If $G$ is a finitely generated free group, then its classifying space $B G$ can be presented as a finite CW complex (a finite bouquet of circles), and therefore is SpanierWhitehead dualizable. Are there any other discrete groups $G$ with the property that $B G$ is SpanierWhitehead dualizable?
5 Answers
Any finitely presented group of type FL admits a finite classifying space. (A group $G$ is of type FL if $\mathbb{Z}$ admits a finite length resolution by finitely generated, free $\mathbb{Z}G$modules.) This is Theorem VIII.7.1 in Brown's book "Cohomology of groups".
More examples are given in section VIII.9 of Brown. These include knot groups, finitely generated nilpotent groups, and certain arithmetic groups.
Also the nonorientable surfaces (other than $\mathbb{R}P^2$) are examples.

$\begingroup$ Thanks; the reference to Brown's book is exactly what I needed. $\endgroup$ Commented Feb 29, 2012 at 5:08
Dualizable is equivalent to $BG$ being finitely dominated (a retract up to homotopy of a finite cell complex). In Wall's "Finiteness conditions on CW complexes, I" Wall shows that finite domination is equivalent to $\pi_1$ being finitely presented and also, the chain complex $C_*(EG)$ being chain finitely dominated over $\Bbb Z[G]$. This means there is a finite resolution of $\Bbb Z$ by finitely generated projective $\Bbb Z[G]$modules. In the group cohomology literature, this last condition is called "FP"
I'll answer a related question: in the $K(n)$local stable category, $BG$ is dualizable for all finite groups $G$, moreover, each is selfdual. You can find this in Hovey and Strickland's 'Morava $K$theories and localisation' Corollary 8.7.
More precisely, the result states that if $G$ is finite, then the $K(n)$ localized norm map: \begin{equation} L_{K(n)}\Sigma^\infty_+ BG\rightarrow F(L_{K(n)}\Sigma^\infty_+ BG, L_{K(n)} S) \end{equation} is a weak equivalence. This is because Tate cohomology lowers chromatic complexity (HoveyStrickland 96, Kuhn 2004), so the cofiber of the localized norm map $L_{K(n)}t_G(L_{K(n)} S)^{G}$ is weakly contractible.

$\begingroup$ I thought that BG wasn't exactly selfdual in the K(n)local sense. Isn't there a twist? $\endgroup$ Commented Feb 28, 2012 at 22:53

$\begingroup$ I expanded the answer to clarify what I meant. If the new answer does not allay your concerns, perhaps you can tell me what you mean by a twist? $\endgroup$ Commented Feb 29, 2012 at 9:42
This property is closed under taking products of groups, which gives you other examples.
Also, any genus $g$ surface (for $g>0$) is a $K(G,1)$, and hence a $BG$, and is clearly a finite CW complex.
This example is undoubtedly covered by John Klein's general response, but is somewhat illustrative in its own right.
By the CartanHadamard theorem, the universal cover of every connected, nonpositively curved manifold $M$ is homeomorphic to $\mathbb{R}^n$. So if $\Gamma = \pi_1(M)$ is the fundamental group of $M$, then $M = \mathbb{R}^n / \Gamma = B \Gamma$ is the classifying space of $\Gamma$. If $M$ is itself compact, it is dualisable.
One tends to consider the case when $\Gamma$ is a discrete subgroup of $Isom(\mathbb{H}^n)$, the group of isometries of hyperbolic space. If $\Gamma$ acts freely and cocompactly on $\mathbb{H}^n$, then $B\Gamma$ is stably dualisable.