Abstract Nonsense

Crushing one theorem at a time

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 \text{MinProj}\left(\mathcal{A}\left(G\right)\right) are either disjoint or equivalent. But, before we prove that actual theorem we need a few lemmas:

 

Lemma: Let p\in\text{MinProj}\left(\mathcal{A}\left(G\right)\right) then \Phi\left(p\right) is equal to a |G|-tuple \left(M_\alpha\right) where M_\alpha=\bold{0} for all but one \alpha and the one non-zero coordinate is a rank one projection.

Proof: From a previous theorem we know that p\in\Lambda^{(\alpha_0)} for some \alpha_0\in\widehat{G}. Then, by prior observation we know that \Phi\left(p^{(\alpha_0)}\right)\in\widetilde{\text{Mat}_{d_{\alpha_0}}\left(\mathbb{C}\right)} and it’s \alpha_0^{\text{th}} coordinate, call it P, satisfies P^2=P and thus by the characterization of projection endomorphism in terms if idempotence we may conclude that P is a projection. Suppose that P projects onto \mathscr{W}\leqslant \mathbb{C}^{d_{\alpha_0}} and \dim\mathscr{W}>1. Then, we may decompose \mathscr{W} into a direct sum \mathscr{X}\oplus\mathscr{Y} where \dim\mathscr{X},\dim\mathscr{Y}>0 and write P as P_\mathscr{X}+P_{\mathscr{Y}} where P_\mathscr{X},P_{\mathscr{Y}} are projections along \mathscr{X},\mathscr{Y} respectively. Thus, if we write \widetilde{P_\mathscr{X}},\widetilde{P_\mathscr{Y}} as the element of \text{Mat}_{d_{\alpha_0}}\left(\mathbb{C}\right) which has \alpha_0^{\text{th}} coordinate P_\mathscr{X} and P_\mathscr{Y} respectively then p=\Phi^{-1}\left(\widetilde{P_\mathscr{X}}\right)+\Phi^{-1}\left(\widetilde{P_\mathscr{Y}}\right). But,

 

\Phi^{-1}\left(\widetilde{P_\mathscr{X}}\right)^2=\Phi^{-1}\left(\widetilde{P_\mathscr{X}}^2\right)=\Phi^{-1}\left(\widetilde{P_\mathscr{X}^2}\right)=\Phi^{-1}\left(\widetilde{P_\mathscr{X}}\right)

 

and so \Phi^{-1}\left(\widetilde{P_\mathscr{X}}\right)\in\text{Proj}\left(\mathcal{A}\left(G\right)\right) and is non-zero, and similarly \Phi^{-1}\left(\widetilde{P_\mathscr{Y}}\right) is a non-zero projection. But, this contradicts that p\in\text{MinProj}\left(\mathcal{A}(G)\right). It follows that since p is non-zero that \dim\mathscr{W}=1 and so P is a rank one projection as claimed. \blacksquare

 

 

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 \displaystyle \bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right) is similar to the definition in \mathcal{A}(G)). Put more formally:

 

 

Lemma: Let p,q\in\text{Proj}\left(\mathcal{A}(G)\right) then p\simeq q if and only if there exists invertible A,B in \displaystyle \bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)  such that A\Phi(p)B=\phi(q) for. Similarly, p\eqsim q if and only if \Phi(p)C\Phi(q)=\bold{0} for every C in \displaystyle \bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right).

Proof: We note that if p\simeq q then there exists u,v\in\mathcal{A}\left(G\right)^\times such that uqv=q and so \Phi(u)\Phi(p)\Phi(v)=\Phi(q) and since \Phi is an algebra isomorphism we know that \Phi(u),\Phi(v) are invertible from where the if part of the first statement follows. Conversely, suppose that A\Phi(p)B=\Phi(q) for some invertible A,B in \displaystyle \bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right) then since \Phi is an algebra isomorphism there exists invertible u,v\in\mathcal{A}\left(G\right)^\times such that A=\Phi(u) and B=\Phi(v) and so \Phi(q)=\Phi(u)\Phi(p)\Phi(v)=\Phi(upv) and thus q=upv so that p\simeq q as desired. The first statement follows.

