direct image of a locally free sheaf
Zhenya
Does anybody know is this statement right or wrong?
f: Y->X - flat projective morphism
E - locally free sheaf on Y
Then f_* (E) is locally free.
Zhenya
David Madore
Zhenya in litteris
<e2557eb2-74a8-42d2...@26g2000hsk.googlegroups.com>
scripsit:
I assume that "locally free" means "locall y free coherent" and that
everything is noetherian.
Then f_*(E) is flat because E is flat and f is (see Hartshorne, *Algebraic
Geometry*, proposition III.9.2(c)).
But f_*(E) is coherent because E is and f is projective
(op. cit. III.8.8(b)).
Now this implies that f_*(E) is locally free (op. cit. III.9.2(e)).
--
David A. Madore
(david....@ens.fr,
http://www.madore.org/~david/ )
Jannick Asmus
On 16.05.2008 11:15, David Madore wrote:
> Zhenya in litteris
> <e2557eb2-74a8-42d2...@26g2000hsk.googlegroups.com>
> scripsit:
>> Does anybody know is this statement right or wrong?
>>
>> f: Y->X - flat projective morphism
>> E - locally free sheaf on Y
>>
>> Then f_* (E) is locally free.
>
> I assume that "locally free" means "locall y free coherent" and that
> everything is noetherian.
>
> Then f_*(E) is flat because E is flat and f is (see Hartshorne, *Algebraic
> Geometry*, proposition III.9.2(c)).
I can see that the cited proposition gives that _E_ is flat over X. I am
wondering how you deduce that f_*(E) is flat over X with that.
Considering the natural morphism of O_Y-modules f^*f_*(E) -> E there
seems more work to be done here. (*)
> But f_*(E) is coherent because E is and f is projective
> (op. cit. III.8.8(b)).
Note that f_*(E) can be seen to be coherent, too, if f is assumed to be
proper. I think that the strong assumption that f be projective needs to
be invested somehow to get a result.
> Now this implies that f_*(E) is locally free (op. cit. III.9.2(e)).
Honestly I doubt that the claim is true as stated although I do not have
any counterexample at hand at the moment.
Note that f_*(E(n)) is locally free for n>>0 where E(n) is the n-th
twist of E with respect to a f-ample line bundle L on Y if X is integral
(cf. http://groups.google.com/group/sci.math/msg/9c4a3dfe05992be0, too).
This appears to be the best result which is possible in the light of
Hartshorne, Thm. III.9.9 and the equivalent assertions in its proof.
... but I am keen on seeing some elaboration on (*). ;)
--
Best wishes,
J.
David Madore
Jannick Asmus in litteris <g0js9i$g4f$1...@dizzy.math.ohio-state.edu>
scripsit:
> On 16.05.2008 11:15, David Madore wrote:
>> Then f_*(E) is flat because E is flat and f is (see Hartshorne, *Algebraic
>> Geometry*, proposition III.9.2(c)).
>
> I can see that the cited proposition gives that _E_ is flat over X. I am
> wondering how you deduce that f_*(E) is flat over X with that.
> Considering the natural morphism of O_Y-modules f^*f_*(E) -> E there
> seems more work to be done here. (*)
By Zeus, you're right. I was confused by Hartshorne's wording of
example III.9.7.1, where he writes that some morphism f:Y->X is not
flat, for "if it were, then f_*(O_Y) would be a flat sheaf of
O_X-modules" [note: I inverted X's and Y's w.r.t. Hartshorne's
notation, for consistency with the original poster's notation]. At
the very least, Hartshorne is horribly confusing here, because never
up to that point is the direct image of a sheaf under a flat morphism
mentioned, and suddenly it pops up without explanation. What did he
have in mind?
In truth, I find it very annoying that this question is not mentioned
in standard texts on algebraic geometry and commutative algebra. If
you have the affine case and quasi-coherent sheaves in mind, the
direct image of a quasi-coherent sheaf by a morphism Spec(B)->Spec(A)
corresponds to viewing a B-module as the induced A-module; so it's
pretty natural to think that if a morphism is flat then direct image
by it will take flat sheaves to flat sheaves (see below). A decent
book on the subject should contain some explanation as to the relation
between "f_*(E) is flat" and "E is f-flat": either my expectations are
too high or there are no decent textbooks on the subject. :-)
> Note that f_*(E) can be seen to be coherent, too, if f is assumed to be
> proper. I think that the strong assumption that f be projective needs to
> be invested somehow to get a result.
