Abstract Nonsense

Crushing one theorem at a time

Complex Conjugate Representation (A Characterization of Self-Conjugate Maps)

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

We now proceed, as promised, to give a necessary and sufficient conditions for an irrep of a space to be self-conjugate real and self-conjugate quaternonic. Namely:

Theorem: Let \mathscr{V} be a vector space, G a finite group, and \rho:G\to\mathcal{U}\left(\mathscr{V}\right) an irrep. Then, \rho is self-conjugate if and only if there exists a J which is antilienar and antiunitary which commutes with \rho.  If there exists such a J it is unique up to a constant of modulus one. Moreover, if there exists such a J then J^2=\pm\mathbf{1} with J^2=\mathbf{1} iff \rho is real and J^2=-\mathbf{1} iff \rho is quaternonic.

Proof: Suppose first that \rho is self-conjugate. Then, for any complex conjugate map J on \mathscr{V} there exists some unitary map U such that WJ\rho_g JW^{-1}=\rho_g for every g\in G. Note though that WJ is antilinear (indeed, W(J(\alpha x+\beta y))=W(\overline{\alpha}x+\overline{\beta}y)=\overline{\alpha}W(J(x))+\overline{\beta}W(J(y))) and antiunitary (indeed, \left\langle WJ(x),WJ(y)\right\rangle=\left\langle W(y),W(x)\right\rangle=\langle y,x\rangle) and since J=J^{-1} it follows from this equation that WJ\rho_g=\rho_g WJ and thus the conclusion follows.

Suppose then that there existed an antilinear, antiunitary map J which commutes with \rho_g for every g\in G. Then, picking some complex conjugate K on \mathscr{V} we see that W=KJ is unitary (we’ve discussed this before) and by assumption J\rho_g=\rho_g J and thus KJ\rho_g=K\rho J=K\rho_g KKJ and thus W\rho_gW^{-1}=K\rho_g K from where it follows that \text{Conj}^K_\rho\simeq\rho and thus \rho is self-conjugate.

Now, to see that if such a J exists then J^2=\pm\mathbf{1}, we merely note that J^2 is unitary and J^2\rho_g=\rho_g J^2 for every g\in G. It follows by the second form of Schur’s lemma that J^2=\alpha\mathbf{1} for some \alpha. But, since J^2 is unitary we must have that |\alpha|=1. Note though that since \alpha\mathbf{1}=J^2 we may apply J to both sides to get that J(\alpha\mathbf{1})=\overline{\alpha}J=J^3=\alpha \mathbf{1}J=\alpha J and thus \alpha=\overline{\alpha}, from where it follows that \alpha=\pm 1.

Now, to see that such a J is unique up to a constant of modulus one, applying the same idea of using Schur’s lemma we can show that if J_1 and J_2 are any such antiunitary,antilinear commuting maps that  J_1J_2 is an intertwinor for \rho and thus by the second form of Schur’s lemma J_1J_2=\alpha\mathbf{1} for some \alpha. Note though that since J_1J_2 is unitary we must have that |\alpha|=1. Note then that by the last part of the proof J_1J_2=\alpha\mathbf{1} implies that J_1=\pm\alpha J_2 from where the conclusion follows.

Now, suppose that J^2=\mathbf{1}. Then, by our previous theorem we have that there exists an orthonormal basis \{x_1,\cdots,x_n\} for \mathscr{V} for which J(x_k)=x_k\;\; k\in[n]. Note though that if g\in G and \displaystyle \rho_g(x_j)=\sum_{i=1}^{n}\alpha_{i,j}x_i then


\displaystyle \left\langle \rho_g(x_j),x_k\right\rangle=\sum_{i=1}^{n}\alpha_{i,j}\langle x_i,x_k\rangle=\alpha_{k,j}


That said,


\begin{aligned}\left\langle \rho_g (x_j),x_k\right\rangle &=\left\langle J(x_k),J\left(\rho_g (x_j)\right)\right\rangle\\ &=\left\langle J(x_k),\rho_g\left(J(x_j)\right)\right\rangle\\ &=\left\langle x_k,\rho_g(x_j)\right\rangle\\ &= \overline{\left\langle \rho_g(x_j),x_k\right\rangle}\end{aligned}


Since j,k\in[n] and g\in G was arbitrary it follows that the matrix with respect to this basis is real for each \rho_g and thus \rho is real. Conversely, suppose that J is real. Then, there exists some basis \{x_1,\cdots,x_n\} for which the matrix of \rho_g is real for each g\in G. Define J by


\displaystyle J\left(\sum_{j=1}^{n}\alpha_j x_j\right)=\sum_{j=1}^{n}\overline{\alpha_j}x_j


it’s easily verified then that J is an antilinear, antinunitary map and J^2=\mathbf{1}. But, it easily follows from the fact that \rho_g has a real matrix with respect to the basis \{x_1,\cdots,x_n\} that J commutes with \rho_g for each g\in G and thus the converse follows.

It follows from this that if \rho is quaternonic then any such J is such that J^2=\pm\mathbf{1}, but the above excludes the case J^2=\mathbf{1} and thus J^2=-\mathbf{1}. Similarly, if J^2=-\mathbf{1} then the above shows that J is not real and thus must be quaternonic. The conclusion follows. \blacksquare



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

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print


February 6, 2011 - Posted by | Algebra, Linear Algebra, Representation Theory | , , ,


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