Abstract Nonsense

Crushing one theorem at a time

Complex Conjugate Representation (Characterization of Complex Conjugates)


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

We’d now like to find necessary and sufficient conditions for an irrep to be a self-conjugate and when this is true whether it is real or quaternionic. But, first we need a theorem which can be used to show that all complex conjugates, given the right basis, have a particularly nice form and which will be integral in this aforementioned characterization of quaternionic and real self-conjugate irreps.

Theorem: Let J be a complex conjugate on a pre-Hilbert space \mathscr{V} of dimension n. Then, \mathscr{V} admits an orthonormal basis \{v_1,\cdots,v_n\} such that J(v_k)=v_k,\;\; k=1,\cdots,n.

Proof: Consider the vector space \mathscr{V}^\mathbb{R} which is \mathscr{V} but considered as a vector space over \mathbb{R} instead of a vector space over \mathbb{C}. It’s an elementary fact that \dim_\mathbb{R} \mathscr{V}^{\mathbb{R}}=2\dim_{\mathbb{C}}\mathscr{V}=2n. We can define an inner product on \mathscr{V}^\mathbb{R}, denoted \langle\cdot,\cdot\rangle by

 

\langle\cdot,\cdot\rangle=\text{Re}\langle\cdot,\cdot\rangle

 

where \langle\cdot,\cdot\rangle is the given inner product on \mathscr{V}. This is indeed a symmetric inner product on \mathscr{V}^\mathbb{R}. To see that it’s linear in the first column note that (recall that this is a \mathbb{R}-space) since

 

\begin{aligned}\langle \alpha x+\beta y,z\rangle_\mathbb{R} &=\text{Re}\langle \alpha x+\beta y,z\rangle\\ &=\text{Re}\left(\alpha\langle x,z\rangle+\beta\langle y,z\rangle\right)\\ &=\alpha\text{Re}\langle x,z\rangle+\beta\text{Re}\langle y,z\rangle\\ &=\alpha\langle x,z\rangle_\mathbb{R}+\beta\langle y,z\rangle_\mathbb{R}\end{aligned}

 

It’s symmetric since

 

\langle x,y\rangle_\mathbb{R}=\text{Re}\langle x,y\rangle=\text{Re}\overline{\langle y,x\rangle}=\text{Re}\langle y,x\rangle=\langle y,x\rangle_\mathbb{R}

 

Lastly, to see that it’s positive definite we merely note that \langle x,x\rangle_\mathbb{R}=\text{Re}\langle x,x\rangle=\langle x,x\rangle since for any inner product \langle x,x\rangle\in\mathbb{R} and the conclusion follows since \langle\cdot,\cdot\rangle is positive definite.

Now, with respect to this new pre-Hilbert space \left(\mathscr{V},\langle\cdot,\cdot\rangle_\mathbb{R}\right) we claim that the map J is linear, involutory, and self-adjoint. This first part of this statement is clear since

J(\alpha x+\beta y)=\overline{\alpha}J(x)+\overline{\beta}J(y)=\alpha J(x)+\beta J(y)

(once again recalling that we are now dealing with an \mathbb{R}-space). The second is also clear by our initial assumptions. The last statement follows since for any x,y\in\mathscr{V}^\mathbb{R} we have that

 

\begin{aligned}\left\langle J(x),y\right\rangle_\mathbb{R} &=\text{Re}\left\langle J(x),y\right\rangle\\ &=\text{Re}\left\langle y,x\right\rangle\\ &=\text{Re}\langle x,y\rangle\\ &=\text{Re}\left\langle J(y),x\right\rangle\\ &=\text{Re}\left\langle x,J(y)\right\rangle\\ &=\left\langle x,J(y)\right\rangle_\mathbb{R}\end{aligned}

 

With this, define \mathscr{X}_1=\text{im }\left(\mathbf{1}+J\right) and \mathscr{X}_2=\text{im}\left(\mathbf{1}-J\right). We claim that \mathscr{V}^\mathbb{R}=\mathscr{X}_1\oplus\mathscr{X}_2. Indeed, to see that \mathscr{V}^\mathbb{R}=\mathscr{X}_1+\mathscr{X}_2  we note that for any x\in\mathscr{V}^\mathbb{R} one has that

x=\frac{1}{4}\left(\mathbf{1}+J\right)^2(x)+\frac{1}{4}\left(\mathbf{1}-J\right)^2(x)

