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With group commutator convention $[a,b]=a^{-1}{b}^{-1}ab$, the displayed group presentation describes the fundamental group of the Borromean rings complement. Its abelianization is ${\mathbb{Z}}^3$, since each relator is a group commutator and hence imposes no abelian relation. It embeds discretely in the Gaussian Bianchi group by sending the meridians to $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, $\left(\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right)$ and $\left(\begin{array}{cc}i& 1\\ 2i& 2-i\end{array}\right)$.

Choose bounded-length substitutions $T$ from the first generating set to the second, and $U$ back. Each substituted relator of the second group presentation has bounded area in the first, and every generator correction $a(UT(a){)}^{-1}$ has bounded area. Substitution into a minimal second-presentation diagram and filling the correction strips givesThe reverse construction gives the opposite inequality. Thus the optimal Dehn functions are coarsely equivalent, and the existence of a recursive isoperimetric bound, hence solvability of the word problem for a group, is independent of the finite group presentation.

A finite group presentation with cyclically reduced symmetrized relators is a Dehn presentation if every nonempty freely reduced null word contains a relator segment longer than half that relator. Replacing the segment by the inverse complementary segment strictly shortens the word. Finiteness of both the alphabet and the relator set is part of this definition; merely permitting all null words as an infinite relator set would make the shortening property vacuous as a finiteness criterion.

A group presentation with a cyclic list of generators $a_i$ and relations $a_i{a}_{i+1}{a}_i^{-1}=a_{i+1}^2$ has no nontrivial finite quotient of a group. In a finite image, all generator orders are odd. Choose the least prime $p$ dividing any nontrivial generator order and let $x$ be the predecessor of a generator $y$ whose order is divisible by $p$. Conjugation by $x$ acts as squaring on $⟨y⟩$, so the multiplicative order of $2$ modulo $p$ divides the order of $x$. That multiplicative order is greater than one and divides $p-1$, giving a prime divisor smaller than $p$ in the order of $x$, a contradiction. This argument applies to any cycle length; it asserts absence of finite quotients, not infinitude of the presented group.

A word in generators and their formal inverses is freely reduced if no consecutive letters are mutual inverses. Deleting such inverse pairs gives the unique reduced representative in a free group. In a general group presentation, free reduction preserves the represented group element but may not produce a shortest representative.

Free reduction removes adjacent inverse-letter pairs. A stack algorithm scans the letters, canceling the top letter when the next is its inverse. The resulting reduced word is unique: deleting an inverse pair before scanning does not change the final stack, so every cancellation sequence gives the same result. In any group presentation these deletions preserve the represented element, and in a free group the reduced word is the identity exactly when it is empty.

Suppose $G=⟨Y\mid R(Y)⟩$ is a finite group presentation, and $X$ is any finite generating set. Write $y_j=u_j(X)$ and $x_i=v_i(Y)$. A finite presentation on $X$ is obtained using the relators $R(u(X))$ and $x_i=v_i(u(X))$. The substitutions in both directions give inverse homomorphisms, proving that finite presentability does not depend on the finite generators chosen.

// Target: geometry-and-topology.bigb

A nondecreasing function bounding the area of every null word of length at most $n$ is an isoperimetric function for a finite group presentation. The Dehn function is the minimal such bound. A subrecursive bound means one dominated by a total computable function. Arbitrarily enlarged bounds are not invariants: even a group presentation with zero area function admits both linear and exponential upper bounds, which are not equivalent under the usual coarse comparison.

The required quasi-isometry is the identity on the group vertices. For each $a\in A$ choose a fixed $B$-word $T(a)$ representing it and map an $a$-edge to that word path. The bounded substitutions in Q2(ii) prove the distortion inequalities.

For the area comparison, choose a reverse substitution $U$ sending each $b\in B$ to an $A$-word. Let $L$ bound the lengths of $T(a)$. Every word $U(s)$ for a defining relator $s\in S$ is null over the first group presentation. Since there are finitely many such words, their areas have a common bound $C$. Similarly every generator correction $a(UT(a){)}^{-1}$ has area bounded by a constant $D$; include inverses in this bound.

