Proof. The following argument also appears in the companion paper (OpenAI 2026, Lemma 3.1); it applies to every finite simple graph. The kernel of the boundary map is spanned by signed simple graph cycles. Indeed, choose a spanning forest, subtract the fundamental cycle associated with each nonforest edge with its appropriate coefficient, and obtain a cycle supported on a forest. Stripping leaves shows that the remaining chain is zero. It therefore suffices to treat two oriented simple graph cycles.

We will replace the first cycle by a closed polygonal walk \(\gamma\) and the second by two closed polygonal walks \(\eta_+,\eta_-\). The average of the signed intersection counts of \(\gamma\) with these two walks will be \(J(z,z')\): crossings outside small vertex disks contribute \(I\), while the averaged contributions inside them give the correction in Equation (2). Each total intersection count is zero, as we prove at the end of the argument.

The local contribution. Consider two oriented chords \(C,E\) of a round disk, with four distinct boundary endpoints. Give a chord’s start coefficient \(-1\) and its end coefficient \(+1\), denoted by \(\epsilon_C\) and \(\epsilon_E\). If \(T(u,v)\) compares boundary points from a counterclockwise cut away from the endpoints, their signed intersection is \[ I(C,E)=\frac12 \sum_{u\in\partial C}\sum_{v\in\partial E} \epsilon_C(u)\epsilon_E(v)T(u,v). \tag{3}\] Moving the cut past one endpoint changes the right side by a multiple of the other chord’s total endpoint coefficient, which is zero. Cutting just before the start of \(C\), the expression becomes minus the sum of \(\epsilon_E\) on the open counterclockwise boundary arc from the start to the end of \(C\). This arc is on the right of the directed chord. The expression is zero for nonalternating endpoints, and is \(+1\) for \(E\) crossing from right to left, or \(-1\) for the reverse. These are exactly the intersection signs. Figure 1 shows this convention and the shared-spoke averaging used below.

Separating shared tracks. Choose disjoint small round vertex disks meeting only the straight spokes. In each vertex disk replace the first cycle’s passage by the chord from its incoming to its outgoing boundary point. Outside these disks retain its edge tracks, obtaining a closed polygonal walk \(\gamma\).

Construct two versions \(\eta_+,\eta_-\) of the second cycle. Outside the vertex disks, displace each track slightly left or right, respectively, relative to its fixed basis-edge orientation. In the vertex disks join its displaced endpoints by chords. Here are the local details needed when the cycles share edges. Place additional disjoint small round disks at crossings and nonstraight bends, meeting only their incident straight pieces. Shift each straight stretch between disks to a parallel one. Across a crossing disk continue along the shifted line: it avoids its own original line and crosses the other original strand once, with the same sign. At a bend the shifted ends lie on the same side of the original bent track. For a left shift they lie clockwise of its incoming outward ray and counterclockwise of its outgoing ray. Join them by a polygonal path in that component of the bend disk; use the other component for a right shift.

The distinct truncated stretches between disks are compact and disjoint. Finiteness therefore permits a common sufficiently small shift introducing no additional intersections. The vertex cuts and comparisons between distinct spokes remain unchanged. At seams the shifted and original endpoints are distinct; chord interiors stay inside their disks, and the intended crossings remain strictly inside crossing disks. These local replacements join to closed polygonal walks \(\eta_+,\eta_-\). Every intersection between \(\gamma\) and either walk is transverse and lies away from their corners; unnecessary straight subdivision points can be removed.

In either version the signed intersection count outside the vertex disks, counting by traversals, is \[\sum_{e,f}z_ez'_f I(e,f).\] This remains true for shared tracks: a shifted edge avoids its own copy, and crossings of different edges give the two ordered contributions when both cycles use both edges.

In a common vertex disk the first chord’s endpoint coefficient on spoke \(e\) is \(\sigma_{we}z_e\): an incoming traversal starts the chord and an outgoing traversal ends it. The second chord has coefficient \(\sigma_{wf}z'_f\). For distinct spokes, the comparison sign in Equation (3) is \(T_w(e,f)\) for both shifts. For a shared spoke the two shifted endpoints are on opposite sides of its original endpoint, so their comparison signs average to zero. The average vertex contribution is therefore \[\frac12\sum_{e,f\text{ incident with }w} z_ez'_f\sigma_{we}\sigma_{wf}T_w(e,f).\] Opposite cycle traversals change the coefficients, not this averaging. In particular, for opposite traversals of a shared edge, the same coefficient multiplies both opposite shared-spoke comparisons, whose average remains zero.

