Let $G$ be a discrete group and let $S$ be a generating set for $G$; assume that $S$ is symmetric (i.e., $g\in S$ iff $g^{-1}\in S$). Let $L=L_S=\frac{1}{|S|}(\sum_{g\in S} g-1)$ be an element of the group algebra. For $F\in \ell^p(G)$ and $g,h\in G$ let $\lambda_p(g) F$ be defined by $\lambda_p(g) F (h) = F(hg^{-1})$; this is the regular representation of $G$ on $\ell^p(G)$. Finally let $L_p = \pi_p(L)$.

It is a classical result of Kesten that $G$ is non-amenable iff $L_2$ has a bounded inverse (i.e., $0$ is not in the spectrum of $L_2$).

What is known in the case of $p=1$? Clearly if $G$ is amenable, then there are functions $f_n\in \ell^1(G)$ with the property that $\Vert L_1 f_n \Vert_1 / \Vert f_n \Vert_1 \to 0$ showing that $0$ is in the spectrum of $L_1$ and so $L_1$ is not invertible. Also, if $G$ is not amenable, then such $f_n$ cannot exist under the additional assumption that $f_n\geq 0$.

Given $G$ non-amenable, can one find a generating set $S$ with the property that $L_1$ is invertible?


2 Answers 2


There is no generating set for which this is possible (actually this is true for any graph with an infinite connected component). Furthermore, your claim that (in non-amenable groups) such $f_n$ cannot exist if $f_n \geq 0$ is false. Here is a detailed proof.

Lemma 1: If $0 \leq v,w \in \ell^1G$ then $\|v+w\|_{\ell^1} = \|v \|_{\ell^1} + \|w\|_{\ell^1}$.

Proof: for all $g \in G$, $|v(g) + w(g)| = v(g) + w(g) = |v(g)| + |w(g)|$ this passes to the $\ell^1$-norm (by summing both sides over all $g$). $\hspace{10.5cm} \square$

Lemma 2: For any function $v \in \ell^1G$ and for any $g \in G$, $\|gv\|_{\ell^1} = \|v\|_{\ell^1}$. If further $v \geq 0$, $gv \geq 0$.

Proof: $gv$ has the same values as $v$ but permuted by the group action, so the norm is the same. The same reasoning shows that $Pv$ is positive when $v$ is. $\hspace{7.5cm} \square$

Definition: Let $P$ be the random walk operator: $P = \frac{1}{|S|} \sum_{g \in S} g$ or $P = L+I$ (where $I$ is the identity).

Lemma 3: for any $0 \leq v \in \ell^1G$, $\|Pv\|_{\ell^1} = \|v\|_{\ell^1}$ and $Pv \geq 0$.

Proof: Note that for any $g \in G$, $gv \geq 0$. $$ \begin{array}{rll} \|Pv\|_{\ell^1} &= \| \tfrac{1}{|S|} \sum_{g \in S} gv \|_{\ell^1} & = \tfrac{1}{|S|} \| \sum_{g \in S} gv \|_{\ell^1} \\ &\overset{Lemma 1}{=} \tfrac{1}{|S|} \sum_{g \in S} \| gv \|_{\ell^1} &\overset{Lemma 2}{=} \tfrac{1}{|S|} \sum_{g \in S} \| v \|_{\ell^1} \\ &= \| v \|_{\ell^1} \end{array} $$ $Pv$ is a strictly positive scalar times a sum of positive function, so it is positive. $\hspace{3cm} \square$

Notation: $P^k$ is $P$ applied $k$ times (e.g. $P^3 v= PPPv$) and $P^0 = I$.

Lemma 4: Let $f_n = \displaystyle \sum_{i=0}^n P^i v$ where $v \geq 0$ and $\|v\|_{\ell^1} =1$. Then $\|f_n\|_{\ell^1}= n+1$.

Proof: $$ \begin{array}{rll} \|f_n\|_{\ell^1} &= \displaystyle \| \sum_{i=0}^n P^i v\|_{\ell^1} & \overset{Lemma 1}{=} \displaystyle \sum_{i=0}^n \| P^i v\|_{\ell^1}\\ & \overset{Lemma 3}{=} \displaystyle \sum_{i=0}^n \|v\|_{\ell^1} & = n+1. \\ \end{array} $$

Lemma 5: Let $f_n = \displaystyle \sum_{i=0}^n P^i v$ where $v \geq 0$ and $\|v\|_{\ell^1} =1$. Then $\displaystyle \frac{\|L f_n\|_{\ell^1}}{\| f_n \|_{\ell^1}} \to 0$.

Proof: Let's compute $Lf_n$ using $L=P-I$: $$ Lf_n = \displaystyle \sum_{i=0}^n LP^i v = \displaystyle \sum_{i=0}^n (P-I)P^i v = \displaystyle \sum_{i=0}^n \big( P^{i+1}v-P^i v \big) = P^{n+1}v -v $$ Now $\|Lf_n\|_{\ell^1} = \|P^{n+1}v -v\|_{\ell^1} \overset{TI}{\leq} \|P^{n+1}v\|_{\ell^1} + \|v\|_{\ell^1} = 2$ where $TI$ stands for the triangle inequality.

Using Lemma 4 $\displaystyle \frac{ \|L f_n\|_{\ell^1}}{\|f_n\|_{\ell^1}} \leq \frac{2}{n+1}$ which obviously tends to 0.

Corollary: 0 is in the spectrum of $L_1$ and the Laplacian cannot be inverted $\ell^1$.

