Abstract Nonsense

Crushing one theorem at a time

A Bijection Between A Subset of the Complex Reps of a Finite Group and the Real Reps (Pt. I)

Point of post: In this post we describe a certain subset of the set of all \mathbb{C}-representations of a finite group G and show that this subset is in bijective correspondence with the set of all \mathbb{R}-representations of G. Moreover, we shall show that this bijection naturally restricts to a subset of the set of all \mathbb{C}-reps of G to the the \mathbb{R}-irreps of G.


In our last post we discussed the notion of \mathbb{R}representations for a finite group G. Naturally our first desire would be to see if we could, in some way, connect \mathbb{R}representations of G to the \mathbb{C}-representations  which has held the center of our attention for so long. We begin this process in this post by showing that there is a natural place for which these \mathbb{R}-representations occur. Namely, we shall see that every \mathbb{C}-representation \rho:G\to\mathcal{U}\left(\mathscr{V}\right) for which there is a complex conjugate J for which J\rho(g)J=\rho(g) for every g\in G naturally admits a \mathbb{R}-representation \rho_{\Re}. We shall show then that in fact the reverse is true–namely that for every \mathbb{R}-representation \psi there is a natural way to produce a \mathbb{C}-representation \rho such that \rho_{\Re}=\psi. Moreover, we’ll show that if \rho:G\to\mathcal{U}\left(\mathscr{V}\right) is a \mathbb{C}-representation satisfying the aforementioned conditions and the added condition that there does not exist \mathscr{W}\leqslant \mathscr{V} such that J\left(\mathscr{W}\right)=\mathscr{W} and \mathscr{W} is \rho-invariant then we shall see that \rho_{\Re} is an irreducible \mathbb{R}-representation.

A Characterization of \mathbb{R}– Representations

Our first goal is to show that a certain class of \mathbb{C}-representations naturally admits a \mathbb{R}representation. Indeed, say that a \mathbb{C}-representation \rho:G\to\mathcal{U}\left(\mathscr{V}\right) satisfies the real condition if there exists a complex conjugate J:\mathscr{V}\to\mathscr{V} such that J\rho(g)J=\rho(g) for every g\in G. We call such a J the ‘realizer’ of \rho. With this in mind:

Theorem: Let G be a finite group and \rho:G\to\mathcal{U}\left(\mathscr{V}\right) be a \mathbb{C}-representation satisfying the real condition with realizer J. Then, the space \mathscr{V}_\Re\overset{\text{def.}}{=}\left\{v\in \mathscr{V}:J(v)=v\right\} is a real vector space with \dim_{\mathbb{R}}\mathscr{V}_\Re=\dim_\mathbb{C}\mathscr{V} and the same inner product as \mathscr{V}. Moreover, the map \rho_\Re:G\to\mathcal{U}\left(\mathscr{V}_\Re\right) given by g\mapsto \rho(g)\mid_{\mathscr{V}_\Re} is a \mathbb{R}-representation of G. Further more, \rho_\Re is irreducible if and only if there does not exist a \rho-invariant \{\bold{0}\}<\mathscr{W}<\mathscr{V} for which J\left(\mathscr{W}\right)=\mathscr{W}.

Proof: We first show that \mathscr{V}_\Re is a real inner product space of real dimension n=\dim_\mathbb{C}\mathscr{V}. The fact that \mathscr{V}_\Re is a real vector space with the same addition as \mathscr{V} and scalar multiplication restricted to \mathbb{R} is clear, thus it remains to prove then that with this definition of operations \dim_\mathbb{R}\mathscr{V}_\Re=n. To see this recall from our past characterizations of complex conjugates that \mathscr{V} admits a basis \left\{x_1,\cdots,x_n\right\} such that J(v_k)=v_k for k=1,\cdots,n. We claim then that \{x_1,\cdots,x_n\} is similarly a basis for \mathscr{V}_\Re. Indeed, the fact that \{x_1,\cdots,x_n\} is linearly independent is evident and thus it suffices to prove that \text{span}_\mathbb{R}\{x_1,\cdots,x_n\}=\mathscr{V}_\Re. To see this let v\in\mathscr{V}_\Re be arbitrary. Then, since \{x_1,\cdots,x_n\} is a basis for \mathscr{V} there exists \zeta_1,\cdots,\zeta_n\in\mathbb{C} such that \displaystyle \sum_{k=1}^{n}\zeta_k x_k=v. Note though that by assumption


