Let $G$ be a locally compact group. Then the space $L^1(G)$ with the convolution $$f\star g = \int_G\ f(xy^{-1}) g(y)\ d\mu(y)$$

is a $*$-algebra ($\mu$ is the left Haar measure). The involution is given by $$f^*(x) = \Delta^{-1}(x) \overline{f}(x^{-1}).$$

Following the paper (page 6, at the bottom):


let $VN^1(G)$ denote the smallest von Neumann algebra generated by $$\bigg\{L_f \ : \ f\in L^1(G)\bigg\},$$

where $L_f(g) = f\star g$. It would be beautiful if $(L_f)^* = L_{f^*}$, but that does not seem to be the case (for non-unimodular groups). Here is my reasoning:

\begin{gather*} \langle g| (L_f)^*h\rangle = \langle L_fg|h \rangle = \int_G\ f\star g(x)\ \overline{h}(x)\ dx \\ = \int_G\ \int_G\ f(xy^{-1})g(y)\ dy\ \overline{h}(x)\ dx \stackrel{Fubini}{=} \int_G\ g(y)\ \int_G\ f(xy^{-1})\ \overline{h}(x)\ dx\ dy \\ = \int_G\ g(y)\ \overline{\int_G\ \overline{f}(xy^{-1})\ h(x)\ dx}\ dy = \int_G\ g(y)\ \overline{\int_G\ \widehat{f}(yx^{-1})\ h(x)\ dx}\ dy \\ = \int_G\ g(y)\ \overline{L_{\widehat{f}}h}\ dy , \end{gather*}

where $\widehat{f} = \overline{f(x^{-1})}$. Hence $(L_f)^* = L_{\widehat{f}}$, which is kind of ugly. My question is: does it have to be that way or am I stubborn making some stupid computational mistake i.e. missing $\Delta^{-1}(x)$? Thank you in advance for any tips!

  • $\begingroup$ Shouldn't the modular function appear after your sixth equality? The relation $(L_f)^*=L_{f^*}$ holds certainly for certain unimodular groups, e.g, discrete groups. I expect it holds for any locally compact groups $G$. $\endgroup$ – user342207 Aug 11 '17 at 8:45
  • $\begingroup$ If I understand correctly, you are referring to the equality where we use $$\widehat{f}(yx^{-1}) = \overline{f((yx^{-1})^{-1})} = \overline{f}(xy^{-1}).$$ Unfortunately, I don't see why the modular function should 'pop out' at this point. Nevertheless, I share your conviction that the relation $(L_f)^* = L_{f^*}$ should hold for all locally compact groups. It seems I just can't work out the details. $\endgroup$ – mateusz Aug 11 '17 at 9:36
  • $\begingroup$ There is the equality $$\int_G f(x^{-1}) \Delta(x^{-1}) \; dx = \int_G f(x) \; dx, $$ which holds for all $f \in L^1 (G)$. Isn't this what you are doing when applying the change of variable $xy^{-1} \mapsto yx^{-1}$? $\endgroup$ – user342207 Aug 11 '17 at 13:22
  • $\begingroup$ I see what you mean, but I don't think I am applying any change of variable. As I have written in the previous post, I am simply relabeling the functions (from $\overline{f}$ to $\widehat{f}$). Besides, using the equality that you mentioned would mess up the function $h$ and I reckon this would complicate the matters even further... $\endgroup$ – mateusz Aug 11 '17 at 17:45

The convolution product $f \ast g$ of $f, g \in L^1 (G)$ is defined as $$ f \ast g (x) = \int_G f(y) g(y^{-1} x) d\mu_G (y). $$ Equipped with this product,the space $L^1 (G)$ forms a Banach $*$-algebra with involution $f^* (x) = \Delta(x^{-1} ) \overline{f(x^{-1})}$. Suitable changes of variables give $$ f \ast g (x) = \int_G f(x y^{-1}) g(y) \Delta (y^{-1}) \; d\mu_G (y). $$ Comparing this latter expression to your definition of $f \star g$, it seems you are missing the essential term $\Delta (y^{-1})$ in the integrand.

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