Given a smooth projective variety $X$ over an algebraically closed field $k$. Now given a another projective variety $T$ and a coherent $O_{X\times T}$-module $F$, which is flat over $T$.

Given $r,s \in T$, and let $F_r$ and $F_s$ be the pullback of $F$ to the fibers over $r$ and $s$.

Do we have $ch(F_r)=ch(F_s)$ in this case? I.e. is the Chern character constant in flat families?

What about the Chern classes $c_i$?

Is there some literature about the behaviour of these characteristic classes in (flat) families?


2 Answers 2


The answer depends on the version of the Chern classes you are using. For example, if you consider Chern classes with values in the Chow ring then the answer is negative. The simplest example is the following, take $X = T$ to be an elliptic curve, and $F$ to be the line bundle corresponding to the diagonal. Then $c_1(F_r)$ is the point $r$ itself, but different points are not rationally equivalent.

On the other hand, if you consider Chern classes with values in cohomology with rational coefficients then the answer is positive (if $T$ is connected!). Indeed, since $X$ is smooth and projective any flat sheaf has a resolution by vector bundles. Because of the flatness its restriction gives a resolution of the restricted sheaf. Since Chern classes are multiplicative in exact seqeunces, the question reduces to the case of a vector bundle. Further, because of the splitting principle we can assume we can assume that $F$ is a line bundle. And in this case we can deduce everything from Riemann--Roch.

Note by the way that the same argument shows that Chern classes are also constant in non-flat families, if you only replace the restriction $F_r$ by derived restriction.

  • $\begingroup$ Okay, $T$ connected and cohomology with rational coefficients is okay for my example. How does the Riemann-Roch argument work? Something like: Because $T$ is connected, the Euler characteristic $\chi(X,L_t)$ of the line bundles is a constant function of $t\in T$. But for all $t\in T$ we have $\chi(X,L_t)=ch(L_t)td(X)$. So $ch(L_t)$ must be constant? $\endgroup$
    – TonyS
    Jan 20, 2011 at 22:18
  • 1
    $\begingroup$ It is better to look at the Hilbert polynomial of $L$, i.e. pick up an ample divisor $H$ on $X$ and consider $f(k) = \chi(X,L_t(kH))$. On one hand it is a constant function since $L$ is flat. On the other, if you compute it by Riemann-Roch you will get $c_1(L_t)H^{n-1}\cdot g(k)$, where $g$ is a certain polynomial and $n = \dim X$. So, it follows that $c_1(L_t)H^{n-1}$ is constant. And this is true FOR ANY ample divisor $H$. It follows that $c_1(L_t)$ is constant. $\endgroup$
    – Sasha
    Jan 21, 2011 at 3:53

As soon as ${\rm Pic}^\circ(X)$ is non-trivial there is a counter-example. Take a non-trivial line bundle $\mathcal L$ that deforms to the trivial one. Then $c_1(\mathcal L)$ goes from non-zero to zero.

  • 1
    $\begingroup$ Sorry Sándor, I don't understand... Is your non-trivial line bundle $\mathcal L$ in $\operatorname{Pic}^0(X)$? In this case then $c_1(L)$ shouldn't be zero? $\endgroup$
    – diverietti
    Jan 20, 2011 at 22:08
  • 2
    $\begingroup$ I am thinking $c_1$ in the Chow group, not in $H^2(X,\mathbb Z)$. The topological $c_1$ is zero, but the algebraic cycle is not. $\endgroup$ Jan 20, 2011 at 22:23

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.