\displaystyle \sum_{k=1}^{n}\zeta_k v_k=v=J(v)=\sum_{k=1}^{n}\overline{\zeta_k}x_k


and thus \zeta_k=\overline{\zeta_k} and thus \zeta_k\in\mathbb{R} it clearly follows then that v\in\text{span}_\mathbb{R}\{x_1,\cdots,x_n\}. Since v was arbitrary it follows that \{x_1,\cdots,x_n\} is a real basis for \mathscr{V}_\Re and thus \dim_\mathbb{R}\mathscr{V}_\Re=n as desired. Thus, it remains to show that the restriction of the inner product \langle\cdot,\cdot\rangle to \mathscr{V}_\Re\times\mathscr{V}_\Re is really a real inner product. But, all the axioms are clear except for possibly that \langle\cdot,\cdot\rangle maps \mathscr{V}_\Re\times\mathscr{V}_\Re into \mathbb{R} (there’s not ostensible reason this must be true). To see this merely note that for any u,v\in\mathscr{V}_\Re one has that


\displaystyle \left\langle u,v\right\rangle=\left\langle J(u),J(v)\right\rangle=\left\langle v,u\right\rangle=\overline{\left\langle u,v\right\rangle}


from where the result follows.

Now, we prove that \rho_\Re as defined in the theorem statement is really a \mathbb{R}-representation. The first thing to show is that for every g\in G one has that \rho(g)\mid_{\mathscr{V}_\Re} really is an element of \mathcal{U}\left(\mathscr{V}_\Re\right). The first, but crucial step in this procedure is to show that, while silly, \rho_g\left(\mathscr{V}_\Re\right)\subseteq\mathscr{V}_\Re since this isn’t absolutely clear. But, this is where the real condition comes into play. Namely, since J^2=\mathbf{1} we may reinterpret the real condition as saying J\rho(g)=\rho(g)J for each g\in G. Thus, in particular for any g\in G and any v\in\mathscr{V}_\Re one has that J(\rho_g(v))=\rho_g(J(v))=\rho_g(v) so that \rho_g(v)\in\mathscr{V}_\Re from where the fact that \rho_g\left(\mathscr{V}_\Re\right)\subseteq\mathscr{V}_\Re follows. Now, the fact that \rho_g is a \mathbb{R}-homomorphism and unitary clearly follow from definition. Thus, to prove that \rho_\Re is a \mathbb{R}-representation it merely suffices to prove that it is a homomorphism, but this is clear since \rho is a representation. Thus, it follows that \rho_\Re is a homomorphism G\to\mathcal{U}\left(\mathscr{V}_\Re\right) from where the fact that it is a \mathbb{R}-representation of G clearly follows.

Finally we prove that \rho_\Re is irreducible if and only if \mathscr{V} doesn’t admit any non-trivial proper subspaces \mathscr{W} for which J\left(\mathscr{W}\right)=\mathscr{W}. Indeed, suppose that \rho_\Re is reducible so that there exists \{\bold{0}\}<\mathscr{W}<\mathscr{V}_\Re such that \mathscr{W} is \rho_\Re invariant. Define then \mathscr{W}_\Im=\text{span}_{\mathbb{C}}\mathscr{W}. Clearly by definition one has that \mathscr{W}_\Im\leqslant\mathscr{V}. Moreover, we claim that J\left(\mathscr{W}_\Im\right)=\mathscr{W}_\Im. Indeed, let u\in\mathscr{W}_\Im then \displaystyle u=\sum_r \alpha_r w_r where w_r\in\mathscr{W} and so


