I have several related questions regarding homological vs. cohomological dimension of a space/group (this is not a duplicate of this).

The standard definition of the cohomological dimension $cd(X)$ of a space $X$ is

definition 1: $cd(X)=$ The minimum over $n\in \mathbb{N}$, such that $H^i(X,M)=0$ for all $i>n$ and all local coefficient systems $M$.

One can similarly define the homological dimension $hd(X)$ (I am not sure it is a standard terminology)

definition 2: $hd(X)=$ The minimum over $n\in \mathbb{N}$, such that $H_i(X,M)=0$ for all $i>n$ and all local coefficient systems $M$.

My main question is

Question1: Do we always have $cd(X)=hd(X)?$ What if one or both assumed to be finite? what if $X$ is a finite CW complex?

For $X = BG$, i.e a classifying space of a group $G$, the definitions reduce to the standard ones in group cohomology. I believe that the $cd(BG)$ coincides with the minimal length of a projective resolution of the trivial $G$-module $\mathbb{Z}$ over the group ring $\mathbb{Z}[G]$ (Namely, the projective dimension of $\mathbb{Z}$).

Question 2: Is it true that the $hd(BG)$ coincides with the flat dimension of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module?

Finally, if $X$ is a connected CW complex with $cd(X)=0$, then $X$ is contractible. One can similarly ask

Question 3: If $X$ is a CW complex with $hd(X)=0$, is it true that $X$ is contractible?

(Edit: The following argument has a gap is wrong. It is not clear true that if $hd(X)=0$ then $X$ is a filtered colimit of finite subcomplexes $X'$ with $hd(X')=0$. An exmaple is given here)

Regardless of the answer to question 1, I think the answer to this is yes by the following argument. Since the universal cover is acyclic and simply connected, it is contractible, so this reduces to a question about group cohomology. If $G$ is finitely generated (if $X$ is a finite CW for example), then $\mathbb{Z}$ is a finitely presented $\mathbb{Z}[G]$-module so if it is flat then it is projective and hence the cohomological dimension is zero as well. Can one reduce the general case to this one since every CW complex is a filtered colimit of its finite subcomplexes. Is there a more direct argument?

  • $\begingroup$ Re the edit, that is not true. See this answer of mine to another question, for instance. $\endgroup$ Jul 15, 2015 at 7:19
  • $\begingroup$ Thanks Eric. I'll make some changes to reflect this fact. $\endgroup$
    – KotelKanim
    Jul 15, 2015 at 7:42
  • $\begingroup$ If $G$ is any non-trivial group and $H$ is a non-trivial abelian subgroup, then $H_1\left(G,\mathbb{Z}[G/H]\right)\cong H_1(H,\mathbb{Z})\neq0$. $\endgroup$ Jul 15, 2015 at 9:38
  • $\begingroup$ @JeremyRickard, right! So a space $X$ with $hd(X)=0$ must be contractible. Could you possibly write this comment as an answer? $\endgroup$
    – KotelKanim
    Jul 15, 2015 at 9:58

3 Answers 3


Q2) You mean $X=BG$, right? Then the answer is yes by standard homological algebra.

Q1) The answer to the first question is no, e.g. $X=M(\mathbb Q,n)=S^n_{\mathbb Q}$ the rational $n$-sphere has homological dimension $n$ but cohomological dimension $n+1$ (since $\mathbb Q$ is flat but not projective as an abelian group). For $X$ finite I don't know the answer in general, but if $X$ is in addition simply connected then yes, since the chain complex of $X$ is quasi-isomorphic to its homology and the projective and flat dimensions coincide on finitely generated abelian groups.

Q3) This is true if $X$ is simply connected (Whitehead's theorem) but false otherwise, e.g. $X=BG$ with $G$ acyclic. It is also positively answered in the question for $X=BG$ with $G$ finitely presented.

  • $\begingroup$ For Q3), I don't think $BG$ with $G$ acyclic is necessarily contractible. Acyclic usually means trivial homology with integer coefficients, so Higman's group is a counter-example. $\endgroup$
    – Mark Grant
    Jul 14, 2015 at 5:59
  • 1
    $\begingroup$ Thanks for the answers, but I don't agree about Q3. As mark says, acyclic means trivial homology with integral coefficients. I am talking about trivial homology with any (twisted/local) coefficients. I don't think there is a non-trivial group with trivial homology as I explained in the question. $\endgroup$
    – KotelKanim
    Jul 14, 2015 at 6:35
  • $\begingroup$ Sorry, it seems that when answering Q3 I suddenly forgot that we're considering local coefficients. $\endgroup$ Jul 14, 2015 at 20:56

As noted in the question, "Question 3" reduces to asking whether, if $H_i(G,M)=0$ for all $i>0$ and all $\mathbb{Z}G$-modules $M$, then $G$ must be trivial.