Total signed intersection. Two closed polygonal walks in the plane that meet transversely away from their corners have total signed intersection zero, with intersections counted by traversals. To see this without a simplicity assumption, cone every oriented segment \([u,v]\) of the first walk to a common point \(o\), forming the oriented triangle \([o,u,v]\). Choose \(o\) outside the finitely many lines that cause degeneracy, a cone segment to pass through a corner of the second walk, or nontransverse intersections. Each triangle boundary has total signed intersection zero: along the closed second walk, entrances and exits have opposite signs. In the sum of the triangle boundaries the extra cone sides cancel, leaving the first walk. This proves the assertion.

Apply it to \((\gamma,\eta_+)\) and \((\gamma,\eta_-)\) and average. The exterior and vertex contributions just computed sum to \(J(z,z')\), proving the lemma. ◻

Proof. The Lagrange formula \[A(t)=\sum_p A(u_p)\frac{G(t)}{(t-u_p)G'(u_p)}\] holds for \(\deg A\le N-1\): both sides have degree at most \(N-1\) and agree at all \(N\) nodes. Comparing coefficients of \(t^{N-1}\) gives the assertion. ◻

Proof. Both spaces and the target have dimension \(s(s-1)\), so it suffices to prove injectivity, simultaneously for the two spaces. For \(s=2\), the two points have distinct first coordinates and distinct second coordinates. They determine an affine polynomial in either one variable.

For larger \(s\), let \(P\in\mathcal P_{s-1,s-2}\) vanish on the pattern. Each of the rows \(p=s,s+1\) contains \(s-1\) points. Thus the polynomials \(P(u_s,v)\) and \(P(u_{s+1},v)\), of degree at most \(s-2\), vanish identically. It follows that \[P(u,v)=(u-u_s)(u-u_{s+1})R(u,v), \qquad R\in\mathcal P_{s-3,s-2}.\] Remove those two indices from the node list. The remaining two triangles are the pattern \(L_{s-1}\); the removed factors are nonzero on it. The induction hypothesis for the second degree rectangle gives \(R=0\).

For \(P\in\mathcal P_{s-2,s-1}\), use instead the columns \(q=1,2s\), each with \(s-1\) points. Factoring \((v-u_1)(v-u_{2s})\) leaves a polynomial in \(\mathcal P_{s-2,s-3}\) on \(L_{s-1}\), zero by the first degree rectangle in the induction hypothesis. This proves both assertions. ◻

Proof. Let \(A\in\mathcal P_{s-1,s-2}\) and \(B\in\mathcal P_{s-2,s-1}\). The product \((u-v)A(u,v)B(u,v)\) has degree at most \(2s-2\) in each variable. Consequently, replacing \(S(p,q)\) by \(S(p,q)+r_p+c_q\) leaves \(M(A,B)\) unchanged: the added \(r_p\) terms vanish after the weighted sum in \(q\), by Lemma 3, and the \(c_q\) terms vanish after the weighted sum in \(p\). Diagonal entries have no effect because \(u_p-u_q=0\) there.

Relabel the nodes in their prescribed order; their numerical values remain arbitrary distinct real numbers. Off the diagonal, \(S(p,q)=\tfrac12-\mathbf1_{\{q<p\}}\). We may therefore replace \(S\) in the pairing by \[K(p,q)= \begin{cases} -1,&1\le q<p\le s,\\ 1,&s<p<q\le2s,\\ 0,&\text{otherwise}. \end{cases}\] Indeed, off the diagonal this is \(S(p,q)-\tfrac12+\mathbf1_{\{p>s\}}\), an allowed change by a row term and a constant. Hence \[M(A,B)=\sum_{(p,q)\in L_s} K(p,q)\frac{u_p-u_q}{G'(u_p)G'(u_q)} A(u_p,u_q)B(u_p,u_q).\] Every displayed weight is nonzero. Lemma 4 identifies each of the two polynomial spaces with \(\mathbb R^{L_s}\) by evaluation. In these two coordinate systems the pairing is diagonal with nonzero entries, so it is nonsingular.

