MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

As I mentioned in my previous post, I am studying the article Moduli of Enriques surfaces and Grothendieck-Riemann-Roch.

The Grothendieck-Riemann-Roch theorem is applied there to show that, for any family of Enriques surfaces $f:Y\longrightarrow T$, the line bundle $$\mathcal{L} := R^0 f_\ast \Big ((\Omega^2_{Y/T})^{\otimes 2}\Big)$$ is a torsion line bundle, i.e., some tensor power $\mathcal{L}^{\otimes n}$ is isomorphic to the structure sheaf on $T$.

To this extent, one applies GRR to the morphism $f$ and the structure sheaf $\mathcal{O}_Y$. The problem I have now is with the "relative tangent sheaf". I am guessing this is the the quotient sheaf $$f^\ast \mathcal{T}_T/\mathcal{T}_Y.$$

Q1. Why is this well-defined? That is, why do we have an injection $ \mathcal{T}_Y \longrightarrow f^\ast \mathcal{T}_T$?

Q2. How can one determine the Chern classes of $\mathcal{T}_f$ by means of the fibres? That is, can one use the structure on the fibres (Enriques surfaces) to determine $c_i(\mathcal{T}_f)$?

[New questions]

Q3. Let $E$ be a fibre of $f:Y\longrightarrow T$ with injection $i:E\longrightarrow Y$. Is the ringmorphism $i^\ast:A^\cdot Y \longrightarrow A^\cdot E$ injective? If not, is it injective after tensoring with $\mathbf{Q}$?

Let $c_i=c_i(T_f)$.

Q4. We have two formulas from the GRR. The first is $1 = \frac{1}{12} f_\ast(c_1^2+c_2).$ This is the degree 0 part. The second comes from the degree 1 part and reads $0 = \frac{1}{24}f_\ast(c_1\cdot c_2).$ Now, why is $f_\ast(c_1^2) = 0$ as is suggested by the article?

share|cite|improve this question
up vote 3 down vote accepted

Q1: It is the other way round. For a smooth family the differential $T_Y \to f^\ast T_T$ is surjective and the relative tangent is the kernel, so you have an exact sequence

$0 \to T_f \to T_Y \to f^\ast T_T \to 0$.

In this way the tangent to $f$ actually restricts to the tangent of the fibers.

Q2: I don't think that the classes $c_i(T_f)$ are determined by the fibers alone; they depend on the family. It does not even make sense to say that $c_i(T_f)$ are determined by the fibers since these classes live in $H^{2i}(Y)$ anyway, so you have to know at least the total space.

But since $T_f$ restricts to the tangent of the fibers, you know, by naturality of the Chern classes, that if $i \colon E \to Y$ is the inclusion of a fiber $i^\ast c_i(T_f) = c_i(T_E)$.

And these you can compute using the fact that $E$ is Enriques. Namely $2 c_1(T_E) = 0$ since twice the canonical is trivial and $c_2(T_E) = \chi(E) = 12$.

Q3: Surely it is not injective in the top degree, for trivial dimensional reasons. I do not see any reason why it should be in other degrees.

Q4: As is written in the article, this follows from $f_\ast c_2 = 12$. This is more or less clear in cohomology. In this case $f_*$ is the integration along fibers, and since $c_2(T_E)$ is $12$ times the fundamental class of $E$ for all fibers $E$ (see Q2), that integral is $12$.

To translate this in the Chow language, I think the folllowing will do. Let $D$ be a cycle representing $c_2(T_f)$. Since $Y$ is smooth, we can compute the intersection number $D \cdot E = c_2(T_f) \cap E = c_2(T_E) \cap E = 12$. So $D$ intersectts the generic fiber in $12$ points, and the morphism $D \to T$ has degree $12$. In follows that $f_\ast D = 12 [T]$, which is what you want.

share|cite|improve this answer
So, clearly $c^2_1(T_E)=\frac{1}{4}(2c_1(T_E))^2=0$ in the Chow ring of $E$ tensored with $\Q$ for every fibre E. Can one conclude from this that $c^2_1(T_f)=0$ in the Chow ring of $Y$ tensored with $\Q$? The reason I'm asking is the following. The author of the article says that Noether's formula implies that $f_\ast(c_2(T_f))=12$ in the Chow ring of $T$. All I can see though is that $$f_\ast(c^2_1(T_f)+c_2(T_f))=12.$$ (I take the degree 2 part of the GRR identity.) So shouldn't $c^2_1(T_f)$ be zero? Another question: is the ringmorphism $i^\ast:A(Y)\rightarrow A(E)$ injective? – Ariyan Javanpeykar Mar 4 '10 at 18:51
that should be "tensored with $\mathbf{Q}$". – Ariyan Javanpeykar Mar 4 '10 at 19:00
Yes, we take $c_i$ to be the $c_i$ of the relative tangent sheaf $\mathcal{T}_f$. Then, for any inclusion $i:E \longrightarrow Y$ of a fibre of $f$, we know that $i^\ast\mathcal{T}_f = \mathcal{T}_E$. Therefore, $i^\ast c_1^2 = 0$ in $A^2(E)\otimes_\mathbf{Z} \mathbf{Q}$. Now, I do not see why this would imply $c_1^2 =0$ or $f_\ast(c_1^2) = 0$... – Ariyan Javanpeykar Mar 5 '10 at 8:07
By the way, I think you should post separate questions, if you have some more. – Andrea Ferretti Mar 5 '10 at 13:12

This is an answer to Q1:

The relative tangent bundle is the vector bundle on $Y$ whose fibre at a point is the tangent space to the fibre of $f$ passing through that point.

How do we tell if a tangent vector at $y$ is pointing along the fibre through $y$? Because it is killed by the derivative mapping $Df_y:\mathrm{T}Y_y \to \mathrm{T}T_{f(y)}.$ We can organize all these maps into a single map $Df:\mathcal{T}_Y \to f^*\mathcal{T}_T,$ and the relative tangent sheaf is then the kernel of this map.

(The dual picture with differentials may be more familiar: in that picture we have $df:f^{\*}\Omega_{T} \to \Omega_Y,$ and the relative differentials are the cokernel of this map.)

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.