A torsion-free counterexample to reduced Baum–Connes injectivity

Proof. The injections exist because the polynomial factor TM and fixed factor 2^b in (2.9) are dominated by the positive exponent gaps r-.40005s and s-.60005s. Choose both injections uniformly. For fixed nonzero \zeta, multiplication identifies the field coefficients bijectively with all output-to-input functionals. On any specified set of n names these are therefore sampled without replacement. Restriction to t input bits has fibers of proportion 2^{-t}\le2^{-b}.

Here is the required tail estimate. For n zero–one samples without replacement, with sum S and mean success proportion p, one has

Indeed the exponential moment before centering is an elementary symmetric mean in the population entries 1,e^u. Replacing two unequal nonnegative entries by their average increases every elementary symmetric sum: the only changing term is their product multiplied by a nonnegative symmetric sum in the other entries. Iteration compares the moment with sampling with replacement. For a centered Bernoulli variable, the second derivative of the logarithmic moment generating function is a tilted variance, at most 1/4; integrating twice proves the displayed bound. Exponential Markov inequality gives

There are at most s2^{2s} choices of an output functional, prefix length, and restricted value. Multiplying by the polynomial number of families and the fixed number of extra slices and sides still gives 2^{O(s)} tests. Each has n\ge d=2^{.40005s}. Their total failure probability tends to zero. Fix a choice for which none fails.□

Proof. Preservation of the common support. At the actual larger endpoint the forced correction cancels its extra bits. Lower correction intervals lie in active coordinates. Higher correction gates vanish: their eight preceding raw blocks are wholly beyond the actual endpoint support, because consecutive marks are separated by K>20 source blocks. Thus corrected streams have exactly the common permitted lengths.

The transformed corrected u streams preserve these lengths. At the last common block the suffix is the identity. The next twenty blocks are identity, and their corrected inputs are zero. Raw u support extends by at most one block beyond the common cutoff. At every later ordinary index the gate window (2.10) is entirely beyond that support, so its gate is closed. The same holds at all higher marked rules. This proves preservation of the padded tail, including an extra interval in the larger endpoint’s u coordinate.

For clarity, the two possible common terminal shapes are

In both rows 1\le t\le s and the output of each block-diagonal K fits in the displayed v length. If an extra interval is in v, it lies beyond this output: its same-side direct block is zero there. Thus computing v from the equations cannot leak a K u term into the prescribed extra interval.

Inversion from either endpoint. Given an A endpoint and slot, its extra value is known, either from the state or from the slot. Its first r evaluation bits determine h_0; there are no marked corrections in the initial blocks. Hence c and the B coefficient name are known. Proceed through the B blocks. To compute its next corrected evaluation from the A endpoint, all correction gates use already known raw blocks by their five-index lag. The B permutation key uses only earlier B blocks and the preceding A block. Invert that known block permutation, then undo the known correction to recover the next raw B u block. The other equation now determines its raw v block. Continue in order. The support argument just given shows that all recovered streams have the required lengths.

Conversely a B slot gives c. Its first s-r evaluation bits give h_0', and hence h_0. Inverting successive A block permutations recovers raw u_A and then v_A. These two procedures solve the same triangular equations and are inverse. The lag is important here: a correction in Q_{B,j} does not read the unknown u_{A,j-1}, and one in Q_{A,j} does not read the unknown u_{B,j}.

Uniform coordinates. First freely specify the raw u blocks on both sides, omitting only the larger endpoint’s fixed extra interval if that interval is in u. Specify h_0 for an A slot, or h_0' for a B slot whose bucket is fixed. Every gate, correction, and permutation value is now determined. Equations (2.12) determine both \bar v streams, and undoing the corrections determines both raw v streams. A forced extra interval in v is exactly e. The number of free coordinates is

respectively. The preceding inversion proves bijectivity, not merely equality of dimensions. Uniform measure therefore becomes product uniform measure in these coordinates.□

Proof. Fix a slot and, where necessary, condition on its actual extra value. Every gate tested at or before the common cutoff, and the gate of the extra interval at most two blocks later, reads eight full free raw u blocks. They are independent uniform by Lemma 2.2. Thus each noninitial gate is closed with probability exactly 2^{-8s}. Initial gates are never closed. Each parity in the third test has at least n_0 full active blocks. Linearity of expectation and Markov’s inequality give probability at most 250\,2^{-8s} for each of its two failures. The first two tests add at most 4\,2^{-8s}. There are at most T2^bD slots at either endpoint, since D\ge d. A union bound proves (2.14), with room in the constant. Averaging over the extra value gives the same bound at a larger endpoint.

The first two tests refer only to raw coordinates obtained before the forced choices can matter. If they pass, the auxiliary and universal rules coincide along this edge; the padding argument of Lemma 2.2 applies to both ends. Restricting to edges with two surviving endpoints therefore gives a subgraph satisfying the universal equations.

For either source parity, the fraction of full active blocks belonging to marked or following twenty-block zones is at most

Indeed these zones have at most eleven indices of each parity, are separated by K interleaved indices, and there is at most one boundary zone in the count. The displayed larger constants cover both endpoints of the counting interval. Together with the separate 1/250 gate bound this is less than .005. A marked partial block is counted as unavailable for a full-s-bit test even if its prefix gate is open. The current cutoff and its preceding two indices are not in a previous marked zone, by (2.8).□

Proof. The label orientation identifies the side, and the complete label identifies the family, both coefficient names, the extras, and every universal rule. At an interior vertex the two incident coefficient names are distinct. Otherwise, at A their family, serial and actual extra agree, and at B their bucket agrees as well; they are the same slot. The unique-edge assertion in Lemma 2.2 would then make the path backtrack.

For either incident name \nu, rearrange its current corrected evaluation block as

If the right sides and correction offsets are known for the two names, their difference recovers the entire raw u block because the difference of the field coefficients is nonzero. Then either equation recovers the raw v block. The argument applies to full padded blocks even when an actual endpoint has only a partial active block.

At an A position of radius one, the two first evaluations are precisely the two label heads, and the initial correction offsets vanish. This recovers A_0. Suppose next that j\ge1. At a B position of radius 2j, each neighbor has A radius at least 2j-1, so its raw streams are known through block A_{j-1} by induction. The central B streams are already known through B_{j-1}. To compute the required right side of the second equation in (2.12) through js bits, one needs only the A transformed stream through

Every key for its last needed source block uses A blocks strictly before A_{j-1} and central B blocks through B_{j-1}. Its argument is the already recovered full corrected A_{j-1} block. Its gates and all correction offsets use still earlier raw blocks by (2.10). Thus the two right sides are known, and (2.17) recovers B_j.

At an A position of radius 2j+1, both neighbors have B radius 2j. The first edge equation through r+js uses the B transformed stream through js. Its last argument is the already recovered corrected B_j block, and its keys use central A blocks only through A_{j-1} and earlier B blocks. Again the correction lag makes every offset known. Subtraction recovers A_j.

This induction is an induction on individual source blocks, not on whole A–B pairs. At each step both table keys and table arguments are known before any current output is used. The universal rules are identical in the two occurrences, including at marked cutoffs lying beyond one actual state’s support. Padding therefore causes no exception to the induction.□

Proof. Uniform raw vertex coordinates are independent bits. The probability of the test for i' is 2^{-q(i,i')}. The elementary inequality

splits its bound into a term depending on the smaller height and one depending on the height difference. Summing the latter over positive multiples of M, and the former even over all nonnegative smaller heights, gives (2.19). Both sums converge uniformly in the larger height.□

Proof. The degree statements follow from Lemma 2.2. At a larger endpoint, a fixed state specifies its extra value; at a smaller endpoint that value ranges over its slots. At an A endpoint equal labels specify equal serial, family, and extra, hence the same slot. At B they also specify the bucket. Uniqueness of the edge at a slot proves foldedness, and restriction to an induced subgraph preserves it. Counting (2.7) gives (2.20); its exponent uses \log d+r=1.00005s, not a coarser word-counting bound. The deletion assertions are Lemmas 2.3 and 2.5.

For fixed b and incidence types, the label count has logarithm at most 1.00005s+O(\log s), whereas \log_2(q^2)=1.0001s. This strict margin persists after fixed finite refinements of the labels. It will make the weighted forest error tend to zero in Lemma 8.4.

It remains to prove global uniqueness. Let N=\lceil1.002m_i\rceil and compare the first N edges of two paths with the same oriented word. Their midpoint radius is at least \lfloor N/2\rfloor. If the first midpoint has height j, then

By Lemma 2.4, their common recovered aligned depth \delta satisfies

The last inequality follows from a\le10^{-6}, H_i\ge(2\cdot10^6+1)s, and b/s=.50005a.

For all heights, (2.2) gives

An aligned A depth \delta recovers \delta/2+.1s bits of each stream, and a B depth recovers \delta/2 bits. Since H_j\ge1000s, the displayed recovery covers the entire first midpoint state.

The labels give equal height residues at the two midpoint positions. If their heights differ, write i_-<i_+ for them. Their own-u length difference is the \Delta of (2.18). The smaller state’s padding is zero on the interval tested there at the larger state. Its endpoint is at most

Because H_{i_-}\le H_j and 2\eta_{\rm sep}<.001, the recovered prefix contains that whole interval, even if i_+ is arbitrarily large. The larger state would therefore have failed its separation test, a contradiction. The two heights are equal. Their entire states agree by the recovery already proved. Foldedness now identifies the paths in both directions from this midpoint, and uniquely continues any longer common word.□

Proof. Fix the serial number, full label head and preceding raw u blocks. If the last common block is B, its remaining B data are exactly the t bits of raw u_B in that block; the known correction merely translates them. Its remaining A data are exactly the corresponding t evaluation bits. All other coordinates are determined by Lemma 2.2. The forced-on current prefix table is an arbitrary bijection between these two 2^t-element sets. Different complete keys specify disjoint clusters of slot incidences. The A-terminal case reverses the roles. The padding proof in Lemma 2.2 shows that higher tables never act on this incidence.□

Proof. Translation of uniform V proves the mean assertion. Fix C,W and write \widehat f_a(\zeta)=2^{-w}\sum_z f_a(C,z)(-1)^{\zeta z}. Orthogonality, first in V and then in U, gives

where \mathcal C ranges over classes with equal \zeta K_a. Cauchy–Schwarz bounds each square by |\mathcal C|\sum_{a\in\mathcal C}|\widehat f_a(\zeta)|^2. The multiplicity assumption and Parseval’s identity prove the result after integration over C,W. In particular, integration over W introduces no variance of the conditional mean.□

Proof. The full blocks. Use the orthogonal difference projections

Equation (3.4) implies E_h^A\mathsf{M}E_j^B=0 unless h\leq j\leq h+1. The band h=j is bounded by (3.5), and the band j=h+1 by the adjoint of (3.6), whenever the blocks in question are full. Within either band the input spaces are mutually orthogonal, as are the output spaces. Each band therefore has norm at most \sqrt\theta, and their sum at most 2\sqrt\theta.

