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I found several papers on a related question called the "quasi-projective dimension" of a group ring $R[G]$. The original paper on this is Groups of finite quasi-projective dimension, by Howie and Schneebeli. Their definition of a quasi-projective resolution is a finite resolution of a module $M$ by projective terms, and at the end a term which is a permutation module. I assume that, certainly for finite groups, it would work just as well to use a free resolution as a projective resolution. Among other results, Howie and Schneebeli establish that if $G$ is a finite group and $R = \mathbb{Z}$, then the quasi-projective dimension of $R[G]$ equals the period of its Tate cohomology. But another theme of the paper is that these questions, both theirs and surely Leonid's also, are perfectly interesting for infinite groups too.

I found several papers on a related question called the "quasi-projective dimension" of a group ring $R[G]$. The original paper on this is Groups of finite quasi-projective dimension, by Howie and Schneebeli. Their definition of a quasi-projective resolution is a finite resolution of a module $M$ by projective terms, and at the end a term which is a permutation module. I assume that, certainly for finite groups, it would work just as well to use a free resolution as a projective resolution. Among other results, Howie and Schneebeli establish that if $G$ is a finite group and $R = \mathbb{Z}$, then the quasi-projective dimension of $R[G]$ equals the period of its Tate cohomology. But another theme of the paper is that these questions, both theirs and surely Leonid's also, are perfectly interesting for infinite groups too.

I found several papers on a related question called the "quasi-projective dimension" of a group ring $R[G]$. The original paper on this is Groups of finite quasi-projective dimension, by Howie and Schneebeli. Their definition of a quasi-projective resolution is a finite resolution of a module $M$ by projective terms, and at the end a term which is a permutation module. I assume that, certainly for finite groups, it would work just as well to use a free resolution as a projective resolution. Among other results, Howie and Schneebeli establish that if $G$ is a finite group and $R = \mathbb{Z}$, then the quasi-projective dimension of $R[G]$ equals the period of its Tate cohomology. But another theme of the paper is that these questions, both theirs and surely Leonid's also, are perfectly interesting for infinite groups too.

The papers that cite this initial paper use the second idea that I propose above. They make CW complexes with an action of the group $G$, and then make chain complexes from these CW complexes. So these CW complexes seem like a main way to understand complexes of permutation modules. In my opinion, the CW complex picture suggests generalizing the original question to include signed permutation modules.

Now that I have thought about the question some more, I can give a better answer. I have a remark about how to search for these resolutions in general, and a construction that leads to many examples.

First, every permutation module is part of many exact sequences of permutation modules in which the subgroup is trivial: $$\cdots \to R[G]^{n_2} \to R[G]^{n_1} \to R[G/H] \to 0.$$ The reason is very simple and standard: Any exact complex of this type is by definition a free resolution. The way that you make a free resolution is that there is some intermediate target or kernel $K$, and you can send the generators $1 \in R[G]$ to some spanning set of $K$. Usually the resolution is infinite, but with standard linear algebra you can search for a finite solution when there is one.

This observation generalizes to other permutation modules. There is a part of induction-restriction reciprocity that holds over any ring. Namely, $$\text{Hom}_G(R[G/H],M) \cong \text{Hom}_H(I,M),$$ where $I$ is the trivial representation. This relation is a generalization of the proof that a free module is projective. So if there is a finite permutation resolution of a module $M$ (which could be the kernel of some incomplete sequence of permutation modules), you can search for it in the same way that you search for free resolutions.

Second (and I suspect that readers will like this answer better), you can obtain many examples from the chain homology complex of a finite CW complex $K$ with an action of $G$. In order to make everything match, let's consider a slight generalization of a permutation module, not just $R[G/H]$ but also a module $R[G/H]_\chi$ induced from a character $$\chi:H \to \{1,-1\}.$$ The basic idea is not hard: Each term $C^n(K)$ is a direct sum of signed permutation module sof $G$, because $G$ acts on the cells. If a cell $c$ has stabilizer $H$, then it makes an orbit equivalent to $G/H$, and we can define $\chi$ by examining which elements of $H$ flip over $c$. If $K$ happens to have the same $R$-homology as a point, then you can augment its chain complex by the trivial module. Or if it is an $R$-homology sphere, you can augment its chain complex at both ends. If you don't like the signed permutation modules, you can subdivide the cell $c$ to get rid of them, or work in characteristic 2.

