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Since the free group $F_2$ of rank $2$ is non-amenable, we have that $l^2$-homology of $F_2$ vanishes in degree $0$, i.e. $H_0(F_2,l^2(F_2))=0$. This means that for any $f\in l^2(F_2)$, the equation $$af_1-f_1+bf_2-f_2=f$$ should have a solution $(f_1,f_2)\in l^2(F_2)\times l^2(F_2)$. Here, $a,b$ are generators of $F_2$, $(af_1)(g)=f_1(ga)$ and $(bf_2)(g)=f_2(gb)$.

Is it possible to write down an explicit solution? For example in the simplest case $f=\delta_1$ where $1\in F_2$ is the unit element?

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The trick is to find an $\ell^2$ function $g\colon F_2\to\mathbb{R}$ whose graph Laplacian $\nabla^2 g$ is equal to $\delta_1$, i.e. $$ ag+a^{-1}g + bg + b^{-1}g - 4g \;=\; \delta_1. $$ If we can find such a function, we can rewrite the above equation as $$ a\bigl(g-a^{-1}g\bigr) \,-\, \bigl(g-a^{-1}g\bigr) \,+\, b\bigl(g-b^{-1}g\bigr) \,-\, \bigl(g-b^{-1}g\bigr) \;=\; \delta_1. $$ and then $f_1 = g-a^{-1}g$ and $f_2 = g-b^{-1}g$ will suffice.

To find the function $g$, consider the Cayley graph of $F_2$ with respect to the $\{a,b\}$ generating set. There is an obvious flow $\gamma$ on this graph whose divergence (or boundary, if you think of $\gamma$ as a $1$-chain) is $\delta_1$. Specifically,

  1. $\gamma(e) = 1/4$ for each of the four edges $e$ that emanate from $1$, with each of these edges directed to point away from $1$.

  2. $\gamma(e) = 1/12$ for each of the twelve outward-pointing edges at the second level.

  3. In general, $\gamma(e) = \dfrac{1}{4\cdot 3^n}$ for any edge $e$ that emanates from the ball of radius $n$ in $F_2$.

Next, we just need to find an $\ell^2$ function $g\colon F_2 \to \mathbb{R}$ whose gradient (or coboundary) is equal to $\gamma$. For any vertex $w$, the value of $g(w)$ should be the negative of the sum of the values of $\gamma(e)$ for the edges $e$ on a directed path from $w$ to infinity. Specifically, $$ g(1) \;=\; - \displaystyle\sum_{n=0}^\infty \frac{1}{4\cdot 3^n} \;=\; -\frac{3}{8}, $$ and it follows that $$ g(w) \;=\; -\frac{1}{8\cdot 3^{|w|-1}} $$ where $|w|$ denotes the word length of an element $w\in F_2$. It's easy to check that $g$ is in fact $\ell^2$, since $$ \sum_{w\in F_2} g(w)^2 \;=\; \sum_{n=0}^\infty \frac{|S_n|}{(8\cdot 3^{n-1})^2} \;=\; \left(\frac{3}{8}\right)^2 + \sum_{n=1}^\infty \frac{4\cdot 3^{n-1}}{(8\cdot 3^{n-1})^2} \;<\; \infty, $$ where $|S_n|$ denotes the cardinality of the $n$-sphere in $F_2$. By construction, the Laplacian of $g$ is equal to $\delta_1$.

Therefore, the desired functions $f_1 = g-a^{-1}g$ and $f_2 = g-b^{-1}g$ are given by $$ f_1(w) \;=\; \frac{1}{8\cdot 3^{|a^{-1}w|-1}} - \frac{1}{8\cdot 3^{|w|-1}} \qquad\text{and}\qquad f_2(w) \;=\; \frac{1}{8\cdot 3^{|b^{-1}w|-1}} - \frac{1}{8\cdot 3^{|w|-1}}. $$ Cheers!

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  • $\begingroup$ Hi Jim, thank you for this nice answer! :) $\endgroup$ – Werner Thumann Jun 5 '15 at 10:27

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