For the consequence, the smaller space for \(A\) is contained in the first nonsingular space, while the given test space for \(B\) contains the second one. Nonsingularity forces \(A=0\). ◻

Proof. For exponents \(a,b\le n-2\), the chain \[z_{a,b}=\sum_{i,p}w_iw_p t_i^a t_p^b[ip]\] has boundary \[\left(\sum_iw_i t_i^a\right)\sum_pw_p t_p^b\mathbf e_p - \left(\sum_pw_p t_p^b\right)\sum_iw_i t_i^a\mathbf e_i=0\] by Lemma 3. For a monomial \(H=x^au^by^cv^h\), Equation (6) is \(J(z_{a,b},z_{c,h})=0\), by Lemma 2. Linearity gives the result. ◻

Proof. Suppose \(E(P)=0\). We first force vanishing at a shared endpoint, then use the resulting polynomial factors to descend in degree.

Localizing at a shared endpoint. In Lemma 6 use \[ H=P(x,u,y,v)(u-v)(u-y)(v-x)U(x)V(y)B(u,v), \tag{9}\] where \(\deg U,\deg V\le s\) and \(B\in\mathcal P_{s-1,s-1}\). The three-factor weight has separate degrees \((1,2,1,2)\); the total degrees are therefore at most \(2s-1=n-2\).

Independent-edge terms vanish either by \(E(P)=0\) or by their zero \(J\). The weight kills the coincidences \(p=q\), \(p=k\), and \(q=i\); loops already have zero edge vector. Only \(i=k\) survives, with \(p,q\) distinct from \(i\) and from each other. Since adjacent edges do not cross and both orientations point out of \(i\), \[J([ip],[iq])=\tfrac12 T_i(ip,iq)=:S_i(p,q).\] Set \(S_i(p,p)=0\) and \(F_i(t)=\prod_{p\ne i}(t-t_p)\). For \(p\ne i\), \((t_p-t_i)w_p=1/F_i'(t_p)\), so Equation (6) becomes \[ 0=\sum_i w_i^2U(t_i)V(t_i) \sum_{\substack{p\ne i\\q\ne i}} S_i(p,q)\frac{t_p-t_q}{F_i'(t_p)F_i'(t_q)} P(t_i,t_p,t_i,t_q)B(t_p,t_q). \tag{10}\]

For each fixed \(B\), products \(U(t)V(t)\) span all polynomials of degree at most \(2s=n-1\): split each monomial’s exponent into two parts of size at most \(s\). Their linear combinations interpolate arbitrary values on the \(n\) nodes. Since \(w_i\ne0\), the inner sum in Equation (10) vanishes separately for every \(i\) and every \(B\). More explicitly, for fixed \(B\), linearity allows \(U V\) to be replaced by the Lagrange polynomial that is one at \(t_i\) and zero at all other nodes. For each \(i\), the inner sum is the order pairing on the \(2s\) nodes different from \(t_i\). Its first polynomial, \(P(t_i,u,t_i,v)\), belongs to \(\mathcal P_{s-2,s-2}\). Lemma 5 gives \[P(t_i,u,t_i,v)=0 \quad\text{as a polynomial in }u,v,\quad\text{for every }i.\]

Dividing the collision factors. The polynomial \(P(x,u,x,v)\) has degree at most \(2s-4<n\) in \(x\). Applying the univariate root bound to each coefficient in \(u,v\) shows that it is identically zero. Hence \(x-y\) divides \(P\). The two within-pair symmetries imply divisibility also by \(x-v\), \(u-y\), and \(u-v\). Each difference is prime: quotienting the polynomial ring by that factor identifies its two variables and leaves a polynomial ring, which has no zero divisors. The four factors are pairwise nonassociate, so their individual divisibility implies divisibility by their product \[ C=(x-y)(x-v)(u-y)(u-v). \tag{11}\]

We have proved that \(C\) divides every polynomial in the kernel. For a nonzero polynomial, this already forces every separate degree to be at least two. If the kernel contains a nonzero polynomial, choose \(P\) of minimum degree in \(x\) and write \(P=CR\) with \(R\ne0\). The product \(C\) is invariant under each within-pair swap; applying either swap and cancelling \(C\) shows that \(R\) has the same two symmetries. Its degree in each variable is two less than that of \(P\), so \(R\in\mathcal S\). Moreover, \(C\) is nonzero at every evaluated independent-edge pair, so \(E(R)=0\). This contradicts the minimal choice of \(P\). ◻

Proof. Define the full difference product \[\Delta=(x-u)(y-v)(x-y)(x-v)(u-y)(u-v).\] For \(P,Q\in\mathcal S\), the polynomial \(PQ\Delta\) has separate degrees at most \(2(s-2)+3=2s-1\), so Lemma 6 applies. Its only nonzero terms come from \(\mathcal T\): every common endpoint kills \(\Delta\).

