# Abstract Nonsense

## 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.

January 18, 2011 -

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

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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 . […]

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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. […]

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