Let $w$ be null over $A$ with length at most $n$. Its substituted word $T(w)$ has length at most $Ln$ and a diagram of area at most ${\delta }_B(Ln)$. Replace each $B$-edge by its $U$-path. Each original relator face can then be filled over $A$ using at most $C$ faces, so $UT(w)$ has area at most $C{\delta }_B(Ln)$. Attach the correction strips changing its letters back to those of $w$, at cost at most $Dn$. ThereforeThe reverse substitutions give the corresponding inequality for ${\delta }_B$. Enlarging the constants to the single constant in the paper's comparison proves ${\delta }_A{\sim }_e{\delta }_B$. This is Dehn functions under a change of finite presentation.

Two qualifications are necessary for the printed continuation. The invariant is the optimal isoperimetric function, namely the Dehn function, not arbitrary upper bounds. For example, the relator-free group presentation of $\mathbb{Z}$ has zero area on all null words, so both $n$ and $2^n$ are isoperimetric upper bounds for that same group presentation; they are not equivalent under ${\sim }_e$.

Also, the comparison supplies no recursive bound unless one was available beforehand. Its actual consequence, by part (ii), isThe unconditional assertion that every finitely presented group has a solvable word problem is false. Finitely presented counterexamples are established in the primary paper The Word Problem. Thus the missing solvability premise cannot be inferred from group presentation equivalence alone.

First construct the bounded-difference automaton. Fix $K$ as above, enlarging it if necessary. Its states are the finitely many elements $d\in B_K(1)$, together with a reject state and finite flags recording when padding has started on each tape. The initial state is $1$. Read the pair of words synchronously, padding their ends with a symbol $\#$ representing the identity. If the current prefixes represent $g,h$, the difference state is $d=g^{-1}h$. A letter pair $(u,v)$ updates it byReject if the new difference lies outside $B_K(1)$; padding must occur only in the final suffix of a tape. The accepting state for the generator $a$ is $d=a$. Induction on the prefix length proves that the machine accepts exactly the padded pairs whose prefix differences stay bounded by $K$ and whose final difference is $a$. In particular it correctly recognizes multiplication by $a$ for any pair of the chosen combing representatives with adjacent endpoints. Taking accepting state $1$ gives the equality machine.

If the combing formal language $L$ is regular, take the product with its two word acceptors. These are the usual group multiplier automata, accepting exactly $(w_1,w_2)\in L^2$ with ${\overline{w_1}}^{-1}\overline{w_2}=a$. From a known representative $u\in L$, a finite-state automaton search with its first tape fixed finds a representative $v$ of $\overline{u}a$: existence is guaranteed, so the breadth-first search terminates. Equality is tested by the identity multiplier. Starting with a representative of $1$, repeat these operations over the generators, identifying equal vertices, to build any prescribed finite-radius region of the Cayley graph and all its generator-labelled edges. Iterating multiplication over an input word and testing equality with the identity gives a word problem algorithm.

A bare geometric combing does not imply that $L$ is a regular language. Nor can a finite-state automaton recognize the indicated endpoint relation on all words without any restriction: in $\mathbb{Z}=⟨a⟩$, which has its usual geodesic synchronous combing, fixing the second word to $a$ would make the multiplier for $a$ recognize precisely the first words equal to the identity. Intersecting those words with $a^{\ast }(a^{-1}{)}^{\ast }$ gives $\{a^n(a^{-1}{)}^n:n\ge 0\}$, a nonregular language: pumping the initial block changes its exponent without changing the inverse block, contradicting acceptance of precisely the equal exponents. Thus the literal unrestricted-pair request is false; the machines above recognize bounded fellow-travelling pairs, or full representative-pair relations under the additional automatic-language hypothesis.

Nevertheless the final decidability assertion is true for a synchronous combable group, without a regular-language assumption. Here is a terminating filling bound, avoiding any assumption that the given combing is computable. Let $D$ be the number of words of length at most $K$ over $\Sigma$. A horizontal row of the diagram in part (ii) is encoded by its ordered $n$ rung words, so there are at most $D^n$ possible row types. If two rows have the same word tuple, delete the band between them and identify corresponding labelled rungs. This preserves the outside boundary and leaves a valid diagram of the same short relators. Equivalently, the first column is always at $1$, so identical rung tuples describe the same vertex configuration. Repeatedly remove such bands. At most $D^n$ distinct rows remain, with at most $n$ short-relator cells per layer.