 

To prove the second statement note that p\eqsim q if and only if puq=\bold{0} for every u\in\mathcal{A}(G) which, since \Phi is an algebra isomorphism, is true if and only if \Phi(p)\Phi(u)\Phi(q)=\bold{0} or every \Phi(u)\in \Phi\left(\mathcal{A}\left(G\right)\right) and since \Phi is surjective the conclusion follows. \blacksquare

 

 

Theorem: Let p,q\in\text{MinProj}\left(\mathcal{A}(G)\right), then either p\simeq q or p\eqsim q.

Proof: We know that since p,q are minimal projections that p\in\Lambda^{(\alpha_0)} and q\in\Lambda^{(\beta_0)} for some \alpha_0,\beta_0\in\widehat{G}. We claim that p\simeq q if \alpha_0=\beta_0 and p\eqsim q if \alpha_0\ne\beta_0. Indeed, by our second lemma we know that \Phi(p) is equal to \widetilde{P_\mathscr{X}}\in\widetilde{\text{Mat}_{d_{\alpha_0}}\left(\mathbb{C}\right)} where P_\mathscr{X} is some rank one projection onto \mathscr{X}  and similarly \Phi(q)=\widetilde{P_\mathscr{Y}}\in\widetilde{\text{Mat}_{d_{\beta_0}}\left(\mathbb{C}\right)}  where P_\mathscr{Y} is some rank one projection onto \mathscr{Y}. If \alpha_0\ne\beta_0 then clearly for any C in \displaystyle \bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right) one has that  \widetilde{P_\mathscr{X}}C\widetilde{P_\mathscr{Y}}=\bold{0} namely since \widetilde{\text{Mat}_{d_{\beta_0}}\left(\mathbb{C}\right)}  is a two-sided ideal we have that C\widetilde{P_\mathscr{Y}}\in\widetilde{\text{Mat}_{d_{\beta_0}}\left(\mathbb{C}\right)} and so clearly \widetilde{P_\mathscr{X}}C\widetilde{P_\mathscr{Y}}=\bold{0}. Thus, we may conclude by our second lemma that p\eqsim q.

 

If \alpha_0=\beta_0 we take the isomorphism T:\mathbb{C}^{d_{\alpha_0}}\to\mathbb{C}^{d_{\alpha_0}} which takes \mathscr{X}\to\mathscr{Y} isomorphically and is the identity on \mathbb{C}^{d_{\alpha_0}}-\mathscr{X} then evidently if L=(T_\alpha) where T_\alpha=T if \alpha=\alpha_0 and \mathbf{1}_\alpha otherwise then \widetilde{P_\mathscr{X}}=L^{-1}\widetilde{P_\mathscr{Y}}L from where it follows by our second lemma that p\simeq q. \blacksquare

 

 

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 \mathcal{A}(G) are…”. It turns out though that what shall be useful to us is knowing the minimal central projections, recall these are the members of \text{MinProj}\left(\text{Cl}(G)\right). Amazingly though, we are able to classify these (how convenient)! Namely:

 

Theorem: Let p\in\text{MinProj}\left(\text{Cl}\left(G\right)\right) then there exists some \alpha\in\widehat{G} such that \displaystyle p=\frac{d_\alpha}{|G|}\chi^{(\alpha)}.

Proof: Using our convolution formula we know that each of these functions is a projection and since they clearly (being a basis for \text{Cl}(G)) can’t be written as the sum of class functions (let alone elements of \text{Proj}\left(\text{Cl}(G)\right)) they must be elements of \text{MinProj}\left(\text{Cl}\left(G\right)\right).

 

Conversely, if p\in\text{MinProj}\left(\text{Cl}(G)\right) then we know (since it itself is a class function) may be rewritten

 

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

 

And, since itself is a projection we have by prior theorem that each m^{(\alpha)}\chi^{(\alpha)} is a projection and a class function we may conclude that m^{(\alpha)}=0 for all but one \alpha, call it \alpha_0. Thus, p=m^{(\alpha_0)}\chi^{(\alpha_0)} but a quick check shows that since \displaystyle m^{(\alpha_0)}\chi^{(\alpha_0)}=\frac{d_{\alpha_0}}{|G|}m^{(\alpha_0)}{}^2\chi^{(\alpha_0)} we may conclude that \displaystyle m^{(\alpha_0)}=\frac{d_{\alpha_0}}{|G|} as desired. \blacksquare

 

Remark: An alternate proof is to prove that if f is a minimal central projection then \Phi(f) is of the form \left(\delta_{\alpha,\alpha_0}\mathbf{1}_{\alpha_0}\right) for some \alpha_0\in\widehat{G} 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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April 9, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

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