The terminal block. There are two possible terminal forms. If the final input block is a B block, write

For the input differences with 1\leq j\leq n, all blocks required in both bands are full, and consequently

Fix a B slot. Inversion through the preceding blocks determines all of u_A and all but the last t bits of v_A using u_B[0,ns) and Q_B[0,(n+1)s). The current prefix permutation uniformly averages the remaining t bits of Q_A. Its coefficient term K_Au_A ends at r+ns, so these are exactly the remaining v_A bits, up to a known correction. The matching key contains only the label and preceding raw inputs. After this average, its slot value has no separate dependence on the last t raw u_B bits. Conditional on P_n^B, it is therefore a function of V+K_aU+o_a, with U\in\F_2^t and V\in\F_2^s. Lemma 3.2 gives

Combining this orthogonally with the adjoint of (3.7) gives \|\mathsf{M}^*g\|^2\leq5\theta\|g\|^2 for centered g.

If the final input block is an A block, the lengths are instead

The full-block argument on E_h^A, 0\leq h\leq n, gives \|\mathsf{M}^*(P_n^A-C_A)\|\leq2\sqrt\theta. The current permutation averages the final t evaluation bits of Q_B, on which K_Bu_B is zero. Conditional on P_n^A, the remaining slot function depends on the final t input bits of A only through their full s-bit evaluation. Hence \|(1-P_n^A)\mathsf{M}\|^2\leq\theta, and the same orthogonal combination proves the assertion. Both arguments include t=s; the initial r-block is never terminal because the tower starts after many complete pairs.

Combining the extra-bit slices. If the larger side is A, its space is A_0\times\F_2^b and the complete operator has components \mathsf{M}_e. For centered f on B, \|\mathsf{M}f\|^2=\mathbb E_e\|\mathsf{M}_ef\|^2\leq\eta^2\|f\|^2. If the larger side is B, use the adjoint version of this argument. Equivalently, write an input as f_e=c_e+f_e^0; its overall mean zero says \mathbb E_ec_e=0, so the constants cancel before Jensen’s inequality is applied to \mathbb E_e\mathsf{M}_ef_e^0.□

Proof. Within each of the finitely many families order the source blocks as A_0,B_1,A_1,B_2,\ldots. Interleave the families by increasing source-block index and a fixed family order. At an unmarked update choose all its active permutation variables uniformly. At a marked update use the finite conditional distribution given avoidance of its current cut failures, separately in the extra-bit slices. Conditional on all earlier source-block indices, the current kernels in different families are independent, since a family’s cut conditions use only its own tables. Thus grouping the families into one source-block update gives the same bound, with k the total number of tested tables. The mark separation ensures at most one actual cutoff per family at an update. Lemma 3.3 and (3.11) show that this distribution exists for every possible preceding history, and (3.12) proves its test bound. Blocks prescribed to be identities have no random variables.

Each update has finitely many possible outcomes. An independent uniform number in [0,1] can choose among them by consecutive intervals of their conditional probabilities. Iterating on the countable sequence of updates constructs the asserted law. A future update does not change an earlier incidence: the prefix and padding properties of Section 2 guarantee this. Therefore all earlier expansion conditions persist.

When a test whose current-table choices and input arguments are measurable from earlier updates is evaluated, all later conditional distributions integrate to one. Thus its probability is governed by the current kernel just described, without an additional conditioning on future expansion successes. This observation is the form of sequential conditioning used in the pruning argument below.□

Proof. The pools are injective on the full finite collection of family, type, slot, and permitted extra-value names, rather than only on the slots at a single vertex. Hence equality of the two coefficient names gives the same family and the same slot data. At an A-endpoint the label head h_{\mathrm{lab}} is the first r bits of its evaluation. At a B-endpoint its suffix is the first s-r evaluation bits and its bucket is slot data. Equality of the raw endpoint prefixes therefore gives the same full outward label. Extra-dependent coefficients cause no exception because those extra values are part of their names.

Remove one edge at each end. By (4.2), the new endpoints agree to depth at least h-s, and the new threshold is (p-2)s/2+M_k. The prefixes are available at the new endpoints by the same propagation statement. The remaining subpath is still in the same band and consists of good vertices. If p>2, this is a proper coherent return, a contradiction. If p=2, the two edges entering its middle vertex have the same outward label there, so foldedness makes the original path an immediate backtrack. This too is impossible.□

Proof. Choose two incident slots at each theta branch vertex and use the two incident slots at every degree-two vertex. At the coherent junction use the two end slots through depth h. At a free end beyond h use its prescribed u-block and its single incident evaluation.

First recover every A-block A_0. Its two evaluation heads are known from the labels. If their coefficients are c_1\ne c_2, subtraction recovers the raw u-block, and then the raw v-block. At later blocks the known correction offsets can be moved to the right-hand sides, so the same two equations have the form

The difference c_1-c_2 is a nonzero field element and is invertible on the full padded block. Thus these equations uniquely give

At a free endpoint, u is already prescribed and the one equation uniquely gives v.

The update of B-block j uses the two values of P_{A,j-1} at the neighboring A-input blocks already recovered. The subsequent update of A-block j uses P_{B,j} at the B-input blocks just recovered. Each key contains raw blocks strictly before its source block; current correction windows end before both coupled blocks. All these data are known before the current table is read. This proves the assertion inductively in the individual source-block order A_0,B_1,A_1,B_2,A_2,\ldots.

The induction is also an unambiguous formal reconstruction when a pattern is invalid: solve on full padded blocks, stop after all blocks that can meet the candidate cutoffs, and then test the required zeros, labels, and vertex conditions. An assignment violating an inactive-coordinate condition need not define a graph vertex, but it causes no ambiguity in earlier query data. Every actual realization is reproduced by the induction, which proves the claimed upper bounds on realizations. Ignoring some final validity conditions will consequently only increase our counts.

It remains to count the free endpoint u-bits. The active interleaving implies

Consequently

At aligned depth h an A-endpoint has specified (h+2r-s)/2 u-bits, and a B-endpoint has specified h/2 u-bits. Adding the two residual lengths on their common side gives (4.6). A theta has two chosen evaluations at every vertex and hence has no freely prescribed coordinate.□

Proof. After a failed test give the process value zero. Otherwise multiply its value at step t by 2^{R_t}\exp(-\epsilon_s k_t) when that test passes. Conditional expectation does not increase, by (4.8). The initial value is one, so the final nonnegative variable has expectation at most one. On A its value is at least 2^R\exp(-\epsilon_sK), proving the claim. Additional final conditions are used only in this last lower bound, not in any conditional probability.□

Proof. Apart from the initial short block, which is never terminal, there are two terminal forms. If the final active input has t bits in its s-bit field, then its evaluation block has all s bits. The map in (4.7) sends an s-bit v and a t-bit u-subspace injectively into two s-bit evaluations. Its image has dimension s+t, so the compatibility codimension is s-t. The nonzero coefficient difference proves injectivity on the full field and therefore on the active subspace.

On the opposite side, the input has ended and the last evaluation has t bits. Its two incoming values are independent uniform t-bit values from the two active-prefix permutations. Since u=0, they must agree after their known offsets are removed. This has codimension t. These two surpluses add to s. The boundary cases t=0,s simply move the one nonzero surplus to the other side. This proves the equal-cutoff statement.

For adjacent incidences the correction extras are fixed label data, and the lagged gates are determined by earlier raw blocks. For incidence a write its known correction as o_a=C_a^v+K_aC_a^u and move the whole vector o_a to the other side of the equation. Its support may fill the output field; translation by it costs no random dimension. On a common active random prefix, with coordinate projection \pi and width w', the map

has affine image of dimension at most w'+t in 2w' bits. Thus its codimension is at least \max\{0,w'-t\}. This uses no invertibility of a truncated multiplier. For the terminal form with no input, replace t in this dimension count by zero.

Here are the active widths, including the boundary between blocks. Write t\in\{b,2b,\ldots,s\} for the width of the last active block in the interleaved raw input stream at the receiver’s height; each incident common cutoff is the receiver cutoff or is b bits earlier. If t>b, both evaluations of the partial-input side occur in the preceding source block and have the full s random bits. Their codimension is at least s-t. On the evaluation-only side the common random prefix has width at least t-b, giving that many compatibility bits. If t=b, the cutoff can cross a block boundary. On the partial-input side the preceding source block still has full random width: it is either ordinary or the full active prefix of a marked block. It therefore still contributes s-b compatibility bits. On the opposite side a lower incidence can already be in its deterministic post-mark block. Its surplus is only b, so we use the certain event and charge zero, as permitted by \max\{0,b-4b\}=0. The same reasoning applies with A and B interchanged; t=s includes the boundary represented by t=0 on the other side. All retained positive-width outputs are actual independent uniform current-table prefixes. These bounds are stronger than the stated conservative loss of 4b, and prove the probability assertion.□

Proof. At a B-block j update the queried table is P_{A,j-1}. Equality of its key identifies the family, full label, A raw inputs through block j-2, and B raw inputs through block j-1. In particular the B u-prefixes through j-1 agree. To compute their B v-prefixes through j-1, the equations use A inputs through j-2, the same earlier keys, and the same coefficient and label offsets. Correction gates use still earlier raw pairs. Thus the evaluation prefixes agree, and v=Q-Ku gives agreement of the v prefixes. At an A-block j update the queried table P_{B,j} identifies A raw inputs through j-1 and B raw inputs through j-1. The A evaluation through j-1 uses only this B data and earlier keys, giving the identical conclusion.

More explicitly, if the current table has source index k, its preceding complete receiver prefix has aligned depth exactly ks: for P_{A,j-1} this is 2(j-1)s, and for P_{B,j} it is (2j-1)s. Write \kappa for the final common source index of an incident edge and t\in\{b,2b,\ldots,s\} for its active length. Its common scale is H_c=\kappa s+t. The two terminal forms test source index \kappa-1 or \kappa, respectively. Thus ks\geq H_c-2s. The common scale is at least the receiving vertex scale minus b, and hence ks\geq H_*-2s-b. If two adjacent incidences straddle a block boundary, restricting both to their earlier common receiver prefix loses at most one more aligned step, of length 2s. This is still greater than H_*-8s for b<s. Every prefix just used ends before the first partial receiving block and is therefore active at both receivers. Repeated keys at the same receiving vertex cannot occur among its two chosen slots: their full label names are different.□

Proof. By the available endpoint depth,

Suppose two keys at terminal receiver positions a<b coincide. Lemma 4.6 gives agreement through depth at least H_*-E_p. The receivers have the same side. Their contiguous path segment is nonbacktracking and lies in the same band. If

it is a proper coherent return, contrary to minimality, unless it is the entire original path. The latter case already has both positions in the guards. Otherwise

It follows that a<4E_p/s and p-b<4E_p/s. Both positions again lie in the guards. This proves the claim using only contiguous subpaths of the original return; the prefix identification does not close a new path.□

Proof. Put H=H_i. The first endpoint has scale H, so (4.4) gives p\leq2(H-M)/s\leq2H/s. There are a fixed number of eligible bands and at most L_s^p label/height patterns of a given length. By Lemma 4.3, their free endpoint coordinates have at most

choices. Each pattern and coordinate choice has at most one actual realization. Apply Lemma 4.4 using (4.11); the event being bounded is that this realization is a minimal return in the good graph. Its probability is at most

Thus its contribution, summed over label patterns and free bits, is at most

Here we have used h\geq ps/2+M, and the larger error constant also absorbs the exponential inflation; for large s, \epsilon_s\leq1, and b\geq1.