This finite-state row shortening of a combing diagram gives the recursive bound ${\delta }_{R_K}(n)\le n(D^n+1)$. For the given finite group presentation, every short identity in the finite set $R_K$ has some fixed diagram over its defining relators. Let $C$ bound their areas. Replacing each short face by its fixed diagram givesPart 3(ii) now supplies a terminating word decider. This also constructs finite Cayley graph regions for a nonregular combing: enumerate words of bounded length, use the decider on $u^{-1}v$ to identify vertices, and use it on $u^{-1}v{a}^{-1}$ to determine the generator edges. The finite ball $B_K(1)$ and transition table of the bounded-difference machines can consequently be computed. The constants depend on a fixed combing and are not a uniform procedure certifying combability of an arbitrary input group presentation. Every finite group presentation of this synchronous combable group has solvable word problem, by the presentation-invariance proof in Q3.

Take an integer $K\ge 1$ bounding synchronous distances at integer times. Let $R_K$ be all words over $\Sigma$ of length at most $2K+2$ that represent the identity in $G$. This is a finite set. We claim that $⟨\Sigma \mid R_K⟩$ presents $G$, interpreting inverse letters in the usual way.

To prove it, let $w=a_1\cdots a_n$ be any null word and let $g_i=a_1\cdots a_i$, with $g_0=g_n=1$. Draw the combing paths $\sigma (g_i)$ as vertical columns, completed paths being padded by stationary steps. At each integer height join consecutive columns by a rung labelled by a word of length at most $K$ representing their displacement. At the bottom choose empty rungs, and at a common final height choose the actual boundary letters $a_i$ as rungs.

Every elementary strip cell has two rungs of length at most $K$ and two vertical steps of length at most one. Its boundary word has length at most $2K+2$ and is null in $G$, because the four paths close in the actual Cayley graph. It is therefore a relator in $R_K$. Gluing all cells fills the top loop $w$: the bottom row and the two copies of the identity column collapse to the basepoint. Thus every null word is a product of conjugates of these finitely many short relators. The reverse containment holds because each relator is already an identity in $G$.

Hence the finite presentation from a synchronous combing provesThis is an existence proof of a finite relator set. It does not assume an algorithm for identifying identities in an arbitrary group presentation before establishing decidability.

We use $[a,b]=a^{-1}{b}^{-1}ab$. To obtain a Wirtinger presentation, choose an oriented regular knot diagram, cut its strands at undercrossings, and give each resulting arc a positively oriented meridian of a knot. The complement of the arcs above the projection plane has a free fundamental group on these meridians. At a crossing, pushing an understrand meridian past the overstrand conjugates it by the overstrand meridian. Thus, if $a$ is the overstrand generator and $b,c$ are the incoming and outgoing understrand generators, the relation is $c=a^{ϵ}b{a}^{-ϵ}$, where $ϵ$ is determined by the crossing orientation. Applying the Seifert-van Kampen theorem while adjoining the crossing neighborhoods gives one such relation per crossing. The remaining three-dimensional region adds no fundamental-group relation. For a connected diagram one crossing relation is redundant. This derives a group presentation from any tame knot diagram, rather than just specifying a recipe without its topological justification.

For the Borromean rings, choose three meridians $x,y,z$ and call the other three arc meridians $x^{\prime },y^{\prime },z^{\prime }$. Base paths and orientations can be chosen so the six crossing relations readEliminate the primed generators with the first row. The second row says respectively that $x$ commutes with $(z^{\prime }{)}^{-1}z=[y^{-1},z]$, that $y$ commutes with $(x^{\prime }{)}^{-1}x=[z^{-1},x]$, and that $z$ commutes with $(y^{\prime }{)}^{-1}y=[x^{-1},y]$. The last relation is redundant, just as one crossing relation is redundant in the Wirtinger presentation. Alternatively the Hall-Witt identity, after conjugating its three factors, shows that any two of these cyclic relations imply the third. Hence the Borromean link group isBoth relators have zero exponent sum in each generator. Passing from the free group to its abelianization therefore imposes no relations beyond commutation, andEquivalently the three linking-number maps to $\mathbb{Z}$ independently detect the three meridians.

