Polygonal normal forms for compact connected surfaces

Apply [L1] to obtain a finite triangulated model of $S$, and then a one-polygon surface schema $P$ with quotient homeomorphic to $S$. Let $W$ be its finite cyclic paired word. Its word has at least one pair by [L2]. All subsequent operations are the finite quotient-preserving operations of [L3], with the local vertex reduction detailed in step 2.

Whenever $W$ has a cyclic adjacent inverse pair $a{a}^{-1}$ and at least one other paired edge remains, use the inverse-pair cancellation of [L3]. It removes one paired edge, so finite repetition stops. If $W$ consists only of $a{a}^{-1}$, stop at this actual digon. Its formal reduced word is empty; no zero-sided polygon is constructed.

Suppose $W$ still has more than one vertex class and no adjacent inverse pair. Choose a vertex class $\alpha$ joined by an edge $b_1$ to a different class; such an edge exists by connectedness of the finite one-skeleton [L4]. Relabel its reference direction so the selected occurrence is $b_1$. In the cyclic oriented-corner list at $\alpha$, let $b_2$ be the oriented edge following $b_1$ locally. Since an immediately returning inverse edge would be an adjacent inverse pair, this list has at least two members. Reading the face from $b_1$, its word has one of the two forms $b_1{b}_2^{-1}A{b}_{2}B$ or $b_1{b}_2^{-1}A{b}_2^{-1}B$, with $A,B$ possibly empty. In either case split the polygon along a fresh diagonal $c$, obtaining face words $b_1{b}_2^{-1}c$ and $c^{-1}A{b}_2^{\epsilon }B$, where $\epsilon =+1$ or $-1$. If $\epsilon =+1$, rotate and merge along the opposite $b_2$ sides to get the cyclic one-face word $B{c}^{-1}Ac{b}_1$. If $\epsilon =-1$, reverse the orientation of the second face, whose word becomes $B^{-1}{b}_2{A}^{-1}c$, then merge it with the first face along the now opposite $b_2$ sides. The cyclic result is $c{b}_1{A}^{-1}c{B}^{-1}$. Reversal of one face before gluing changes only its presentation, and each split/merge preserves the quotient by [L3]. This equal-sign branch is necessary for nonorientable words such as $abab$, which has two vertex classes and no adjacent inverse pair.

For the exact vertex-class calculation, use oriented edge germs: $b$ and $d$ lead to the same vertex when a boundary occurrence of $d^{-1}$ succeeds an occurrence of $b$, and take the transitive closure. This is the side-endpoint pairing rule [L4] expressed with directed sides. Before either move of step 3.1, $b_1,b_2$ belong to the selected cyclic class $\alpha$ because $b_2^{-1}$ succeeds $b_1$. In the split face $b_1{b}_2^{-1}c$, the new $c^{-1}$ germ follows the same old $b_2^{-1}$ side. In the opposite-sign branch, the merged word $B{c}^{-1}Ac{b}_1$ identifies $c$ with the old $b_1^{-1}$ successor; in the equal-sign branch, reversing the second oriented face gives $c{b}_1{A}^{-1}c{B}^{-1}$ and the corresponding reversed successor relation. For a direct corner check, number the input corners $v_0,v_1,\dots$ starting before $b_1$, and write $m=\mathrm{\mid }A\mathrm{\mid }$. The input corner $v_1$ just after $b_1$ is paired by the two $b_2$ sides with $v_{m+3}$ in the opposite-sign case, and with $v_{m+2}$ in the equal-sign case. The displayed merged word has one corner after $b_1$ representing exactly that paired pair. The vertex class containing input $v_0$ (before $b_1$, outside $\alpha$) gains one occurrence: in the equal-sign output the corners just before $b_1$ and just after the second $c$ both represent $v_0$; in the opposite-sign output the corner after $B$ and the corner before $b_1$ both represent $v_0$. Every remaining old corner has one output representative. Thus the total number of boundary corners stays fixed, the selected class shrinks and the other class gains one occurrence. Thus the chosen class $\alpha$ loses one corner occurrence in either sign case. In oriented-germ language, if $b_2^{-1}\mathrm{\notin }\alpha$, $b_2$ leaves $\alpha$; if $b_2^{-1}\in \alpha$, both $b_2$ and $b_2^{-1}$ leave and one $c$ orientation replaces them. All other members stay in the same cyclic order. The split and inverse merge preserve $E,F$ and the quotient surface, hence preserve $V$ by [L5] and the finite-cell Euler count; equivalently direct endpoint tracing gives the same result. This is Gallier–Xu Lemma 6.1 Step 2's oriented-corner calculation applied to a one-face schema, including its inverse-face option. Repeat at most $\mathrm{\mid }\alpha \mathrm{\mid }-1$ times. At length one its incident side returns immediately as an adjacent inverse pair, which step 2.1 cancels; then both $V$ and $E$ decrease by one. Restart with a remaining nontrivial vertex class. Each outer restart decreases finite $V$ and each inner operation decreases finite selected-class length, so the process ends at one vertex class or at the sphere digon.

