If $A$ is an algebra over a field $k$ and $M$ is a finite-dimensional $A$-module, then Alperin showed in a paper [Diagrams for modules, JPAA, 1980] how to associate a diagram to $M$ with the vertices being the composition factors, putting all simple submodules (the socle) on the bottom, all simple quotients (the head) on the top and connecting vertices in the middle if there is a subquotient in which the two vertices form an indecomposable module; that is, there is a non-trivial element of $\operatorname{Ext}^1$ creating a non-split extension between them. I would guess that there are various issues with the well-definedness of this process for general $A$ and $M$, but one can make sense of it in various situations.

My question is: Suppose the number of composition factors in $M$ is a fixed $n$, so that the number of vertices of the Alperin diagram of $M$ is $n$. Are there any conditions on the algebra $A$ (homological or otherwise) which impose bounds on the number of edges there can be in the Alperin diagram of $M$? For example, do we know anything if $A=kG$ is the group algebra of a finite group $G$?

For a more specific question, is it true that for each $n$ there is a finite group $G$ and a $kG$-module $M$ such that the Alperin diagram is a complete graph?

  • $\begingroup$ Alperin's definition of module diagrams appears to rule out the possibility of triangles ($K_3$'s) as subgraphs: ``if $y_1,\ldots,y_n$ are nodes, $n>2$, and there is an edge from $y_i$ to $y_{i+1}$, $1 \leq i <n$, then there is no edge from $y_1$ to $y_n$.'' So complete graphs on $n>2$ vertices cannot occur. $\endgroup$
    – Alex Dugas
    Apr 10 '19 at 23:19

I don't have an answer to your main question but I can speak to some of your other points.

Firstly, Alperin diagrams are not so much graphs as lattices. They are in fact an ordering on the composition factors of $M$ such that $N_1 \ge N_2$ iff there is a submodule with head $N_1$ and socle $N_2$. As such, triangles (as @AlexDugas notes) are not possible — $N_1 \ge N_2 \ge N_3$ and $N_3 \ge N_1$ are incompatible.

Secondly, a very important restriction on $M$ that must hold before we can construct Alperin diagrams is that $M$ must be multiplicity free. That is, each factor must appear in a composition series exactly once.

To see why this is important, consider two identical modules with composition factors $[N_1, N_2]$ (so the head is $N_1$). Then the direct sum of these modules, quotiented out by a suitable diagonal image of $N_2$ will have head $N_1\oplus N_1$ and socle $N_2$. What is its Alperin diagram? At first we may say

N_1   N_1
   \ /

but in fact a suitable change of basis gives that $N_1$ falls out as a direct summand! Thus a "better" diagram is the forest

N_1  +    |

Which should we use? This is a simple case, but you can imagine that in a module with a dozen factors some of which are repeated multiple times, this becomes a serious question.


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