\displaystyle J(u)=\sum_r \overline{\alpha_r}J(w_r)=\sum_r \overline{\alpha_r}w_r\in\mathscr{W}_\Im


Conversely, if \displaystyle \sum_r\alpha_r w_r\in\mathscr{W}_\Im where w_r\in\mathscr{W} then \displaystyle \sum_{r}\overline{\alpha_r}w_r\in\mathscr{W}_\Im and evidently


\displaystyle J\left(\sum_r\overline{\alpha_r}w_r\right)=\sum_r \alpha_r J(w_r)=\sum_r \alpha_r w_r


and so \displaystyle \sum_r\alpha_r w_r\in J\left(\mathscr{W}_\Im\right) from where the fact that J\left(\mathscr{W}_\Im\right)=\mathscr{W}_\Im follows. We now claim that \mathscr{W}_\Im is \rho-invariant. Indeed, let g\in G and \displaystyle \sum_{r}\alpha_r w_r\in\mathscr{W}_\Im be arbitrary (where w_r\in\mathscr{W}). Then, one has that


\displaystyle \left(\rho(g)\right)\left(\sum_r \alpha_r w_r\right) =\sum_r \alpha_r \left(\rho(g)\right)\left(w_r\right)= \sum_r \alpha_r \left(\rho_\Re(g)\right)(w_r)


but by assumption one has that \left(\rho_\Re(g)\right)(w_r)\in\mathscr{W} so that we may conclude that \displaystyle \left(\rho(g)\right)\left(\sum_r \alpha_r w_r\right)\in\text{span}_\mathbb{C}\mathscr{W}=\mathscr{W}_\Im. Since g\in G and \displaystyle \sum_r \alpha_r w_r\in\mathscr{W}_\Im was arbitrary we may conclude that \mathscr{W}_\Im is \rho-invariant as claimed. Lastly, noting that since \{\bold{0}\}<\mathscr{W}<\mathscr{V}_\Re we clearly have that \text{span}_\mathbb{C}\{\bold{0}\}<\mathscr{W}_\Im<\text{span}_\mathbb{C}\mathscr{V}_\Re which (recalling that \mathscr{V}_\Re contains a complex basis for \mathscr{V}) implies that \{\bold{0}\}<\mathscr{W}_\Im<\mathscr{V} from where it follows that \mathscr{W}_\Im is a \rho-invariant, non-trivial proper subspace of \mathscr{V} for which J\left(\mathscr{W}_\Im\right)=\mathscr{W}_\Im. It follows then that if \mathscr{V} does not admit such a subspace then \rho_\Re must be irreducible.


Conversely, suppose that \{\bold{0}\}<\mathscr{W}<\mathscr{V} is \rho-invariant and J\left(\mathscr{W}\right)=\mathscr{W}. Define then \mathscr{W}_\Re=\left\{w\in\mathscr{W}:J(w)=w\right\}. Evidently, it’s true then that \{\bold{0}\}<\mathscr{W}_\Re<\mathscr{V}_\Re. We claim that \mathscr{W}_\Re is \rho_\Re invariant. Indeed, let g\in G and g\in G and w\in\mathscr{W}_\Re then \left(\rho_\Re(g)\right)(w)=\left(\rho(g)\right)(w)\in \mathscr{W} and moreover J\left(\rho(g)\right)(w)=\left(\rho(g)\right)\left(J(w)\right)=\left(\rho(g)\right)(w) so that in fact \left(\rho(g)\right)(w)\in\mathscr{W}_\Re. Since g\in G and w\in\mathscr{W}_\Re were arbitrary we have that \mathscr{W}_\Re is \rho_\Re-invariant and thus \rho_\Re is reducible. Thus, if \rho_\Re is irreducible then \mathscr{V} cannot admit a non-trivial proper subspace which is invariant under both \rho and J.

Putting this all together the conclusion follows. \blacksquare


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


March 29, 2011 - Posted by | Algebra, Representation Theory | , , , ,


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