Here is an explicit representation, followed by a faithfulness argument:Each determinant is one. Direct multiplication verifies the two relators. For example $[Y^{-1},Z]=\left(\begin{array}{cc}-1& 2i\\ 0& -1\end{array}\right)$, which commutes with $X$. Merely checking those relators would not prove faithfulness, so we identify the image's complete group presentation.

Write $B=PS{L}_2(\mathbb{Z}[i])$, the Gaussian Bianchi group, and use its four generators $T,U,L,A$ from the allowed presentation. In $B$ the three proposed generators areThe last equality is in $PS{L}_2$, so an overall minus sign in matrix multiplication is immaterial. Reduce matrices modulo $4$ in $\mathbb{Z}[i]$, and then quotient by $\{\pm I\}$. The finite image $Q$ has order $1536$: the residue field is ${\mathbb{F}}_2$, $S{L}_2({\mathbb{F}}_2)$ has order six, and each of the three subsequent local-ring layers has a kernel of order $2^3$, before the final quotient by $\{\pm I\}$. Elementary elimination over this local ring shows that the images of $T,U,A$ generate it. The subgroup $\bar{H}=⟨\bar{X},\bar{Y},\bar{Z}⟩$ has order $64$, as multiplication in this finite ring verifies. Let $K$ be its inverse image in $B$; then $[B:K]=24$.

For completeness the following gives the coset computation needed to check the subgroup presentation explicitly. Cosets are right cosets $Kr$; columns record right multiplication by $T,U,L,A$. Representatives are words in the allowed generators.Apply the Reidemeister–Schreier theorem: for a transition $j{⟶}^{g}k$, use generator $r_{j}g{r}_k^{-1}$, omitting the 23 tree transitions. Rewrite each of the eight allowed ambient relators starting at each of the 24 rows. There are 73 remaining Schreier generators. Adjoin the names $x=T$, $y=A{U}^{-2}{A}^{-1}$, $z=A{T}^{-2}{A}^{-1}L{U}^{-1}$ and eliminate the Schreier generators using relations in which they occur once. The remaining relations, with an uppercase letter denoting the inverse of the corresponding lowercase one, are the wordsThis table and the rewriting rule determine every relation without an appeal to a guessed presentation. In this notation cyclically reduced versions of the two desired relators are $r=ZYzXZyzYxy$ and $q$. Elementary substitutions of relator segments, with cyclic reduction at each step, giveusing $r,q$; conversely $r⟶p⟶1$ using $p,q$. These substitutions show equality of the two normal closures. For example the first substitution replaces $xyX$ in the cyclic rotation of $p$ by $zxZyzXZ$, as allowed by $q$, and then the initial $ZYz$ is replaced by $YXyZYzx$ using $r$. Thus $K$ has exactly the presentation of $G$, with $x,y,z$ represented by $X,Y,Z$. In particular these three matrices generate $K$, and

To identify the actual link complement as a hyperbolic three-manifold, rather than only on a manifold with an isomorphic group, cut along the projection-plane regions, making the usual small crossing detours above and below. The octahedral projection graph has eight triangular regions. The upper and lower pieces are ideal octahedra; corresponding faces are matched with one-third-turn corner permutations, with opposite directions on adjacent checkerboard faces. Reversing these cuts reconstructs the link complement. Give both pieces the regular ideal hyperbolic space octahedron metric. All triangular face maps are isometries and each edge cycle has four dihedral angles $\pi /2$, so the metric has no edge singularity. Equal horospherical truncations match in Euclidean squares and assemble into three flat tori, giving complete torus cusps. The compact truncated remainder and these cusps show completeness and finite volume. Thus the Borromean complement is a complete finite-volume hyperbolic three-manifold. This also supplies a geometric discrete faithful holonomy, independently of the subgroup calculation.

