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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?

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2 Answers 2

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.

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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? –  Ari Mar 4 '10 at 18:51
    
that should be "tensored with $\mathbf{Q}$". –  Ari 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 $f_\ast(c_1^2) = 0$... –  Ari 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.)

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