44.2 Group von Neumann Algebras

Proof. Firstly, for each $f \in L^{1}(G; \complex)$ and $g, h \in L^{2}(G; \complex)$, by the Fubini-Tonelli Theorem,

By Theorem 32.2.2, $\ol{\lambda(G)}^{\text{\small SOT}}= \ol{\lambda(L^1(G;\complex))}^{\text{\small SOT}}$.$\square$

Proof. By Young’s Inequality, for each $g \in L^{2}(G; \complex)$, $\norm{f * g}_{L^2(G; \complex)}\le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^2(G; \complex)}$.$\square$

Proof. (1): Let $\seq{g_n}\subset L^{2}(G; \complex)$, $\seq{f * g_n}\subset L^{2}(G; \complex)$, and $g, h \in L^{2}(G; \complex)$ such that $g_{n} \to g$ and $f * g_{n} \to h$ in $L^{2}(G; \complex)$ as $n \to \infty$. After taking a subsequence, assume without loss of generality that:

1. (i)

   $f * g_{n} \to h$ almost everywhere.

2. (ii)

   $\sum_{n \in \natp}\norm{g_n - g}_{L^2(G; \complex)}< \infty$.

In which case, for almost every $x \in G$,

where $\int_{G}|f(y)|\sum_{n \in \natp}|(g_{n} - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the Dominated Convergence Theorem, $f * g_{N} \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the Closed Graph Theorem implies that $L_{f}$ is continuous.

(2): Let $U \in \cn_{G}(1_{G})$ be relatively compact with $U = U^{-1}$, then $\one_{U} \in L^{2}(G; \complex)$, and $\int_{G} |f(y)|\one_{U}(y^{-1}x)dy < \infty$ for almost every $x \in G$. For each $x_{0} \in G$, by Proposition 31.2.7, there exists $x \in G$ such that $x_{0} \in xU$ and $\int_{G} |f(y)|\one_{U}(y^{-1}x)dy < \infty$. As such, for any $K \subset G$ compact, there exists $F \subset G$ finite such that $\int_{G} |f(y)|\one_{U}(y^{-1}x)dy < \infty$ for each $x \in F$ and $K \subset \bigcup_{x \in F}xU$. In which case,

(3): By Lemma 44.2.4, $f$ is locally integrable. Let $K \subset G$ be relatively compact. Using Lemma 5.20.2, assume without loss of generality that $K$ is also open.

Let $g \in C_{c}(G; \complex)$ with $\supp{g}\subset K^{-1}$, then there exists $V \in \cn_{G}(1_{G})$ such that $\supp{g}^{-1}V \subset K^{-1}$. In which case, for each $x \in V$,

so $f * g$ is continuous on $V$ by Young’s Inequality. In particular, $f * g|_{V}= 0$ almost everywhere implies that $\int_{G} f(y)g(y^{-1}) dy = 0$. As such, $\int_{G} fg = 0$ for all $g \in C_{c}(G; \complex)$ with $\supp{g}\subset K$. By uniqueness from the Riesz Representation Theorem, $f|_{K} = 0$ almost everywhere. Therefore $f = 0$ locally almost everywhere.$\square$

Proof, [Proposition 1.3.5, AP]. (1): $f * \delta_{e} = f \in l^{2}(G; \complex)$.

(2): Let $x, y \in G$, then

(3): By (1), $f * g \in l^{2}(G; \complex)$ is defined. For any $h \in l^{2}(G; \complex)$, by associativity of convolution,

(4): Let $T \in \ol{LC(G)}^{\text{\small SOT}}$ and $\angles{f_\alpha}_{\alpha \in A}$ be left convolvers such that $T = \sotlim_{\alpha \in A}L_{f_\alpha}$. Let $f = T\delta_{e} \in l^{2}(G; \complex)$, then

For any $g \in l^{2}(G; \complex)$,

by Young’s Inequality. Therefore $Tg = f * g$, $T = L_{f}$, and $LC(G)$ is strong-operator closed. By von Neumann’s Bicommutant Theorem, $LC(G)$ is a von Neumann algebra.$\square$

Proof, [Theorem 1.3.6, AP]. ($\subset$): For any $x, y, z \in G$ and $f \in l^{2}(G; \complex)$,

so $\lambda(G) \subset R(G)'$. By Definition 44.2.5, $L(G) \subset LC(G) \subset R(G)'$.

($\supset$): Let $T \in R(G)'$ and $f = T\delta_{e} \in l^{2}(G; \complex)$, then for each $x \in G$,

As such, $Tg = f * g$ for all $g \in l^{2}(G; \complex)$, and $T = L_{f}$.$\square$

Proof, [Proposition 1.3.9, AP]. (2) $\Rightarrow$ (1): Let $T \in Z(L(G))$, then for each $x \in G$,

so $T\delta_{e}$ is invariant under conjugation, and hence constant on conjugacy classes of $G$.

Since $G$ is ICC and $T\delta_{e} \in l^{2}(G; \complex)$, there exists $\lambda \in \complex$ with $T\delta_{e} = \lambda \delta_{e}$. As such, $T = \lambda I$, and $Z(L(G)) = \complex I$.

$\neg$ (2) $\Rightarrow$ $\neg$ (1): Let $C \subset G$ be a non-trivial finite conjugacy class, then for each $x \in G$, $\delta_{x} * \one_{C} = \one_{xC}= \one_{Cx}= \one_{C} * \delta_{x}$, so $L_{\one_C}$ is a non-trivial element of $Z(L(G))$.$\square$

Proof, [Proposition 1.3.10, AP]. For each $g \in H \wr_{I} \Gamma$ and subgroup $K \subset H \wr_{I} \Gamma$, let $g^{K} = \bracsn{kgk^{-1}|k \in K}$.

Let $g = (\xi, \gamma) \in H \wr_{I} \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^{G} \supset g^{\Gamma}$ is infinite.

Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_{f}$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_{f}$. By (b), there exists $i_{0} \in I$ with $\gamma i_{0} \ne i_{0}$. As $H$ is non-trivial, there exists $\xi_{0} \in H^{(I)}\setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_{0} = (\xi_{0}, e_{\Gamma})$, then

and hence $g$ has infinite conjugacy class because $\xi_{0}^{-1}\gamma\xi_{0} \ne e_{H^{(I)}}$. Therefore $g^{G}$ is also infinite.$\square$