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\title{Math 8120 - Homework Assignment 2}

\author{Due Wednesday, January 22 by 4:00pm}

\date{\today}

\begin{document}



\maketitle

Let $\Gamma$ be a discrete abelian group. A character on $\Gamma$ is a homomorphism $\chi: \Gamma \to \mathbb T$. We let $\hat{\Gamma}$ denote the set of characters, and we endow $\hat{\Gamma}$ with a group structure by defining $(\chi \psi)(s) = \chi(s) \psi(s)$. 


\begin{exer}
Show that $\hat{\Gamma}$ can naturally be identified with the spectrum $\sigma( C^* \Gamma)$ and that with the induced topology we have that $\hat{\Gamma}$ is a compact group. 
\end{exer}

The Gelfand transform gives an isomorphism $C^* \Gamma \cong C(\hat{\Gamma})$ and we have that $\ell^1 \Gamma$ is a dense subspace in $C^*\Gamma$, thus the Gelfand transform gives a map from $\ell^1 \Gamma$ onto a dense subspace of $C(\hat{\Gamma})$ which maps $\xi \in \ell^1 \Gamma$ to the function $\hat{\Gamma} \ni \chi \mapsto \sum_{s \in \Gamma} \xi(s) \chi(s)$.  This is closely related to the Fourier transform $\mathcal F: \ell^1 \Gamma \to C(\hat{\Gamma})$ given by $\mathcal F( \xi) (\chi) = \sum_{s \in \Gamma} \xi(s) \overline{\chi(s)}$. 

\begin{exer}[Plancherel's theorem]
Let $\lambda$ denote the Haar measure on $\hat{\Gamma}$ such that $\lambda(\hat{\Gamma}) = 1$. Show that the Fourier transform extends to a unitary operator between $\ell^2 \Gamma$ and $L^2(\hat{\Gamma}, \lambda)$.
\end{exer}

We denote by $\mathcal F_0: \ell^2 \Gamma \to L^2(\hat{\Gamma}, \lambda)$ the unitary from the previous exercise. (Which is also called the Fourier transform).

The left regular representation of a group $\Gamma$ is $\lambda: \Gamma \to \mathcal U(\ell^2 \Gamma)$ defined by $(\lambda(s) \xi)(t) = \xi(s^{-1} t)$. The reduced group $C^*$-algebra of $\Gamma$ is the $C^*$-algebra generated by $\lambda(\Gamma)$ and is denoted by $C^*_r(\Gamma)$. 

We now consider the natural inclusions $C(\hat{\Gamma}) \subset L^\infty(\hat{\Gamma}) \subset \mathcal B(L^2(\hat{\Gamma}))$. Where the latter inclusion associates each function $f \in L^\infty(\hat{\Gamma})$ to the operator $L^2(\hat{\Gamma}) \ni \xi \mapsto f \xi \in L^2(\hat{\Gamma})$.

\begin{exer}
Let $\Gamma$ be a discrete abelian group. Show that $\mathcal F ( C^*_r \Gamma ) \mathcal F^* = C(\hat{\Gamma})$ and $\mathcal F ( L \Gamma ) \mathcal F^* = L^\infty( \hat{\Gamma} )$. In particular, for abelian discrete groups, the canonical $*$-homomorphism from $C^* \Gamma$ onto $C^*_r \Gamma$ given by $\ell^1 \Gamma \ni \xi \mapsto \lambda(\xi) \in C^*_r \Gamma$, is a $*$-isomorphism. 
\end{exer}




\begin{exer}
Let $\Gamma$ be a discrete group. Consider the following conditions:
\begin{enumerate}
\item The canonical $*$-homomorphism from $C^* \Gamma$ onto $C^*_r \Gamma$ is a $*$-isomorphism.
\item For any unitary representation $\pi: \Gamma \to \mathcal U(\mathcal H)$, and any $\xi \in \ell^1 \Gamma$, we have $\| \pi(\xi) \| \leq \| \lambda(\xi) \|$. 
\item For any finite set $F \subset \Gamma$ we have $\| \sum_{s \in F} \lambda(s) \| = | F |$. 
\item For any $\varepsilon > 0$ and $F \subset \Gamma$ finite, there exists a unit vector $\xi \in \ell^2 \Gamma$ such that $\sum_{s \in F} \| \lambda(s)\xi - \xi \|^2 < \varepsilon$. 
\end{enumerate}
Prove the implications (1) $\Leftrightarrow$ (2) $\Rightarrow$ (3) $\Leftrightarrow$ (4).\footnote{In fact, all four of these conditions are equivalent.}

Hint: To prove (3) $\Leftrightarrow$ (4) you may want to use the fact that if $\{ \xi_i \}_{i = 1}^n$ is a family of unit vectors in a Hilbert space then we have $\sum_{i, j = 1}^n \| \xi_i - \xi_j \|^2 = 2n^2 - 2 \sum_{i, j = 1}^n \langle \xi_i, \xi_j \rangle = 2n^2 - 2 \| \sum_{i = 1}^n \xi_i \|^2$.
\end{exer}



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