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

**Point of post: **In this post we discuss the notion of the group algebra, in preparation for our eventual discussion about the representation of symmetric groups.

*Motivation*

We’ve seen in past posts that the group algebra is isomorphic in all the important ways to the direct sum of matrix algebras. We’ll use this fact to study projections in the group algebra which are functions generalizing the notion of projections on an endomorphism algebra. Namely, projections are elements of the group algebra which are idempotent under convolution. These shall prove to be very important when we attempt, at a later date, to classify the representations of the symmetric group.

*Projections Into the Group Algebra*

* *

Let be a finite group and be an element of the group algebra we call (mimicking somewhat the notion of projections in the endomorphism algebra) a *projection *if is idempotent–i.e. that . For a subalgebra (or more generally a subset) of we denote the set of all projections in by . We call an element *minimal on *if whenever where the either or . We denote the set of all minimal projections on by . We call a projection which is also a class function, in other words a *central *projection and an element of a *minimal central *projection.

We can define an equivalence relation , verbalized “equivalent”, on by saying if there exists (where the notation means is invertible in the algebra) such that . We call two elements of *disjoint*, denoted (to accentuate the dual notions of equivalence and disjointness), if for every one has that . We claim that is a symmetric relation– it may not be. But we shall prove that any two elements of are either equivalent or disjoint. That said, before we prove this we prove two very interesting, on their own.

**Theorem:** *Let have the decomposition, in terms of the *

* *

* *

* *

*Then, if and only if for each .*

**Proof: **We recall that

So, suppose then that for each , then evidently we see from that so that . Conversely, we know that since is an algebra that for each . So, assume that , then it’s clear from that

but since is equal to this implies that or for every .

And,

**Theorem: ***Let with decomposition with respect to equal to . Then, if and only if for all but at most one and for each .*

**Proof: **Suppose first that and . Then, we must have that for some . Clearly then for every by the definition of minimality, noting that by the previous problem every . Now, we must clearly then have that since if where then and so by assumption of ‘s minimality we may conclude that either or as desired.

The converse is clearly true.

**References:**

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

[…] Point of Post: This post is a continuation of this one. […]

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