Note the argument is true in any graph. If $v$ is supported on a finite component, then $\tfrac{1}{n+1} f_n$ tends to the constant function, which is indeed in the kernel. Otherwise $\tfrac{1}{n+1} f_n$ tends weak$^*$ to 0 but not in norm.

As a complementary remark the image of the Laplacian is not dense (in a graph with an infinite connected component). By taking $v = \delta_x$, the Dirac mass at some vertex $x$, the sequence $f_n$ above shows its image is weak$^*$ dense. It's easy to check that its image lies in $\ell^1_0X = \lbrace f \in \ell^1X \mid \sum_{x \in X} f(x) =0 \rbrace$. One can further check that $\overline{\mathrm{Im} \Delta} = \ell^1_0 X$ if and only if the graph has no non-constant bounded harmonic functions. Non-amenable groups always have non-constant bounded harmonic functions so that the (norm) closure of the image of $\Delta$ is strict subspace of $\ell^1_0 X$.


The Laplacian is never invertible on $\ell^1(G)$.

There is already a correct answer by ARG. I am writing this answer as there is a demand for more information. I will thus try to provide a more conceptual point of view. But I should remark that my answer is essentially the same as ARG's answer.

Let $\mu$ be a probability measure on $G$ and assume the Laplacian $\Delta=1-\mu$ is an automorphism of $\ell^1(G)$. By taking the double dual, $\Delta$ extends to an automorphism of $\ell^\infty(G)^*$. Recall that a mean on $G$ is a norm 1 positive element of $\ell^\infty(G)^*$ and let $M$ be the space of all means on $G$. It is easy to see that $M$ is convex and compact for the w*-topology. Note that $G$ acts on $M$ preserving its structure. In particular, $M$ is $\mu$-invariant. By Markov–Kakutani fixed-point theorem there is a $\mu$-invariant mean $m\in M$. Thus $\ker \Delta\neq 0$ as $$ \Delta m=m-\mu m =0. $$ This is a contradiction.

Note that, by the same argument, $\Delta$ is never an automorphism of $\ell^\infty(G)$. For completeness let me add that for $1<p<\infty$ and $\mu$ with a generating support, $\Delta$ is an automorphism of $\ell^p(G)$ iff $G$ is not amenable. For this recall that the Mazur map $f \mapsto \text{sgn}(f)|f|^{2/p}$ is a $G$-equivariant uniformly continuous homeomorphism from the unit sphere of $\ell^2(G)$ to the unit sphere of $\ell^p(G)$. It follows that there exist almost invariant vectors in $\ell^p(G)$ iff there exist almost invariant vectors in $\ell^2(G)$, and this happens iff $G$ is amenable (by Kesten). Thus, if $G$ is amenable $\Delta$ is not an automorphism (and this works also for $p=1$). If $G$ is not amenable, by the uniform convexity of the unit ball of $\ell^p(G)$ (which fails for $p=1$), we get that $\|\mu\|<1$, thus the sum $\sum_{n=0}^\infty \mu^n$ exists in the space of bounded operators, and since $(1-\mu)\sum_{n=0}^\infty \mu^n=1$, $\Delta$ is indeed an automorphism.

  • $\begingroup$ On a non-mathematical note: it appears that the OP is no longer active on this site $\endgroup$
    – Yemon Choi
    Commented Dec 10, 2017 at 21:09
  • 1
    $\begingroup$ Oh... I didn't realize that the bounty was suggested by ARG. I thought it was offered by the OP. I was indeed surprised, as ARG's answer is perfect. $\endgroup$
    – Uri Bader
    Commented Dec 10, 2017 at 21:51
  • 1
    $\begingroup$ @Yemon Thanks for asking. Indeed, I was very brief at this part. Yes, the point is that the Mazur map intertwines the action. It follows that the existence of almost invariant vectors is a phenomena independent of $1\leq p <\infty$. This statement was recorded as Remark 4.3 of arxiv.org/pdf/math/0506361.pdf, and further explanation could be found there. But the thing I wanted to emphasize here, is that not having almost invariants does not automatically imply invertability. It does, under further uniform convexity assumption, which is what fails for $p=1$. $\endgroup$
    – Uri Bader
    Commented Dec 10, 2017 at 22:07
  • 1
    $\begingroup$ By the way, another method to do the $\ell^p$ case is as follows (it can be found in a very short paper of Lohoue). Check that the operator norm of $P$ is $\leq 1$ for $\ell^\infty \to \ell^\infty$ and for $\ell^1 \to \ell^1$. Now a group is amenable iff $\|P\|_{\ell^2 \to \ell^2} < 1$. By the Riesz-Thorin interpolation, $\|P\|_{\ell^p \to \ell^p} \leq \|P\|_{\ell^2 \to \ell^2}^r \|P\|_{\ell^1 \to \ell^1}^s$ or $\|P\|_{\ell^p \to \ell^p} \leq \|P\|_{\ell^2 \to \ell^2}^r \|P\|_{\ell^\infty \to \ell^\infty}^s$ ($r$ and $s$ depend on $p$, as well as the bound by $\ell^1$ or $\ell^\infty$. $\endgroup$
    – ARG
    Commented Dec 11, 2017 at 16:14
  • 2
    $\begingroup$ Let me also add that, by my personal experience, I am not expecting people to upvote good answers. It does happen some time, but usually a good question with a simple good answer are just being ignored by most of us. This is compatible with Cunningham's Law, see meta.wikimedia.org/wiki/Cunningham%27s_Law. I have more to say here, but probably this is not the right forum. $\endgroup$
    – Uri Bader
    Commented Dec 13, 2017 at 7:09

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