Abstract Nonsense

Crushing one theorem at a time

Representation Theory: Definitions and Basics (Pt. II Equivalent Representations)


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

Equivalent Representations

Let G be a finite group and \rho:g\to\mathcal{U}\left(\mathscr{V}\right) and \rho':G\to\mathcal{U}\left(\mathscr{W}\right) be two representations of G. We say that \rho is equivalent to \rho' if there exists some unitary map U:\mathscr{V}\to\mathscr{W} such that \rho'_g=U\rho_g U^{-1} for every g\in G. We denote this relation symbolically by \rho\simeq \rho' We now prove that since each \rho_g,\rho'_g is unitary this is equivalent to stating that there just exists some isomorphism T:\mathscr{V}\to\mathscr{W} such that \rho'_g=T^{-1}\rho_gT. Indeed:

Theorem: Let \rho_g:G\to\mathcal{U}\left(\mathscr{V}\right) and \rho'_g:G\to\mathcal{U}\left(\mathscr{W}\right) be two unitary representations. Suppose then that there existed some isomorphism T:\mathscr{V}\to\mathscr{W} such that \rho'_g=T^{-1}\rho_gT for each g\in G. Then, there exists some unitary map U:\mathscr{V}\to\mathscr{W} such that \rho_g=U^{-1}\rho'_gU for each g\in G.

Proof: Note that since \rho_g=T^{-1}\rho'_g T we clearly have that \rho_g^\ast=T^\ast\rho\prime_g^\ast \left(T^{\ast}\right)^{-1} and then taking the inverse of both sides (recalling that \rho_g,\rho'_g are unitary and thus \rho_g^\ast=\rho_g^{-1} etc.) we find that \rho_g=T^\ast \rho'_g\left(T^\ast\right)^{-1} using the fact though that \rho'_g=T\rho_gT^{-1} we may put these two together to conclude that

T^\ast T\rho_g \left(T^\ast T\right)^{-1}=T^\ast \rho'_g\left(T^\ast\right)^{-1}=\rho_g

or by rearranging

\rho_g^{-1}\left(T^\ast T\right)\rho_g=T^\ast T

Note that since T^\ast T>0 (this notation means that T^\ast T is positive-definite) by general principles we know that T^\ast T has some unique positive-definite square root \sqrt{T^\ast T}. Noting then that \rho_{g}^{-1}\sqrt{T^\ast T}\rho_g is positive-definite  and

\left(\rho_g^{-1}\sqrt{T^\ast T}\rho_g\right)^2=\rho_g^{-1}\left(T^\ast T\right)\rho_g=T^\ast T

 

we may conclude by the uniqueness of the square root that

\rho_g^{-1}\sqrt{T^\ast T}\rho_g=\sqrt{T^\ast T}

 

Note then that we may use the polar decomposition of T to write T=U\sqrt{T^\ast T} where U is unitary. Note then that

\begin{aligned}U\rho_gU^{-1} &= T\sqrt{T^\ast T}^{-1}\rho_g \sqrt{T^\ast T}T^{-1}\\ &= T\rho_g T^{-1}\\ &= \rho'_g\end{aligned}

from where the conclusion follows. \blacksquare

Both these examples show why we so much desire the unitarity of our representations. It makes everything easier. We’ll see that this is carried further in the next post.

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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January 18, 2011 - Posted by | Algebra, Representation Theory | , , , ,

5 Comments »

  1. […] Indeed, by Schur’s lemma we know that must be an isomorphism such that , and thus by earlier theorem it follows that as […]

    Pingback by Representation Theory: Schur’s Lemma (First and Second Forms) « Abstract Nonsense | January 27, 2011 | Reply

  2. […] that  (since is -invariant) or for every –and by an earlier theorem we may conclude that is self-conjugate. Thus, there exists some unitary such that for every . […]

    Pingback by Representation Theory: A Way of Creating C-representations Satisfying the Real Condition With No (rho,J)-invariant Subspaces (Pt. II) « Abstract Nonsense | April 2, 2011 | Reply

  3. i understand that ρg commutes with the square root of T*T (= P).

    but shouldn’t UρgU^-1 be TP^-1ρgPT^-1?

    unless T*T = I i don’t see how we can conclude P^-1 = P.

    Comment by David Wheeler | April 11, 2011 | Reply

    • Dear David,

      I agree made a typo, but I believe I am missing what you’re saying. We have that T=U\sqrt{T^\ast T} and so U^{-1}=\sqrt{T^\ast T}T^{-1} and U=T\sqrt{T^\ast T}^{-1} and so U\rho_gU^{-1}=T\sqrt{T^\ast T}^{-1}\rho_g\sqrt{T^\ast T}T^{-1}=T\rho_gT^{-1}=\rho'_g. Which is what we wanted. Is there something I’m missing?

      Best,
      Alex

      P.S. Thanks for catching the typo! Please let me know if you catch any more!

      Comment by drexel28 | April 11, 2011 | Reply

  4. […] And, since was arbitrary we may appeal to a previous theorem. […]

    Pingback by Another Way of Looking at Induced Representations (Pt. II) « Abstract Nonsense | April 24, 2011 | Reply


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