
Sheaves on Graphs, Their Homological Invariants, and a Proof of the Hanna Neumann Conjecture
Here
is a remarkable and much shorter
proof of the Hanna Neumann Conjecture,
which simplifies, generalizes, and clarifies aspects of
my sheaf theoretic proof,
by Warren Dicks, written entirely in the language of skew group rings.
remarkable simplifications, by Warren Dicks.
You may want to read Dicks' version before mine, although my long
article here also proves that the maximum excess is an L^2 Betti number.
 Postscript version.
 Dvi version.
 PDF version.

Abstract:
In this paper we establish some foundations regarding sheaves of vector
spaces on graphs and their invariants, such as homology groups and their
limits.
We then use these ideas to prove the Hanna Neumann
Conjecture of the 1950's; in fact,
we prove a strengthened form of the conjecture.
We introduce a notion of a sheaf of vector spaces on a graph,
and develop the foundations of homology theories for such sheaves.
One sheaf invariant,
its ``maximum excess,'' has a number of remarkable
properties.
It has a simple definition, with no reference to homology theory,
that resembles graph expansion.
Yet it is a ``limit'' of Betti numbers, and hence has a
short/long exact sequence theory and resembles the $L^2$ Betti numbers
of Atiyah.
Also, the maximum excess is defined via a
supermodular function, which gives the maximum excess
much stronger properties than
one has of a typical Betti number.
Our sheaf theory can be viewed as a vast generalization of algebraic graph
theory: each sheaf has invariants associated
to itsuch as Betti numbers and Laplacian matricesthat generalize
those in classical graph theory.
We shall use ``Galois graph theory'' to reduce the Strengthened
Hanna Neumann Conjecture to showing that certain sheaves, that we
call $\rho$kernels,
have zero maximum excess.
We use the symmetry in Galois theory to argue that if the
Strengthened Hanna Neumann Conjecture is false, then the maximum excess
of ``most of'' these $\rho$kernels must be large.
We then give an inductive argument to show that this is impossible.
