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From a relation to a planar surface

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson proves Theorem 2.1 of Equations over groups and the Kervaire conjecture: for a unimodular word \(w\), no element \(g\ne1\) of \(A\) lies in the normal closure of \(w\) in \(A*\langle t\rangle\) [OpenAI-K, Section 5]. A product of conjugates of \(w^{\pm1}\) equal to \(g\) gives a map of a disk with holes into a space whose fundamental group is \(A*\langle t\rangle\). The circle coordinate of this map, made piecewise affine, has a level set made of arcs that join the occurrences of \(t\) in pairs; thickening these arcs gives bands, and with the holes they form disk-and-band surfaces. Their components are removed from the inside out: each time, all inner boundary words are trivial, so the planar boundary theorem of The planar boundary theorem makes the outer one trivial as well, and the map extends over the whole component. At the end the loop reading \(g\) bounds a disk in a space with fundamental group \(A\), so \(g=1\).

We use Theorem 2.1 of Disks, bands and boundary words, proved in the planar boundary lesson, with the statement (SURF) and Lemma 4.2 there; the statement (PI1) of Cone pictures and asphericity, Section 2, proved in Fundamental groups of graphs and two-complexes, Corollary 2.4 and Theorem 3.3; and the following standard statement, which is not proved in the programme and is recorded as an open obligation of this course.

(REL) (relations as maps of punctured disks; [Chen, Example 2.5 and Figure 3], [Hatcher, Section 1.1 and Proposition 1.26]). Let \(X\) be a path-connected space with a basepoint \(x_0\), let \(\gamma\) and \(\omega\) be loops at \(x_0\), and suppose that \([\gamma]=\prod_{j=1}^r b_j[\omega]^{\epsilon_j}b_j^{-1}\) in \(\pi_1(X,x_0)\), with \(r\ge1\), \(b_j\in\pi_1(X,x_0)\) and \(\epsilon_j=\pm1\). Let \(\Delta\) be a closed polygonal disk in the plane, \(H_1,\ldots,H_r\) pairwise disjoint closed polygonal disks in its interior, and \(P=\Delta\setminus(\operatorname{int}H_1\cup\cdots\cup\operatorname{int}H_r)\). Then there is a continuous map \(f:P\to X\) whose restriction to \(\partial\Delta\), read from a point of \(\partial\Delta\) in the boundary orientation of \(\Delta\), is \(\gamma\), and whose restriction to each \(\partial H_j\), read from a point of \(\partial H_j\) in the boundary orientation of \(H_j\), is \(\omega^{\epsilon_j}\).

The standard proof joins a point of \(\partial\Delta\) to a point of each \(\partial H_j\) by arcs that meet only at their common end, maps the \(j\)-th arc to a path representing \(b_j\), observes that cutting \(P\) along the arcs leaves a disk whose boundary loop is null-homotopic by the relation, extends over that disk, and glues back.

We also use the following elementary fact. A map \(\phi:\partial D\to Y\) from the boundary circle of a closed disk \(D\) extends to a map \(D\to Y\) if and only if the loop \(\phi\) is null-homotopic. Indeed, identify \(D\) with the unit disk; an extension \(F\) gives the homotopy \((z,s)\mapsto F((1-s)z)\) to a constant, and a homotopy \(G\) from \(\phi\) to a constant is constant on \(\partial D\times\{1\}\), so \(F(\rho z)=G(z,1-\rho)\) for \(|z|=1\), \(0\le\rho\le1\), is a continuous extension. A loop at a point \(y_0\) is null-homotopic, as a free loop, exactly when its class in \(\pi_1(Y,y_0)\) is trivial.

1. Spaces for \(A\) and \(A*\langle t\rangle\)

Let \(A\) be a group, possibly uncountable. Let \(Y\) be the CW complex with one vertex \(y_0\), one loop \(x_a\) for each \(a\in A\), and a closed disk attached along \(x_1\) and along \(x_ax_bx_{ab}^{-1}\) for all \(a,b\in A\). Let \(\mathbb T=\mathbb R/\mathbb Z\), with basepoint \(0\), and let \(X=Y\vee\mathbb T\), identifying \(y_0\) with \(0\).

