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. II)


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

The Complexification of a Real Inner Product Space

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Before we get to the main part of this post we must discuss the complexification of a real inner product space. Namely, we define the complexification of the real inner product space \left(\mathscr{R},\langle\cdot,\cdot\rangle\right), denoted \mathscr{R}^{\mathbb{C}},  to be the set \mathscr{R}\times\mathscr{R} with addition defined coordinatewise, scalar multiplication defined (a+bi)(u,v)=(au-bv,av+bu), and inner product defined by \left\langle (u,v),(u',v')\right\rangle^\mathbb{C}=\langle u,u'\rangle+\langle v,v'\rangle+i\left(\langle v,u'\rangle-\langle u,v'\rangle\right). It is easy to verify that \mathscr{R}^\mathbb{C} is indeed a pre-Hilbert space and moreover that \dim_\mathbb{C}\mathscr{R}^\mathbb{C}=\dim_\mathbb{R}\mathscr{V}. What we now claim is that the map \mathbf{1}\oplus -\mathbf{1} is a complex conjugate. Indeed:

 

Theorem: Let \left(\mathscr{R},\langle\cdot,\cdot\rangle\right) be a real inner product space. Then the map \mathbf{1}\oplus -\mathbf{1} is a complex conjugate on \mathscr{R}^\mathbb{C}.

Proof: It’s clear that \left(\mathbf{1}\oplus\mathbf{-1}\right)^2=\left(\mathbf{1}^2\right)\oplus\left(\left(-\mathbf{1}\right)^2\right)=\mathbf{1}\oplus\mathbf{1}=\mathbf{1}_{\mathscr{R}\times\mathscr{R}}. Next we next claim that \mathbf{1}\oplus -\mathbf{1} is antilinear. Indeed, for any a+bi\in\mathbb{C} and (u,v)\in\mathscr{V}^\mathbb{C} we have that

 

\displaystyle \left(\mathbf{1}\oplus-\mathbf{1}\right)\left((a+bi)\left(u,v\right)\right)=\left(\mathbf{1}\oplus-\mathbf{1}\right)\left(au-bv,av+bu\right)=\left(au-bv,-av-bu\right)

 

and

 

\overline{\left(a+bi\right)}\left(\mathbf{1}\oplus-\mathbf{1}\right)(u,v)=\left(a-bi\right)\left(u,-v\right)=\left(au-bv,-av-bu\right)

 

and thus \left(\mathbf{1}\oplus-\mathbf{1}\right)\left(z(u,v)\right)=\overline{z}\left(\mathbf{1}\oplus-\mathbf{1}\right)(u,v) and since \mathbf{1}\oplus-\mathbf{1} is trivially linear antilinearity follows. Lastly, we claim with the proposed inner product that \mathbf{1}\oplus-\mathbf{1} is antiunitary. Indeed, let (u,v),(u',v')\in\mathscr{V}^\mathbb{C} be arbitrary then

 

\displaystyle \begin{aligned}\left\langle \left(\mathbf{1}\oplus-\mathbf{1}\right)(u,v),\left(\mathbf{1}\oplus-\mathbf{1}\right)(u',v')\right\rangle^\mathbb{C} &= \left\langle (u,-v),(u',-v')\right\rangle^\mathbb{C}\\ &= \langle u,u'\rangle+\langle -v,-v'\rangle+i\left(\langle-v,u'\rangle-\langle u,-v'\rangle\right)\\ &=\langle u,u'\rangle+\langle v,v'\rangle+i\left(\langle u,v'\rangle-\langle v,u'\rangle\right)\\ &= \langle u',u\rangle+\langle v',v\rangle+i\left(\langle v',u\rangle-\langle u',v\rangle\right)\\ &= \left\langle (u',v'),(u,v)\right\rangle^\mathbb{C}\end{aligned}

 

and since (u,v) and (u',v') were arbitrary antiunitarity follows. Thus, combining these three attributes let us conclude. \blacksquare

 

Inverse Map

In this post we define an inverse (in a sense soon made to be rigorous) to our map \rho\mapsto \rho_\Re discussed previously. Namely:

 

Theorem: Let G be a finite group and \rho:G\to\mathcal{U}\left(\mathscr{R}\right) be a \mathbb{R}-representation on the real inner product space \left(\mathscr{R},\langle\cdot,\cdot\rangle\right). Then the mapping \rho_\Im:G\to\mathcal{U}\left(\mathscr{R}^\mathbb{C}\right) by \rho_\Im(g)=\rho(g)\oplus\rho(g) is a \mathbb{C}-representation of G which satisfies the real condition with realizer \mathbf{1}\oplus-\mathbf{1}. Moreover, \mathscr{R}^\mathbb{C} has no non-trivial  subspaces \mathscr{S} such that \left(\mathbf{1}\oplus-\mathbf{1}\right)\left(\mathscr{S}\right)=\mathscr{S} which are \rho_\Im-invariant  if and only if \rho is an irreducible \mathbb{R}-representation.

