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Let $\mathcal{A}$ be an abelican category with enough injectives, let $K^\bullet \in Kom^+(\mathcal{A})$ be a complex, where $Kom^+(\mathcal{A})$ is the category of cochain complexes over $\mathcal{A}$ bounded on the left. I read in Weibel's Homological algebra book that the Hyper(co)homology $\mathbb{H}^n(K^\bullet)$ of an exact complex $K^\bullet$ is $0$ for all $n$, I also read somewhere else that the hyper(co)homology of an acyclic complex is $0$, but I didn't find a proof for both cases.

Now let's say that I got a complex $K^\bullet$ and that the cohomology of this complex $H^n(K^\bullet) = 0$ for $n \geq m$, for some $m$. I want to know if this implies that $\mathbb{H}^n(K^\bullet) = 0$ for $n \geq m$? I couldn't find a reference for either case (exact and acyclic), can anybody help me?

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Hypercohomology is with respect to a functor on the category $A$. What is that functor in this case ? The fact that hypercohomology of a complex with respect to a functor will vanish when the complex is exact is a consequence of the existence of the first hypercohomology spectral sequence. See for instance Griffiths and Harris, Principles ..., p446. The $E_2$ page of the spectral sequence is $0$ so the same must be true of its abutment. – Damian Rössler Sep 6 '12 at 16:23
Thank you, I found the book, very clear exposition, thanks – Mario Carrasco Sep 7 '12 at 18:10

Hypercohomology (with respect to any functor) of an exact=acyclic complex is zero, because it is defined on the derived category and an exact complex is quasi-isomorphic to the zero object.

The question in your last paragraph is only wishful thinking. If this were true, there would not be any cohomology theories.

For an example to see that this fails take an arbitrary (say finite type over an algebraically closed field) scheme $X$ with a coherent sheaf $\mathscr F$ that has nontrivial higher cohomology groups. (Say the sheaf $\mathscr O(-2)$ on $\mathbb P^1$). Then take $\mathcal A$ to be the category of coherent sheaves on $X$, $K^\bullet=\mathscr F$ (that is $K^0=\mathscr F$ and $K^i=0$ for $i\neq 0$) and consider the hypercohomology corresponding to the global section functor. By construction $h^n(K^\bullet)=0$ for $n\neq 0$, but $\mathbb H^n(K^\bullet)\neq 0$ for some $n>0$. (In the case of $\mathscr O(-2)$ on $\mathbb P^1$, we have $\mathbb H^1(K^\bullet)\neq 0$).

Obviously there are many similar examples.

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You mean even for an "acyclic complex" $K^\bullet$, i.e. one with $H^n(K^\bullet) = 0$ for $n \geq 1$?? I mean a complex $K^\bullet$ whose cohomology (as a complex) vanishes for $n \geq 1$, I think that's the wrong terminology. I read in Shafarevich's book about algebraic geometry, that the hypercohomology of a complex of sheaves that is 'acyclic' (NOT a complex of 'acyclic sheaves') is 0. I'm confused now – Mario Carrasco Sep 7 '12 at 18:34
I believe a complex $K^\bullet$ is acyclic if $h^n(K^\bullet)=0$ for all $n\in\mathbb Z$. – Sándor Kovács Sep 7 '12 at 19:18
Perhaps you are thinking of replacing an object with its resolution, i.e., a complex that is exact everywhere except at degree $0$? That is not an acyclic complex and usually has (hyper)cohomology. – Sándor Kovács Sep 7 '12 at 19:21
You know I was confused by the whole exact or acyclic thing, these guys say that EXACT is not the same as ACYCLIC: According to these guys my complex is acyclic, so I guess I was being misled the entire time? Shafarevich wrote in his book that an ACYCLIC complex of sheaves has hypercohomology 0, but didn't provide proof, when I read that (thinking that my complex is acyclic) I thought that I was done, now I guess he's using the same definition as you and not the one in the discussion in the link, I feel so bad right now – Mario Carrasco Sep 7 '12 at 19:57
I added a competing answer to that question on math.stackexchange – Sándor Kovács Sep 7 '12 at 20:41

EDIT: reading your question again, I'm a bit confused about both terminology and what you are actually asking. Below by an exact complex I mean one with vanishing cohomology (as a complex!), whereas by an acyclic complex I mean a complex of ($\Gamma$-)acyclic objects. In general, the hypercohomology of an acyclic complex is easy to compute (see first part of my answer), and the hypercohomology of an exact complex is zero (see second half).

