Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 7 down vote accepted

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.

share|improve this answer
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? –  TonyS Jan 20 '11 at 22:18
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. –  Sasha Jan 21 '11 at 3:53
add comment

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.

share|improve this answer
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? –  diverietti Jan 20 '11 at 22:08
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. –  Sándor Kovács Jan 20 '11 at 22:23
add comment

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.