I don't think the distinction between "projective" and "proper" is the
issue here, but I may be wrong, of course.
> Honestly I doubt that the claim is true as stated although I do not have
> any counterexample at hand at the moment.
I also thought this statement was fishy, but I couldn't find a
counterexample, so I was confused as explained above. But this
confusing being cleared away, I agree with you that the statement is
probably wrong without further hypotheses.
However, I can see more clearly the reason why finding a
counterexample is not obvious. Let f:Y->X be a morphism of finite
type between noetherian schemes, and assume X = Spec A is affine
(which we can do, because our problem is local on the target).
Consider a finite covering of Y by affine subschemes V_i = Spec(B_i),
each B_i being an A-algebra (of finite type). If E is a
quasi-coherent sheaf on Y, the restriction E|V_i of E to each V_i is
some (M_i)~ with M_i a B_i-module; the direct image f_*(E|V_i) is
([A]M_i)~ where [A]M_i is M_i seen as an A-module; if we assume that E
and f are flat (or more generally that E is flat over X), then the
[A]M_i are flat, so the f_*(E|V_i) are (this is essentially what
Hartshorne says around II.5 and III.9). Now this does not imply that
f_*(E) is flat as I was too quick to believe, but it is true that
f_*(E) injects in the direct sum sum of the f_*(E|V_i) which are flat
(so f_*(E) is, e.g., torsion free). Now it is not true in general
that a subsheaf (submodule) of a flat sheaf (module) is flat, but it
is true if, say, X is a smooth curve. So my attempts to find a
counterexample with X = A^1 were doomed.
I need to think further about this...
Jannick Asmus
On 17.05.2008 16:15, David Madore wrote:
> Jannick Asmus in litteris <g0js9i$g4f$1...@dizzy.math.ohio-state.edu>
> scripsit:
>> On 16.05.2008 11:15, David Madore wrote:
>>> Then f_*(E) is flat because E is flat and f is (see Hartshorne, *Algebraic
>>> Geometry*, proposition III.9.2(c)).
>> I can see that the cited proposition gives that _E_ is flat over X. I am
>> wondering how you deduce that f_*(E) is flat over X with that.
>> Considering the natural morphism of O_Y-modules f^*f_*(E) -> E there
>> seems more work to be done here. (*)
>
> By Zeus, you're right. I was confused by Hartshorne's wording of
> example III.9.7.1, where he writes that some morphism f:Y->X is not
> flat, for "if it were, then f_*(O_Y) would be a flat sheaf of
> O_X-modules" [note: I inverted X's and Y's w.r.t. Hartshorne's
> notation, for consistency with the original poster's notation]. At
> the very least, Hartshorne is horribly confusing here, because never
> up to that point is the direct image of a sheaf under a flat morphism
> mentioned, and suddenly it pops up without explanation. What did he
> have in mind?
The subtle point in example III.9.7.1 is that the morphism stems from
the normalization process, so it is affine - and everything is fine
since things can be easily considered under one single affine chart such
that the tool-kit of commutative algebra can be applied. But our
morphism in the original problem is complete or projective. They are not
affine in general.
The direct image is thoroughly examined in terms of the cohomology
groups H^i(X,F) or higher direct images R^i(f_*)(F). This might be a bit
disguised, but basically direct images are considered by cohomological
tools noted before.
> In truth, I find it very annoying that this question is not mentioned
> in standard texts on algebraic geometry and commutative algebra.
I understand. I would expect some remarks on that, too, together with
examples in chapters on flat complete morphisms.
> If
> you have the affine case and quasi-coherent sheaves in mind, the
> direct image of a quasi-coherent sheaf by a morphism Spec(B)->Spec(A)
> corresponds to viewing a B-module as the induced A-module; so it's
> pretty natural to think that if a morphism is flat then direct image
> by it will take flat sheaves to flat sheaves (see below). A decent
> book on the subject should contain some explanation as to the relation
> between "f_*(E) is flat" and "E is f-flat": either my expectations are
> too high or there are no decent textbooks on the subject. :-)
Note that things get a bit more complicated if a fibre is not contained
in a single _affine_ subset. In general this is the case when complete
morphisms are considered.
>> Note that f_*(E) can be seen to be coherent, too, if f is assumed to be
>> proper. I think that the strong assumption that f be projective needs to
>> be invested somehow to get a result.
>
> I don't think the distinction between "projective" and "proper" is the
> issue here, but I may be wrong, of course.
