Or equivalently, if $G$ is a group, do the projective and injective dimension of $Z$ (viewed as a $ZG$-module) agree?
Thanks!
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$\begingroup$ You probably mean to ask «If $G$ is a group, do the projective and the injective dimensions of $\mathbb Z$ as a $\mathbb ZG$-module aggree?» $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 13, 2010 at 16:15
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$\begingroup$ Yes. They are equivalent. $\endgroup$– HaoCommented Sep 13, 2010 at 16:24
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10$\begingroup$ The answer is that it is false. For example if $G$ is the trivial group then the projective dimension of $\mathbb{Z}$ is zero but the injective dimension is one. $\endgroup$– Simon WadsleyCommented Sep 13, 2010 at 16:30
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1$\begingroup$ @Mariano: I think the "it is equivalent to say" is pointing out the equivalence between the question in the title and the question in the body. Editing to make more clear. $\endgroup$– Cam McLemanCommented Sep 13, 2010 at 19:39
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5$\begingroup$ Judging by the title, my guess is that the body of the question should read "Or equivalently, if G is a group, do the projective and the flat dimension of Z (viewed as ZG-module) agree?" $\endgroup$– Leonid PositselskiCommented Sep 13, 2010 at 21:33
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3 Answers
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The cohomological and homological dimension of a group do not agree in general. For example, the homological dimension of the group $Z[\frac{1}{2}]$ is one, while its cohomological dimension is 2. However, if |G| is countable, then cohomological dimension is either equal to the homological dimension or one dimension greater, see Proposition 2.4 in the following paper:
R. Bieri, Normal subgroups in duality groups and in groups of cohomological dimension 2. J. Pure Appl. Algebra 7 (1976), no. 1, 35–51.
Also , cohomological dimension of a group is defined by the highest dimension n such that $H^n(G,M)$ is nonzero, where M is a G-module. It is important to use a G-module instead of the usual cofficient. For example, the cohomological dimension for any nontrivial knot group is 2, while its cohomology with $Z$ cofficient, or any G-module with trivial group action is always the same as the cohomology of $S^1$.
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I like to think of group (co)homology topologically, so I would say the integral (co)homological dimension of G is the integral (co)homological dimension of BG. Thinking this way there are lots of geometric examples in which the integral homological dimension is less than the integral cohomological dimension: for example, non-orientable surfaces are classifying spaces for their fundamental groups. Their integral homological dimension is 1, and their integral cohomological dimension is 2.
Of course for other coefficient systems, this will no longer be the case (as Tom points out in his comment below). Note, for example, that the rational cohomological dimension of a non-orientable surface is 1.
[Okay, I'm going to admit some confusion in regards to the comments on the original question. Am I thinking of the flat or the injective dimension here, when I take homology of BG? From the comments, it sounds like this must correspond to the flat dimension? I don't have Brown's book in front of me to un-confuse myself...]
answered Sep 14, 2010 at 5:17
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2$\begingroup$ Cohomological dimension is usually defined to be the supremum over all coefficient systems (in particular, this is necessary for many theorems and applications, including the fact that virtual cohomological dimension is well-defined). Taking the orientation character as your coefficients rules out your proposed counterexample. $\endgroup$ Commented Sep 14, 2010 at 6:35
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$\begingroup$ @Tom: Since the question and the comments seemed to be talking about integral coefficients in particular, I was just talking about integral (co)homological dimensions. I edited to make this clearer. $\endgroup$ Commented Sep 14, 2010 at 19:00
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So, Simon Wadsley’s comment clearly answers this question but a hypothetical future user will not have his/her eye drawn to that answer. That’s why I’m posting this, which is all Simon’s idea, in the hope that the OP will come back on at some point and accept this answer (thus preventing this problem from being bumped up to the front-page by the Mathoverflowbot). If the OP is reading this, please click the check mark next to this box so it'll count as being answered. I'm giving a CW answer so I don't get rep (in line with the recommended procedure on meta)
The homological and cohomological dimensions of a group do NOT have to agree. As you point out, if they did agree then the projective and injective dimensions of $\mathbb{Z}$ as a $\mathbb{Z}[G]$ module would agree (this is basically just the definition of Ext). An example that this could fail, take $G$ to be the trivial group. Then the projective dimension of $\mathbb{Z}$ as a $\mathbb{Z}$-module is zero because any ring is projective over itself. But the injective dimension is not zero because $\mathbb{Z}$ is not a divisible abelian group. Indeed, the injective dimension is 1, as can be seen from the fact that $\mathbb{Z}$ is a PID and hence has global dimension 1. Or you can just write down an injective resolution. Or you can read Dummit-Foote for their treatment.
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2$\begingroup$ I'm a little confused by this example. Thinking topologically, the classifying space of the trivial group is a point, and a $G$-module is just an abelian group, so I don't see how the first cohomology can be nonzero? $\endgroup$ Commented Dec 1, 2011 at 7:16
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1$\begingroup$ Cohomology is given by Ext, but homology is given by Tor. The problem is whether the flat dimension of $\mathbb Z$ over $\mathbb Z[G]$ agrees with its projective dimension. For the trivial group they are both 0, of course. $\endgroup$– AngeloCommented Dec 2, 2011 at 11:21