## Projections Into the Group Algebra (Pt. II)

**Point of Post: **This post is a continuation of this one.

We now return to our goal of showing that any two elements of are either disjoint or equivalent. But, before we prove that actual theorem we need a few lemmas:

**Lemma: ***Let then is equal to a -tuple where for all but one and the one non-zero coordinate is a rank one projection.*

**Proof: **From a previous theorem we know that for some . Then, by prior observation we know that and it’s coordinate, call it , satisfies and thus by the characterization of projection endomorphism in terms if idempotence we may conclude that is a projection. Suppose that projects onto and . Then, we may decompose into a direct sum where and write as where are projections along respectively. Thus, if we write as the element of which has coordinate and respectively then . But,

and so and is non-zero, and similarly is a non-zero projection. But, this contradicts that . It follows that since is non-zero that and so is a rank one projection as claimed.

Our next lemma claims that checking that projections are disjoint or equivalent is equivalent to checking that it’s image is equivalent or disjoint (where of course the notion of equivalence and disjointness in is similar to the definition in ). Put more formally:

**Lemma: ***Let then if and only if there exists invertible in such that for. Similarly, if and only if for every in .*

**Proof: **We note that if then there exists such that and so and since is an algebra isomorphism we know that are invertible from where the if part of the first statement follows. Conversely, suppose that for some invertible in then since is an algebra isomorphism there exists invertible such that and and so and thus so that as desired. The first statement follows.

To prove the second statement note that if and only if for every which, since is an algebra isomorphism, is true if and only if or every and since is surjective the conclusion follows.

**Theorem: ***Let , then either or .*

**Proof: **We know that since are minimal projections that and for some . We claim that if and if . Indeed, by our second lemma we know that is equal to where is some rank one projection onto and similarly where is some rank one projection onto . If then clearly for any in one has that namely since is a two-sided ideal we have that and so clearly . Thus, we may conclude by our second lemma that .

If we take the isomorphism which takes isomorphically and is the identity on then evidently if where if and otherwise then from where it follows by our second lemma that .

We are now prepared to prove what shall become most useful to us in the future. Namely, even with all of this effort we have not been able to classify minimal projections in a nice closed form way in the sense that we haven’t been able to say “the minimal projections of are…”. It turns out though that what shall be useful to us is knowing the minimal central projections, recall these are the members of . Amazingly though, we are able to classify these (how convenient)! Namely:

**Theorem: ***Let then there exists some such that .*

**Proof: **Using our convolution formula we know that each of these functions is a projection and since they clearly (being a basis for ) can’t be written as the sum of class functions (let alone elements of ) they must be elements of .

Conversely, if then we know (since it itself is a class function) may be rewritten

And, since itself is a projection we have by prior theorem that each is a projection and a class function we may conclude that for all but one , call it . Thus, but a quick check shows that since we may conclude that as desired.

*Remark: *An alternate proof is to prove that if is a minimal central projection then is of the form for some and check that the preimage of this is precisely what we claimed.

**References:**

1. Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Math. Soc., 1996. Print.

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