Fix representative orientations for the edges. For each ordered pair in \(\mathcal T\), its four choices of endpoint orientations give equal contributions. Reversing either edge changes the signs of both \(J\) and \(\Delta\), while leaving \(P,Q\) and the node-weight product unchanged. After division by four, \[\sum_{\alpha\in\mathcal T}a_\alpha E(P)_\alpha E(Q)_\alpha=0, \quad a_{(\{i,p\},\{k,q\})} =J([ip],[kq])w_iw_pw_kw_q \Delta(t_i,t_p,t_k,t_q).\] Every \(a_\alpha\) is nonzero, so this is a nondegenerate symmetric diagonal form on \(\mathbb R^\mathcal T\). In particular it is not the skew form \(J\). Exchanging the two whole edges gives a different coordinate with opposite coefficient, a block \(\operatorname{diag}(a,-a)\); this does not destroy nondegeneracy.

The image \(W=E(\mathcal S)\) is contained in its orthogonal complement. To justify the dimension count, let \(V=\mathbb R^\mathcal T\). The nondegenerate form identifies \(V\) with its dual. Restricting a linear functional on \(V\) to \(W\) is onto \(W^*\), since any basis of \(W\) extends to a basis of \(V\). The resulting map \(V\to W^*\) has kernel \(W^\perp\); hence \(\dim W^\perp=|\mathcal T|-\dim W\), and therefore \(2\dim W\le|\mathcal T|\). By Lemma 7 and Equations (7) and (8), \[c(D)\ge\frac{|\mathcal T|}{2}\ge\dim W=\binom{s}{2}^{\!2}.\] Cancellation of signs at repeated crossings can only remove coordinates from \(\mathcal T\), and so cannot invalidate this inequality. ◻

Proof. Proposition 8 proves the odd cases \(n\ge5\). For \(n=2s>4\), delete each vertex in turn from a normalized minimum drawing. Every crossing involves four distinct vertices, so survives exactly \(n-4\) deletions. Each remaining drawing of \(K_{n-1}\), whether minimum or not, has at least \(\binom{s-1}{2}^{2}\) crossings. Thus \[(n-4)c(D)\ge n\binom{s-1}{2}^{\!2}, \qquad c(D)\ge\frac{s(s-1)^2(s-2)}4.\] For \(n=3,4\), nonnegativity gives the claimed zero lower bound. ◻

Proof. Each four-set of vertices has one alternating edge pair, its diagonals. Count cyclic tuples starting at a label \(a\), with successive positive integer gaps \(g_1,g_2,g_3,g_4\) summing to \(n\). The four vertices are \[a,\quad a+g_1,\quad a+g_1+g_2,\quad a+g_1+g_2+g_3 \pmod n.\] Each four-set occurs four times, once for each starting vertex. The diagonal endpoint-sum residues are \[r=2a+g_1+g_2,\qquad r+k,\qquad k=g_1+g_3.\] For \(2\le k\le n-2\), put \(\ell=\min(k,n-k)\), and let \(R_k\) be the residues \(r\) for which \(r,r+k\) have the same color. Shifting by \(\ell\le m\) changes color on \(\ell\) residues from each block, so \[|R_k|=n-2\ell.\] For each \(k\) there are \((k-1)(n-k-1)\) gap choices: choose \(1\le g_1\le k-1\) and \(1\le g_2\le n-k-1\); the other gaps are determined.

We claim that, including all \(a\), the number of same-color tuples for this \(k\) is \[ (k-1)(n-k-1)|R_k|. \tag{12}\] For odd \(n\), multiplication by two permutes the residues, so each fixed gap choice already contributes \(|R_k|\). For even \(n=2m\), as \(a\) varies, \(r=2a+g_1+g_2\) visits each residue in the parity class of \(g_1+g_2\) twice. We must check that this restriction still gives Equation (12) after summing over the gap choices. There are three cases. If \(m\) is odd, \(R_k\) is invariant under adding \(m\), which switches parity, so its parity classes have equal sizes. If \(m,k\) are both even, \(R_k\) consists of two cyclic intervals of even length \(m-\ell\), and again has equal parity counts. Finally, if \(m\) is even and \(k\) odd, the \(k-1\) choices for \(g_1\), for each fixed \(g_2\), give both parities of \(g_1+g_2\) equally often. Averaging their doubled parity-class counts proves Equation (12) in this case too.