The identified involution is $z=g^2=h^5$. It is central: it commutes with $g$ because it is a power of $g$, and with $h$ because it is a power of $h$. We use the normal form theorem for an amalgamated free product: the two factors embed, an alternating product of elements outside the amalgamated subgroup is nonidentity, and fixed coset representatives give a unique reduced expression. Here choose representatives $1,g$ in the first cyclic group and $1,h,h^2,h^3,h^4$ in the second.

An effective expression from the normal form theorem for an amalgamated free product iswhere each $x_j$ is a nonidentity representative and successive representatives come from different factors. Reduce powers by $g^u=z^{\lfloor u/2\rfloor }{g}^{u\mathrm{mod}2}$ and $h^v=z^{\lfloor v/5\rfloor }{h}^{v\mathrm{mod}5}$, using these identities for negative integers too. Move the central $z$ factors left and reduce their total exponent modulo two. Whenever adjacent representatives come from one factor, multiply them, split off their $z$ contribution, and delete a resulting identity. Each such combination reduces the number of representatives, so the procedure terminates. By the normal form theorem for an amalgamated free product, the input is trivial exactly when $\boxed{r=0\text{ and }\epsilon =0}$. In particular $z\ne 1$ because each factor embeds. This gives the soluble word problem in finite cyclic amalgams algorithm without merely asserting that a normal form exists.

In the specified quotient, the extra relator makes $g,h$ commute. Put $t=g{h}^{-2}$. Its abelian group presentation givesConversely $g\mapsto s^5$, $h\mapsto s^2$ satisfy every quotient relator in $C_{20}=⟨s\mid s^{20}=1⟩$, and send $t$ to $s$. Thus the quotient is exactly $C_{20}$, rather than just a cyclic group of order dividing twenty. If $M,N$ are the exponent sums of $g,h$ in the input, its quotient word problem is

A closed connected three-manifold has a handle decomposition with one zero-handle and one three-handle. To arrange this, cancel the zero-handles along a spanning tree of connecting one-handles; perform the corresponding construction in the dual handle decomposition to consolidate the three-handles. This does not require the three-manifold to be orientable.

If there are $r$ one-handles and $s$ two-handles, the Euler characteristic isMod-two Poincare duality pairs the Betti numbers of a closed odd-dimensional manifold, giving $\chi (Y)=0$. Thus $r=s$. The one-handles give $r$ generators of a group of the fundamental group, the attaching circles of the two-handles give $r$ relators, and the three-handle does not change the fundamental group. Therefore ${\pi }_1(Y)$ admits a balanced presentation.

Now letbe any finite group presentation. Start with a four-dimensional zero-handle and $r$ one-handles. Its fundamental group is the free group on the $x_i$. Represent the finitely many relators by disjoint embedded circles in its three-dimensional boundary; a small perturbation makes the circles disjoint. Choose framings of an embedded sphere and attach $s$ two-handles. The resulting compact connected oriented four-manifold $W$ has ${\pi }_1(W)\cong G$ by the Seifert-van Kampen theorem.

The map ${\pi }_1(\partial W)\to {\pi }_1(W)$ is surjective. One way to see this is to turn the handle decomposition upside down: the relative handles based on $\partial W$ have indices $2,3,4$, and so add no fundamental group generators. Also $\partial W$ is connected, since surgery on embedded circles in a connected three-manifold preserves connectedness.

Take the double of a manifoldIt is a closed connected smooth oriented four-manifold. The Seifert-van Kampen theorem givesBoth maps from ${\pi }_1(\partial W)$ are the same surjection, under the natural identification of the two copies. The pushout identifies the two copies of every element of $G$ and imposes no new relations: the fold homomorphism is inverse to either inclusion. ThusThis is four-manifold realization of finitely presented groups.

Orient $S^3$ as the boundary of $B^4$. Push the interior of an oriented Seifert surface $F$ slightly into $B^4$, and let $X_F$ be the double branched covering of $B^4$ over this pushed-in surface. Its boundary is the two-fold branched cover of a knot. Define the knot signature byThe signature of a possibly degenerate intersection form means the number of positive eigenvalues minus the number of negative eigenvalues; its radical contributes zero.

