The end of the previous Research Log suggested a way towards showing that can be associated to an idempotent state
. Over night I thought of another way.
Using the Pierce decomposition with respect to (where
),
.
The corner is a hereditary
-subalgebra of
. This implies that if
and for
,
.
Let . We know from Fagnola and Pellicer that
and
.
By assumption in the background here we have an irreducible and periodic random walk driven by . This means that for all projections
, there exists
such that
.
Define:
.
Define:
.
The claim is that the support of ,
is equal to
.
We probably need to write down that:
.
Consider for any
. Note
that is each is supported on
. This means furthermore that
.
Suppose that the support . A question arises… is
? This follows from the fact that
and
is hereditary.
Consider a projection . We know that there exists a
such that
.
This implies that , say
(note
):
By assumption . Consider
.
For this to equal one every must equal one but
.
Therefore is the support of
.
Let . We have shown above that
for all
. This is an idempotent state such that
is its support (a similar argument to above shows this). Therefore
is a group like projection and so we denote it by
and
!
But is it a subgroup? But is it a normal subgroup? Can we construct a normal quasi-subgroup?
My feeling at the moment is that is probably not a subgroup. One idea I had was to look at an element
such that
, and see can I get anything from that.
A good idea would be to look for a counterexample. Perhaps dual groups might be a good place to look? After all, we can construct non-Haar idempotents for any non-abelian finite using
. Let
be non-normal and consider
. If it is a state, it is an idempotent state. We need to find a unitary representation
of
and a unit vector
such that:
Bekka, de la Harpe, and Vallette say this is possible (Exercise C. 6. 7).
Now… what does periodicity look like in . Is it even possible?
Suppose that are cyclic for
. Let
.
It is straightforward to show that for all :
.
Letting shows that all
are
-th roots of unity:
.
This means that is a characteristic function.
What we need to invoke now is irreducibility. Every subgroup induces a projection:
.
Every element is in some subgroup of
(say,
).
We are assuming that the random walk given by is irreducible. Therefore there exists a
such that
…
this seems promising but does not go anywhere.
My feeling is that periodicity is not compatible with irreducibility for random walks on . We will see can we show this on St Patrick’s Day! Perhaps assuming irreducibility will be faster (and showing it implies aperiodicity?)
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