To elaborate on Damian's comment (as Sándor Kovács points out, the second half is wishful thinking):

One way to define hypercohomology of a complex $K^\bullet$ is to take a Cartan-Eilenberg resolution $K^{\bullet, \bullet}$ and then take the cohomology of the total complex of $K^{\bullet, \bullet}$. Cartan-Eilenberg resolutions are pretty awesome (they have basically all nice properties you could ask for), but in particular every columns is an injective resolution of the respective term of $K^\bullet$.

Now given any left-bounded double complex $K^{\bullet, \bullet}$ there exist two spectral sequences converging to the cohomology of the total complex (see Weibel). The $E_2$ pages are obtained by taking first horizontal and then vertical cohomology, and vice versa.

In what follows, I shall write $h^p(K^\bullet)$ for the cohomology of the complex $K^\bullet$, and $H^p(K)$ for the derived functor of global sections on $K$.

In our case, taking vertical cohomology first, we get an $E_1$ page $H^q(K^p)$. By your acyclicity assumption, we have $H^q(K^p) = 0$ for $q>0$, i.e. the $E^1$ page just consists of $K^\bullet$ again in the lowest row, and zeros else. Thus the spectral sequence degenerates at the $E_2$ page, leaving you with $h^p(\Gamma(K^\bullet))$ for the cohomology of the total complex, i.e. hypercohomology (see Weibel again for how the separate pages relate to the cohomology of the total complex).

This is the statement you were asking for (to the extent that I am aware of a correct formulation): the hypercohomology of an acyclic complex is just the cohomology of its complex of global sections. In particular an exact acyclic complex has vanishing hypercohomology.

More generally, any exact complex has vanishing hypercohomology: this time look at the spectral sequence where we take horizontal cohomology first. Another awesome property of cartan-eilenberg resolutions: the horizontal boundaries and cycles are themselves injectives, and hence taking horizontal cohomology of the cartan-eilenberg resolution yields an injective resolution of the cohomology of $K^\bullet$. There is a small difficulty: we are not supposed to compute the hypercohomology of $K^{\bullet,\bullet}$ but of $\Gamma(K^{\bullet,\bullet})$. However, since injectives are acyclic, "taking horizontal cohomology" and "taking global sections" actually commutes in our case, so we are good. In summary: there exists a convergent spectral sequence with $E_2$ page $H^q(h^p K^\bullet)$, which is evidently zero if $K^\bullet$ is exact.

Even more generally (and by a very similar argument), hypercohomology factors through quasi-isomorphism.

Note: again as Damian points out, "hypercohomology on some abelian category" does not really make sense. You can in general define hyperderived functors, and then hypercohomology is usually the name for the hyperderived functor of global sections. My above statements hold for the hyperderived functors of an arbitrary left-exact additive functor $\Gamma$.

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No, by "acyclic complex" I meant a complex $K^\bullet$ with vanishing cohomology for $n \geq 1$, not a complex of "acyclic objects", I think I'm using the wrong terminology here, the thing is that my complex $K^\bullet$ fails to be exact by one degree. I read in Shafarevich's algebraic geometry book 2 things: 1) That the hypercohomology of a complex of acyclic sheaves is the cohomology of the complex of global sections (as you wrote) and 2) That the hypercohomology of a complex that is 'acyclic' (which I'm assuming acyclic as a complex, i.e. the thing that I have) is 0. Thank you so much BTW – Mario Carrasco Sep 7 '12 at 18:28
I see; my use of the terminology may be wrong as well. In any case, just to have "almost all" cohomology of $K$ vanish is obviously not enough. – Tom Bachmann Sep 7 '12 at 18:53
Then Shafarevich's book is wrong? – Mario Carrasco Sep 7 '12 at 19:10
@Mario: that's not what an acyclic complex is. – Sándor Kovács Sep 7 '12 at 19:50
Thank YOU. Another thing I was so confused about is: You know how in an abelican category $\mathcal{A}$, you have the derived functors $R^nF(A)$ of a left exact functor $F$ of an object $A$. To say that the object $A$ is acyclic is to say that the $R^nF(A) = 0$ for $n \geq 1$ right? But that means that the complex $0 \rightarrow F(I^0) \rightarrow F(I^1) \rightarrow F(I^2) \cdots$ is exact except at 0 right?? I get confused because I've seen some people use $0 \rightarrow F(I^1) \rightarrow F(I^2) \rightarrow F(I^3) \cdots$ – Mario Carrasco Sep 7 '12 at 21:26

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