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](http://en.wikipedia.org/wiki/Polycyclic_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](http://en.wikipedia.org/wiki/Heisenberg_group)?




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



$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat (right) $\mathbb{Z}G$-module?
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](http://arxiv.org/pdf/1103.1567v3.pdf). 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.