I agree, but I think the assumption on f to be projective needs to be
thoroughly used at this point to get a decent result. As it is the
properness of f which guarantees that the direct image of a coherent
sheaf is coherent I was wondering where some ample line bundle should
show up in the argument.
>> Honestly I doubt that the claim is true as stated although I do not have
>> any counterexample at hand at the moment.
>
> I also thought this statement was fishy, but I couldn't find a
> counterexample, so I was confused as explained above. But this
> confusing being cleared away, I agree with you that the statement is
> probably wrong without further hypotheses.
>
> However, I can see more clearly the reason why finding a
> counterexample is not obvious. Let f:Y->X be a morphism of finite
> type between noetherian schemes, and assume X = Spec A is affine
> (which we can do, because our problem is local on the target).
> Consider a finite covering of Y by affine subschemes V_i = Spec(B_i),
> each B_i being an A-algebra (of finite type). If E is a
> quasi-coherent sheaf on Y, the restriction E|V_i of E to each V_i is
> some (M_i)~ with M_i a B_i-module; the direct image f_*(E|V_i) is
> ([A]M_i)~ where [A]M_i is M_i seen as an A-module; if we assume that E
> and f are flat (or more generally that E is flat over X), then the
> [A]M_i are flat, so the f_*(E|V_i) are (this is essentially what
> Hartshorne says around II.5 and III.9). Now this does not imply that
> f_*(E) is flat as I was too quick to believe, but it is true that
> f_*(E) injects in the direct sum sum of the f_*(E|V_i) which are flat
> (so f_*(E) is, e.g., torsion free). Now it is not true in general
> that a subsheaf (submodule) of a flat sheaf (module) is flat, but it
> is true if, say, X is a smooth curve. So my attempts to find a
> counterexample with X = A^1 were doomed.
Basically your argument can be extended by having a look at the Cech
complex C^i of E and some affine covering of Y. By assumption C^i are
flat A-modules, but C^i is not exact in general. Taking some twist of E
high enough (w.r.t. an ample line bundle L on Y) then C^*(E(n)) is exact
such that it is a flat resolution of finite length of M:=H^0(Y,E(n)).
This implies that the A-module M is flat. By properness the A-module is
finitely generated, hence locally free. (This is the idea of one of the
implications proven in Hartshorne, Thm. III.9.9.)
> I need to think further about this...
... looking forward to seeing a counterexample on this. :-) If I can
find some more time I will try to find a counterexample as well.
--
Best wishes,
J.
Jurgen Bohm
If I am not mistaken in the following conclusions, the statement is
wrong. To find a counterexample I refer to the article
D. Eisenbud, F.-O. Schreyer, Relative Beilinson Monad and Direct Image
for Families of Coherent Sheaves,
arXiv:math.AG/0506391 v2 27 Sep 2006
The articles cites the following lemma, which is also in
Hartshorne(III,Lemma 12.3):
If f:X->Y, Y=spec A, is a projective morphism and F a Y-flat sheaf on X,
then there exists a complex AA^*
AA^*: 0 -> A^h0 -> A^h1 -> A^h2 -> ... -> A^hn -> 0
whose cohomologies are h^i(AA^*) = R^i f_* F(Y)
The abovementioned article gives a kind of converse of this lemma as
Theorem 0.4:
Let A be a noetherian ring and let AA^* a complex like the above. Then a
vector bundle F on P^n_A exists, such that h^i(AA^*)=R^i f_*(Y)
Therefore f_*F can correspond to an arbitrary kernel ker(A^h0 -> A^h1)
and need not be a flat (=locally free) sheaf on Y.
I constructed a counterexample (more or less by trial and error) with
Macaulay2 by choosing A=k[a,b,c,d] and S=A[x,y], therefore X=Proj
S=P^1_A, and a certain matrix m : S^5(-1) -> S^7. So there is a sequence
S^5(-1) -> S^7 -> M' -> 0
Let M be M' ** S^(-2)
Let I3 the ideal in k[a,b,c,d], where there are points x,y, so that
the matrix m doesn't have the full rank 5, it can be computed by taking
the 5x5 minors of m in k[a,b,c,d,x,y] and eliminating x, setting y=1
intersect eliminate y, setting x=1. Above Y-V(I3) F=M^\tilde is a vector
bundle.
