I am reading the book “Fourier-Mukai transforms in algebraic geometry” by Daniel Huybrechts. Before stating my question, let me state some definitions. Here, $X$ is a smooth projective variety.

Definition: In a k-linear triangulated category $\mathcal{D}$, an object $P\in \mathcal{D}$ is called simple if $Hom(P,P)$ is a field.

Definition: The support of a complex $\mathcal{F}^{\bullet}\in D^b(X)$ is the union of all its cohomology sheaves, i.e. it is the closed subset $$supp(\mathcal{F}^{\bullet}):=\cup \ supp(H^i(\mathcal{F}^{\bullet}))$$.

And for support of complexes, we have the following lemma (it is stated on page 65 as Lemma 3.9 in Huybrecht’s book):

Lemma: Suppose $\mathcal{F}^{\bullet}\in D^b(X)$ and $supp(\mathcal{F}^{\bullet})=Z_1\cup Z_2$ where $Z_1,Z_2$ are disjoint closed subsets of $X$. Then, $\mathcal{F}^{\bullet}\cong \mathcal{F}_1^{\bullet}\oplus\mathcal{F}_2^{\bullet}$

Now, my question is about the proof of Lemma 4.5, on page 92, in Huybrecht’s book. This lemma is as follows:

Lemma: Suppose $\mathcal{F}^{\bullet}$ is a simple object in $D^b(X)$ with zero dimensional support. If $Hom(\mathcal{F}^{\bullet},\mathcal{F}[i])=0$ for $i<0$, then $$\mathcal{F}^{\bullet}\cong k(x)[m]$$ for some closed point $x\in X$ and some integer $m$.

In its proof, to show $\mathcal{F}^{\bullet}$ is concentrated in exactly one closed point, he uses Lemma 3.9, which I stated before, and he get the contradiction if the assumption of Lemma 3.9 holds and then he conclude that we may assume that the support of all cohomology sheaves $H^i(\mathcal{F}^{\bullet})$ consists of the same closed point $x\in X$. My question is that why is this true?

  • $\begingroup$ Maybe I am overlooking something, but to me it seems that the following is going on: If Huybrechts shows that the complex is supported at a single point and the support is by definition the support of all the cohomology sheaves, then certainly all of these must be supported at exactly this single point or have no support at all. The relevant ones are of course the ones with non-empty support, so that we will just assume that all of their supports are non-empty instead of just considering the ones that are relevant to us (which would require a lot more tedious phrasing). $\endgroup$
    – Con
    Commented Nov 18, 2021 at 10:02
  • $\begingroup$ @Con Yes, I agree with you and my question is your first sentence. He conclude it as I said. $\endgroup$ Commented Nov 18, 2021 at 10:25
  • $\begingroup$ As I said, this is just applying the definition of the support of a complex. If $\bigcup \text{supp}(H^i(\mathcal{F}^{\bullet})) = \text{supp}(\mathcal{F}^{\bullet}) = \lbrace * \rbrace$, then each of the members of the union have to be that point or empty. $\endgroup$
    – Con
    Commented Nov 18, 2021 at 10:52
  • $\begingroup$ @Con I mean I know this. My question is that how can we conclude that this union is just a point. $\endgroup$ Commented Nov 18, 2021 at 10:54
  • $\begingroup$ Oh, I must have missunderstood your actual question. I will have to leave for a few minutes, but afterwards I will write an answer. $\endgroup$
    – Con
    Commented Nov 18, 2021 at 10:58

1 Answer 1


Assume on the contrary, that the support of $\mathcal{F}$ consists of multiple closed points. Then by the Lemma you stated, there exist non-trivial $\mathcal{F}_1$ and $\mathcal{F}_2$ with $\mathcal{F} \cong \mathcal{F}_1 \oplus \mathcal{F}_2$. Now we can consider the following morphism:

$$\mathcal{F} \cong \mathcal{F}_1 \oplus \mathcal{F}_2 \twoheadrightarrow \mathcal{F}_1 \hookrightarrow \mathcal{F}_1 \oplus \mathcal{F}_2 \cong \mathcal{F},$$ where the two morphisms in the middle are the projection onto the first factor and the inclusion as first factor. Since this morphism is non-zero and not invertible, its existence contradicts $\mathcal{F}$ being simple. Hence our assumption was wrong, so that the support only consists of a single closed point.

  • 1
    $\begingroup$ Why the support consists only closed points? Maybe the support is some union of closed subset and these subsets have non-empty intersections? Thank you. $\endgroup$
    – Jian
    Commented Nov 18, 2021 at 11:22
  • 1
    $\begingroup$ Before Huybrechts shows this, he shows in a previous step that the support is $0$-dimensional, i.e. consists of finitely many closed points. $\endgroup$
    – Con
    Commented Nov 18, 2021 at 11:25
  • $\begingroup$ Thanks. I missed the assumption in the Lemma. $\endgroup$
    – Jian
    Commented Nov 18, 2021 at 11:29
  • $\begingroup$ @Con Many thanks for your answer. May you also write in your answer that being zero dimensional implies that the support consists of finitely many points and why is this true? $\endgroup$ Commented Nov 18, 2021 at 12:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .