Abstract Nonsense

Crushing one theorem at a time

Representation Theory: A Way of Creating C-representations Satisfying the Real Condition With No (rho,J)-invariant Subspaces (Pt. II)

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


Conversely, let \rho:G\to\mathscr{U}\left(\mathscr{V}\right) be a complex or quaternionic irreducible \mathbb{C}-representation and suppose that \{\bold{0}\}<\mathscr{W}<\mathscr{V}\oplus\mathscr{V} is \left(\rho\oplus\text{Conj}_\rho^J,\widetilde{J}\right)-invariant. Let then \mathscr{W}_2=\mathscr{W}\cap\left(\{\bold{0}\}\times\mathscr{V}\right) if \pi_2:\mathscr{W}_2\to\mathscr{V}:(0,w)\mapsto w we claim that \pi_2\left(\mathscr{W}_2\right) is \text{Conj}_\rho^J-invariant. Indeed, let g\in G and w\in\pi_2\left(\mathscr{W}_2\right) be arbitrary, then (0,w)\in\mathscr{W}_2 and so by assumption, since \mathscr{W} is \rho\oplus\text{Conj}_\rho^J-invariant, we have that \left(\rho(g)\oplus\text{Conj}_\rho^J(g)\right)(0,w)\in\mathscr{W} but evidently it’s also a member of \{\bold{0}\}\times\mathscr{V} since \left(\rho\oplus\text{Conj}_\rho^J\right)(0,w)=\left(0,\left(\text{Conj}_\rho^J(g)\right)(w)\right). It then follows that \left(0,\left(\text{Conj}_\rho^J(g)\right)(w)\right)\in\mathscr{W}_2 and thus \left(\text{Conj}_\rho^J(g)\right)(w)\in\pi_2\left(\mathscr{W}_2\right) as desired. It follows then by assumption (since \rho is an irrep and thus \text{Conj}_\rho^J is an irrep) that \pi_2\left(\mathscr{W}_2\right)=\{\bold{0}\} or \pi_2\left(\mathscr{W}_2\right)=\mathscr{V}–or equivalently \mathscr{W}_2=\{(0,0)\} or \mathscr{W}_2=\{\bold{0}\}\times\mathscr{V}. Suppose that the latter is true, then let v\in\mathscr{V} be arbitrary, then by assumption \left(0,J(v)\right)\in\{\bold{0}\}\times\mathscr{V}=\mathscr{W}_2\subseteq\mathscr{W} and since \mathscr{W} is \widetilde{J}-invariant we have that \widetilde{J}\left(0,J(v)\right)=\left(J(J(v)),J(0)\right)=(v,0)\in\mathscr{W}. Thus, since v\in\mathscr{V} was arbitrary we have that \mathscr{V}\times\{\bold{0}\}\subseteq\mathscr{W}. Thus, we must have (since both of the following are contained in the subspace \mathscr{W}) that \mathscr{V}\times\{\bold{0}\}+\{\bold{0}\}\times\mathscr{V}=\mathscr{V}\oplus\mathscr{V}\subseteq\mathscr{W} so that \mathscr{W}=\mathscr{V}\oplus\mathscr{V} contradictory to assumption. Thus, we may assume that \mathscr{W}_2=\{(0,0)\}. Using the exact same methodology we can show that \left\{w:(w,0)\in\mathscr{W}\right\} is a \rho-invariant subspace and thus by the same reasoning we must have that \mathscr{W}\cap\left(\mathscr{V}\times\{\bold{0}\}\right) is either \{(0,0)\} or \mathscr{V}\times\{\bold{0}\}. We clearly must have though that it is the latter.



Thus, for every v\in\mathscr{V} we know that there exists some w\in\mathscr{V} such that (v,w)\in\mathscr{W}. Moreover, we know that w\in\mathscr{V} is the only element of \mathscr{V} which appears as a second coordinate after v. Indeed, if (v,u)\in\mathscr{W} then (v,w)-(v,u)=(0,w-u)\in\mathscr{W} but by the first part of the above paragraph we know that w-u=\bold{0} or w=u. Thus, we can unambiguously define the map A:\mathscr{V}\to\mathscr{V} by A(v) is the unique element of \mathscr{V} such that \left(v,A(v)\right)\in\mathscr{W}. Note that evidently A is linear since (\alpha v,\alpha A(v))+(\beta w,\beta A(w))=\left(\alpha v+\beta w,\alpha A(v)+\beta A(w)\right) and so by definition A\left(\alpha v+\beta w\right)=\alpha A(v)+\beta A(w). Note though that \ker A is \rho-invariant since if v\in\ker A then (v,0)\in\mathscr{V} and so \left(\rho_g\oplus\text{Conj}_\rho^J(g)\right)\left(v,0\right)=\left(\rho_g(v),0\right)\in\mathscr{W} so that A\left(\rho_g(v)\right)=0 or \rho_g(v)\in\ker A. But, evidently \ker A\ne\mathscr{V} otherwise \mathscr{W}=\left\{(v,A(v)):v\in\mathscr{V}\right\}=\left\{(v,0):v\in\mathscr{V}\right\} but a quick check shows then that \mathscr{W} is not \widetilde{J}-invariant in this case. Thus, \ker A=\{\bold{0}\} and thus \text{im} A=\mathscr{V}. Note though that for any g\in G and v\in\mathscr{V}




so that  (since \mathscr{W} is \rho\oplus\text{Conj}_\rho^J-invariant) A\rho_g=J\rho_g(v)JA or A\rho_gA^{-1}=J\rho_g J for every g\in G–and by an earlier theorem we may conclude that \rho is self-conjugate. Thus, there exists some unitary U\in\mathcal{U}\left(\mathscr{V}\right) such that UJ\rho_g JU^{-1}=\rho_g for every g\in G. Thus, we have that


J\rho_g J=A\rho_gA^{-1}=AUJ\rho_gJU^{-1}A^{-1}


for every g\in G and thus (since \rho and so \text{Conj}_\rho^J is an irrep) by Schur’s lemma we may conclude that AU=\alpha\mathbf{1} and so A=\alpha U^{-1}. Note though that since \mathscr{W} is \widetilde{J}-invariant one has that for every v\in\mathscr{V}




so that AJA=J. Or using the fact that A=\alpha U^{-1} we have that (\alpha U^{-1})J(\alpha U^{-1})=|\alpha|^2 U^{-1}JU^{-1}=J so that |\alpha|=1 and so A itself is unitary. Finally, let K=JA=A^{-1}J then K is antilinear, antiunitary, and K^2=KK=JAA^{1}J=J^2=\mathbf{1} and so K is a complex conjugate. Note though that


K\rho_g K=A^{-1}J\rho_g J=A^{-1}\left(A\rho_g A^{-1}\right)A=\rho_g


for every g\in G. Thus, by one of our previous characterizations of real irreps we may conclude that \rho is real–contradictory to assumption. The conclusion follows. \blacksquare




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



April 2, 2011 - Posted by | Uncategorized

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