Proof: We first show that g\mapsto \rho(g)\oplus \rho(g) really is a homomorphism G\to\mathcal{U}\left(\mathscr{R}^\mathbb{C}\right). We first verify that \rho(g)\oplus \rho(g) is actually a unitary linear transformation on \mathscr{R}^\mathbb{C} for each g\in G. Indeed, since it’s clear that \rho(g)\oplus\rho(g) is linear it suffices to show that for each g\in G \rho_g\oplus\rho_g is unitary. So, let  (u,v),(u',v')\in\mathscr{R}^\mathbb{C}, then one has that

 

\begin{aligned}\left\langle (\rho_g\oplus\rho_g)(u,v),\left(\rho_g\oplus\rho_g\right)(u',v')\right\rangle^\mathbb{C} &=\left\langle \left(\rho_g(u),\rho_g(v)\right),\left(\rho_g(u'),\rho_g(v')\right)\right\rangle^\mathbb{C}\\ &=\left\langle\rho_g(u),\rho_g(u')\right\rangle+\left\langle\rho_g(v),\rho_g(v')\right\rangle+i\left(\left\langle\rho_g(v),\rho_g(u')\right\rangle-\left\langle\rho_g(u),\rho_g(v')\right\rangle\right)\\ &= \left\langle u,u'\right\rangle+\left\langle v,v'\right\rangle+i\left(\left\langle v,u'\right\rangle-\left\langle u,v'\right\rangle\right)\\ &=\left\langle (u,v),(u',v')\right\rangle^\mathbb{C}\end{aligned}

 

from where unitarity of \rho(g)\oplus\rho(g) follows, and since g\in G was arbitrary it follows that \rho_\Im really is a mapping G\to\mathcal{U}\left(\mathscr{R}^\mathbb{C}\right).

 

We now show that \rho_\Im satisfies the real condition with realizer \mathbf{1}\oplus-\mathbf{1}. But, this is evident since

 

\left(\mathbf{1}\oplus-\mathbf{1}\right)(\rho_g\oplus\rho_g)=(\mathbf{1}\rho_g)\oplus(-\mathbf{1}\rho_g)=\left(\rho_g\mathbf{1}\right)\oplus\left(\rho_g(-\mathbf{1})\right)=\left(\rho_g\oplus\rho_g\right)\left(\mathbf{1}\oplus-\mathbf{1}\right)

 

 

Therefore, it remains to verify our last claim. Suppose first that \rho wasn’t reducible so that there existed some \{\bold{0}\}<\mathscr{S}<\mathscr{R} such that \mathscr{S} is \rho-invariant. Note then that \mathscr{S}\times\mathscr{S} is a subspace of \mathscr{R}^\mathbb{C} and clearly it is \rho_\Im and \mathbf{1}\oplus-\mathbf{1} invariant. Conversely, suppose that \{\bold{0}\}<\mathscr{W}<\mathscr{R}^\mathbb{C} then by assumption for every (u,v)\in\mathscr{W} we have that (u,-v)\in\mathscr{W} and so (u,\bold{0})\in\mathscr{W} and for every a\in\mathbb{R} we have that (a+0i)(u,\bold{0})\in\mathscr{W} and similarly for every (u,0),(u',0)\in\mathscr{W} one has that (u+u',\bold{0})\in\mathscr{W}. Thus, we may conclude that \mathscr{S}=\left\{w:(w,\bold{0})\in\mathscr{W}\right\} is a subspace of \mathscr{R} which is clearly \rho-invariant by assumption since for every g\in G and w\in\mathscr{S} we have that (w,\bold{0})\in\mathscr{W} and thus by assumption (\rho_g\oplus\rho_g)(w,\bold{0})=(\rho_g(w),\bold{0})\in\mathscr{W} so that \rho_g(w)\in\mathscr{S}. The conclusion follows. \blacksquare

 

References:

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

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March 30, 2011 - Posted by | Algebra, Representation Theory | , , ,

1 Comment »

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

    Pingback by Representation Theory: A Bijection Between A Subset of The Complex Reps of a Finite Group and the Real Reps (Pt. III) « Abstract Nonsense | March 30, 2011 | Reply


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