Let $\Gamma$ be some group, and let $$\ell^2(\Gamma) := \{f : \Gamma \to \Bbb{C} \mid \sum_{\gamma \in \Gamma} |f(\gamma)|^2 < \infty\}.$$ Given $\gamma \in \Gamma$, define $u_\gamma : \ell^2(\Gamma) \to \ell^2(\Gamma)$ by $u_\gamma(f)(\eta) := f(\gamma^{-1} \eta)$. I am reading an expository paper in which in claims that $u_\gamma$ is a unitary operator and that $\gamma \mapsto u_\gamma$ defines a unitary representation (called the left regular representation; see page 7 of this). However, this seems to be a mistake. Shouldn't $u_\gamma$ actually be $u_\gamma (f)(\eta) := f(\gamma \eta)$? Defined the former way, I wasn't able to show that $\gamma \mapsto u_\gamma$ is a homomorphism, but I was able to show that it is a homomorphism if defined the later way. Am I misunderstanding something?


Given the second way of construing the left regular representation, it is easy to show that $u_\gamma$ is invertible with inverse $u_{\gamma^{-1}}$. However, when I try to show that $u_{\gamma}^{\ast} = u_{\gamma^{-1}}$ (i.e., that $u_{\gamma}$ is unitary), I run into some difficulties. I could use some help with showing that $u_{\gamma}$ is unitary.

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    $\begingroup$ $$u_{\gamma\delta}f(\eta) = f((\gamma\delta)^{-1}\eta) = f(\delta^{-1}\gamma^{-1}\eta) = u_\delta f(\gamma^{-1}\eta) = u_\gamma(u_\delta f)(\eta),$$hence $u_{\gamma\delta} = u_\gamma\circ u_\delta$. Note that for $\Gamma = (\mathbb R,+)$ you have $\ell^2(\Gamma) = L^2(\mathbb R)$ and $u_\gamma f(x) = f(x-\gamma)$, so $u_\gamma$ is translation by $\gamma$. So, the $u_\gamma$ you have here is just a generalization of the concept of translation. $\endgroup$ – amsmath Jun 5 at 13:32
  • $\begingroup$ The computation by @amsmath shows why the definition using $\gamma^{-1}$ works. You should also check why your proposed definition doesn't work; an analogous computation gives $u_\gamma(u_\delta(f))=u_{\delta\gamma}(f)$. $\endgroup$ – Andreas Blass Jun 5 at 16:03

The comments explain why the representation defined by $\gamma\mapsto u_\gamma$ with $(u_\gamma f)(\eta)=f(\gamma^{-1}\eta)$ is indeed a homomorphism.

To see that this representation is unitary, it suffices to show that $\langle u_\gamma f,u_\gamma g\rangle=\langle f,g\rangle$ for all $f,g\in \ell^2\Gamma$ (this shows that $u_\gamma$ is an isometry, and thus a unitary since it is invertible). And to show this, it suffices to consider $f=\delta_\mu$, $g=\delta_\nu$ for some $\mu,\nu\in\Gamma$, where $\delta_\mu(\gamma)=1$ if $\gamma=\mu$ and $=0$ otherwise (since $\{\delta_\gamma\mid\gamma\in\Gamma\}$ is an orthonormal basis of $\ell^2\Gamma$). And seeing this is easy: \begin{align*} \langle u_\gamma\delta_\mu,u_\gamma\delta_\nu\rangle&=\langle\delta_{\gamma\mu},\delta_{\gamma\nu}\rangle\\ &=\left\{\begin{array}{lcl} 1&:&\gamma\mu=\gamma\nu,\\ 0&:&\text{otherwise}, \end{array}\right. \\ &=\langle\delta_\mu,\delta_\nu\rangle. \end{align*}


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