Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for a reference for the fact that the top cohomology $H^n(X;A)$ of an $n$-dimensional manifold $X$ is non-trivial precisely when $X$ is compact.

I tried to ask this question on Math.Stackexchange, but there was some issue regarding orientability.

If $X$ is orientable, this should follow from standard Poincaré duality, but in my understanding the assumption of orientability can be removed by using "twisted" Poincaré duality. In this case, the coefficients must be local for $H^n$ to be non-trivial. (Edit: They don't need to be local, see the answer by Johannes Ebert below.) But if the coefficients respect the orientation (i.e. the coefficients are a constant sheaf, which is twisted by the orientation character), then $H^n\neq0$ if and only if $X$ is compact.

I vaguely remember hearing about an isomorphism $H^n\cong H_c^0$, where $H_c^\ast$ denotes compactly supported cohomology. The argument then goes that $H_c^0$ is non-trivial precisely when $X$ is non-compact, which I think of by using covers in Čech cohomology. $0$th cohomology with values in the sheaf of locally constant functions is just constant functions on all open sets of the cover $\{U_i\}$ and these functions must agree on single intersections $U_i\cap U_j$. The cohomology classes are thus given by globally constant functions, which, for compactly supported cohomology can only be the zero function when $X$ is non-compact, and can be any constant function when $X$ is compact.

It seems this argument makes no use of orientability, provided, there exists a twisted version of $H^n\cong H_c^0$...

share|improve this question
You use the sheaf cohomology tag. Is $A$ an abelian sheaf on $X$? Does it vary? –  Martin Brandenburg Jul 7 '12 at 2:04
Yes, I think of $A$ as a locally constant sheaf on $X$, which "varies" with the orientation of $X$. –  Earthliŋ Jul 7 '12 at 2:11
$A$ is an Abelian sheaf. I'm not quite know what you mean by "vary", though, unless you just mean the sheaf is not constant. –  Earthliŋ Jul 7 '12 at 2:24
If by $A$ you mean the constant sheaf $\mathbb Z$ twisted by the orientation character $\pi_1(X)\to\{\pm 1\}$, then the answer is yes. But in any case, this is not research level. –  John Pardon Jul 7 '12 at 2:43
Do you have a reference for this? –  Earthliŋ Jul 7 '12 at 2:59
show 4 more comments

2 Answers

up vote 4 down vote accepted

A connected $n$-manifold is compact iff $H^n (M;\mathbb{Z}) \neq 0$, see Bredon, Topology and Geometry, Corollaries 7.12 and 7.14, page 346 ff.

share|improve this answer
Thank you. I got confused by the fact that $H^n(\mathbb RP^n;\mathbb R)=0$. In your reference the top cohomology is non-trivial even though the coefficients are not twisted, which is even better. –  Earthliŋ Jul 7 '12 at 12:41
This statement is false over the integers, unless the manifold can be oriented, but true over the field with two elements for all manifolds. –  Fernando Muro Jul 7 '12 at 17:20
How about "A connected $n$-manifold is compact iff $H^n(M;A^w)\neq0$"? (For any non-trivial constant Abelian sheaf $A$ and $w$ the orientation character of $M$.) –  Earthliŋ Jul 8 '12 at 0:55
Don't you like the easier statement that it is comact iff $H^n(M,\mathbb{F}_2)\neq 0$? –  Fernando Muro Jul 8 '12 at 21:52
Why the downvote? @Fernando: what is wrong with my statement?? –  Johannes Ebert Jul 18 '12 at 19:03
add comment

For algebraic varieties, there is an analogue in terms of cohomology of coherent sheaves. It reads as follows: if $X$ is an irreducible quasi-projective algebraic variety over a field $k$, of dimension $n$, then $X$ is projective if and only if $H^n(X,F)\ne 0$ for some coherent sheaf $F$ on $X$. This was conjectured by Lichtenbaum, a first proof was given by Grothendieck (see thm 6.9 in Hartshorne, Local cohomology, Springer LNM), another one by Kleiman (On the Vanishing of $H^n(X,F)$ for an $n$-dimensional variety, Proc. AMS 1967).

share|improve this answer
Nice, thank you. –  Earthliŋ Jul 7 '12 at 12:44
add comment

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.