# Left Regular Representation of a Group

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?

EDIT:

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.

• $$u_{\gamma\delta}f(\eta) = f((\gamma\delta)^{-1}\eta) = f(\delta^{-1}\gamma^{-1}\eta) = u_\delta f(\gamma^{-1}\eta) = u_\gamma(u_\delta f)(\eta),$$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. – amsmath Jun 5 at 13:32
• 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 $u_\gamma(u_\delta(f))=u_{\delta\gamma}(f)$. – 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*}