Indeed, expansion merely gives

\frac{1}{4}\left(\mathbf{1}+J+J^2+\mathbf{1}-J+J^2\right)=\frac{1}{4} 4\mathbf{1}=\mathbf{1}

from where the rest is trivial. To see that \mathscr{X}_1\cap\mathscr{X}_2 is trivial, assume that x\in\mathscr{X}_1\cap\mathscr{X}_2 then x=(\mathbf{1}+J)(y)=(\mathbf{1}-J)(z) for some y,z\in\mathscr{V}^\mathbb{R}. But,

\left\langle (\mathbf{1}+J)(y),(\mathbf{1}-J)(z)\right\rangle_\mathbb{R}=\left\langle y,\left(\mathbf{1}+J\right)\left(\mathbf{1}-J\right)(y)\right\rangle_\mathbb{R}=0

from where it follows that x=\bold{0} (note we made use of the fact that (\mathbf{1}+J)(\mathbf{1}-J)=\mathbf{1}-J^2=\bold{0}).

 

Next note that if x\in\mathscr{X}_1 then J(x)=x since x=(\mathbf{1}+J)(y) for some y\in\mathscr{V}^\mathbb{R} and thus

J(x)=J(\mathbf{1}+J)(y)=\left(J+J^2\right)(y)=\left(\mathbf{1}+J\right)(y)=x

similarly one shows that if x\in\mathscr{X}_2 then J(x)=-x. With this we claim that the map I:\mathscr{X}_2\to\mathscr{X}_1 by x\mapsto ix is an isomorphism. Indeed, I is clearly linear. To see that I is bijective we note that JI(x)=J(ix)=\bar{i}K(x)=-IJ(x) and thus I is invertible. It follows then that I is an isomorphism and thus by a counting argument \dim_\mathbb{R}\mathscr{X}_1=\dim_\mathbb{R}\mathscr{X}_2=n.  Let \{x_1,\cdots,x_n\} be an orthonormal basis for \mathscr{X}_1.  But, recalling the simple identity -\text{Re}(iz)=\text{Im}(z) one sees that

 

\begin{aligned}\langle x,y\rangle &=\text{Re}\langle x,y\rangle+i\text{Im}\langle x,y\rangle\\ &=\text{Re}\langle x,y\rangle-i\text{Re}\left(i\langle x,y\rangle\right)\\ &=\text{Re}\langle x,y\rangle-i\text{Re}\langle ix,y\rangle\\ &= \langle x,y\rangle_\mathbb{R}-i\langle ix,y\rangle_\mathbb{R}\end{aligned}

 

So, note then that if i\ne j then by assumption of the way the basis was chosen \langle x_i,x_j\rangle=0. That said, this implies that \text{Re}\langle x_i,x_j\rangle=0 so that \langle x_i,x_j\rangle=ir for some real number r. It follows then that \text{Re}i\langle x_i,y_j\rangle=-r and thus ir=-r from where it follows that r=0 and thus \langle x_i,x_j\rangle=0. That said, note that since \langle x_i,x_i\rangle\in\mathbb{R} we have that \text{Re}\left(i\langle x_i,x_i\rangle\right)=0 and thus

\langle x_i,x_i\rangle=\langle x_i,x_i\rangle_\mathbb{R}=1

by assumption. It follows that \{x_1,\cdots,x_n\} is orthonormal in \left(\mathscr{V},\langle\cdot,\cdot\rangle\right) and since \dim\mathscr{V}=n it follows that \{x_1,\cdots,x_n\} is an orthonormal basis for \left(\mathscr{V},\langle\cdot,\cdot\rangle\right).

That said, since x_1,\cdots,x_n\in\mathscr{X}_1 by prior discussion we know that K(x_j)=x_j,\;\; j=1,\cdots,n from where the theorem follows. \blacksquare

From this we get the following interesting corollary:

Corollary: Let \mathscr{V} be a pre-Hilbert space and J a complex conjugate on \mathscr{V}. Then, there exists some orthonormal basis \{x_1,\cdots,x_n\} such that


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

 

References:

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


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

5 Comments »

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

    Pingback by Representation Theory: Complex Conjugate Representation (A Characterization of Self-Conjugate Maps) « Abstract Nonsense | February 6, 2011 | Reply

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  4. […] it remains to prove then that with this definition of operations . To see this recall from our past characterizations of complex conjugates that admits a basis such that for . We claim then that is similarly a basis for . Indeed, the […]

    Pingback by Representation Theory: A Bijection Between A Subset of the Complex Reps of a Finite Group and the Real Reps (Pt. I) « Abstract Nonsense | March 29, 2011 | Reply

  5. […] Indeed, for any we have that there exists such that where is the basis guaranteed by our characterization of complex conjugates such that . Clearly then and since we have that and evidently . Thus, is an epimorphism but […]

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