Lemma 1.1. \(a\mapsto[x_a]\) is an isomorphism \(A\to\pi_1(Y,y_0)\), and together with the loop \(t\) that runs once around \(\mathbb T\) it gives an isomorphism \(A*\langle t\rangle\to\pi_1(X,y_0)\).

Proof. By (PI1) 1, the fundamental group of the graph \(Y^1\), a wedge of circles, is free on the classes of the loops \(x_a\); by (PI1) 2, \(\pi_1(Y,y_0)\) is its quotient by the normal subgroup \(N\) generated by \(x_1\) and the \(x_ax_bx_{ab}^{-1}\). In the quotient \(x_ax_b=x_{ab}\) and \(x_1=1\), so \(x_a^{-1}=x_{a^{-1}}\), and every element is a single \(x_a\). The homomorphism from the free group to \(A\), \(x_a\mapsto a\), kills \(N\), and the induced map sends \(x_a\) to \(a\); so the classes \(x_a\) are pairwise distinct, and \(a\mapsto x_a\) is an isomorphism. The same argument for \(X\), whose \(1\)-skeleton has the additional loop \(t\) and the same disks, presents \(\pi_1(X,y_0)\) by the generators \(x_a\) and \(t\) and the relations of \(A\), which is a presentation of \(A*\langle t\rangle\). \(\square\)

The collapse maps \(c_{\mathbb T}:X\to\mathbb T\), sending \(Y\) to \(0\), and \(c_Y:X\to Y\), sending \(\mathbb T\) to \(y_0\), are continuous; \(c_Y\) induces the homomorphism \(A*\langle t\rangle\to A\) that kills \(t\).

2. A punctured disk for a relation

Let \(w\) be unimodular. Conjugating, we may assume (Exercise 6.1 of the disk lesson) that \[ \begin{gathered} w=t^{s_1}a_1t^{s_2}a_2\cdots t^{s_n}a_n,\\ s_i\in\{1,-1\},\quad\sum_is_i=\pm1 , \end{gathered} \] since conjugate words have the same normal closure. Represent \(w\) by the loop \(\omega\) in \(X\) that runs, in order, once around \(\mathbb T\) in the direction \(s_1\), with constant speed, then along \(x_{a_1}\), then around \(\mathbb T\) in the direction \(s_2\), and so on. Call the parameter intervals of the first kind the \(t\)-intervals and the others the coefficient intervals.

Suppose that \(g\in A\) lies in the normal closure of \(w\): \(g=\prod_{j=1}^rb_jw^{\epsilon_j}b_j^{-1}\) in \(A*\langle t\rangle\), with \(b_j\in A*\langle t\rangle\) and \(\epsilon_j=\pm1\). If \(r=0\), then \(g=1\). Otherwise apply (REL) with \(\gamma=x_g\) and the loop \(\omega\): we obtain \(f:P\to X\) on a disk \(\Delta\) with holes \(H_1,\ldots,H_r\). Put \[ \begin{gathered} u=c_{\mathbb T}\circ f:P\to\mathbb T,\\ v=c_Y\circ f:P\to Y . \end{gathered} \] On \(\partial\Delta\), \(u=0\) and \(v\) reads \(x_g\). On \(\partial H_j\), read in the orientation of \(H_j\), the map \(u\) is constant \(0\) on the coefficient intervals and runs once around \(\mathbb T\), monotonically, on each \(t\)-interval, forward for a letter \(t\) and backward for a letter \(t^{-1}\) of \(w^{\epsilon_j}\); the map \(v\) runs along \(x_{a}\) on the coefficient interval of a letter \(a\) and is constant on the \(t\)-intervals. We call \(H_j\) positive if \(\epsilon_j=1\) and negative if \(\epsilon_j=-1\).

3. A piecewise affine circle coordinate and its level arcs

We identify points of \(\mathbb T\) with angles in \(\mathbb R/\mathbb Z\) and use the circle distance \(|\cdot|_{\mathbb T}\).

Lemma 3.1 (approximation). There are a continuous map \(u':P\to\mathbb T\) with \(|u'-u|_{\mathbb T}<\frac1{12}\) everywhere, and a finite triangulation of \(P\) by straight triangles on each of which \(u'\) lifts to an affine function to \(\mathbb R\), such that on \(\partial P\): \(u'=0\) on \(\partial\Delta\) and on the coefficient intervals, and on each \(t\)-interval \(u'\) runs once around \(\mathbb T\), monotonically in the same direction as \(u\).

