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

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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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