ag.algebraic geometry - 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.

• Next nt.number theory - Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?
• Previous dna - Do mitochondria simply automatically convert glucose to ATP?