Let $A_\ast$ and $F_\ast$ be the functors $\textrm{Var}_\mathbb C\to \textrm{Ab}$ of Chow groups and constructible functions, respectively, with respect to proper maps. Then the Chern-Schwartz-MacPherson class is the unique natural transformation $$c_{\textrm{SM}}:F_\ast\to A_\ast$$ taking the value $c_{\textrm{SM}}(\textbf{1}_X)=c(TX)\cap [X]\in A_\ast X$ on $\textbf{1}_X\in F_\ast X$ for nonsingular $X$, and commuting with proper pushforwards. It can be a tool for computing the Euler characteristic of singular varieties.

The Chern-Schwartz-MacPherson class of $X$ is by definition the class $c_{\textrm{SM}}(\textbf{1}_X)$. It respects the excision relation exactly as (generalized) Euler characteristics, meaning that if $Z\subset X$ is closed an $U=X\setminus Z$, then $$c_{\textrm{SM}}(\textbf{1}_X)=c_{\textrm{SM}}(\textbf{1}_Z)+c_{\textrm{SM}}(\textbf{1}_{U})$$

Question. When is $c_{\textrm{SM}}(\textbf{1}_X)$ computable in practice for singular $X$? Is it true that (at least the degree zero part of) $c_{\textrm{SM}}(\textbf{1}_X)$ vanishes if $\chi_{top}(X)=0$?


There are some special cases where the CSM classes of singular varieties can be computed easily.

For example there is Ehler's formula for $c_{SM}(X)$ where $X$ is any complete toric variety. Let $\Sigma$ be the fan of $X$ with torus orbits $B_\sigma$ for $\sigma \in \Sigma$, then the CSM class of $X$ is given by

$$ c_{SM}(X) = \sum_{\sigma \in \Sigma}[\overline{B}_\sigma] \in A_*(X) $$

where $\overline{B}_\sigma$ are closures of the torus orbits.

Paolo Aluffi has done a lot of work on computing these things. For example (see here) he has constructed a generalization that lives in proChow groups instead of the usual chow groups. The proChow group of a variety $X$ agrees with the Chow group when $X$ is proper and the proCSM class of a not necessarily complete toric variety still satisfies the same formula as above. This also generalizes to other varieties with stratification by smooth locally closed subvarieties but the formula gets more complicated.

In another direction, if we have an embedding $i : X \subset M$ where $M$ is smooth, then we can compute $i_*c_{SM}(X)$ in $M$ by the following formula.

$$ i_*c_{SM}(X) = c(TM) \cap [M] - \pi_*\left(c(\Omega^1_{\tilde{M}}(log X')^\vee)\cap [\widetilde{M}]\right) \in A_*(M) $$

where $\pi : \widetilde{M} \to M$ is an isomorphism away from $X$ and $X' = \pi^{-1}(X)_{red}$ is an snc divisor. See this paper.

When $X$ is itself a hypersurface in $M$ there is a much more computable formula:

$$ c_{SM}(X) = c_F(X) + c(TM) \cap \left(\frac{1}{c(\mathcal{O}(X))}\cdot (s(Y,M)^\vee \otimes \mathcal{O}(X))\right) $$

where $c_F$ is the Chern-Fulton class and $s$ is the Segre class and $Y$ is the singular locus of $X$. Each of the pieces of this formula are in general much more computable then $c_{SM}$ itself in the sense that there are algorithms and Macaulay2 codes for doing this. For more on this aspect and especially techniques and examples of these computations, check out this survey.

As for your second question, it is true in general that the degree of the zero dimensional piece of $c_{SM}(X)$ is the Euler characteristic,

$$ \int c_{SM}(X) = \chi(X). $$

So the answer to your second question is yes.

  • $\begingroup$ Thank you for this very rich answer! I was wondering, is it also true that if a constructible function $f\in F_\ast X$ is supported on a locus $Y\subset X$ with $\chi(Y)=0$ then $\int c_{SM}(f)=0$? I would say no, but don't know enough examples... Thanks again! $\endgroup$ – Brenin May 13 '15 at 19:51
  • $\begingroup$ You're welcome! No I don't think that's true. Here is a counterexample. Take $X = \mathbb{C}^*$ and fix a point $p \in X$. Take the constructible function $f = \mathbb{1}_X + \mathbb{1}_p$. This function is supported on $X$ which has Euler characteristic $0$ but $\int c_{SM}(f) = \chi(X) + \chi(p) = 1$. $\endgroup$ – Dori Bejleri May 15 '15 at 18:10

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.