Using the command directImageComplex(M) one computes a complex of A
modules of the abovementioned kind, whose cohomologies are the (higher)
direct image sheaves of M^\tilde. In the case I considered, it took the form
0 -> A^6 -> A^3 -> 0
with 0 -> K -> A^6 -> A^3 -> 0 exact and
phi
0 -> A^1 -----> A^4 -> K -> 0
exact (and K=f_*(M^\tilde)(Y))
Let I4 be the ideal of k[a,b,c,d], where the map A^1 -> A^4 vanishes. It
can be checked, that rad(I4) \not\supseteq rad(I3), so there are points
P in spec A, above which F=M^\tilde is a vector-bundle, but on tensoring
with k(P) the map phi ** k(P) vanishes, so Tor_1^(A_P)(k(p),K_P) <> 0,
so K_P = (f_* M^\tilde)_P is not projective.
If there is interest, I can post the script doing the calculation for
further experimenting.
Greetings
Jurgen
--
Jurgen Bohm www.aviduratas.de
"At a time when so many scholars in the world are calculating, is it not
desirable that some, who can, dream ?" R. Thom
Ilya Zakharevich
=?ISO-8859-1?Q?Jurgen_Bohm?=
<jbo...@gmx.net>], who wrote in article <g1fmj8$827$1...@dizzy.math.ohio-state.edu>:
> If I am not mistaken in the following conclusions, the statement is
> wrong.
> Macaulay2 by choosing A=k[a,b,c,d] and S=A[x,y], therefore X=Proj
> S=P^1_A, and a certain matrix m : S^5(-1) -> S^7. So there is a sequence
I tried to reduce your arguments to a little bit more (IMO) pedestrian form...
==== Dictionary of linear algebra [with O(d) a line bundle on P1]:
a pencil is a pair of MxN matrices (A,B), or a family xA + yB;
isomorphism of pencils is (A,B) |--> (QAR,QBR) with invertible Q,R;
a pencil is "a Jordan block" if isomorphic to one with A or B equal to
id, and the other one a Jordan block;
a pencil is "decreasing Kronecker" if isomorphic to (d/dx,d/dy) acting
Pol_n(x,y) --> Pol_{n-1}(x,y); "increasing Kronecker" is transposed;
over algebraically closed fields, any pencil decomposes to a direct sum
of Jordan blocks, and increasing or decreasing Kronecker blocks; the
blocks appearing in the decomposition are uniquely defined;
Ker and Coker(xA + yB) induce sheaves on P1 (consisting of ratios x:y);
Jordan blocks correspond to skyscrapers in Coker, increasing
Kronecker blocks to O(n) in Coker, decreasing to O(-n) in Ker;
a family of pencils has pencils with Jordan blocks on a closed subset;
a generic pencil with M=N breaks into 1x1 Jordan blocks; with M>N into
M-N decreasing Kronecker blocks; n's of these blocks differ at most
by one (which determines the list uniquely);
a generic family of pencils has principal degeneration on:
M=N: 2x2 Jordan block appears on hypersurface Disc(det(xA+yB))=0;
M=N+1: 1x1 Jordan block appears on hypersurface HyperDet(A,B)=0
for N+1 x N x 2 flavor of GKZ hyperdeterminant;
M>N: 1x1 Jordan block appears in codimension M-N;
M=N+2, M=2m+1, m>1: a sum of two Kronecker blocks with n=m,m+1
degenerates into one with n=m-1,m+2 on a subfamily of codimension 3;
M=N+2, M=2m, m>1: a sum of two Kronecker blocks with n=m degenerates
into a sum with n=m-1,m+1 on a subfamily of codimension 2;
[I vaguely rememeber seeing calculations of stabilizers published
which would imply the last two subcases, but do not recollect
where; I needed to calculate stabs explicitly... When m=1,
codim decreases by 1.
The M=N+1 case was observed by I.M.Gelfand/Valery???]
To translate to the cases N>M, transpose.
=============================== The last two sub-statements suggest:
Conjecture: a generic family of rk=2 vector bundles on P1 of degree
2m+1 degenerates into O(m-1) (+) O(m+2) on codimension 3; of degree
2m degenerates into O(m-1) (+) O(m+1) on codimension 2.
[These families are generically O(m) (+) O(m+1) and O(m) (+) O(m).]
================================ Comparing with your example:
Conjecture: consider a family of rk=2 vector bundles on P1 parameterized
by S which is isomorphic to O(n) (+) O(n') outside of subvariety S'
of S, and degenerates into O(N) (+) O(N') on S'; N<n,n'. Then the
direct image to S is not flat if codim S' >= 2 and n,n' >= -1, N < -1.
