# convolution, $L^1(G)$ and von Neumann algebra $VN^1(G)$

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):

https://dmitripavlov.org/scans/terp-fourier.pdf

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!

• 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 $G$. – user342207 Aug 11 '17 at 8:45
• 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. – mateusz Aug 11 '17 at 9:36
• 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 $xy^{-1} \mapsto yx^{-1}$? – user342207 Aug 11 '17 at 13:22
• 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... – 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.