Finite C prime(1/6) presentations satisfy a linear isoperimetric inequality

Freely reduce $w$ to a word $u$. Because free reduction does not change the represented group element, $u$ is still null, and $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$. If $u$ is the empty word, then the null diagram with no faces has area $0$, so the claim is immediate. Otherwise let $D$ be a minimal-area van Kampen diagram for $u$. By [L2], the diagram $D$ is reduced, so [L1] applies.

By [L1], some face $f$ of $D$ contributes an outer boundary arc $p$ with $\mathrm{\mid }p\mathrm{\mid }>\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}2$. Let $q$ be the complementary boundary arc of $f$, so $\mathrm{\mid }q\mathrm{\mid }<\mathrm{\mid }p\mathrm{\mid }$. Replacing $p$ by $q^{-1}$ and freely reducing gives a null word $u^{\mathrm{\prime }}$ with $\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }-1$. Conversely, attach one $f$-cell along the occurrence of $q^{-1}$ in any minimal diagram for the unreduced replacement word and add the free-cancellation strips. This constructs a diagram for $u$ with one more face, so $\mathrm{Area}(u)\le \mathrm{Area}(u^{\mathrm{\prime }})+1.$

Induct on the freely reduced boundary length. Step 1.1 gives the base case $\mathrm{\mid }u\mathrm{\mid }=0$. For $\mathrm{\mid }u\mathrm{\mid }>0$, step 2.1 yields a shorter freely reduced null word $u^{\mathrm{\prime }}$. By the induction hypothesis, $\mathrm{Area}(u)\le \mathrm{Area}(u^{\mathrm{\prime }})+1\le \mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }+1\le \mathrm{\mid }u\mathrm{\mid }\mathrm{.}$ Because $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$, this is a linear isoperimetric inequality.

Finally, [F1] identifies van Kampen area with algebraic relator area, so the same linear bound holds in the algebraic formulation.