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

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.

• Just a comment - any group $G$ is canonically isomorphic to the opposite group $G^{op}$, where multiplication is reversed (i.e. $g\cdot h := hg$). So $\mathbb{Z} G$ is left-Noetherian iff it is right-Noetherian. But I don't know how the $l^1$ condition interacts with standard homological algebra. Can you prove what you want for $G = \mathbb Z$? – Peter Samuelson Jul 21 '13 at 0:51
• I am not sure it has been proved when $G=\mathbb{Z}$, but perhaps you may find it helpful to read theorem 3.1(4) and corollary 3.5 in the following paper:arxiv.org/abs/1103.1567 – Jiang Jul 21 '13 at 2:36
• And "how the $l^1$ condition interacts with standard homological algebra" is more or less what I want to know. – Jiang Jul 21 '13 at 2:37

Here are a few observations that are too long for a comment but don't provide a full answer by any means.

As Peter Samuelson observed left-right questions aren't significant in this setting since $\mathbb{Z}G$ is isomorphic to $\mathbb{Z}G^{op}$.

Also, note that $\mathbb{C}G$ is a flat $\mathbb{Z}G$-module since $\mathbb{C}$ is a flat $\mathbb{Z}$ module. Thus, by standard base change arguments, to prove that $l^1(G)$ is flat over $\mathbb{Z}G$ it would suffice to prove that it is flat over $\mathbb{C}G$. At least in the case $G$ is polycyclic-by-finite, $\mathbb{C}G$ will also be (left and right) Noetherian.

Now I'm going to restrict to the case $G=\mathbb{Z}$, so $\mathbb{C}G\cong \mathbb{C}[x,x^{-1}]$, the ring of Laurent polynomials in $x$. This ring is a principal ideal domain.

Now I refer to Weibel's Introduction to homological algebra (CUP) Proposition 3.2.4. A left $R:=\mathbb{C}[x,x^{-1}]$-module $B$ is flat as such if and only if $\mathrm{Tor}_1^R(R/I,B)=0$ for every right ideal $I$ of $R$.

But we've already observed that every (right) ideal $I$ is of the form $fR$ in this case, and (for $f\neq 0$), we can compute $\mathrm{Tor}_1^R(R/fR,B)$ explicitly as $[f]B :=\{ b\in B\mid fb=0\}$.

In other words, to answer your question positively for $G=\mathbb{Z}$ it suffices to prove that $l^1(G)$ has no elements killed by $f$ for any non-zero $f\in \mathbb{C}G$. I think it should be possible to prove this by elementary arguments although I haven't even begun to check.

• It occurs to me that in the final paragraph one might as well assume that $f$ is irreducible in $R$. Since (by the fundamental theorem of algebra) such an $f$ is a unit times $(x-\lambda)$ for $\lambda\in \mathbb{C}$ non-zero it should be very easy to complete the case $G=\mathbb{Z}$ by hand. – Simon Wadsley Sep 17 '13 at 10:54
• in this paper ams.org/journals/proc/1998-126-03/S0002-9939-98-04025-8/… the author mentioned at the last line of page 721 that $0\neq f\in \mathbb{C}(\mathbb{Z})$ is a uniform nonzero divisor. – Jiang Sep 19 '13 at 2:01
• How to modify your reduction to the case such that $G$ is a non-commutative polycyclic-by-finite group, say the discrete Heisenberg group? In particular, what does a general right ideal $I$ look like in $\mathbb{C}G$? – Jiang Sep 19 '13 at 2:06
• I'm not even sure that it is easy to extend my strategy to $G=\mathbb{Z}^n$ for $n>1$, although it may be easier than I think. Classifying right ideals in $\mathbb{C}G$ for $G$ a discrete Heisenberg group is probably hard. To get an idea why, see arxiv.org/abs/math/0102190. The first sentence of section 7 of your reference suggests that $Tor^1_{\mathbb{C}G}(\mathbb{C}G/f\mathbb{C}G,l^1(G))=0$ for all non-zero $f\in \mathbb{C}G$ whenever $G$ is torsionfree polycyclic. However, it isn't clear to me how hard this result was to prove already in that generality. – Simon Wadsley Sep 19 '13 at 13:27

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.