-2
$\begingroup$

In general, we know that the higher direct images of finite morphisms are zero.

Now, suppose we have a morphism $f: S\rightarrow B$ which is finite over a Zariski open subset (say $Z$) of the projective variety $B$. But, the fibers over the points of the complement of $Z$ are not finite.

In the above situation can we still say $Rf_* = f_*$?

$\endgroup$
  • 2
    $\begingroup$ just consider a birational map $\endgroup$ – Chen Jiang Mar 6 '18 at 6:53
  • $\begingroup$ Can you explain? $\endgroup$ – Mohsen Karkheiran Mar 6 '18 at 6:57
1
$\begingroup$

No. For example, if you blow up a regular point on a smooth affine variety, you get non-affine variety, so it must have some sheaf with non-vanishing higher cohomologies by the cohomological criterion for affine-ness. But then the blow-down morphism can not be acyclic on this sheaf.indeed, by the identity $R((fg)_*) = Rf_* Rg_*$ for $f$ the blow-down and $g$ the projection to a point, if $Rf_* = f_*$ then $R((fg)_*)= (fg)_*$ and there are no higher cohomologies.

$\endgroup$
  • $\begingroup$ Thanks. Let me be more specific. Suppose the fibers on the complement of $Z$ are an elliptic curve. Then is it possible to blow-up $B$ (with $P^1$) to $\tilde{B}$? In this case, $f=h g$, where $g: S\rightarrow \tilde{B}$ is now a finite morphism (because elliptic curve is a double cover of $P^1$), and $Rg_*=g_*$, and $h$ is the blow down. So now we get $Rf_*$ = $h_*$ $ Rg_*$=$(hg)_*$=$f_*$. Is this true? $\endgroup$ – Mohsen Karkheiran Mar 6 '18 at 7:31
  • $\begingroup$ Probably you disagree with me on $Rh_*=h_*$. But the whole question coming from a paper by Donagi (arxiv.org/pdf/hep-th/0405014.pdf eq.157), and he used $Rf_*=f_*$, in the situation I explained... $\endgroup$ – Mohsen Karkheiran Mar 6 '18 at 7:46
  • 1
    $\begingroup$ This can not be true, because if the fiber is an elliptic curve, by pushing the sheaf $\mathcal{O}_S / \mathcal{I}_F$ for $F$ the elliptic fiber, you get the cohomology of an elliptic curve, supported at the image of the elliptic curve. This is not concentrated in degree 0. Besides, it seems that in eq. 157 in the paper they use it for a finite Galois cover. Anyway, the higher direct image as a functor can not vanish for non-affine morphisms in general, because the cohomology of the pre-image of at least one affine piece with at least one sheaf must have components higher than 0. $\endgroup$ – S. carmeli Mar 6 '18 at 8:14
  • $\begingroup$ Thanks. I wish I could discuss more clearly. The Variety S I mentioned, is called "spectral cover" which parametrizes the "S-equivalence" classes of stable $SU(N)$ bundles over a Weierstrass elliptic fibration with base B. The spectral cover S is originally defined as a finite cover of an open dense subset of B. But then they can extend it to every point. The blow up idea came from the following paper arxiv.org/pdf/hep-th/9701162.pdf (page 59 footnote). Maybe in Donagi's paper, they are restricting on the open dense subset, but it's not quite satisfactory from the physics point of view $\endgroup$ – Mohsen Karkheiran Mar 6 '18 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.