Proof. By uniform continuity of \(u\), there is \(\rho>0\) such that \(u\) maps every subset of \(P\) of diameter less than \(\rho\) into an arc of length less than \(\frac1{24}\). The polygonal region \(P\) has triangulations by straight triangles; choose one with triangles of diameter less than \(\rho\), whose vertices include the endpoints of all coefficient intervals and \(t\)-intervals. On each triangle, the values of \(u\) lie in an arc \(I\) of length less than \(\frac1{24}\); lift the vertex values to real numbers in a lift of \(I\), interpolate affinely, and reduce modulo \(1\). Lifts on two triangles sharing an edge differ there by an integer, so the circle values agree, and \(u'\) is continuous. Inside a triangle, \(u'\) and \(u\) both lie in \(I\), so \(|u'-u|_{\mathbb T}<\frac1{24}\). On \(\partial\Delta\) and on the coefficient intervals the vertex values are \(0\), so \(u'=0\) there. Along a \(t\)-interval the vertex values of \(u\) move monotonically once around the circle, by steps shorter than \(\frac1{24}\), and \(u'\) interpolates them affinely in lifted angles; so \(u'\) moves monotonically in the same direction, once around. \(\square\)

Choose an angle \(\zeta\in(\frac13,\frac23)\) that is not the value of \(u'\) at any vertex of the triangulation, and put \(\Lambda=u'^{-1}(\zeta)\).

Lemma 3.2 (level arcs).

  1. \(\Lambda\) is a finite disjoint union of polygonal circles and polygonal arcs, the arcs meeting \(\partial P\) exactly in their endpoints. Every \(t\)-interval of every \(\partial H_j\) contains exactly one endpoint of an arc, and there are no other endpoints; in particular \(\Lambda\) misses \(\partial\Delta\) and the coefficient intervals.
  2. At every point of \(\Lambda\), \(u\) lies within \(\frac14\) of \(\frac12\), so \(f\) maps an open neighbourhood \(U\) of \(\Lambda\) into \(\mathbb T\setminus\{0\}\), and \(v\) is constant, equal to \(y_0\), on \(U\).
  3. Read in the orientations of the disks \(H_j\), the two endpoints of each arc lie on \(t\)-intervals of opposite letters \(t\) and \(t^{-1}\).

Proof. (1) On a triangle, \(u'\) lifts to an affine function \(\tilde u\), and \(\Lambda\) is the set where \(\tilde u\) takes one of the values \(\zeta+m\), \(m\in\mathbb Z\). Since \(\tilde u\) varies by less than \(\frac1{24}\) on the triangle, at most one such value occurs. If \(\tilde u\) is constant, the level set is empty, because no vertex value is \(\zeta\). Otherwise it is the intersection of the triangle with a line that avoids the vertices, so it is empty or a segment joining interior points of two edges. So \(\Lambda\) meets each closed triangle in at most one segment, and each interior edge in at most one point, which is an endpoint of exactly the two segments of the two adjacent triangles. Following segments from triangle to triangle gives polygonal circles and arcs; an arc can only end at a point of a boundary edge. On \(\partial P\), \(u'\) is \(0\) on \(\partial\Delta\) and on the coefficient intervals, and takes the value \(\zeta\) exactly once on each \(t\)-interval, because it runs once around the circle monotonically there (Lemma 3.1).

(2) On \(\Lambda\), \(u'=\zeta\in(\frac13,\frac23)\) and \(|u-u'|_{\mathbb T}<\frac1{12}\), so \(u\in(\frac14,\frac34)\), an open condition. A point mapped by \(f\) into \(Y\) has \(u=0\). So the open set \(U=u^{-1}((\frac14,\frac34))\) contains \(\Lambda\) and \(f(U)\subseteq\mathbb T\setminus\{0\}\), where \(c_Y\) is constant.

