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

Let $R$ be a commutative ring, like the ring of integers $\mathbb Z$ or the ring of $p$-adic integers $\mathbb Z_p$. Let $G$ be a finite group; let us consider permutational representations of $G$ over $R$, i.e., $R[G]$-modules of the form $R[G/H]$, where $H\subset G$ is a subgroup, and direct sums of such modules. These are free $R$-modules where $G$ acts so that there exists a basis of the module preserved (as a whole) by the action.

I am interested in finite exact sequences of representations of the above type, particularly in those of them that are not very long. There are some beautiful examples, e.g., for $G=\mathbb Z/2$ and $R=\mathbb Z/2$ there is an exact sequence $$ 0\rightarrow R\rightarrow R[G]\rightarrow R\rightarrow 0, $$ while for $G=\mathbb Z/n$ and $R=\mathbb Z$ there is an exact sequence $$ 0\rightarrow R\rightarrow R[G]\rightarrow R[G]\rightarrow R\rightarrow 0. $$ For the fourth symmetric group $G=\mathbb S_4$ and $R=\mathbb Z$ there is an exact sequence $$ 0\rightarrow R\rightarrow R[\mathbb X_4]\oplus R\rightarrow R[\mathbb X_6]\oplus R\rightarrow R[\mathbb X_3]\rightarrow0, $$ where $\mathbb X_4$ is the four-element set that $\mathbb S_4$ permutes, $\mathbb X_6$ is the set of all two-element subsets of $\mathbb X_4$, and $\mathbb X_3$ is the quotient set of $\mathbb X_6$ by the obvious involution. Dihedral groups also have some four-term exact sequences of permutational representations.

Where is one supposed to get such exact sequences? There are some obvious ways, like e.g. one can take cones of morphisms of exact sequences of this type, or one can do restriction or induction from one group to another one. Are there any other constructions?

For constructions to be interesting, they should of course be removed far enough from the trivial case when $|G|$ is invertible in $R$. E.g., to have $R=\mathbb Z_p$ and $|G|$ a large $p$-group would be perhaps most highly nontrivial.

EDIT: One of the commenters asked where does the sequence for $\mathbb S_4$ come from, so let me say a few words about this. Not that I really understand it, but there is a geometric construction using a CW complex, and not quite of the kind that Greg suggests in his second answer below.

Represent the group $\mathbb S_4$ as the group of rotations of the $3$-dimensional cube. Consider the quotient CW complex of the cube's surface by the central symmetry involution. The group $\mathbb S_4$ still acts on the quotient. The set of vertices of the quotient is the $\mathbb S_4$-set $\mathbb X_4$, the set of edges is the $\mathbb S_4$-set $\mathbb X_6$, and the set of faces is the $\mathbb S_4$-set $\mathbb X_3$.

Now consider the map $R[\mathbb X_4]\rightarrow R[\mathbb X_6]$ assigning to a vertex the sum of the three edges ending in it (without any signs!) and also the map $R[\mathbb X_6]\rightarrow R[\mathbb X_3]$ assigning to an edge the sum of the two faces bordering on it (also without any signs!). The composition of these two maps is not zero, of course; what it does is taking every vertex to twice the sum of all the three faces. One somehow transforms this pair of arrows into a four-term exact sequence by adding the trivial $\mathbb S_4$-module direct summands $R$ in several degrees.

share|improve this question
1  
This seems like very strange question. I've thought a reasonable amount about permutation modules over finite fields, but I just have no idea what sort of answer you really want from this question. –  Ben Webster Dec 10 '09 at 18:00
    
The bounded homotopy category of permutational representations maps to the bounded derived category of arbitrary representations and this functor has some kernel, which is a triangulated category. I want to acquire some understanding or at least feeling of what this kernel consists of. I am also very much interested in any nice examples of objects in the kernel. –  Leonid Positselski Dec 10 '09 at 18:49
1  
Ah. That's a much more answerable question, and one I may actually have thoughts on. –  Ben Webster Dec 11 '09 at 15:13
    
So, please share them. –  Leonid Positselski Dec 11 '09 at 16:21
    
Where does the sequence for $\mathbb S_4$ come from? –  Mariano Suárez-Alvarez Dec 13 '09 at 5:38
show 1 more comment

3 Answers

up vote 5 down vote accepted

Here is a preprint by Boltje and Hartmann that constructs a conjectured resolution of Specht modules of $S_n$ (over $\mathbb{Z}$) by Young modules. This is presumably a related tool.

(An earlier version of this answer had some out-of-step comments about resolutions that were either not helpful or already addressed in the original question.)

share|improve this answer
1  
Thank you for the reference! As to the first paragraph of your answer, how does one extend a permutation module to a chain complex with semisimple homology? –  Leonid Positselski Dec 4 '09 at 20:07
    
You always catch my worst mistakes. Thanks again. –  Greg Kuperberg Dec 4 '09 at 20:16
1  
A brief remark on the last paragraph of your answer - it seems that some info on resolutions of Specht modules over Z (actually, maybe all possible information) can be extracted from arxiv.org/abs/0803.4382 –  Vladimir Dotsenko Dec 5 '09 at 1:34
add comment

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.


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.

share|improve this answer
    
I don't want to extend the "answer itself" further, but let me briefly add this. There is a tiling of the projective plane by three squares with an $S_4$ action that comes from the surface of the cube. We can obtain Leonid's third example over $\mathbb{Z}/2$ by modifying the CW cohomology of this projective plane. I concede that I do not see a natural CW model for this example with coefficients in $\mathbb{Z}$. It may not be hard to contrive a CW model over $\mathbb{Z}$ for this case, but that is not the best answer. –  Greg Kuperberg Dec 16 '09 at 7:09
add comment

[Katsura, Takeshi. Permutation presentations of modules over finite groups. J. Algebra 319 (2008), no. 9, 3653--3665.] proves that if $G$ is finite, then every $\mathbb ZG$-module $M$ has a permutation presentation (which is a short exact sequence of the form $0\to F\to F\to M\to0$ with $F$ a permutation module) iff $G$ has all its Sylow subgroups cyclic. For such a group, if you pick $M$ itself to be a permutation module, then you get one of the sequences you want.

share|improve this answer
    
Thank you, this is a helpful reference. But aren't all short exact sequences of permutational representations obtained in this way contractible, perhaps? –  Leonid Positselski Dec 4 '09 at 23:58
add comment

Your Answer

 
discard

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.