If $K$ is a line segment and $G = C_2$ acts by reflecting it, then result is Leonid's first example.

If $K$ is a polygon with $n$ sides and $G = C_n$ acts by rotation, the result is Leonid's second example.

If $K$ is a polygonal tiling of the 2-sphere and $G$ is a rotation group that acts on $K$ without flipping over any edges, the result is a new example. For instance you can take a dodecahedron graph and divide each edge into two edges. Again, the point of splitting the edges is just to get rid of the signed permutation modules.

Every finite group $G$ acts faithfully on a sphere of some dimension, because $G$ has a faithful linear representation. So there are many sphere examples for every finite group.

At first glance, the second answer is a type of construction. In many cases, it is also an interpretation of a chain complex $C$, because if you have $C$ you can try to build a CW complex to represent it. In order to be an augmented chain complex, $C$ needs to end in the trivial module $I$. If it ends in something else, you can concatenate with $$0 \to I \to I \to 0.$$ The differentials of $C$ also need to have integer matrices.

Now that I have thought about the question some more, I can give a better answer. I have a remark about how to search for these resolutions in general, and a construction that leads to many examples.

First, every permutation module is part of many exact sequences of permutation modules in which the subgroup is trivial: $$\cdots \to R[G]^{n_2} \to R[G]^{n_1} \to R[G/H] \to 0.$$ The reason is very simple and standard: Any exact complex of this type is by definition a free resolution. The way that you make a free resolution is that there is some intermediate target or kernel $K$, and you can send the generators $1 \in R[G]$ to some spanning set of $K$. Usually the resolution is infinite, but with standard linear algebra you can search for a finite solution when there is one.

This observation generalizes to other permutation modules. There is a part of induction-restriction reciprocity that holds over any ring. Namely, $$\text{Hom}_G(R[G/H],M) \cong \text{Hom}_H(I,M),$$ where $I$ is the trivial representation. This relation is a generalization of the proof that a free module is projective. So if there is a finite permutation resolution of a module $M$ (which could be the kernel of some incomplete sequence of permutation modules), you can search for it in the same way that you search for free resolutions.

Second (and I suspect that readers will like this answer better), you can obtain many examples from the chain homology complex of a finite CW complex $K$ with an action of $G$. In order to make everything match, let's consider a slight generalization of a permutation module, not just $R[G/H]$ but also a module $R[G/H]_\chi$ induced from a character $$\chi:H \to \{1,-1\}.$$ The basic idea is not hard: Each term $C_n(K)$ is a direct sum of signed permutation modules of $G$, because $G$ acts on the cells. If a cell $c$ has stabilizer $H$, then it makes an orbit equivalent to $G/H$, and we can define $\chi$ by examining which elements of $H$ flip over $c$. If $K$ happens to have the same $R$-homology as a point, then you can augment its chain complex by the trivial module. Or if it is an $R$-homology sphere, you can augment its chain complex at both ends. If you don't like the signed permutation modules, you can subdivide the cell $c$ to get rid of them, or work in characteristic 2.

If $K$ is a line segment and $G = C_2$ acts by reflecting it, the result is Leonid's first example.

If $K$ is a polygon with $n$ sides and $G = C_n$ acts by rotation, the result is Leonid's second example.

If $K$ is a polygonal tiling of the 2-sphere and $G$ is a rotation group that acts on $K$ without reversing edges, the result is a new example. For instance you can take a dodecahedron graph and divide each edge into two edges. Again, the point of splitting the edges is just to get rid of the signed permutation modules.

Every finite group $G$ acts faithfully on a sphere of some dimension, because $G$ has a faithful linear representation. So there are many sphere examples for every finite group.

At first glance, the second answer is a type of construction. In many cases, it is also an interpretation of a chain complex $C$, because if you have $C$ you can try to build a CW complex to represent it. In order to be an augmented chain complex, $C$ needs to end in the trivial module $I$. If it ends in something else, you can concatenate with $$0 \to I \to I \to 0.$$ The differentials of $C$ also need to have integer matrices.