Since M\geq.03H and sp\leq2H, the exponent is at most

Choose, for example, s/b\geq10^{12} and H_0/s\geq10^{12}, in addition to the earlier requirements. The error here is less than .001H. The finite sum over lengths and bands is bounded by a fixed polynomial in H, which is absorbed by another .002H on increasing H_0 if necessary. This proves the stated bound. These numerical choices are only convenient sufficient choices; no degree grows with tower height.□

Proof. Such a path lies in one of the interior bands chosen above. Its endpoints are the same vertex, and its available prefix depth is at least H_i-2s. Its required depth is at most

The least aligned depth above this number adds less than 2s. For sufficiently large H_0/s it is still available. Hence this path is a coherent return for that band’s margin and contains a minimal return. At least one of its vertices, in fact all vertices of that minimal return, was deleted. This is a contradiction.□

Proof. A repeated terminal key gives receiver-prefix agreement through H-be-8s by Lemma 4.6. Join the receivers by a shortest theta path. It has length below .76m_i and lies in an interior band containing the theta. Its coherent threshold, including its margin, is at most

This is smaller than the available agreement depth for sufficiently small b/s and large H_0/s. The path is a coherent return, contradicting its survival in X^{(1)}.

For a shared key at the branch vertices, the last sentence of Lemma 4.6 identifies their preceding complete receiver prefixes. A source update at aligned depth at least .65H identifies them to depth at least .65H-4s. Their shortest branch has length below .51m_i, whose coherent threshold is less than

Again the available margin is much larger than the block and height errors. This gives the same contradiction. Equality of unordered triples in particular would give a shared key.□

Proof. The three slots at one branch vertex have different labels and hence different keys. At their already known input arguments, the three fresh table values are independent uniform field elements. Equation (4.12) can be solved for any one of them and has probability 2^{-s}.

Two distinct three-element key sets each contain a key absent from the other set. Condition on all the other whole tables, leaving these two exclusive tables unexposed. Each exclusive table is used at one known argument in its branch-vertex equation. Its value is uniform and has a nonzero coefficient; it does not enter the other branch equation. The two equations therefore have probability 2^{-2s}. If other formal recovery queries use those same tables, their results have not been prescribed and are integrated out. With identical triples retain only one of the two necessary equations, giving the last assertion.□

Proof. Only theta patterns need be considered. For a fixed total length e, choose its three ordered positive branch lengths, the position and orientation of its root, and its bipartite side. The number of these choices is at most 100(e+1)^4. For example there are fewer than (e+1)^2 length triples, at most 2e oriented root positions, two side choices, and a fixed finite number of choices of the first branch. Give its edges orientations for enumeration and choose at most L_s^e label/height patterns. Assigning heights along a spanning tree determines the remaining heights; any inconsistency on a remaining edge causes rejection.

Lemma 4.3 has no free coordinates for a theta. For each fixed pattern use the predictable terminal and branch tests above. Every actual theta in X^{(1)} passes their key and gate rejections and has the stated accumulated ranks. The current-table count is at most 2(e-1) for the terminal tests plus six per branch update. Since e>.9m_i, the latter count is O(H_i/s)=O(e); a bound of 20e suffices after omitting the fixed initial updates. Lemma 4.4 bounds its probability by

After summing patterns, the exponent is at most

For e<1.52H_i/(.50005s) its main coefficient of H_i is

The error and the polynomial sum over e fit below .007H_i by the allowed parameter choices. This proves the bound. Embeddedness and survival in X^{(1)} are final predicates on the reconstructed assignment; they were not used to condition any current table law.□

Proof. We first use one-variable substitutions to balance every edge. We then use two-variable substitutions to isolate a balanced closed path supported on edges traversed at least four times. Its support will contain the required bicycle.

For an oriented edge e, let n_e(V) denote the number of its positive traversals minus the number of its negative traversals in an edge word V. Assigning t_e=x and all other variables the identity gives x^{n_e(W)}=1, so

Thus every used edge has positive even traversal multiplicity. Call an edge paired if its multiplicity in W is exactly two, and call it heavy if its multiplicity is at least four.

We first examine a paired edge e. Its two signs are opposite by (4.15). Starting the cyclic word at its first occurrence, write

Cyclically rotating a law preserves lawhood, since a cyclic rotation of a word is a conjugate of it under every assignment. Fix an edge f\ne e and assign t_e=x, t_f=y, and all remaining variables the identity. Put n=n_f(U). Equation (4.15) gives n_f(V)=-n, and the assigned word becomes

The infinite order of z implies n_f(U)=0. The edge e does not occur inside U, so we have proved the following interval property:

Thus the open interval between the occurrences of a paired edge is balanced on every edge. The two-variable substitution gives the required pairwise area condition directly.

Fix the original linear ordering of the occurrences in W, and join the two positions of each paired edge by an interval. These intervals cannot cross. Indeed, if the intervals for e and f crossed, the open interval for e would contain exactly one occurrence of f, contradicting (4.16). If paired edges exist, choose an inclusion-minimal paired interval, and let U be its interior edge path. Every edge occurring in U is heavy: if a paired edge occurred there, the interval property would force both of its occurrences to lie there, giving a smaller paired interval.

Moreover, U is a closed edge path, because the inside endpoints of the two opposite traversals of the same edge coincide. It is nonempty, since an empty interior would give an immediate reversal in W. It is reduced as a based path, being a contiguous subpath of W, and it is balanced by (4.16). If there are no paired edges, take U=W; these same properties follow from the hypotheses and (4.15), and every edge of U is again heavy in W.

We claim that the connected support S of U has first Betti number at least two. Here it matters that U need not be cyclically reduced at its basepoint. Repeatedly remove its first and last traversals whenever they are mutually inverse. This expresses U as P C P^{-1}, where C is a cyclically reduced closed path. The path C is nonempty: if this process reached the empty path, its last removal would have removed two adjacent inverse traversals from the original reduced path. Each removal subtracts opposite traversals of the same edge, so C is still balanced.

Deleting a vertex of valence one and its incident edge from a finite graph cannot affect a cyclically reduced closed path: a visit to that vertex would require an immediate reversal. Such deletions also preserve first Betti number as long as an edge remains. If S had first Betti number zero, this deletion process would leave no edges, contradicting the existence of C. If S had first Betti number one, the remaining nonempty core would be a single cycle. To see the last assertion, the core is connected with minimum valence two and satisfies |E|=|V|; hence all its valences are exactly two. A nonempty cyclically reduced path on a cycle traverses it a nonzero number of times in one fixed direction. Its exponent sum on each cycle edge is therefore nonzero, contradicting the balance of C. This proves the claim, including when the basepoint of U lies on a path leading to its cyclic core.

Finally, choose a spanning tree of S and two distinct edges outside that tree. Let T' be the smallest subtree containing the endpoints of those two edges, and let B be their union with T'. The subgraph B is connected and has first Betti number two. Every leaf of T' is an endpoint of one of the added edges, so B has no vertices of valence one; the case where T' is a single vertex gives the same conclusion directly. Thus B is a bicycle contained in S. Each of its edges is heavy in W, and positivity of the edge lengths gives

The asserted lower bound follows.□

Proof. For a realization, let Z_i^A,Z_i^B be the additional deleted fractions. Bounds on rooted objects imply bounds on all their vertices, with only polynomial and small scale factors, as follows. An object of length \ell rooted at height j can meet layer i only if |i-j|\leq\ell. Every counted object satisfies \ell\leq4H_j/s. From H_j=H_i+b(j-i) and s\geq20b we obtain

Its vertex count is at most \ell+1\leq5H_i/s+1. There are at most 10H_i/s+1 possible root layers and at most a linear number of lengths. Lemmas 4.8 and 4.12 therefore imply, after division by |\mathscr B_i|=2^{H_i} or |\mathscr A_i|=2^{H_i+2r-s},

for sufficiently large s/b and H_0/s. More explicitly, the larger rooted bound contributes 2^{-.025H_i+.975(5bH_i/s)} times a fixed polynomial in H_i. The coefficient remains below -.02 for small b/s, and that polynomial is absorbed by another .01H_i after increasing H_0. This proves (4.17), including the cost of deleting vertices of an object whose qualifying root is in a different layer.

Since H_i=H_0+bi,

Choose H_0 still larger so the last expression is less than \varepsilon/2. Markov’s inequality then gives positive probability that \sum_i(Z_i^A+Z_i^B)<\varepsilon. On that event each individual deleted fraction is below \varepsilon. Every table realization in the sequential law already has the undeleted expansion properties, so choose one in this positive-probability event. The earlier good-vertex deletion budget is separately supplied by Proposition 2.6 and can be made as small as required there.

Passing to an induced subgraph preserves foldedness and uniqueness of a word that occurred at most once before deletion. Lemma 4.9 proves the second assertion, and the definition of the second deletion proves the third. It remains to deduce the nonlaw assertion, allowing for the variation of scale along the closed path.

Let W be a nonempty cyclically reduced closed path of length L\leq6.04m_i based at height i in X_{\mathrm{old}}. Every vertex v of its support has scale m(v)\geq m_i-aL\geq(1-6.04a)m_i. If W were a \HH-law, Lemma 4.13 would give a bicycle B in its support with L\geq4|E(B)|. The third assertion, already proved by the deletion, gives

Consequently

where the last inequality holds for a\leq10^{-6}. This contradicts the choice of W. Independent edge variables here include every underlying edge, regardless of any scalar weight assigned later; reverse traversals use the inverse of the same variable.□

Proof. The relation [x,y]=z gives y^jxy^{-j}=xz^{-j}, so conjugating the torus rectangle changes the \xi-coordinate by -j\vartheta. If 0<\abs{j-k}\leq2J_i and \abs{\vartheta-\alpha}\leq R/Z_i,

This exceeds 2R/N_i at all sufficiently large heights because N_i=J_i^4 and Z_i=J_i^{10}. The translated supports are disjoint. Expansion of E_iT_i^*T_iE_i now leaves only its diagonal terms, proving the first identity. All subsequent product and norm identities follow by inserting the auxiliary support projections. The trace on the torus algebra is Haar measure: it vanishes on every nonidentity character \lambda(z)^c\lambda(x)^n, which determines its integral on continuous functions and hence on Borel spectral projections. The trace property and E_iT_i^*T_iE_i=E_i give \tau(P_i)=\tau(E_i), and the rectangle has area at most 4R^2/(Z_iN_i). In particular the number of translates does not multiply the trace of this isometric image.□

Proof. We first bound the central coefficients at fixed n and l, and then sum the resulting horizontal majorant. Fourier transform only in \xi. With t=Z(\vartheta-\alpha), its n-th coefficient is

The function \Phi is supported in one fixed compact t-interval and is Schwartz in v, uniformly with all t-derivatives and all profile parameters. For example integration by parts in u bounds \norm{\partial_t^a\Phi(t,v)} by C_{a,M}(1+\abs v)^{-M} for every fixed a,M.

The Heisenberg commutation formula gives the coefficient at y-power l, before central Fourier transformation, as

The ratios J_k/J_i are fixed. Thus b_{j,l} is supported on O_b(J) indices, is zero unless \abs l\leq C_bJ, and its r-th discrete differences satisfy

These statements follow by applying the fundamental theorem of calculus r times to the two fixed bump factors, whose normalization is comparable to J.

It is essential to retain the central residual phase. Treat e^{-2\pi i n jt/Z} as part of the amplitude and sum by parts against e^{-2\pi i n j\alpha}. For all fixed nonnegative integers r,a, the discrete product rule and \abs j\leq C_bJ give

The trivial sum bound and r summations by parts therefore imply

This also covers n=0 by the trivial bound. To see the factor 1+J\norm{n\alpha} precisely, summation by parts divides by \abs{1-e^{2\pi in\alpha}}^r, which is bounded below by a constant times \norm{n\alpha}^r; taking the minimum with the trivial bound gives the stated expression.

