MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f: X\rightarrow Y$ be a morphism of varieties. Let $0\rightarrow F\rightarrow E\rightarrow G\rightarrow 0$ be a short exact sequence of locally free sheave of finite rank. If direct images of above sheaves are locally free, then is it true that it induces a short exact sequence $0\rightarrow f_*F\rightarrow f_*E\rightarrow f_*G\rightarrow 0$?

share|cite|improve this question
No. You should learn about derived functors, in particular $R^if_*$. Hartshorne's book is a good place to start. – Francesco Polizzi Dec 9 '10 at 15:35
@Francesco: Higher direct images does not have to vanish in general, but john has added some asssumptions. But perhaps this just makes it harder to find counterexamples. – Martin Brandenburg Dec 9 '10 at 15:49
@Martin: the counterexample is not too difficult. I wrote it as an answer, in order to be clearer... – Francesco Polizzi Dec 9 '10 at 16:20
Take $f$ to be a constant map, then the sequence is not exact. – J.C. Ottem Dec 9 '10 at 16:33
up vote 3 down vote accepted

Let $X,Y$ be defined over the field $k$ and take $f$ to be the structure map $f:X\to {\rm Spec}\, k$. Then let $E\to G$ be a surjective morphism of sheaves that is not surjective on global sections, e.g., $$E=\mathcal O_{\mathbb P^1}(-1)\oplus \mathcal O_{\mathbb P^1}(-1)\to G=\mathcal O_{\mathbb P^1}.$$ Then $f_*$ is just $H^0$ and the desired statement is false.

EDIT: (to have an example mapping to an arbitrary scheme) Consider the base change of $f$ via $Y\to {\rm Spec}\, k$: $$g: X\times_{{\rm Spec}\, k} Y \to Y.$$ and let $\mathcal E:=p^*E$ and $\mathcal G:=p^*G$ where $p:X\times_{{\rm Spec}\, k} Y \to X$ is the projection to $X$. Then $g_*\mathcal E\simeq H^0(X, E)\otimes_k \mathcal O_Y$ and $g_*\mathcal G\simeq H^0(X, G)\otimes_k \mathcal O_Y$, so again the desired statement is false.

share|cite|improve this answer
I guess Torsten beat me to this example while I was typing it (and stopped in mid-sentence).... :) – Sándor Kovács Dec 9 '10 at 17:20
But, I am trying to find counter example in the where $X$ and $Y$ are projective varieties and $f$ is surjective. – john Dec 9 '10 at 18:18
John, this is a counterexample (a point is a projective variety...). – Donu Arapura Dec 9 '10 at 19:13
John, as Donu pointed out, this is such an example, but I added an addition to show that you can actually choose your $Y$. – Sándor Kovács Dec 9 '10 at 19:35

Two simple examples where $Y=\mathrm{Spec}k$ a point:

Consider $0\to\mathcal{O}(-2)\to\mathcal{O}(-1)^2\to\mathcal{O}\to 0$ on $\mathbb P^1$, where the two maps $\mathcal{O}(-1)\to\mathcal{O}$ are given by multiplication by $x$ and $y$ respectively, then the sequence of global sections is $0\to0\to0\to k$ and hence is not exact.

On an elliptic curve there is a non-split exact sequence $0\to\mathcal{O}\to\mathcal E\to\mathcal{O}\to0$ as $\mathrm{Ext}^1(\mathcal{O},\mathcal{O})$ is $1$-dimensional. Then $H^0(\mathcal E)\to H^0(\mathcal O)$ can not be surjective as otherwise the sequence would be split.

share|cite|improve this answer

The answer is no, as the following examples shows. It is inspired by Sandor's answer to this question of mine.

Let $Y$ be an elliptic curve, $X=Y \times Y$ and $f \colon X \to Y$ the projection onto the first factor.

Since $f$ has connected fibres, we have $f_* \mathcal{O}_X= \mathcal{O}_Y$. On the other hand, since $X$ is a product, for all $p \in Y$ we may identify canonically $H^1(f^{-1}(p), \mathcal{O}_{f^{-1}(p)})$ with $H^1(Y, \mathcal{O}_Y) \cong \mathbb{C}$, hence $R^1f_* \mathcal{O}_X=\mathcal{O}_Y$, and by projection formula $R^1f_* \mathcal{O}_X(-p)=\mathcal{O}_Y(-p)$.

Set $E_p:=f^{-1}(p)$, and apply the functor $f_*$ the exact sequence

$0 \to \mathcal{O}_X \to \mathcal{F} \to \mathcal{O}_X(-E_p) \to 0$,

where $\mathcal{F}$ is the unique non-trivial extension of $\mathcal{O}_X(-E_p)$ by $\mathcal{O}_X$ (one can check that $\mathcal{F}$ is an indecomposable rank $2$ vector bundle on $X$).

We obtain

$0 \to \mathcal{O}_Y \to f_* \mathcal{F} \to \mathcal{O}_Y(-p) \stackrel{\delta}{\to} \mathcal{O}_Y \to R^1 f_* \mathcal{F} \to \mathcal{O}_Y(-p) \to 0$.

The sheaf $f_* \mathcal{F}$ is reflexive on a smooth curve, hence locally free. On the other hand, if $\delta$ were the zero map then $f_* \mathcal{F}=\mathcal{O}_Y \oplus \mathcal{O}_Y(-p)$ and so, by funtoriality of $f_*$, the vector bundle $\mathcal{F}$ would be decomposable, contradiction.

share|cite|improve this answer
This does not appear to be a counterexample. – Steven Landsburg Dec 9 '10 at 16:21
There are simple counterexamples though. For instance, if $f$ is the map to a point, the local freeness condition is vacuous, but the sequence of direct images (which are basically the spaces of global sections) does not need to be exact. (Of course, you can't construct a counterexample with a split exact sequence.) – t3suji Dec 9 '10 at 16:28
@Steven: Actually it was not a counterexample :-) I was too concentrated to the non-vanishing of $R^1f_∗$ and I did not think about the original question. I've edited the post, now it's ok... – Francesco Polizzi Dec 9 '10 at 16:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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