But this is true, since if $G$ is non-trivial, we can choose a non-trivial abelian subgroup $A$, and then, by Shapiro's Lemma, $$H_1\left(G,\mathbb{Z}[G/A]\right)\cong H_1\left(A,\mathbb{Z}\right)\cong A\neq0.$$

  • 1
    $\begingroup$ This actually answers Q3 not only for $X=BG$ but for any $X$ since the canonical map $X\rightarrow B\pi_1(X)$ induces an isomorphism on $H^1$ with coefficients in any $\pi_1(X)$-module. $\endgroup$ Jul 15, 2015 at 21:54
  • $\begingroup$ What does the notation $\mathbb{Z}[G/A]$ mean here? Since $A$ is not necessarily normal I’m guessing it refers to something other than the quotient. $\endgroup$ Apr 21, 2023 at 18:04
  • 1
    $\begingroup$ @Carl-FredrikNybergBrodda $G/A$ is the set of left cosets of $A$ in $G$, and $\mathbb{Z}[G/A]$ is the free abelian group with basis $G/A$, which is a left $\mathbb{Z}G$-module via the action of $G$ on $G/A$ by left multiplication. $\endgroup$ Apr 21, 2023 at 20:40

If $X$ is a finite CW complex then $hd(X) = cd(X)$. To prove this it is clearly enough to assume that $X$ is connected. Let $n$ be the dimension of $X$, let $G$ be the fundamental group of $X$ and let $\widetilde{X}$ be the universal cover of $X$. Let $$ 0 \longrightarrow C_n \stackrel{\partial_n}{\longrightarrow} C_{n-1} \stackrel{\partial_{n-1}}{\longrightarrow} \dots \stackrel{\partial_{1}}{\longrightarrow} C_0 $$ be the cellular complex of $\widetilde{X}$ (i.e. $C_n = H_n(\widetilde{X}_n,\widetilde{X}_{n-1})$) considered as a complex of $G$-modules. The homological and cohomological dimensions of $X$ are the same as the flat and projective dimensions respectively of the complex of $G$-modules $C_\bullet$ (lets call these latter two guys $pd(C_\bullet)$ and $fd(C_\bullet)$). Since $X$ is finite each $C_n$ is a finitely generated free $G$-module.

Given a complex of $G$-modules $D_\bullet$, indexed by non-negative integers, we will call the minimal $i$ such that $D_j = 0$ for $j > i$ the dimension of $D_\bullet$. Now let $D_\bullet$ be a complex of finitely generated projective $G$-modules which is quasi-isomorphic to $C_\bullet$ and has minimal dimension among all such complexes (since $C_\bullet$ is made of free finitely generated $G$-modules and has finite dimension such a $D_\bullet$ exists). Then $hd(X) = fd(C_\bullet) = fd(D_\bullet)$ and $cd(X) = pd(C_\bullet) = pd(D_\bullet)$. Let $m$ be the dimension of $D_\bullet$. We next claim that $fd(D_\bullet) = pd(D_\bullet) = m$. First it is clear that $fd(D_\bullet),pd(D_\bullet) \leq m$. Now if the map $d_m: D_m \longrightarrow D_{m-1}$ is not injective then $Tor^m(D_\bullet,\mathbb{Z}G) \neq 0$ and $Ext^m(D_\bullet,D_m) \neq 0$ and so $fd(D_\bullet)=pd(D_\bullet)=m$. Hence we may assume that the map $D_m \longrightarrow D_{m-1}$ is injective. Let $M$ be the cokernel of $d_m$. Then $M$ is finitely generated and by the minimality of $D_\bullet$ we see that $M$ cannot be projective. It follows that there exists a $G$-module $N$ such that the map $Hom_G(D_{m-1},N) \longrightarrow Hom_G(D_m,N)$ is not injective, and hence $Ext^m(D_\bullet,N) \neq 0$. This gives us $pd(D_\bullet) = m$. Now let $D'$ be a finitely generated projective $G$-module such that $D' \oplus D_{m-1}$ is a free $G$-module. Then $M$ is the quotient of $D' \oplus D_{m-1}$ by the finitely generated $G$-module $D' \oplus D_m$ and hence $M$ is finitely presented. It follows that $M$ cannot be flat, since any finitely presented flat $G$-module is projective. This means that there exists a $G$-module $N$ such that the map $N \otimes D_m \longrightarrow N \otimes D_{m-1}$ is not injective, and hence $Tor^m(D_\bullet,N) \neq 0$. We may now conclude that $fd(D_\bullet) = m$ as desired.

Regarding Q3, it seems very likely that $hd(X) = 0$ implies that $X$ is contractible, but the argument you present seems to have a problem. Indeed, $X$ is the filtered colimit of its finite sub complexes, but to use your argument we need to know that if $hd(X) = 0$ then $X$ can be written as a filtered colimit of finite complexes, each of whom has homological dimensional $0$.

  • $\begingroup$ Nice :) And you are right that my argument has a problem. I'm not sure it's fixable... $\endgroup$
    – KotelKanim
    Jul 15, 2015 at 6:27
  • $\begingroup$ In don't understand this argument. What's the projective/flat dimension of a complex? How can Tor be non-trivial if the second argument is projective? $\endgroup$ Jul 15, 2015 at 22:00
  • $\begingroup$ @FernandoMuro. Maybe my terminology is not standard. Let us work with non-negatively graded complexes. By $Tor_n(C_\bullet,D_\bullet)$ I mean the $n$'th homology group of the derived tensor product of $C_\bullet$ and $D_\bullet$. If either $C_\bullet$ or $D_\bullet$ are flat than this will coincide with the $n$'th homology group of the ordinary tensor product. Similarly, by $Ext^n(C_\bullet,D_\bullet)$ I mean the $n$'th cohomology group of the derived hom complex. -> $\endgroup$ Jul 16, 2015 at 10:36
  • $\begingroup$ -> The projective dimension of a complex $C_\bullet$ is the minimal $n$ such that $Ext^m(C_\bullet,M) = 0$ for every $m > n$ and every $G$-module $M$ (considered as a complex concentrated in degree $0$). Similarly, one can defined the flat dimension using $Tor$. These dimensions can be characterized as the minimal lengths of a projective/flat replacement of $C_\bullet$, respectively. The main point of the argument is that if $C_\bullet$ is a finite complex of finitely generated $G$-modules then it is enough to take the shortest finitely generated projective replacement. $\endgroup$ Jul 16, 2015 at 10:42

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