Apply (5.7) to g(t)=N^{-1}\Phi(t,n/N)S_{n,l}(\alpha+t/Z), using r=3 in (5.10). Since

the polynomial factors introduced by every fixed number of central derivatives are absorbed by choosing the Schwartz exponent of \Phi larger. This proves (5.8), for arbitrarily large n as well as for the principal x-frequency range.

It remains to obtain the compensating factor J^{-1} when summing over n. On any interval of M consecutive integers the points n\alpha\bmod1 are separated by at least 1/(3M), by (5.5). Hence at most 1+6Mu lie at distance at most u from any fixed point of the circle. Integrating the identity

against this counting bound gives

For u beyond half the circle the same upper bound still bounds the count, so the integral is valid as written. Split the integers into the central interval \abs n\leq N and dyadic annuli of lengths comparable to 2^vN. Since N\geq J, (5.11) yields

Summation over the O_b(J) possible l proves the one-norm assertion. Using a larger exponent P gives arbitrarily rapid polynomial tails when \abs n>L N; using a larger A gives the same statement for \abs c>L Z. Their sum proves (5.9).□

Proof. Truncate the Fourier series to (5.9), with error at most \eta/2, and denote its matrix coefficients by A_{n,l,c}. Set

By Lemma 5.3, R_{n,l}:=\sum_c\norm{A_{n,l,c}}\leq C_1w_n, while \sum_nw_n\leq C_0/J_i. On the finite allowed n-interval begin a probability measure with masses J_iw_n/(2C_0), whose total is at most one half. Distribute the remaining mass uniformly on \{-N_i,\ldots,N_i\}. Call the result \mu_i. Let \nu_i be uniform on the integer interval containing all allowed l, of length comparable to J_i. Then (5.13) holds and

for a height-independent constant. The finite cutoff L and fixed neighboring ratios are harmless in this inequality.

Draw n and l independently from \mu_i and \nu_i. If R_{n,l}>0, draw c conditionally with probabilities \norm{A_{n,l,c}}/R_{n,l} on the nonzero coefficients, and set

If R_{n,l}=0, set W=0 and choose c=0. Thus zero-weight draws still have exactly the same diffuse horizontal product law. Directly summing the expectation recovers every truncated coefficient, and \norm W\leq M_\eta.

Choose a finite net in the closed matrix ball of radius M_\eta, include zero, close it under adjoint, and round W within \eta/2, keeping zero fixed. The resulting expectation changes in one-norm by at most \mathbb E\norm{W-W_{\rm rounded}}\leq\eta/2. For a reversed edge use the adjoint weight and inverse voltage rather than an independent draw. The net cardinality depends on the accuracy and the fixed matrix dimension, not on height.

The coordinate bounds follow from (5.9). In the chosen normal form,

The product \abs{nl} is bounded by a constant times 2^{250H_i}, much smaller than 2^{500H_i}. Every fixed prefactor is absorbed by raising the initial height, giving the stated exponent for both voltages and inverses.□

Proof. The polynomial estimate is the maximal-atom version of the usual polynomial zero bound [34, Lemma 1]; we include its short induction. If independent real-valued variables have atoms at most \kappa, a nonzero polynomial of total degree d vanishes with probability at most d\kappa. Induct on the number of variables. If its degree in the last variable is s, its leading coefficient is a nonzero polynomial in the preceding variables of degree at most d-s. By induction that coefficient vanishes with probability at most (d-s)\kappa; otherwise the one-variable polynomial has at most s roots and costs at most s\kappa. Variables absent from the polynomial are discarded. A nonzero constant has zero vanishing probability, starting the induction.

By the normal-form multiplication law just displayed, the horizontal coordinates of any word are integral linear combinations of the edge coordinates n_e,l_e, with coefficients equal to signed occurrence counts. If any such count is nonzero, one horizontal polynomial is nonzero. If all counts vanish, every central coordinate c_e cancels, and the central coordinate of the word is a polynomial of degree at most two in the n_e,l_e. Were that polynomial identically zero, the word would vanish for every assignment in \HH, contrary to the hypothesis. Thus a nonzero polynomial of degree at most two is tested on independent horizontal variables. Their maximal atoms are bounded by C_b2^{-50H_{\min}}, since J_i/N_i=2^{-150H_i}. Apply the polynomial bound. No conditioning on the central coordinate or on the matrix weight is performed. This proves the assertion also for an edge whose realized weight is zero.□

Proof. For an integer T let

If v has coordinates bounded by R_0 and T\geq R_0, left or right multiplication by v changes a horizontal coordinate by at most R_0 and the central coordinate by at most R_0+R_0T. This follows directly from the normal-form multiplication law. Points outside the corresponding boundary strips remain in the box. The total strip fraction is at most C R_0/T. Multiplication is a bijection, so the same bound, with a factor two, controls the symmetric difference. Take T=\lceil2^{(a_0+B+2)H}\rceil. The box has (2T+1)^2(2T^2+1)=2^{O(H)} elements, proving the first assertion uniformly over all possible translations.

For the norm estimate, let Q_t retain the deck coordinates in the right translate \mathcal B_Ht. Right translation commutes with left convolution, so every Q_tAQ_t, acting on its range, has the same norm as Q_HAQ_H. For a finitely supported vector \xi,

Average \ip{AQ_t\xi}{Q_t\xi} over t and divide by \abs{\mathcal B_H}. In this average the coefficient of \lambda(v) is multiplied by \abs{\mathcal B_H\cap v\mathcal B_H}/\abs{\mathcal B_H}, which differs from one by at most C2^{-BH}. The averaged quadratic form has absolute value at most \norm{Q_HAQ_H}\norm\xi^2. The difference from A has norm at most C2^{-BH}R: to verify this last assertion, apply Cauchy–Schwarz to a matrix of nonnegative entry-norm bounds whose row and column sums are at most that number. Density and the selfadjoint quadratic form formula prove (5.14). None of these estimates separates coordinates of the matrix amplification. Doubling a rectangular matrix gives the final statement.□

Proof. Set a=g/10, and choose a maximal a-separated set P in the intrinsic metric. It is an a-net. If it has at least two points, the open balls of radius a/2 around them are disjoint and each contains a path interval of length a/2 pointing toward another net point. Hence \abs P a/2\leq Kg, and \abs P\leq\max\{1,\lceil20K\rceil\}, also covering a singleton net. Join each pair of distinct net points at distance at most 3a by one edge of an abstract graph L, mapped to a chosen shortest path in S. This map is surjective on first homology. To check this, subdivide a path of S into pieces of length at most a, choose net points within a of the subdivision points, and replace each piece and its two endpoint connectors by the chosen edge path. The replacement bounds a closed path of length at most 6a<g, which is null-homotopic: cancellation of immediate reversals in a nontrivial closed graph path leaves a path containing a simple cycle no longer than itself. Connectors cancel consecutively around a loop. Thus L is connected and its mapped loops span first homology. Finally \dim H_1(S;\mathbb Q)\leq\dim H_1(L;\mathbb Q)\leq\abs P^2.□

Proof. A support has at most h edges. Prune leaves other than the starting vertex, retaining its rooted core K. A tree leaves just that vertex. Otherwise K has no degree-one vertex except possibly the root and has the same cycle rank as the support. The identity

gives \deg_K(v)\leq2R+2. The removed pieces are trees attached at single core vertices, since neither a cycle nor a path between two core vertices could be removed by this pruning.

There are at most 4^e\Delta^e possible rooted connected subgraphs K with e edges. For clarity this includes cyclic K. Fix orders on ambient incident edges and explore K by marking edges. When an unmarked K-edge is available, mark and traverse it, push it onto a stack, and recursively explore at its endpoint, even if that vertex was already visited. When the recursive call has exhausted its unmarked edges, return along the pushed edge and pop it. Every edge is pushed and popped exactly once. The balanced push/pop word has at most 4^e possibilities, and at each push an ambient edge has at most \Delta choices. These data recover the traversal and its edge set.

For fixed K, classify each walk step as a core step, a tree descent away from the core, or a forced tree return. There are at most 3^{2h} type words. Each core step has at most 2R+2 choices. If t denotes the number of descents and c the number of core steps, closure gives 2t+c=2h. Every core edge is traversed at least twice, so c\geq2e and t\leq h-e. Descents have at most \Delta choices and returns are determined by their stack. Thus the count is bounded by

using h+1\leq2^h. Invalid formal encodings only increase this upper bound. No assertion that a short backtracking walk is a simple path is used.□

Proof. An edge entry is W_e\lambda(v_e)-\mathbb E[W_e\lambda(v_e)] and has norm at most 2M_\eta. Double the block off the diagonal to make it selfadjoint. Its absolute row and column coefficient sums are at most C_*M_\eta D.

We first make precise why a fixed finite-test exponent is available before choosing s. Require H_0\geq s, so

After raising H_0 to absorb fixed constants, the row bound is at most 2^{2H_i}, and the number of base and matrix indices is at most 2^{3H_i}. The coordinates of every possible voltage and inverse have the fixed exponent of Lemma 5.4. Apply Lemma 5.6 with a boundary exponent larger than four. Its box error tends to zero exponentially even after the row loss, and its total finite compression dimension is at most 2^{K_1H_i} for a fixed K_1. Enlarge H_0 further so the box error is at most \eta q_{\tau,ik}/2. These enlargements do not change K_1.

Now fix C>2(K_1+6), increasing it strictly if needed, and put

Every no-singleton support in the 2h-moment has at most h base edges. It lies in the two fixed layers of this block. Their base girth is at least .9H_i/\log_2q after increasing H_0, since adjacent H-values differ by only b. Because \log_2q/\log_2d<1.25, the ratio of support length to this girth is at most 2C, again increasing H_0 for the ceiling. Lemma 5.7 gives a rank bound depending on C alone. This is a use of base girth, not the as-yet unproved voltage girth.

Let \widehat X denote the finite selfadjoint compression. Expand \mathbb E\operatorname{Tr}(\widehat X^{2h}) along base vertices and individual edges. Any summand with a singleton unoriented edge has expectation zero: after conditioning on the other edge draws it is a fixed left factor, one centered compressed entry or its adjoint, and a fixed right factor. Compression does not change this independence. For every other summand its absolute trace on the deck and matrix indices is at most their dimension times (2M_\eta)^{2h}. Lemma 5.8, with \Delta\leq C_*D, proves

where C_1 is fixed after C and the coefficient data, but before s. Since \widehat X is selfadjoint, its largest absolute eigenvalue to the power 2h is at most this nonnegative trace. Markov’s inequality and (5.1) give

Choose s large enough that C_2/\eta^2\leq d^{1/2}. The last expression is at most 2^{(K_1-C/2)H_i}\leq2^{-6H_i}, by the choice of C. The potentially large dependence of C_2 on C is not circular: s is chosen after both constants. This proves the claim.□

Proof of Proposition 5.2. For a block put a_{\tau,ik}=q/q_{\tau,ik}. This scalar lies in a fixed compact subinterval of (0,\infty). Apply Lemma 5.4, with smaller tolerances, to the symbol a_{\tau,ik}S_{\tau,ik}. Multiplication of the profile family by these bounded scalars preserves every uniform estimate. Round these already rescaled weights to one finite net, and normalize the weighted incidence by the common denominator q. Within a block use the same law at every edge, independently over all underlying unoriented edges. Its expectation is

whose distance from U_{\tau,ik}\otimes S_{\tau,ik} is bounded by a fixed constant times the assigned one-norm error. Lemma 5.9, with its tolerance reduced using q_{\tau,ik}/q\leq C_*, supplies the remaining error. This common-q normalization avoids any need to recover an unbounded height from a label.