Here is why the definition is independent of $F$. Given two choices $F_0,F_1$, glue their branched covers along the common boundary to obtainThis is the double branched covering of $S^4$ over the closed oriented surface obtained by gluing the pushed-in $F_0$ to $-F_1$. Every closed oriented surface in $S^4$ admits a Seifert hypersurface: its complement has an integral meridional class, represent it by a map to the circle, and take a regular value. Near the surface choose the angular map in its trivial normal disk bundle. The closure of a suitable regular fiber is a compact oriented three-manifold bounding the surface. Its trivial normal framing can be chosen to match this construction.

Push the interior of this Seifert hypersurface into $B^5$. The double branched covering of $B^5$ over the resulting properly embedded three-manifold has boundary $X$. The existence of the cover follows from the meridian homomorphism to $\mathbb{Z}/2$; smoothness near the branch set is the local map $(z,u)\mapsto (z^2,u)$. By the hypothesis about boundaries of five-manifolds, $\mathrm{sign}X=0$. Novikov additivity for gluing along an entire closed three-dimensional boundary givesThus the knot signature is independent of the Seifert surface. An isotopy of the knot carries the construction to an equivalent one, so it is also a knot invariant.

To calculate its intersection form, express $F$ as a disk with $2g$ bands. The double branched covering of $B^4$ over the pushed-in disk is again $B^4$. Each band lifts to a two-handle. The capped lifted cores give a basis of $H_2(X_F;\mathbb{Z})$ corresponding to the band-core basis of $H_1(F;\mathbb{Z})$. The framing of the $i$th lifted attaching circle is $2\mathrm{lk}(a_i^{+},a_i)=2V_{ii}$. For distinct bands, the two sheets contribute the two push-off linkings $V_{ij}$ and $V_{ji}$, so the mutual linking number is $V_{ij}+V_{ji}$. HenceA different integral basis changes this matrix by matrix congruence and leaves its signature unchanged.

In Figure 7 the upper knot is a connected sum of knots $T\#T$, while the lower one is $T\#\overline{T}$, where $T$ is the positive trefoil knot. Both knot groups have the same presentationTo see this, the Seifert-van Kampen theorem expresses the knot group of a connected sum of knots as the amalgamated free product of the two summand knot groups, amalgamating their meridians of a knot. A trefoil knot has presentation $⟨a,b\mid aba=bab⟩$ with $a$ a meridian. Reflection gives the same abstract group for the mirror. If it reverses the chosen meridian, send both generators to their inverses; the inverse braid relation is again the same relation. Thus the meridian amalgamation yields the displayed group presentation in both cases.

A band basis for the positive trefoil knot giveswhose symmetrization is negative definite. Therefore $\sigma (T)=-2$ and $\sigma (\overline{T})=2$. Boundary-connected-summing Seifert surfaces makes the Seifert matrix block diagonal, so the knot signature is additive under connected sum of knots. ConsequentlyThe upper knot and lower knot are therefore not isotopic, even though their knot groups are isomorphic. With the opposite global knot signature convention the first value is $+4$, and the distinction is unchanged.

The group presentation gives the quotient group ${\mathbb{Z}}^3$ by the subgroup generated by the rows ofFor the Smith normal form, the first determinantal divisor is the greatest common divisor of all entries, namely $3$. The nonzero two-by-two matrix minors are $\pm 27$, so the second determinantal divisor is $27$. The third row is the second minus the first, and a nonzero minor shows the rank is exactly two. Consequently the nonzero invariant factors are $3$ and $27/3=9$, with one free generator. The Fundamental theorem of finitely generated abelian groups givesAn element killed by $3$ has zero free coordinate, arbitrary coordinate in $\mathbb{Z}/3\mathbb{Z}$, and coordinate $0,3$ or $6$ in $\mathbb{Z}/9\mathbb{Z}$. There are nine such elements. Removing the identity leaves eight elements of order three.