Twists
Motivation
A previous REU project demonstrated that taking twisted contractive
similarities instead of the standard ones in the definition of the
3-gasket can greatly simplify the problem of determining explicit
solutions to the renormalization problem. Alex and I attempted to work
with a partial twist on SG4 with the reducibility problem, but found
solutions which were close to trivial, and possibly non-extensible.
What works for twists on SG3 is the uneven degree of the renormalization
eigenequations. In each eigenequation, there was one variable which was
linear in that equation, permitting the system to be solved
algebraically. We were hoping for a similar stroke of luck with
variable degrees in higher dimensions.
Group Theory
At its core, twisting is an exercise in the theory of binary operators
on finite sets. We have a set of n vertices which must be twisted. A
twist is characterized by the property that for each pair of boundary
vertices a and b there exists a unique vertex c
such that Fa(c)=Fb(c). This can be characterized in terms of the
binary twist operator f:VxV->V which takes a vertex pair
(a,b) and returns a vertex c as before. There are three
properties that f must satisfy:
f(a,a)=a (fixity)
f(a,b)=f(b,a) (commutativity)
and f(a,b)=f(a,c) => b=c (uniqueness)
These properties imply that the operator has a unique fixed point, and
we believe it can be shown that it is an order 2 operator, implying that
the twist map can only be applied to odd vertex sets. We also conjecture
a fourth condition:
f(f(a,b),f(c,d))=f(f(a,c),f(b,d))
This condition seems to amount to invariance of the twist under
permutation of indices.
If we assume an abelian group structure on the vertices, then f becomes a
homomorphism from VxV->V, although it is non-associative so doesn't
determine a ring structure on the vertices. Under this assumption it is
easy to see that |G| must be odd. If it weren't odd, then there would
have to be an element of order 2, say a, so f(a,a)=a and f(0,a)=f(a,0)=b
for some b equal to neither a nor 0. This gives 2b=a so 4b=0 and b is
of order 4. We continue as above to get the existence of an element of
order 2^k for all k, contradicting the finiteness of G.
It can be easily seen that f is surjective, and that its kernel consists
of the vertex pairs between which the vertex 0 is mapped, so cosets of
the kernel give vertex pairs between which the same point is mapped.
We can give an explicit map for f on Zn: f maps (a,b)->k (a+b) mod n
where n=2k-1. It is easy to verify the above properties for this map.
Given any abelian group structure on the vertices we can write it as a
direct product of Zi for various i and use this homomorphism on each
part to give a twist homomorphism for the entire vertex set. Thus,
there is a distinct (up to re-indexing) twist homomorphism for each
factorization of n.
Computation
To test the hypothesis that twists lead to a certain imbalance in
degrees of variables in the eigen-equations, we stripped down our
harmonic extension matrix to have the minimum amount of information from
which some variable degree can be recovered. This meant substituting
large-ish primes (starting at 29 in our case) for all variables in the
harmonic extension matrix except for the xis and c01. I
then computed the harmonic extension and plugged it into the
appropriately twisted level 1 energy form and found the new conductances
by looking at coefficients of cross-terms and dividing by 2.
Unfortunately this approach did not give the desired result for
n=5, 7, and 9 (2 types of twists on 9), and we were
back to the drawing board.
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