We verify the voltage-girth probability for the same law. A forbidden closed lift projects to a cyclically nonbacktracking closed base walk. Test all such base walks of length at most \lceil6.03m_i\rceil from height i. For large m_i, this length is below 6.04m_i; hence the local bound in Proposition 4.14 says that each is a non-law. This also removes any rounding issue at exactly 6.03m_i. Along such a walk,

after raising the initial height. Taking b/\log_2q\leq10^{-6}, Lemma 5.5 bounds its identity probability by 2^{-40H_i}, with all fixed factors absorbed by the initial height.

The number of possible rooted walk instructions is bounded by

for sufficiently large s and then H_0/s. Indeed its leading exponent divided by H_i is

the finite type factor, the side-size offset .2s, and rounding fit the remaining margin. The union bound for height i is therefore at most 2^{-31H_i}. It counts every edge, even those with realized weight zero. Triviality of holonomy is independent of the chosen point in the initial deck fiber, so no additional union over that infinite fiber is needed.

There are finitely many block types at each height. Since H_i=H_0+bi, the sum over all heights of their norm-failure bounds and the girth-failure bounds can be made less than one by raising H_0. No conditioning on the girth events has entered the centered moment calculation. For completeness, one can obtain a simultaneous choice without invoking an infinite-product measure. Enumerate all the required block inequalities and walk tests. Any finite list involves finitely many edge variables and, by the same summed probability bound, has a satisfying assignment. Extend those assignments arbitrarily to all other edges. Each edge has finitely many possible voltage/net-weight pairs, so successive subsequences stabilize the choice at the first edge, the second edge, and so on. The diagonal limit satisfies every condition, because each condition depends on finitely many edge choices. Fix such an assignment.

Finally append the finite matrix-weight index to each old label, and use the inverse label with adjoint weight on its reverse. Each refined positive label now specifies its matrix weight, and every type uses the common normalizer q^{-1}. These are the coefficients of the finite Cayley polynomial in Equation (8.15). Its comparison with the sampled blocks on retained sheet domains, and the estimate on the remaining edges, are proved in Section 8. Equality of refined words implies equality of the original words, so all original uniqueness conclusions persist. The original leading label count is d2^r=2^{1.00005s}; prescribed types, the weight net and extras contribute height-independent factors, and the residue flags contribute 2^{O(\log s)}. Large voltage coordinates are separate edge data and do not enter this alphabet. Since \log_2(q^2)=1.0001s, the displayed sharp margin follows. The ratio is more precisely 2^{-.000025s+O(\log s)+O_{b,\eta,\mathrm{profiles}}(1)}, not a claimed constant-prefactor exact decay. This proves all assertions.□

Proof. Choose an old vertex in each component of X_{\mathrm o} and enumerate the components. Join the chosen vertices in a chain by new subdivided paths; if there are finitely many components use a finite chain. The chain has no new branching vertices. For each joining path with endpoints v,w, choose its integer length n to satisfy

Interpolate m linearly along the path. All of its voltages may be the identity. At one old basepoint v_* attach two further subdivided based loops, each of length at least 20m(v_*), with new interior vertices. Keep m=m(v_*) on these loops. Assign their voltages so that their ordered products are respectively x and y; for example use identity voltages except on one positively traversed edge of each loop. Give every added positive edge a distinct new letter. This preserves foldedness. The chain adds at most two incidences at any old vertex and the two loops add four at v_*. Their lengths are greater than two, so all new vertices have degree two. The interpolation proves the second assertion.

The completed base graph is connected. A path from v_* to a given base vertex v has some voltage h_0. By first traversing based words in the two added loops, one can obtain any prescribed voltage in \HH, since x,y generate \HH. Following the chosen path thereafter reaches any specified lift (v,h). Thus the cover in (6.1) is connected. Its displayed left action is free on vertices and transitive on each fiber.

Consider a simple cycle that uses a private edge. Since every private interior vertex has degree two, the cycle traverses an entire lift of a joining path or one of the two added subdivided loops. Let v be an old endpoint of that lifted path and let \ell be the length of the cycle. Then \ell\ge20m(v). At any other vertex w of the cycle the Lipschitz bound gives m(w)\le m(v)+a\ell. Consequently

A cycle using only old edges satisfies the required inequality by the input hypothesis. This proves the fourth assertion, and in particular the cover has neither loops nor two-edge circuits.

A private letter in a common word identifies its unique oriented base edge. Foldedness then propagates equality of the two projected paths forwards and backwards from that edge, without any assumption that the paths are immersed. This proves the first statement of the last assertion. When different projected occurrences are immersed they therefore consist of old edges, and the old recovery bound applies. Reversing the paths gives the bounds at their other endpoints as well.□

Proof. Only the finitely many letters in the displayed expression matter. Let B be their finite bouquet of oriented circles. The elementary free-cancellation rule constructs a null homotopy in B for a freely trivial word: delete adjacent inverse traversals one pair at a time. Take an oriented sphere with one disk removed for each of the words w and r_1,\ldots,r_n, reading w with the opposite boundary orientation. Choose a cutting tree joining these boundary circles, and prescribe the words u_j on its cuts. Cutting along the tree gives a disk whose boundary reads the freely trivial word supplied by the displayed identity. The cancellation homotopy extends its boundary map to B. Regluing the cuts gives a map from the punctured sphere to B with the specified word maps on all boundary circles.

Choose one interior point of each circle of B. A piecewise linear approximation relative to the boundary can be made transverse to these points as follows: triangulate the source sufficiently finely so each triangle maps into an interval of a circle or into a small tree around the bouquet vertex; choose the interior points outside the images of source vertices. Inside each triangle the inverse image of a selected point is a line segment or is empty. These segments fit to disjoint arcs and circles. Their boundary endpoints are precisely the occurrences of the selected letters on the word boundaries, and each arc joins opposite signed occurrences of the same letter. Closed inverse-image circles may be ignored. Thicken each remaining arc to a narrow disjoint band, cap the word boundaries by disks, and contract each cap to a vertex and each band to an edge. This is an embedded ribbon multigraph in the sphere; its cyclic order at a word vertex is exactly the cyclic order of that word’s letters.

If w is nonempty, retain the component containing its word vertex and take the planar dual, specifying that vertex’s dual face as exterior. Every bounded dual face corresponds to one of the retained r_j word vertices, with its original source walk, so the perimeter budget follows. This argument permits loops, multiple edges, and nonsimple dual face contours. If w is empty, the one-vertex empty diagram suffices. A freely trivial but nonempty w is covered by the same construction, even when there are no relator word vertices in its component.

For the converse, take a thin regular neighborhood of the embedded graph. Its inner boundary circles correspond to the bounded facial contours and its outer boundary to the exterior contour. Cut between these boundary components along a tree of arcs. The cut surface is a disk, and the neighborhood retracts to the labeled graph. Reading the disk boundary in B gives the exterior word as a product of conjugates of the bounded face words, each occurring once with its compatible orientation. The same conclusion holds for the contour of each simple cycle of the graph, by taking the portion on its bounded side. Consequently every closed walk in the graph is trivial in \Gamma: split a walk at repeated vertices until it consists of simple cycles and immediate edge reversals. Labels of paths from a chosen vertex to any other vertex are therefore independent of the chosen path in \Gamma. This is the asserted vertex map; each edge maps to its formal Cayley cell.□

Proof. We argue by a minimal counterexample. Coincident projected edge occurrences will permit reductions that lower the total face perimeter. The remaining comparisons are short by recovery, so each bounded face needs at least seven arcs; the final Euler count will rule out such a diagram.

Suppose a counterexample exists. Among all such source paths and all finite diagrams for their words, minimize first the filling cost c, then the exterior path length, and finally the number of diagram edges. All three quantities are nonnegative integers. The exterior path has one marked end-to-start join. Its source endpoints are required to be distinct, but may change when considering a competing counterexample. This global choice is needed for the bridge argument below.

Each bounded source face may be assumed nonempty and cyclically immersed. To justify this without assuming anything about diagram regularity, use the free-group face expression from Lemma 6.2. Foldedness turns every free cancellation in a source walk into an actual edge reversal. A freely trivial closed face word can be omitted from that expression. Otherwise delete its backtracks and its cyclic conjugating tail, absorbing the tail into its conjugator in the expression. The remaining word is read on a closed source walk with smaller perimeter. The construction part of Lemma 6.2 gives a diagram for the unchanged exterior word with strictly smaller filling cost. Minimality excludes these possibilities. In the same way the exterior source path is linearly immersed: deleting any backtrack preserves its two source endpoints, does not increase the least filling cost, and decreases its length.

We next compare the two source occurrences of each diagram edge incident to a bounded face, orienting them to read the same signed letter. There are three cases, according to the two face incidences. In each case, coincident projections can be eliminated; a bridge on one bounded face can be eliminated without even requiring coincident projections.

First, if two different bounded faces project their common edge to the same base edge of X, a unique deck transformation aligns the two lifted edges. Delete this diagram edge. It is not a bridge, so the remaining graph is connected and the faces merge. Splice their aligned closed source walks, omitting the two opposite edge traversals. The new source walk is closed, and its perimeter is the sum of the old two perimeters minus two. Other shared edges are retained and may occur twice on the new contour. The new diagram is permitted and has smaller filling cost, a contradiction.

Second, suppose an edge has the same bounded face F on both sides. It is a bridge. The exterior face of the original diagram, including its marked join, lies entirely on one side of that bridge. Let D_1 be the other component after deleting the bridge. In the source walk assigned to F, the excursion into D_1 is a subpath \chi reading the bridge, the exterior contour of D_1, and the reverse bridge. All bounded faces of D_1 are former bounded faces other than F. Lemma 6.2 shows that \operatorname{lab}(\chi)=1 in \Gamma and gives a filling of cost strictly less than c, because the positive perimeter of F is absent. If the source endpoints of \chi differed, it would be a smaller counterexample to the globally chosen one. They therefore agree. Delete this source excursion, the bridge, and D_1. The remaining walk assigned to F still closes, the original exterior source path is unchanged, and the filling cost has decreased by at least the two bridge incidences. This is again impossible. Notice that this argument did not assume embeddedness and did not need the two projected bridge occurrences to coincide.

Third, if a bounded face and the exterior path project their common edge to the same edge of X, align their lifts by a deck transformation. Replace this step of the exterior source path by the complementary path of the bounded face, oriented to have the same lifted endpoints. Delete the shared diagram edge, thereby absorbing that bounded face into the exterior. The source endpoints of the exterior path remain distinct, and its filling cost drops. Any subsequent source backtrack deletions preserve that failure. This also contradicts minimality.

These reductions leave only different projected occurrences along edges incident to bounded faces. We now collect them into arcs. Retain the marked join and suppress all other vertices of degree two. There is no unmarked vertex of degree one: it would force an immediate reversal in a cyclically immersed bounded face or within the linearly immersed exterior path. The marked vertex has degree at least one if any edge is present. Each resulting edge, called an arc, has constant face incidences and a constant exterior-path incidence along its interior. An arc incident to a bounded face reads an immersed word on both its occurrences. By the preceding reductions those projected occurrences differ. Lemma 6.1 therefore bounds its length by Lm at the initial point of either occurrence.