[*] [I do not know what happens if one drops n,n' >= -1, N < -1.]
[Heuristically, I would expect that if direct image is flat, so is the
higher direct image (to "preserve Euler characteristic"). If so, then
since higher direct image vanishes at generic point, it vanishes
everywhere; hence the fiber of direct image is global sections of the
vector bundle. But dimension of global sections jumps up on S'.]
================================= Going back to your example:
You have N=M+2, M=5, and take Coker; generically, it would be
O(2) (+) O(3). You twist by O(-2); generically, the result would
be O(0) (+) O(1), which in codimension 3 would degenerate into
O(-1) (+) O(2) (on a closed subset it may also acquire Jordan blocks;
one can ignore this). This is not in context of the conjecture above,
but of [*], which for me is an unchartered territory.
Following my argument, the minimal example to check would be a pencil
with M=3, N=M+2: take a direct sum of Kronecker blocks with n=0,3, and
take a generic 2-parameter deformation (it is enough to require that
for a deformed pencil, Ker A and Ker B do not intersect). So, I would
take Ker(xA + yB) with matrices A,B depending on a,b:
1 0 0 0 0 0 1 0 0 0
0 1 0 0 a 0 0 1 0 0
0 0 1 0 b 0 0 0 1 0
- and I would twist by O(1) - while to closely follow your example,
the twist would be by O(2). (Since I prefer Ker to Coker, the picture
is kinda dual to yours.) So O(-2) (+) O(1) would deform into O(-1) (+) O(0).
Should not one be able to do it without computer algebra? :-(
Hope this helps,
Ilya
Ilya Zakharevich
Ilya Zakharevich
<nospam...@ilyaz.org>], who wrote in article <g5ouam$153g$1...@agate.berkeley.edu>:
> Conjecture: consider a family of rk=2 vector bundles on P1 parameterized
> by S which is isomorphic to O(n) (+) O(n') outside of subvariety S'
> of S, and degenerates into O(N) (+) O(N') on S'; N<n,n'. Then the
> direct image to S is not flat if codim S' >= 2 and n,n' >= -1, N < -1.
>
> [*] [I do not know what happens if one drops n,n' >= -1, N < -1.]
>
> [Heuristically, I would expect that if direct image is flat, so is the
> higher direct image (to "preserve Euler characteristic")
Apparently, it was a kind of circular reasoning...
> take Ker(xA + yB) with matrices A,B depending on a,b:
>
> 1 0 0 0 0 0 1 0 0 0
> 0 1 0 0 a 0 0 1 0 0
> 0 0 1 0 b 0 0 0 1 0
>
> - and I would twist by O(1) - while to closely follow your example,
> the twist would be by O(2). (Since I prefer Ker to Coker, the picture
> is kinda dual to yours.) So O(-2) (+) O(1) would deform into O(-1) (+) O(0).
>
> Should not one be able to do it without computer algebra? :-(
It turns out that the calculation is very simple indeed - and bye-bye
the conjecture! (It looks like the direct image is flat for both
twistings...) One hope which remains (to use the machinery described
in my preceeding message to understand the original counterexample) is
that a deformation of O(-2) (+) O(1) into O(-1) (+) O(0) obtained via
Ker is not isomorphic to one obtained via Coker...
============================================ Calculation:
Twist by O(1): to calculate ( Ker(xA+yB) )(1), one needs to find
linear functions V(x,y) = x v1 + y v2 of (x,y) such that (xA + yB)V = 0.
When (a,b) != (0,0), one gets that V is proportional to V0 with
v1 = -b^2 e4 + a (e5 - a e2 - b e3)
v2 = a^2 e1 + b e5.
So the direct image is {P(a,b) (x v1 + y v2)} with a regular scalar
function P; so it is iso to the structure sheaf.
Twist by O(2): to calculate ( Ker(xA+yB) )(2), one needs to find
quadratic functions Q(x,y) of (x,y) such that (xA + yB)V = 0. When
(a,b) != (0,0), one gets x V0(x,y), y V0(x,y), and
Q0 = x^2 (e5 - a e2 - b e3) + xy (a e1 + b e2) + y^2 b e2;
all 3 quadratic functions are linearly independent (at least when a,b
!= 0), so the direct image is {x P1(a,b) V0 + y P2(a,b) V0 + P3(a,b) Q0}.
So AFAICS, the direct image is 3 copies of the structure sheaf...
Puzzled,
Ilya