As I mentioned in my previous post, I am studying the article *[Moduli of Enriques surfaces and Grothendieck-Riemann-Roch](http://arxiv.org/abs/math/0701546)*.


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?



Family of Enriques surfaces and GRR, Part 2
**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.