Assume a nonterminal one-vertex word with no adjacent inverse pair. If a letter $a$ occurs twice in the same boundary direction, write the cyclic word $aXaY$. The same-direction extraction of [L3] gives $bb{Y}^{-1}X$, an adjacent square block followed by a paired residual word. It replaces one side pair by one side pair, keeps one face and preserves the quotient surface, so [L5] and the finite-cell Euler count keep the vertex count equal to one. Every previously extracted square block consists of two consecutive occurrences of a different label, so neither occurrence of the selected $a$ can cut between its two sides. Such a block lies wholly in $X$ or $Y$; the rewrite may reverse its order and reference direction, but its two equal-sign occurrences remain consecutive. Declare these intact blocks processed and repeat on the other paired letters. Each extraction consumes one previously unprocessed side pair, so this stage is finite.

If no unprocessed pair has equal exponents, every remaining pair is of the form $a,a^{-1}$. The one-vertex nonterminal schema has no adjacent inverse pair: in a cyclic string $b{b}^{-1}Y$ with $Y$ nonempty, the corner between $b$ and $b^{-1}$ is paired only with itself by those two sides and is separate from the corners of $Y$, contradicting one vertex. Now take any unprocessed inverse pair $b,b^{-1}$. If no other pair crosses it, rotate the word to $bX{b}^{-1}Y$. Both $X$ and $Y$ are nonempty by the preceding observation. Every other paired label has both occurrences wholly in $X$ or wholly in $Y$; in particular an earlier contiguous square or handle block cannot straddle the two arcs. The opposite-direction $b$ pairing joins the two end corners of $X$ to each other and the two end corners of $Y$ to each other, while every other paired side joins corners within its own arc. Thus the corner classes carried by $X$ and $Y$ stay distinct, again contradicting the one-vertex hypothesis. Consequently an unprocessed inverse pair has a crossing partner, and that partner is unprocessed because each processed block is contiguous. Orient and cyclically rotate the two interlaced pairs, reversing the polygon if needed, to obtain $aUbV{a}^{-1}X{b}^{-1}Y$. By [L3] its interlaced pair extracts one commutator block $cd{c}^{-1}{d}^{-1}$, leaving residual word $YXVU$. This move replaces two paired labels by two and keeps one face and the quotient homeomorphism type, so [L5] and the Euler cell count keep one vertex for the next repetition. Any earlier contiguous square or handle block involves labels different from $a,b$ and therefore lies wholly in one of $U,V,X,Y$; the rewrite only moves those strings, so it never splits an earlier block. Reversal of a whole block, if needed by a preceding square extraction, still gives a square block or a commutator block after relabelling. Repeat on the shorter unprocessed residual word. The number of unprocessed pairs drops by two at each step, so this too terminates.

After steps 5.1–6.1, the word is a finite product of handle and square blocks. When both kinds occur, cyclically put one square block next to one handle block and apply the mixed move of [L3], replacing that pair by three square blocks. The handle-block count drops by one. Iterate until either all blocks are handles or all are squares. All intermediate cyclic words have a positive even number of sides and remain surface schemas by [L2, L3]. Consequently the terminal form is one of the three forms in the statement. The only AC use is the one inherited from the triangulation in [L1]; the word manipulations and their termination use finite choices.