The length bounds force every bounded face to have at least seven arc incidences. Otherwise apply (6.4) to its at most six arcs. Its nonempty cyclically immersed closed source walk would violate (6.3) and (6.6).

If there are no bounded faces, the diagram is a tree. A nonempty finite tree has at least two leaves, whereas only the marked vertex could be a leaf. The empty diagram has empty word and cannot represent a counterexample. Thus at least one bounded face is present. Suppression is also defined when the graph is a lone cycle, because its marked join is retained. Write V,E,F for the compressed vertex, edge, and total face counts. Euler’s formula and the vertex degrees give

On the other hand this expression equals \sum_f(6-\deg(f)). Each bounded-face term is at most -1, and the one exterior term is at most 6. Their sum is less than 8, a contradiction. The retained exceptional join is the reason for using 8, rather than assuming the stronger bound 12.□

Proof. Assign to a source vertex the element represented by any path from (v_*,1) to that vertex. Two choices differ by a closed source walk, whose word is a relator, so the assignment is well defined. If two vertices have the same image, a source path between them has null label; Lemma 6.3 identifies its endpoints. This proves vertex injectivity. Edges with the same formal Cayley image have the same initial image and the same signed label. Their initial vertices agree by injectivity and their source edges agree by foldedness. Hence formal edges also embed.

For f\in\HH, let \phi(f)=j(v_*,f). A source path from (v_*,1) to (v_*,f_1), followed by the left deck translate by f_1 of a path to (v_*,f_2), ends at (v_*,f_1f_2) and reads the product of their words. Thus \phi(f_1f_2)=\phi(f_1)\phi(f_2). It is injective by vertex injectivity. Following a path to (v_*,h) and then the left deck translate of a path to (v,1) proves (6.9). More generally, connectedness and label preservation show

Fix one of the private positive base edges. Its letter appears on no other base edge. A label-preserving source automorphism must send a chosen lift of this edge to a lift of the same base edge. Compose the automorphism with a deck transformation to fix that edge’s initial vertex. An automorphism of a connected folded labeled graph fixing a vertex fixes every path from that vertex, hence every vertex and edge. It is the identity. Thus every source automorphism is a deck transformation.

If a left translation by g stabilizes j(\widetilde X), its restriction, conjugated by j, is a label-preserving source automorphism. It is deck multiplication by some f. Evaluating at (v_*,1) gives g=\phi(f). Conversely (6.9) shows that every \phi(f) stabilizes the image. The remaining stabilizer and coset assertions follow by conjugation and translation.□

Proof. The concatenation of any chosen sheet paths is a closed walk in the Cayley graph and has null word. Lemma 6.2 supplies a finite diagram. Over all choices with the prescribed contacts fixed, minimize lexicographically the filling cost, the total side length, and the number of edges. Retain every boundary vertex carrying a junction mark; there are at most k such vertices, even if several junctions coincide.

The source-path and bounded-face reductions at the beginning of Lemma 6.3 apply unchanged: bounded source faces are nonempty and cyclically immersed, while each side is linearly immersed. Coincident projected edges between two bounded faces can be merged after deck alignment, decreasing filling cost. A bridge with both incidences on one bounded face can be removed as in that proof. Here there is no preliminary injectivity issue: its source excursion has null label by the cut-off component, and Lemma 6.3 makes it close. The component containing the exterior retains all marked contacts.

If a bounded face and a boundary side have coincident projected occurrences along an edge, align the closed face lift to that side and replace the one side step by the complementary face path. The two source endpoints of the replaced step agree with the old ones, and therefore both prescribed endpoints of the full side remain fixed. Deleting the separating edge absorbs the face and decreases filling cost. This is impossible by minimality.

It remains to analyze edges whose two sides are exterior. Such an edge is a bridge. If its two traversals occur in the same side, the interval between them within that side is an excursion with no junction in its interior. Its endpoints have the same Cayley vertex, and hence the same source vertex by Proposition 6.4. Delete the excursion from that side and remove the bridge and its cut-off component. No prescribed junction is lost. Filling cost does not increase and total side length decreases, again contradicting minimality. If instead the two traversals belong to different sides, their sheets are distinct. Coincident projected occurrences would be deck-aligned and agree at a physical vertex; the rigidity observation after Proposition 6.4 would identify the two sheets. Thus these projected occurrences differ too.

Suppress every unmarked degree-two vertex. An unmarked leaf would force a backtrack in one side or one bounded face, so all unmarked vertices have degree at least three. Marked vertices have degree at least one when edges occur. The face and side incidences of each resulting arc are constant along its interior: a side change occurs at a retained mark, and a face change requires branching. Both of its occurrences therefore lie within immersed source paths. The preceding reductions show they have different projections. Apply Lemma 6.1 to obtain all the assertions about arc lengths and private edges. The proof using (6.3) and (6.6) again gives at least seven arcs on every bounded face.

For completeness the Euler count uses the exterior degree itself, rather than the coarser estimate by all edges. Let f be the number of bounded faces and let V,E count the compressed vertices and edges. When edges occur,

Every edge has two face incidences, so

Consequently B\le2E-6f\le4k-6. If the diagram has no edges, all of its marked vertices coincide and every side is constant, giving the stated separate case.

Every edge incident to a private source vertex is private. If a side with such an endpoint were nonconstant, its initial or terminal arc would contain a private edge, contradicting the arc conclusion already proved. Finally, (6.5) bounds a side of at most three arcs by 3.014m_Q(v) at its initial endpoint. The contraposition proves the last assertion.□

Proof. Let m=m_Q(v) and R=\lfloor\kappa m\rfloor. All vertices of the induced ball B_Q(v) can be joined to v by paths inside that ball; choosing a predecessor at distance one less gives a shortest-path tree. If the ball had a non-tree edge, that edge and its tree path would contain a simple circuit of length at most 2R+1. If an outside vertex had two edge incidences into the ball, these incidences and a path in the tree would contain a simple circuit of length at most 2R+2. This includes the possibility of parallel edges. Either circuit has a vertex u in the ball, at which m_Q(u)\geq m-aR\geq(1-\kappa a)m. Both possibilities contradict the local girth bound, because

For the declared parameters, the difference between the first two expressions is at least (0.002-6.03\kappa\cdot10^{-6})10^6-2>0. Thus the ball is a tree and each outside vertex has at most one edge incidence into it. Its incidence neighborhood is a tree with possible additional leaves, hence a forest. The incidence neighborhood of a singleton is a star: loops or repeated neighbors would give a circuit of length one or two, which the same local girth bound excludes.

The constants in K_{b,\epsilon} are fixed before s. Since \log_2D=.60005s, we may arrange \log_2(\Delta+1)\leq .61s. The ball has at most (\Delta+1)^{R+1} vertices. If an old vertex of layer i lies in it, Equation (7.1) gives m\leq m_i/(1-\kappa a). Therefore

The displayed strict estimate follows from a\leq10^{-6} and m_i\geq10^6. This proves Equation (7.3).□

Proof. Suppose that no sheet has one of the three properties. Every sheet then has a contact. We will construct a polygon in distinct sheets with all but one side forced to use either a private edge or at least four comparison arcs. The exterior-arc bound will exclude this polygon.

Enter a sheet at a contact u. If u is old, choose a contact w\notin B_Q(u); its integer distance from u is strictly greater than \kappa m_Q(u). If u is private, choose any contact w\ne u, which is possible because the contact set is not that private singleton. Transfer at w to another sheet containing it and repeat. Since the family is finite and consecutive sheets are different, a sheet eventually repeats. At the first repetition, retain the intervening k\geq2 distinct sheets and close the path inside the repeated sheet. The resulting polygon has k-1 chosen nonconstant sides; only the closing side is unrestricted.

Apply Lemma 6.5 with all these contact vertices fixed. If a chosen side has a private endpoint in its sheet, any nonconstant path with those endpoints starts or ends with a private edge. Some private edge is therefore unavoidable in every replacement path with the same distinct endpoints. Yet every arc of the reduced side is a comparison arc, none of which contains a private edge. This is impossible. Otherwise all k-1 chosen sides have old endpoints and have intrinsic endpoint distance greater than \kappa m_Q(u) at their incoming contacts. By Equation (7.2), each such side requires at least four exterior arc incidences. There are therefore at least 4(k-1)>4k-6 exterior incidences, again a contradiction. The argument does not require the closing side to be nonconstant or long.

Remove the sheet just obtained and apply the same assertion to the remaining finite family. At each removal take its whole contact ball, its private singleton, or the empty set, respectively. This gives Equation (7.5).□

Proof. First, every finite cycle chain in a graph is a finite linear combination of circuit chains. To see this, take a spanning forest of its finite support graph. For each edge outside that forest, subtract its coefficient times the circuit formed with the forest path between its endpoints. This removes all edges outside the forest. The remainder is a cycle supported on a forest and is zero. Loops and pairs of parallel edges are included in this argument.

Let a circuit in \mathcal C start at g and read the word w. The definition of \Gamma means that, in the free group on the formal labels, w is a finite product of conjugates of words read on closed walks of \widetilde X, and their inverses. Follow that product from g in \mathcal C. Each conjugating stem is traversed once in each direction, so its oriented chain cancels. Each relator walk lies in a translated sheet. Free cancellation of adjacent inverse letters also cancels the corresponding oriented edge chains in the formal Cayley graph. Thus the circuit chain is a finite sum of sheet cycle chains. The preceding circuit decomposition proves surjectivity of \Phi_R.

For injectivity, suppose that \sum_{j=1}^n c_{Q_j}=0. Order the finitely many sheets in this relation by Lemma 7.2. An edge of Q_1 with a nonzero coefficient in c_{Q_1} must occur in another nonzero coordinate, since the total coefficient is zero. Both endpoints of that edge are contacts with a later sheet and belong to R_1. Hence c_{Q_1} is supported on \operatorname{Star}_{Q_1}(R_1), a forest by Lemma 7.1, and is zero. Remove this coordinate and repeat. Every coordinate is zero.□

Proof. We first describe the possible limits of allowed removed sets. We then express disjointness and the residual-forest requirement by conditions on finitely many membership coordinates, verify them on finite families by peeling, and take a diagonal limit.

For a fixed sheet Q, let \mathcal K_Q be the collection of all allowed removed sets in the statement, represented by their characteristic functions in \{0,1\}^{V(Q)}. Every fixed vertex p belongs to only finitely many allowed old-centered balls. Indeed, if p\in B_Q(v), then

The last radius is finite, and Q is locally finite. Also p belongs to at most one allowed private singleton. It follows that \mathcal K_Q is closed in the characteristic-function topology. For completeness, if a pointwise limit is nonempty, choose a vertex p in it. All sufficiently late removed sets contain p, so they belong to a fixed finite collection. A subsequence has one constant value, which must equal the pointwise limit. If the limit is empty, it is an allowed value already. This also explains why the empty choice must be retained.

All vertices, sheets, finite circuits, and their incidences are countable. Choose an orientation for each finite circuit C of \mathcal C and let c_C be its integral circuit chain. Using \Phi_{\Z} from Lemma 7.3, fix its finite decomposition

We impose the following countable collection of finite-coordinate conditions on the removed sets:

1. 1.

