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\title{The Sharp Commutator Bound for the Polymath14 Inequality}

\begin{document}

\begin{center}
    {\Large \bfseries The Sharp Commutator Bound for the Polymath14 Inequality}\\[0.5cm]
	We present a simplified proof of the Polymath14 inequality, along with a new sharpness result.
\end{center}

Let $G$ be a group. Write $\cl_G(g)$ for the commutator length of $g$, with $\cl_G(g)=\infty$ if $g\notin[G,G]$, and set $\scl_G(g)\coloneqq\inf_{n\ge1}\cl_G(g^n)/n$.

\begin{theorem}
	Let $f\colon G\to\mathbb R$ satisfy
	\[
		f(g^2)\ge 2f(g)-1
		\qquad\text{and}\qquad
		f(gh)\le f(g)+f(h)
	\]
	for all $g,h\in G$. Then $f(g)\le 1+2\scl_G(g)$ whenever $\scl_G(g)<\infty$. In particular, $f([g,h])\le2$ for all $g,h\in G$.

	Moreover, there exists $F\colon G\to\mathbb R$ satisfying the same two hypotheses and $F(g)=1+2\scl_G(g)$ whenever $\scl_G(g)<\infty$.
\end{theorem}

We call the two hypotheses the \emph{doubling inequality} and \emph{subadditivity}.

\begin{lemma}\label{lem:power}
	For every $g\in G$ and $k\ge0$,
	\[
		f(g)\le \frac{f(g^{2^k})}{2^k}+1.
	\]
\end{lemma}

\begin{proof}
	Iterate $f(g)\le \frac12f(g^2)+\frac12$.
\end{proof}

\begin{lemma}\label{lem:splitting}
	Assume that $f$ is conjugation invariant. If $g$ is conjugate to $h$, then,
	for every $x\in G$,
	\[
		f(g)=f(h)\le \frac{f(x^{-1}g)+f(hx)}2+1.
	\]
\end{lemma}

\begin{proof}
	Choose $y\in G$ such that $g=yhy^{-1}$. Since
	\[
		g^{n+1}yh^{n+1}
		\sim (x^{-1}g)g^nyh^n(hx),
	\]
	repeated use of conjugation invariance and subadditivity gives
	\[
		f(g^Nyh^N)\le f(y)+N\bigl(f(x^{-1}g)+f(hx)\bigr).
	\]
	Also $g^{2N}=g^Nyh^Ny^{-1}$. Taking $2N=2^k$, applying \cref{lem:power}, and letting $k\to\infty$ gives the result.
\end{proof}

\begin{lemma}\label{lem:commutator}
	Assume that $f$ is conjugation invariant. Then $f([g,h])\le3$ for all
	$g,h\in G$.
\end{lemma}

\begin{proof}
	This is the random-walk argument of \cite[Proposition~2.1]{polymath}. Fix $g,h\in G$ and put $q=[g,h]$. We have
	\begin{align*}
		f(g^mq^k)
		 & \le \frac{f(g^{-1}g^mq^k)+f(h^{-1}g^mq^khg)}2+1
		 &                                                 & (\Cref{lem:splitting})           \\
		 & =\frac{f(g^{m-1}q^k)+f(h^{-1}g^mq^{k-1}gh)}2+1                                     \\
		 & =\frac{f(g^{m-1}q^k)+f(g^{m+1}q^{k-1})}2+1
		 &                                                 & (\text{conjugation invariance}).
	\end{align*}

	Let $S_{2n}$ be the sum of $2n$ independent random signs. Iterating the inequality above from $(m,k)=(0,n)$ gives
	\[
		f(q^n)\le
		\E f\left(g^{S_{2n}}q^{-S_{2n}/2}\right)+2n.
	\]
	By subadditivity, there is a constant $C=C(g,h)$ such that $f(g^sq^{-s/2})\le C(\abs{s}+1)$ for every even integer $s$. Since $\E\abs{S_{2n}}\le \sqrt{\E(S_{2n}^2)}=\sqrt{2n}$, we obtain $f(q^n)\le C(\sqrt{2n}+1)+2n$. Taking $n=2^k$, applying \cref{lem:power}, and letting $k\to\infty$ gives $f(q)\le3$.
\end{proof}

\begin{proof}[Proof of the theorem]
	We first reduce to the conjugation-invariant case. By \cref{lem:power} and subadditivity, $f(xgx^{-1})\le f(g)+(f(x)+f(x^{-1}))/2^k+1$, so $f(xgx^{-1})\le f(g)+1$. Thus $\bar f(g)\coloneqq\sup_{x\in G}f(xgx^{-1})$ is finite, dominates $f$, is conjugation invariant, and satisfies the same two hypotheses. Replacing $f$ by $\bar f$, we may assume that $f$ is conjugation invariant.