(3) Coorient each arc \(\alpha\) by the direction in which \(u'\) increases through \(\zeta\). Take a thin closed neighbourhood \(N\) of \(\alpha\) in \(P\) that is a rectangle \(\alpha\times[-1,1]\), with \(\alpha\times\{1\}\) on the side where \(u'>\zeta\), meeting \(\partial P\) in the two short segments \(\{p_1\}\times[-1,1]\) and \(\{p_2\}\times[-1,1]\) at the endpoints. The boundary orientation of \(N\) runs along one of these short segments from the side \(u'<\zeta\) to the side \(u'>\zeta\), then along a long side, then along the other short segment from \(u'>\zeta\) to \(u'<\zeta\). Near the endpoints, \(N\) and \(P\) induce the same orientation on \(\partial P\). So along \(\partial P\), \(u'\) crosses \(\zeta\) upwards at one endpoint and downwards at the other; by Lemma 3.1, \(u\) moves in the same direction as \(u'\) along each \(t\)-interval. The boundary orientation of \(H_j\) is the opposite of that of \(P\) along \(\partial H_j\), so in the orientations of the disks one endpoint lies on a \(t\)-interval where \(u\) increases, a letter \(t\), and the other on one where it decreases, a letter \(t^{-1}\). \(\square\)

4. Bands and the surface they form

Ignore the circles of \(\Lambda\). The arcs of \(\Lambda\) are pairwise disjoint polygonal arcs in the plane, with interiors outside the disks \(H_j\) and endpoints on their boundaries; by (SURF), applied in the sphere obtained by adding a point at infinity to the plane, they have pairwise disjoint thin bands. A polygonal arc has rectangular neighbourhoods inside any given neighbourhood, so we take the bands inside the set \(U\) of Lemma 3.2(2); they meet the circles \(\partial H_j\) in short intervals, the slots, inside the \(t\)-intervals.

Each \(H_j\) is now a word disk for \(w\) in the sense of Disks, bands and boundary words, Section 1. Its slots are the attachment intervals, indexed by the occurrences of \(t^{\pm1}\) in \(w\). Between two consecutive slots, the boundary of \(H_j\) consists of the coefficient interval of one letter \(a_i\) together with pieces of the two adjacent \(t\)-intervals, on which \(v\) is constant; so the class of the corner under \(v\) is \(a_i\) if \(H_j\) is positive, and \(a_i^{-1}\) if it is negative, in the order of the definition, since a negative disk reads \(w^{-1}\). The traversal sign of a slot is the exponent of its letter in the reading of \(H_j\), and by Lemma 3.2(3) every band joins opposite traversal signs.

Let \(\mathcal S\) be the union of the disks \(H_j\) and the bands. Each component of \(\mathcal S\) contains a disk, every slot carries exactly one band end, and each component is a connected disk-and-band surface for \(w\), oriented by the plane. Along a boundary circle of \(\mathcal S\), the map \(v\) runs along the corners and is constant on the band sides, which lie in \(U\); so the loop \(v\) along a boundary circle, read in its boundary orientation, represents in \(\pi_1(Y,y_0)=A\) the boundary word of that circle, up to conjugation.

5. Removing components from the inside out

Proof of Theorem 2.1 of Equations over groups and the Kervaire conjecture. We prove by induction on the number of components the following statement. Let \(\mathcal S\subseteq\operatorname{int}\Delta\) be a finite union of pairwise disjoint connected disk-and-band surfaces for \(w\) in the plane, and let \(v:\Delta\setminus\operatorname{int}\mathcal S'\to Y\) be continuous, where \(\mathcal S'\) is the union of the word disks of \(\mathcal S\), such that \(v\) reads the corner labels along the corners and is constant on the bands. Then the loop \(v|_{\partial\Delta}\) is null-homotopic.

If there are no components, \(v\) is defined on all of \(\Delta\), and the loop \(v|_{\partial\Delta}\) extends over the disk \(\Delta\), so it is null-homotopic.

Otherwise, for each component \(C\), Lemma 4.2 of the disk lesson, applied in the sphere, shows that \(C\) is the complement of the interiors of disjoint closed disks bounded by its boundary circles; exactly one of these disks contains the point at infinity, and its boundary is the exterior circle of \(C\). Let \(D(C)\) be the closed disk in the plane bounded by the exterior circle; then \(C\subseteq D(C)\), and the other boundary circles of \(C\) bound the inner disks of \(C\), contained in \(D(C)\). Choose a component \(C\) whose \(D(C)\) is minimal under inclusion. No inner disk \(D_i\) of \(C\) contains another component \(C'\): otherwise the unbounded complementary region of \(C'\) would contain the connected set \(\mathbb R^2\setminus D_i\), so \(D(C')\subseteq D_i\subsetneq D(C)\), against minimality.

