MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've recently started to learn about Chern and Segre classes, and it seems to me that they are very similar, sharing the same important properties and having closely related definitions.

Fulton's book starts by defining Segre classes, and then defines Chern classes. My question is: what are the advantages of Chern classes over Segre classes (or vice-versa)? Are they really very similar, or is there some conceptual difference?

share|cite|improve this question
In brief, Segre classes make sense even for singular spaces, the Chern classes are derived. – Leo Alonso May 8 '14 at 9:41
up vote 4 down vote accepted

Chern classes of a rank $r$ vector bundle vanish in degree greater that $r$ while Segre classes do not. On the other hand the definition of Segre classes easily generalizes to singular vector bundles such as cones ( Normal cones are cental objects in deformation theory. For instance the generalization of the concept of normal cone to what is called intrinsic normal cone led to the definition of virtual fundamental class for Deligne-Mumford stacks (

Even if one works with smooth varieties Segre classes are very useful in order to expresses intersection products.

For instance if $Y\subset\mathbb{P}^n$ be a smooth variety, and $\epsilon:X = Bl_Y\mathbb{P}^n\rightarrow\mathbb{P}^n$ is the blow-up of $\mathbb{P}^n$ along $Y$, $\widetilde{H}$ be the pull-back of the hyperplane section $H$ of $\mathbb{P}^n$, and $E$ is the exceptional divisor, $H_Y =H\cdot Y$ we have $$\widetilde{H}^{h-i}E^i = p^*H_Y^{n-i}\cdot i^*E^{i-1} = H_Y^{n-i}\cdot p_*i^*E^{i-1}.$$ Now, $E = \mathbb{P}(N_{Y/\mathbb{P}^n})$, and $i^*E = -e$, where $e = c_1(\mathcal{O}_E(1))$. Let use denote by $s_j$ the Segre classes of $N_{Y/\mathbb{P}^n}$, and let $c = codim_{\mathbb{P}^n}(Y)$. We have the following intersection numbers:

  • $\widetilde{H}^n = 1$;
  • $\widetilde{H}^{n-i}\cdot E^i = 0$ for $i < c$;
  • $\widetilde{H}^{n-i}\cdot E^i = (-1)^{i-1}s_{i-c}H_Y^{n-i}$ for $i\geq c$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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