Let $H=L^2(\mathbb{T})$, where $\mathbb{T}$ is the Torus. Consider a multiplication operator with a sufficiently nice function $f$.
Is there somehow a formula like $$\mathrm{tr} M_f = C \int_{\mathbb{T}} f(x) dx\;\;\;\;\text{ or }\;\;\;\; \mathrm{tr}|M_f| = C \|f\|_{L^1(\mathbb{T})}?$$
Or is there another way to compute the trace of a multiplication operator?
Note $\mathbb T = S^1 \times S^1$. The case on $\mathbb T$ and $S^1$ is not much different:
There is a standard ONB on $S^1$, namely $e^{i kx}$ for $k \in \mathbb Z$. The point is that that $\langle e^{ikx}| M_f e^{ikx}\rangle =\frac1V\int f$, and if you would sum over all $k$ this diverges unless $\int f =0$, and you have that $M$ is not trace-class.
On the other hand if $\int f = 0$ then you have to note that $M$ is trace class iff $$\sum_i \frac1V \int \overline{e_i}\ |M_f|(e_i) \tag{1}$$ converges for any ONB $e_i$. Now $|M|$ is the unique positive root of $M^*M^\mathstrut$. In this case you can see that $|M_f|=M_{|f|}$. But the result for the basis chosen before shows that the sum (1) diverges unless $\int |f| =0$, that is $f=0$ almost everywhere and $M_f=0$.
This was for $S^1$, but note that $e^{ik x_1}$ are orthonormal on $\mathbb T$ (after rescaling the scalar product maybe), and after extending to an ONB the only way to get that $M_f$ is trace class is once again if $M_f=0$.