	Set $M\coloneqq\sup_{g,h\in G}f([g,h])$. By \cref{lem:commutator}, $M\le3$. We claim that $f(g)\le1+M\scl_G(g)$ whenever $\scl_G(g)<\infty$. Let $r$ be the order of the image of $g$ in $G^{\rm ab}$. \cref{lem:power} and subadditivity give
	\[
		f(g)\le
		1+\frac{\max_{0\le a<r}f(g^a)+M\max\{1,\cl_G(g^{r\lfloor2^k/r\rfloor})\}}{2^k}.
	\]
	Letting $k\to\infty$ proves the claim.

	For a commutator $q$, Culler's identities\cite{culler} give $\cl_G(q^N)\le\lfloor N/2\rfloor+1$, hence $\scl_G(q)\le1/2$. The claim gives $f([g,h])\le1+M/2$, so $M\le1+M/2$ and hence $M\le2$. Applying the claim again gives $f(g)\le1+2\scl_G(g)$ whenever $\scl_G(g)<\infty$.


	For the second assertion, choose $(g_i)_{i\in I}$ in $G$ whose images form a basis of $G^{\rm ab}\otimes_{\mathbb Z}\mathbb Q$, and let $\mathcal Q$ be the set of homogeneous quasimorphisms $\phi\colon G\to\mathbb R$ with defect $D(\phi)\le1$ such that $\phi(g_i)=0$ for all $i\in I$, where
	\[
		D(\phi)\coloneqq\sup_{g,h\in G}
		\abs{\phi(gh)-\phi(g)-\phi(h)}.
	\]
	
	Set $F(g)\coloneqq1+\sup_{\phi\in\mathcal Q}\abs{\phi(g)}$. Since the images of the $g_i$ form a basis of $G^{\rm ab}\otimes_{\mathbb Z}\mathbb Q$, there exist $m\ge1$, indices $i_1,\ldots,i_k$, and integers $n_1,\ldots,n_k$ such that $z\coloneqq g^m g_{i_1}^{-n_1}\cdots g_{i_k}^{-n_k}\in[G,G]$. Writing $\ell=\cl_G(z)$, every $\phi\in\mathcal Q$ satisfies
	\[
		m\abs{\phi(g)} \le \abs{\phi(z)}+k \le 2\ell+k.
	\]
	Thus $F(g)<\infty$.

	The defect condition gives $F(gh)\le F(g)+F(h)$, while homogeneity gives $F(g^2)=2F(g)-1$. Suppose $\scl_G(g)<\infty$, and let $r$ be the order of the image of $g$ in $G^{\rm ab}$. Bavard duality~\cite[Theorem~2.70]{calegari-book} for $g^r$, together with homogeneity, gives
	\[
		2\scl_G(g)=\sup_{D(\phi)\le1}\abs{\phi(g)}.
	\]
	Given such a $\phi$, let $\chi\colon G\to\mathbb R$ be the homomorphism determined by $\chi(g_i)=\phi(g_i)$ for all $i\in I$. Since $g$ has torsion image in $G^{\mathrm{ab}}$, we have $\chi(g)=0$. Thus $\phi-\chi\in\mathcal Q$ has the same defect and value at $g$ as $\phi$. Hence the supremum may be restricted to $\mathcal Q$, giving $F(g)=1+2\scl_G(g)$.
\end{proof}

\begin{thebibliography}{9}

	\bibitem{polymath}
	D.H.J. Polymath,
	\emph{Homogeneous length functions on groups},
	Algebra Number Theory \textbf{12} (2018), 1773--1786;
	\href{https://arxiv.org/abs/1801.03908}{arXiv:1801.03908}.

	\bibitem{culler}
	M. Culler,
	\emph{Using surfaces to solve equations in free groups},
	Topology \textbf{20} (1981), 133--145.

	\bibitem{calegari-book}
	D. Calegari,
	\emph{scl},
	MSJ Memoirs, vol.~20,
	Mathematical Society of Japan, Tokyo, 2009.

\end{thebibliography}

\end{document}