Abstract Nonsense

Crushing one theorem at a time

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.


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 G be a finite group and f be an element of the group algebra \mathcal{A}\left(G\right) we call (mimicking somewhat the notion of projections in the endomorphism algebra) f a projection if f is idempotent–i.e. that f^2=f. For a subalgebra \mathscr{A} (or more generally a subset) of \mathcal{A}\left(G\right) we denote the set of all projections in \mathscr{A} by \text{Proj}\left(\mathscr{A}\right). We call an element f\in\text{Proj}\left(\mathscr{A}\right) minimal on \mathscr{A} if whenever f=p+q where p,q\in\text{Proj}\left(\mathscr{A}\right) the either p=\bold{0} or q=\bold{0}. We denote the set of all minimal projections on \mathscr{A} by \text{MinProj}\left(\mathscr{A}\right). We call a projection which is also a class function, in other words  f\in\text{Proj}\left(\mathcal{Z}\left(\mathcal{A}(G)\right)\right)=\text{Proj}\left(\text{Cl}\left(G\right)\right) a central projection and an element of \text{MinProj}\left(\text{Cl}\left(G\right)\right) a minimal central projection.

We can define an equivalence relation \simeq, verbalized “equivalent”, on \text{Proj}\left(\mathcal{A}(G)\right)  by saying p\simeq q if there exists u,v\in\text{Proj}\left(\mathcal{A}(G)\right)^\times (where the notation ^\times means is invertible in the algebra) such that p=uqv. We call two elements of p,q\in \text{Proj}\left(\mathcal{A}\left(G\right)\right) disjoint, denoted \eqsim (to accentuate the dual notions of equivalence and disjointness), if for every f\in\text{Proj}\left(\mathcal{A}(G)\right) one has that pfq=\bold{0}. We claim that \eqsim is a symmetric relation– it may not be. But we shall prove that any two elements of \text{MinProj}\left(\mathcal{A}(G)\right) are either equivalent or disjoint. That said, before we prove this we prove two very interesting, on their own.

Theorem: Let f\in\mathcal{A}(G) have the decomposition, in terms of the \Lambda^{(\alpha)}

\text{ }

\displaystyle \sum_{\alpha\in\widehat{G}}f^{(\alpha)}

\text{ }

Then, f\in\text{Proj}\left(\mathcal{A}(G)\right) if and only if f^{(\alpha)}\in\text{Proj}\left(\mathcal{A}(G)\right) for each \alpha\in\widehat{G}.

Proof: We recall that


\displaystyle f^2=\sum_{\alpha\in\widehat{G}}f^{(\alpha)}{}^2\quad\mathbf{(1)}


So, suppose then that f^{(\alpha)}{}^2=f^{(\alpha)} for each \alpha\in\widehat{G}, then evidently we see from \mathbf{(1)} that f^2=f so that f\in\text{Proj}\left(\mathcal{A}(G)\right). Conversely, we know that since \Lambda^{(\alpha)} is an algebra that f^{(\alpha)}{}^2\in\Lambda^{(\alpha)} for each \alpha\in\widehat{G}. So, assume that f^2=f, then it’s clear from \mathbf{(1)} that

\text{ }

\displaystyle \sum_{\alpha\in\widehat{G}}f^{(\alpha)}=\sum_{\alpha\in\widehat{G}}f^{(\alpha)}{}^2


but since \displaystyle \mathcal{A}(G) is equal to \displaystyle \bigoplus_{\alpha\in\widehat{G}}\Lambda^{(\alpha)} this implies that f^{(\alpha)}=f^{(\alpha)}{}^2 or f^{(\alpha)}\in\text{Proj}\left(\mathcal{A}(G)\right) for every \alpha\in\widehat{G}. \blacksquare

\text{ }


\text{ }

Theorem: Let f\in\text{Proj}\left(\mathcal{A}(G)\right) with decomposition with respect to \displaystyle \bigoplus_{\alpha\in\widehat{G}}\Lambda^{(\alpha)} equal to \displaystyle f=\sum_{\alpha\in\widehat{G}}f^{(\alpha)}. Then, f\in\text{MinProj}\left(\mathcal{A}(G)\right) if and only if f^{(\alpha)}=\bold{0} for all but at most one \alpha and f^{(\alpha)}\in\text{MinProj}\left(\Lambda^{(\alpha)}\right) for each \alpha\in\widehat{G}.

Proof: Suppose first that f\in\text{MinProj}\left(\mathcal{A}(G)\right) and f\ne\bold{0}. Then, we must have that f^{(\alpha_0)}\ne\bold{0} for some \alpha_0\in\widehat{G}. Clearly then f^{(\alpha)}=\bold{0} for every \alpha\ne\alpha_0 by the definition of minimality, noting that by the previous problem every f^{(\alpha)}\in\text{Proj}\left(\mathcal{A}(G)\right). Now, we must clearly then have that f^{(\alpha_0)}\in\text{MinProj}\left(\Lambda^{(\alpha_0)}\right) since if f^{(\alpha_0)}=p^{(\alpha_0)}+q^{(\alpha_0)} where p^{(\alpha_0)},q^{(\alpha_0)}\in\text{Proj}\left(\Lambda^{(\alpha_0)}\right)\subseteq\text{Proj}\left(\mathcal{A}(G)\right) then f=p^{(\alpha_0)}+q^{(\alpha_0)} and so by assumption of f‘s minimality we may conclude that either p^{(\alpha_0)}=\bold{0} or q^{(\alpha_0)}=\bold{0} as desired.

\text{ }

The converse is clearly true. \blacksquare


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


April 9, 2011 - Posted by | Algebra, Representation Theory | , , , , , , , , ,


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

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