TITLE: $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat (right) $\mathbb{Z}G$-module?
QUESTION [14 upvotes]: Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by-finite group. 
Denote the (Banach) space (in fact an algebra under convolution) $$\ell^1(G)=\left\{\sum_{g\in G}\lambda_g g\;\Big|\; \sum_{g\in G}|\lambda_g|<\infty, \,\lambda_g\in \mathbb{C}\right\}.$$ 
Note that $\mathbb{Z}G\subset \ell^1(G)$ and we can consider $\ell^1(G)$ as a right $\mathbb{Z}G$-module. My question is:

$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat $\mathbb{Z}G$-(right) module?

If it is true, any reference (maybe for related topics)? Otherwise, any counterexample? Especially, is it true for the Heisenberg group?

RK: I have asked a general one in MSE but no answer appeared, so I asked here a more specific one.

REPLY [4 votes]: The general answer is no for my question. 
For example, for $G=\mathbb{Z}^2\rtimes\mathbb{Z}$ with exponential growth rate, $\ell^1(G)$ is not flat over $\mathbb{Z}G$, the proof is based on rather elementary calculation. But it is too long to be present here..
The point is to find $f\in\mathbb{Z}G\cap\ell^1(G)^{\times}$, find $h\in\mathbb{Z}G$, and do calculation to show for any $t\in\mathbb{Z}G$ with $th\in\mathbb{Z}Gf$, $t\not\in\ell^1(G)^{\times}$. Then we have $0\to\frac{\mathbb{Z}Gf+\mathbb{Z}Gh}{\mathbb{Z}Gf}\to\frac{\mathbb{Z}G}{\mathbb{Z}Gf}$ fails the flatness of $\ell^1{G}$.
It is open when $G$ is the Heisenberg group.
Motivation behind this question is proposition 2.1, theorem 3.1 and conjecture 3.6 in the paper here. Note that conjecture 3.6 is false ingle general, but for the most interesting case, i.e., $G$ is Heisenberg group, it is still open.