   For each physical vertex x and distinct sheets Q,R containing it, x is not retained in both D_Q and D_R.

2. 2.

   For each circuit C, there are Q\in\mathcal F_C and e\in E(C)\cap E(Q) such that both endpoints of e lie in D_Q.

Each condition is a finite Boolean condition, hence is preserved by coordinatewise limits. In the second condition both the list of edges and the list of candidate sheets are finite. Replacing that list by all sheets would not give the same closedness argument.

Any finite collection of these conditions can be satisfied. Take the finite union of all sheets appearing in the selected first-type conditions and of all sets \mathcal F_C for the selected second-type conditions. Peel this finite family by Lemma 7.2 and use its allowed removed sets. Assign the empty removed set to every sheet outside the family. The selected first-type conditions hold by Equation (7.5).

Suppose that a selected circuit condition failed. In its decomposition (7.7), take the earliest peeled sheet with a nonzero coordinate, after discarding earlier zero coordinates. An edge of this coordinate whose two endpoints lie outside its removed set cannot occur in any later coordinate: if it did, its endpoints would be contacts in the removed set. If this edge belonged to C, it and the present sheet would witness the selected condition, contrary to its failure. Hence it lies outside C, and its coefficient in c_C, and therefore in the present coordinate, is zero. Thus the coordinate is supported on the incidence neighborhood of its removed set and is a cycle in a forest. It is zero. Induction removes all its coordinates, contradicting c_C\ne0. This proves finite satisfiability.

Here one can finish by an explicit diagonal argument. Enumerate all the conditions, and for each n choose a full assignment satisfying the first n. Enumerate the countably many membership coordinates (Q,v). Pass successively to an infinite subsequence on which the first coordinate is constant, one on which the second is constant, and so on, and take the diagonal subsequence. It converges pointwise. The preceding closedness argument shows that each sheet’s limiting removed set belongs to \mathcal K_Q. Every one of the finitely supported conditions holds in the limit.

The first conditions give disjoint domains. An edge internal to two domains would have an endpoint in their intersection, so its sheet is unique. A circuit in the residual graph would violate its own second-type condition. The residual graph is therefore a forest. The last assertions follow from Lemma 7.1.□

Proof. Normalize the embedded reference sheet as j(v,f)=f t_v, with f\in\HH and v\in V(X). The deck group is identified with its embedded image in \Gamma. A sheet is hQ_0 for a coset h\HH, because its stabilizer is \HH. At a physical vertex u and specified base type v, the equality u=hf t_v forces and is achieved by the unique coset h\HH=u t_v^{-1}\HH. Embeddedness makes its source vertex unique. This proves the bijection. The displayed cardinality follows from the definitions of the two old layer state sets. It counts only old types; private singleton removals add no old incidence or active old coordinate.□

Proof. For an edge chain b in a sheet Q, let c_Q*b denote the same linear combination of its triangles, and let \sigma_Q(t) denote the spoke chain with coefficients given by a vertex chain t in that sheet. Equation (7.10) reads

Any finite 2-chain determines a finite family of sheet edge chains b_Q. If its boundary vanishes, its components on the distinct spoke sets give \partial b_Q=0 for every Q. Its original-edge component is then \sum_Q b_Q=0. Injectivity of \Phi_R in Lemma 7.3 forces every b_Q=0, proving injectivity at C_2.

Let b+\sum_Q\sigma_Q(t_Q) be a finite 1-cycle, with b on \mathcal C. Its coefficient at c_Q says \varepsilon(t_Q)=0. Since Q is connected, t_Q is the boundary of a finite edge chain a_Q: join the finitely many supporting vertices to one chosen vertex by finite paths and use their coefficients. Adding \sum_Q\partial(c_Q*a_Q) removes all spoke terms. The remaining 1-cycle lies on \mathcal C, and surjectivity of \Phi_R writes it as a finite sum of sheet cycles. Each such cycle is the boundary of the identical combination of its cone triangles. Hence the original 1-cycle was a boundary.

Finally \mathcal C is connected and every cone is attached to it. A finite vertex chain of total coefficient zero is the boundary of paths to a fixed vertex. This proves exactness at C_0 and at the augmentation. There are no cells of dimension greater than two. Every argument involved finite-support chains only.□

Proof. Choose the image of each free basis vector of P_0 to lift its prescribed image under augmentation. Having defined the map through degree n-1, the desired image of the boundary of a basis vector of P_n is a cycle in Q_{n-1}. Exactness supplies a preimage in Q_n. Choose it and extend linearly. This constructs the chain map in every degree.

For two lifts, let f be their difference. Its augmentation is zero. Inductively suppose that h_{n-1} has been chosen with the earlier homotopy identities. The map f_n-h_{n-1}\partial_n takes values in the cycles of Q_n; this follows on applying the boundary and using the chain-map identity and the preceding homotopy identity. Choose preimages on the free basis of P_n to obtain h_n. Then f_n=\partial_{n+1}h_n+h_{n-1}\partial_n. For n=0 the same argument uses the zero augmentation. Thus the lifts are homotopic. Lifts of the identity in both directions give the stated chain homotopy equivalence. Tensor and Hom preserve the displayed homotopy equations. No step requires a finite basis.□

Proof. The integral Heisenberg group is torsion-free. For example, in matrix coordinates (\alpha,\beta,\gamma)\in\Z^3, the normal-form element x^n y^l z^c from Section 5 has coordinates (\alpha,\beta,\gamma)=(n,l,c+nl). Its law is

If a positive power is the identity, its first two coordinates force \alpha=\beta=0, and its third then forces \gamma=0.

Suppose P\leq\Gamma has prime order p. The stabilizer of a cone vertex is a conjugate of \HH, so it intersects P trivially. The other cell stabilizers are already trivial by Equation (7.12). Thus P acts freely on the set of cells of every dimension. This is freeness of setwise cell stabilizers, not just absence of pointwise-fixed edges. In particular, no involution can invert a formal positive Cayley edge. Choosing orbit representatives gives free modules over A_p=\F_p[P]. Lemma 7.6, with coefficients \F_p, would give a free resolution of its trivial module of length two.

Let t=u-1 for a generator u of P. Then A_p\cong\F_p[t]/(t^p), and the complex

is exact. In the basis 1,t,\ldots,t^{p-1}, the kernels of multiplication by t and t^{p-1} are respectively (t^{p-1}) and (t). Tensoring this resolution with the trivial right module \F_p makes every differential zero; its homology is \F_p in every nonnegative degree. Tensoring the alleged length-two resolution gives zero homology in every degree above two. This contradicts Lemma 7.7. Hence \Gamma has no subgroup of prime order. A nonidentity element of finite order n would have a power of order p for any prime divisor p of n, which proves the assertion.□

Proof. The binomial expression \binom{\alpha}{2}=\alpha(\alpha-1)/2 is an integer also for negative \alpha. Normalization follows immediately from Equation (7.15). Using Equation (7.13), the identity

gives, for all g,h,k\in\HH,

Explicitly, the only mixed term added to the central coordinate of hk is \alpha'\beta''; multiplying it by \alpha agrees with the mixed term \alpha\alpha'\beta'' in the binomial identity. All other terms on the two sides coincide separately. This proves the cocycle equation in the inhomogeneous bar complex. Equivalently, the multiplication

on \Z\times\HH is associative, with central kernel \Z\times\{1\}, and defines the corresponding integral central extension. For the canonical lifts X=(0,x), Y=(0,y), Z=(0,z) and the positive central kernel generator T=(1,1), one has [X,Y]=Z, [X,Z]=T, and Y,Z,T commute. This is the unipotent shear extension that lengthens the Heisenberg block.

The subgroup generated by z=(0,0,1) and x=(1,0,0) is freely abelian of rank two: z^u x^v=(v,0,u) in these coordinates. For its oriented torus, the bar 2-chain

represents the oriented fundamental class. To check the sign and the normalization directly, subdivide the fundamental square with successive vertices 1,z,zx,x by the diagonal from 1 to zx. The two oriented triangles give (1,z,zx)-(1,x,xz), which are the displayed inhomogeneous bar chains. Sending grid vertices (u,v) to z^u x^v supplies the equivariant map from the triangulated universal cover of the torus into the bar model. Evaluation is therefore

□

Proof. Put A=\Z\Gamma, B=\Z\HH, M=\Z, and C_0=C_0(\mathcal Z;\Z). By Lemma 7.6 and Equation (7.12), the module K=\ker(C_0\to M) has a length-one free resolution

This is the vanishing of \operatorname{Ext}^2_A(K,M) that makes the degree-two restriction map surjective. The proof has two parts: we construct the resulting surjection in Ext using a free resolution, and then identify it with restriction by comparing the induced Heisenberg resolution with the bar resolution of \Gamma.

As a right B-module, A is a direct sum of copies of B, using representatives of the cosets \Gamma/\HH. Therefore

is an exact free A-resolution of A\otimes_B M. Let Q_\bullet resolve C_0=A\oplus A\otimes_B M by taking Q_0=A\oplus I_0, Q_n=I_n for n\geq1, with the extra A summand concentrated in degree zero. Lift K\hookrightarrow C_0 to a chain map f:P_\bullet\to Q_\bullet as follows. Choose f_0 on a free basis to lift that inclusion. The elements f_0d_P(P_1) have zero augmentation in C_0, so exactness of Q_\bullet permits a lift f_1:P_1\to Q_1 with d_Qf_1=f_0d_P. There are no higher P terms.

Define a free augmented complex T_\bullet\to M by

where P_j=0 for j\geq2, with boundary

The augmentation is Q_0\to C_0\to M. The identities defining f give d_T^2=0. This complex is exact. In degree two, a cycle (q,p) has d_Pp=0, hence p=0, and then q bounds in Q_\bullet. The same is immediate in degrees at least three. In degree one, if d_Qq+f_0p=0, augmentation in C_0 implies that p=d_Pp_1 for some p_1. Then q+f_1p_1 is a Q-cycle and equals d_Qq_2; thus (q,p)=d_T(q_2,-p_1). In degree zero, an element mapping to zero in M has augmentation in K; subtract a suitable f_0p and then lift the remainder through d_Q. Surjectivity of the augmentation is immediate. Hence T_\bullet is a free resolution of M.

The inclusion Q_\bullet\to T_\bullet lifts C_0\to M. Any degree-two cocycle \phi:Q_2\to M extends to \psi:T_2=Q_2\oplus P_1\to M by \psi(q,p)=\phi(q). It is still a cocycle because T_3=Q_3 and its boundary has zero P_1 component. Therefore the induced map

is surjective. This is explicitly the degree-two portion of the Ext sequence for 0\to K\to C_0\to M\to0.

There is a cochain isomorphism

Its inverse sends \eta to a\otimes b\mapsto a\eta(b); this is well-defined by B-linearity and is A-linear. The maps commute with the boundaries. The extra free A summand in Q_0 contributes no cohomology in degree two, so Equation (7.18) identifies the target of Equation (7.17) with H^2(B\HH;\Z). This proves the needed instance of Shapiro’s isomorphism at the cochain level.