So each inner disk \(D_i\) of \(C\) contains no word disk, and \(v\) is defined on all of \(D_i\). Its boundary loop extends over \(D_i\), so it is null-homotopic, and the boundary word of that circle is trivial in \(A\) by the last paragraph of Section 4. By the planar boundary theorem applied to the connected surface \(C\), the boundary word of the exterior circle is trivial as well, so the loop \(v|_{\partial D(C)}\) is null-homotopic, and it extends to a map \(D(C)\to Y\). Replace \(v\) on \(D(C)\) by such an extension, and remove the word disks and bands of \(C\). Nothing changes outside \(D(C)\), and no other component meets \(D(C)\) (a component inside \(D(C)\) would lie in an inner disk of \(C\)). The hypotheses hold for the remaining components, and the loop \(v|_{\partial\Delta}\) is unchanged; by induction it is null-homotopic.

Apply this to the surface \(\mathcal S\) of Section 4 and the map \(v=c_Y\circ f\), extended over nothing. The loop \(v|_{\partial\Delta}\) is \(x_g\), whose class is \(g\) by Lemma 1.1. So \(g=1\) in \(A\). Hence the coefficient map \(A\to(A*\langle t\rangle)/\langle\!\langle w\rangle\!\rangle\) has trivial kernel. \(\square\)

With Corollary 2.2 of the first lesson this proves the Kervaire conjecture: a nontrivial group cannot be killed by adding one generator and one relation.

What the course proves and what it assumes. The lessons prove every step of the argument, with these inputs from outside the programme, each stated where it is used and recorded as an open obligation: the statement (REL) on relations and punctured disks (this lesson), and the statement (SURF) on surfaces in the sphere, stated in the course on the torsion-free group algebra with zero divisors. The statement (PI1) on fundamental groups of \(2\)-complexes is proved in the course on cell complexes, fundamental groups and covering spaces. The Borel functional calculus, the degree modulo two and the inverse function theorem come from other courses of the programme.

6. Exercises

6.1 (easy). Where does the proof use that the holes \(H_j\) of \(P\) and the bands lie in the plane, and not on a surface of higher genus?

6.2 (medium). Let \(w=ta\). Show directly that the surface \(\mathcal S\) of Section 4 consists of components, each made of one positive and one negative disk joined by one band, and that the induction of Section 5 then shows \(g=1\) without the planar boundary theorem.

6.3 (easy). Explain why the level \(\zeta\) is chosen in \((\frac13,\frac23)\) and the approximation error below \(\frac1{12}\).

6.4 (medium). Explain why the bands of Section 4 are taken inside the set \(U\) of Lemma 3.2(2). Which step of Section 5 would fail if \(v\) were not constant on some band?

7. Solutions

6.1. Twice: the planar boundary theorem uses the count \(q=nk-2k+2\), which holds for surfaces in the sphere (Exercise 5.2 of the planar boundary lesson shows what is lost in higher genus); and the induction uses the Jordan disks \(D(C)\) and the inner disks of a component, which exist in the plane by Lemma 4.2 of the disk lesson.

6.2. Each disk has one slot, so a band joins two disks, one positive and one negative by Lemma 3.2(3), and each component is such a pair. Its single boundary circle reads \(a\,a^{-1}\) up to conjugation, which is trivial. So in the induction the exterior word of every minimal component is trivial directly, and the extension step goes through.

6.3. The level must stay away from the angle \(0\), where \(f\) can enter \(Y\): with \(\zeta\in(\frac13,\frac23)\) and error below \(\frac1{12}\), the original coordinate \(u\) lies in \((\frac14,\frac34)\) along the level, so \(f\) is in the circle part there and \(v\) is constant near the bands.

6.4. The boundary word of a circle of \(\mathcal S\) is computed from the corners only; if \(v\) varied along a band side, the loop \(v\) along the circle would differ from the corner product by the band contributions, and the triviality of the loop over an inner disk would say nothing about the boundary word. The construction in Section 4 keeps the bands inside \(U\), where \(v\) is constant, precisely to avoid this.

References