34

1 Answer 1

Reset to default
1
$\begingroup$

Unfortunately Segre class doesn't share very good inclusion-exclusion property. For example, Let $Y$ ve irreducible smooth, and let $X=\cup_t C_t\subset Y$ is a union of curves. Assume that the intersection of curves are only nodal type point (no three points collapse together, and no tangent points). Then we have the following $$ s(X,Y)=\sum_t s(C_t,Y)+\sum [C_a\cap C_b] . $$ For general case, it will get only more complicated. For a detailed discussion, you can read this paper by Aluffi (https://www.math.fsu.edu/~aluffi/archive/paper165.pdf)

Improve this answer
$\endgroup$

You must log in to answer this question.

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