Dividing the total by four, and pairing \(k\) with \(n-k\), gives \[\begin{align*} c(D) &=\frac14\sum_{k=2}^{n-2} (k-1)(n-k-1)\bigl(n-2\min(k,n-k)\bigr)\\ &=\frac12\sum_{k=1}^{m} (k-1)(n-k-1)(n-2k). \end{align*}\] The added \(k=1\) term is zero, as is the possible central term \(k=n/2\). Set \[B_k=k(k-1)(n-k-1)(n-k-2).\] A direct subtraction gives \[B_k-B_{k-1}=2(k-1)(n-k-1)(n-2k).\] Since \(B_0=0\), the crossing count is \(c(D)=B_m/4\). For \(n=2m+1\) this is \(m^2(m-1)^2/4\); for \(n=2m\) it is \(m(m-1)^2(m-2)/4\). These are \(Z(n)\), including the zero values at \(n=3,4\). ◻

Proof. Drawings with finitely many crossings exist, so their crossing counts have a least nonnegative integer value, attained by a drawing. Start with such a minimum drawing, and call its vertices and crossings marks. Parameterize every edge injectively on \([0,1]\). On each edge choose pairwise disjoint relative parameter neighborhoods, with disjoint closures, of its marked parameters, one-sided at its endpoints.

Choose pairwise disjoint small closed round disks centered at the marks. They can be made so small that every visit of an edge to a disk lies in the corresponding parameter neighborhood, and nonincident edges avoid it. Indeed, the image of the compact parameter set outside that neighborhood avoids the mark, so has positive distance from it; if the mark is not on the edge, use the entire compact edge. There are finitely many such conditions.

For an interior mark, remove the edge passage from its first to its last visit to the closed disk, including all intermediate excursions. At an initial vertex remove through the last visit; at a terminal vertex remove from the first visit. These intervals are disjoint along each edge. The retained middle pieces are finitely many pairwise disjoint compact arcs with distinct boundary endpoints and interiors outside all closed disks. This step does not require finitely many visits to a circle.

Choose pairwise disjoint open neighborhoods of the middle pieces. At a boundary endpoint take a short radial spur out of its disk. The spur tip can be joined to a nearby interior point of the original middle piece within the chosen neighborhood and the open exterior of all disks: a sufficiently small exterior collar sector at that endpoint contains both points, and radial paths with a short concentric arc join them. Do this at both ends and use the middle piece between the selected interior points.

A compact path in an open subset of the plane has a polygonal replacement there with the same endpoints: cover its image by small balls inside the open set, subdivide its parameter interval so that each subarc lies in one ball, and use chords in those balls. Apply this between the spur tips and append the spurs. Subdivide the finite union of segments at intersections and endpoints, also splitting and identifying collinear overlaps. A simple path in the resulting finite embedded graph gives a simple polygonal replacement. The disjoint neighborhoods ensure that the different exterior pieces do not meet.

In each crossing disk join its respective endpoint pairs by straight chords. Their four boundary endpoints are distinct, so the chords have at most one crossing, transverse if present. In each vertex disk use straight spokes to its center; distinct boundary endpoints give distinct rays, so the spokes meet only at the vertex. Reassembling in the original edge order gives simple edges: an edge uses at most one strand in each disk, and all exterior pieces are disjoint. Every crossing lies in an old crossing disk, with at most one there. The result is polygonal and has no greater crossing count; hence it is minimum.

Suppose two edges \(vu,vw\) cross. Around one such crossing take a small disk containing only their straight crossing strands. Exchange the prefixes from \(v\) up to that disk and join them to the original suffixes by disjoint chords. The new pairing is nonalternating. We obtain two walks with the required endpoints and with that crossing removed; outside the disk the two old tracks are used exactly once in total.

Subdivide at bends and remaining intersections, and erase closed subwalks between repeated subdivision vertices until both walks are simple. At an old crossing of these two edges, either its two tracks belong to different new walks, in which case each retained passage remains straight, or both belong to one walk. Only in the latter case can loop erasure make a turn, and neither the other walk nor a third edge is present at that point. A crossing with a third edge lies on a track visited at most once by the two walks together, so any retained passage there also remains straight. No new intersection, overlap, or improper contact is introduced; all interiors still avoid graph vertices. This gives an admissible drawing with fewer crossings, a contradiction. Thus the polygonal minimum has no adjacent-edge crossings. ◻