Why is this a $C^*$ algebra? (twisted convolution algebra of $\mathbb{R}^2$) Consider the Lie group $\mathbb{R}^2$ and for $(x_1,y_1)\,,(x_2,y_2)\in\mathbb{R}^2$ let $\sigma((x_1,y_1),(x_2,y_2))=x_1y_2-x_2y_1\,.$ Then $\sigma$ is a 2-cocycle on $\mathbb{R}^2$. We can define an involution by $f^{*}(x)=\bar{f}(-x).$ We get an associative product by defining
$$(f\star g)(x,y)=\int_{\mathbb{R}^2}f(x',y')g(x-x',y-y')e^{i(x'(y-y')-y'(x-x'))}dx'dy'.$$
We can define the norm $\vert\vert f\vert\vert$ to be the operator norm of $f$ acting on $L^2(\mathbb{R}^2)$ with respect to the above operation. This is supposed to be a $C^*$ algebra (after performing a completion), see example 1 in https://math.berkeley.edu/~rieffel/papers/deformation.pdf.
In particular, for this to be a $C^{*}$ norm it must be the case that $*$ is an isometry. Now in order to show it's an isometry I've computed that
$$\lvert\lvert(f^{*}\star g^{*})\lvert\lvert_{L^2}=\lvert\lvert\int_{\mathbb{R}^2}f(x',y')g(x-x',y-y')e^{-i(x'(y-y')-y'(x-x'))}dx'dy'\lvert\lvert_{L^2}$$
but I can't see why $\vert\vert f^{*}\vert\vert=\vert\vert f\vert\vert.$ The integral on the right is almost just $(f\star g)(x,y),$ except that there is a minus sign in the exponential.
 A: Taking the adjoint of a Hilbert space operator is isometric with respect to the operator norm, so it suffices to check that
$$
\langle f\star g,h\rangle_{L^2}=\langle g,f^\ast\star h\rangle_{L^2}
$$
for all $f,g,h\in \mathcal S(\mathbb R^2)$.
This can be done as follows. I will use the notation from Rieffel's paper to make the calculations a little more compact (I just write $\hbar$ for $2\pi^2\hbar$ in Rieffel's paper).
\begin{align*}
\langle f\star g,h\rangle_{L^2}&=\int\int \overline{f(r)g(t-r)\sigma_{\hbar}(r,t-r)}h(t)\,dr\,dt\\
&\overset{s=t-r}{=}\int \overline{g(s)}\int \overline{f(r)\sigma_{\hbar}(r,s)}h(s+r)\,dr\,ds\\
&\overset{u-r}=\int \overline{g(s)}\int f^\ast(r)\overline{\sigma_{\hbar}(-r,s)}h(s-r)\,dr\,ds.
\end{align*}
Since
$$
\overline{\sigma_{\hbar}(-r,s)}=e^{i\hbar(-r_1s_2+r_2 s_1)}=e^{i\hbar(r_2(s_1-r_1)-r_1(s_2-r_2))}=\sigma_{\hbar}(r,s-r),
$$
we have
$$
\int f^\ast(r)\overline{\sigma_{\hbar}(-r,s)}h(s-r)\,dr=\int f^\ast(r)h(s-r)\sigma_{\hbar}(r,s-r)\,dr=(f^\ast\star h)(s)
$$
Thus
$$
\langle f\star g,h\rangle_{L^2}=\int \overline{g(s)}(f^\ast\star h)(s)\,ds=\langle g,f^\ast\star h\rangle_{L^2}.
$$