It remains to identify the map, rather than just its two groups. Let \mathcal B_\bullet(\Gamma) be the preceding bar resolution. Choose a comparison map T_\bullet\to\mathcal B_\bullet(\Gamma) lifting the identity of M. The composite

lifts the augmentation A\otimes_B M\to M. Another such lift is the explicit map

where the tuple on the right is viewed in \Gamma. Lemma 7.7 makes these lifts chain homotopic. After Equation (7.18), precomposition with \theta restricts a \Gamma-cochain to tuples from \HH. As observed in the bar model, this is exactly pullback along B\HH\to B\Gamma. Thus the surjection in Equation (7.17) is the map in Equation (7.16).□

Proof. Lemma 7.8 proves torsion-freeness. The deck embedding of Proposition 6.4 and the coordinate calculation in Lemma 7.9 prove injectivity of i_\Gamma. Choose c_\Gamma mapping to c_{\HH} under the surjection of Lemma 7.10. Its map was identified there with actual restriction in integral cohomology. Equation (7.19) now follows from the explicit evaluation in Lemma 7.9.□

Proof. For a projection P and a finite coordinate set B_0, the ordinary Hilbert–Schmidt norm gives

In a matrix amplification one sums over the fixed matrix basis as well. Every regular deck translate of a given old vertex has the same diagonal trace. If the removed set in Q is an old-centered ball, then Proposition 7.4 bounds its entire cardinality by 2^{10H_i} whenever it meets height i. Equation (8.7) therefore gives the first estimate, with all repeated deck lifts counted separately. An empty removed set costs nothing. A private singleton also costs nothing, since E_{Q,i} has zero private coordinates. This statement is about indexed sheet coordinates even if the same physical vertex is old in another sheet.

Both D and E preserve the height decomposition. The norm of their orthogonal direct sum is the supremum of the block norms, which gives the bound on \delta_\nu. Finally, using U^*U=D, the difference in (8.9) is UD A(D-1)B D U^*. Its norm is at most \|A\|\|B\|\,\|E(1-D)E\|, and \|E(1-D)E\|=\|(1-D)E\|^2.□

Proof. For a physical vertex u and a specified old base vertex v, the only sheet coset that can contain u at base type v is u t_v^{-1}\HH, by (8.1). This also follows from the incidence bijection in Lemma 7.5. Thus at height i each physical vertex receives at most n_i indexed coordinates. Pointwise Cauchy–Schwarz gives the first inequality in (8.10), initially on the stated dense domain. The layer cardinalities give n_i\le C_s2^{H_i}; Lemma 8.1 gives the second inequality. The bounded extensions at individual heights may be summed in operator norm, because H_i=H_0+bi and the square-root estimates are summable. On their common dense domain \mathcal J E=UDE+\mathcal J(1-D)E, proving (8.11) and the claimed extension.□

Proof. For one incidence let a,b be the old normalized constants and let P_A,P_B be the surviving coordinate projections. Their normalized restrictions a',b' differ from a,b, after extension by zero, by at most C\sqrt{\varepsilon_{\mathrm{del}}}. Put Q_A=1-|a'\rangle\langle a'| and Q_B=1-|b'\rangle\langle b'| on the surviving spaces and define

Write U^0=|a\rangle\langle b|+U_0. Then Q_AP_Aa=0 and Q_BP_Bb=0, so \|Q_AUQ_B\|\le\|U_0\|\le\rho. The two displayed summands of \widehat U have orthogonal source and range spaces, proving its asserted gap and norm. The estimates \|Ub'-a'\|+\|U^*a'-b'\|\le C\sqrt{\varepsilon_{\mathrm{del}}} bound the remaining three blocks in U-\widehat U.□

Proof. First choose k so that the gap term in (8.13) is sufficiently small. Include the finitely many cutoff and symbol blocks V and W(f) as designated incidence types in Proposition 5.2. That Proposition supplies a finite weight net. We first compare each sampled factor with its compressed sheet mean, and then control the products in (8.12).

For a type \tau define the actual finite polynomial

where the positive augmented label \sigma determines its matrix weight W_\sigma. These weights already incorporate the bounded factor q/q_{\tau,ik} in Proposition 5.2, before finite-net rounding; the polynomial uses the single common normalizer q^{-1}. The set \mathcal S_\tau is finite. On a sheet Q, sum the polynomial’s contributions over the prescribed source edges of type \tau in Q. Equation (8.3) identifies this source-edge sum, on each basis vector, with the sampled \mathscr B-to-\mathscr A convolution for that type, with forward voltage equal to the sampled group element. The sampled operator differs in full regular norm from its prescribed mean by an arbitrarily small uniform amount.

Compressing to the disjoint domains and taking the orthogonal sum does not increase the sampling error. It remains to estimate the edges outside that sum. The physical Cayley polynomial and this domain sum differ precisely on the remaining formal edges, with their specified weights. These include nonsource Cayley chords and unused weight-tag edges, even when both endpoints lie in one retained domain. An edge is removed from the remainder only when it is an edge of its retained source sheet. Disjointness of the domains prevents any ambiguity. The remaining edges form a forest by Proposition 7.4. A matrix adjacency on a forest of degree at most L and edge weight norm at most w has norm at most 2w\sqrt L: root each component, take the operator sending parent coordinates to child coordinates, use that each child has a unique parent, and add its adjoint. For a rectangular or directed block, self-adjoint doubling has support in the bipartite double cover of this forest, which is again a forest. Consequently the residual error here is at most

where \mathcal S_{\mathrm{wt}} is the finite weighted old alphabet. All private letters have coefficient zero. Each normalized nonzero weight has norm at most C/q, with C fixed before s. The sharp alphabet bound is

which tends to zero after the finite type and net choices. Thus every factor of (8.12) has a finite Cayley formula close to its \Psi_\nu image in full norm.

We now use those finite formulas in the same finite word (8.12). Equation (8.9) controls replacing the product of compressed means by the compression of their product. Ordinary telescoping controls perturbing the factors: for a product of r factors of norm at most L, per-factor error \theta costs at most r\theta(L+\theta)^{r-1}. The numbers r,L are fixed before the degree choice, so the sampling and forest errors can be chosen small enough. Choose s and then H_0 also to make the deletion term and the trimming term as small as required. Equation (8.13) now proves (8.14).

There is no physical projection selecting a height in this argument. Instead, each designated symbol has its prescribed value, including zero, at each block. On a zero block Proposition 5.2 uses zero scalar weights with diffuse voltages. The finite type and weight tags record everything needed for the finite Cayley polynomial. A parity pattern or an initial-height exception changes this finite prescription but introduces no family of new labels indexed by height.□

Proof. Take any sequence \eta_\nu\downarrow0. At stage \nu, first choose the extraction length from the fixed gap. Next choose finite meshes of the two uniform model paths, with mesh error at most \eta_\nu. Collect their finitely many scalar matrix profiles, the four endpoint profiles, and the larger cutoffs. Apply Lemma 8.4 to that finite family, with errors sufficiently small compared with \eta_\nu. Its finite type and weight choices precede s_\nu; choose s_\nu and then H_{0,\nu} to make all errors tend to zero. In particular require both \delta_\nu and the right-hand side of (8.11) to be at most \eta_\nu. The value of b remains fixed.

Each endpoint family now has an actual bounded sequence in a fixed matrix algebra over A_\nu with the same \mathcal Q_{\mathrm{op}} class as its \Theta image. For an arbitrary path parameter, linearly interpolate the physical mesh values. The interpolated sequence differs from the \Theta image of the model path by a norm-null sequence, uniformly in the parameter. The uniform model modulus proves that this is a continuous path in the quotient. Thus every generating path value has image in \mathcal Q_r. This latter algebra is a closed C^*-subalgebra of \mathcal Q_{\mathrm{op}}, so it contains \Theta(\mathcal B).

Only a finite profile family was sampled at each stage. Uniform approximation of the specified compact parameter spaces supplies every path value in the quotient. No claim about arbitrary coordinatewise continuous paths is used.□

Proof. The adjoint relation is av^*v=a. Put c=\cos t, s=\sin t and

It is self-adjoint. With W_t=(cI_m,sv)^T and E_{\mathrm{sc}}=\operatorname{diag}(e,e), this is E_{\mathrm{sc}}+W_taW_t^*. The scalar-factor assumption gives E_{\mathrm{sc}}W_t=W_te. Moreover aW_t^*W_ta=a^2 and ea+ae+a^2=a. Multiplication now gives P_t^2=P_t. The endpoints are \operatorname{diag}(e+a,e) and \operatorname{diag}(e,e+vav^*), proving the equality after one fixed stabilization. All nonscalar entries belong to M_{2m}(B) and the path is norm continuous. Its modulus depends only on \|a\| and \|v\|. In particular it is uniform for bounded coordinate sequences.□

Proof. The definition of K_0 supplies a projection R in a finite matrix algebra over the quotient and, after finite zero padding, a partial isometry w with

Lift R to a bounded self-adjoint sequence r_\nu. Its idempotence defect tends to zero. On a tail its spectrum avoids 1/2, and spectral cutting gives projections R_\nu with \|R_\nu-r_\nu\|\to0. Form the corresponding E_\nu,F_\nu and lift w to a bounded sequence b_\nu. Set z_\nu=F_\nu b_\nu E_\nu. Then

For large \nu these products are invertible in the respective corners. The element z_\nu(z_\nu^*z_\nu)^{-1/2} is a partial isometry with initial projection E_\nu and final projection F_\nu; the latter assertion uses invertibility in the F_\nu corner. Thus [P_\nu]+[R_\nu]=[Q_\nu]+[R_\nu], and cancellation in K_0(B_\nu) proves the result. Every stabilization used is finite and fixed after the quotient witness has been chosen.□

Proof. Apply Lemma 8.7 to (8.25). At every coordinate (8.4) identifies the difference with the physical torus Bott generator. Equation (8.2) transfers the equality to the usual left reduced algebra without changing the subgroup elements or their functional-calculus projections.□

Proof. Choose one finite stabilization projection and a partial isometry witnessing the equality over \Cr(\Gamma). Approximate their finitely many entries by group polynomials, using self-adjoint approximations for the projection. Adjoin all group elements in these finite supports to generators of H; they generate a finitely generated subgroup \Ginj. The reduced inclusion \Cr(\Ginj)\to\Cr(\Gamma) is isometric, since restriction of the regular representation of \Gamma to \Ginj is a direct sum of regular representations of \Ginj. Thus the required small projection and partial-isometry defects have the same bounds in \Cr(\Ginj). Spectral cutting gives the stabilizing projection in \Cr(\Ginj); compressing the approximate partial isometry to the two corners and polar-normalizing it gives an exact witness there, just as in Lemma 8.7. Functional calculus stays inside \Cr(\Ginj). The original projections P,Q were retained exactly throughout.□

Proof of Theorem 1.1. Make the sequence of compatible constructions just described. Proposition 6.4 embeds \HH in every \Gamma_\nu. Proposition 7.11 proves torsion-freeness and provides an integral class whose evaluation on the ordered torus (z,x) is -1. Proposition 8.8 proves that, for all sufficiently large \nu, the rank-zero Bott class of this torus is zero in K_0(\Cr(\Gamma_\nu)).

Fix one such \nu. Apply Lemma 8.9 with H=\HH and the physical Bott projections to obtain a finitely generated subgroup \HH\leq \Ginj\leq\Gamma_\nu in which that equality already holds. The subgroup \Ginj is torsion-free. Restrict the negative of the detector class to B\Ginj, and denote it by c. Let i:\Z^2\hookrightarrow \Ginj be the inclusion with ordered generators (z,x). The detector evaluation is unchanged by restriction, so

It remains to detect the corresponding geometric class and identify its assembly image. We use compactly supported K-homology throughout. On finite CW complexes, the natural comparison of geometric cycles with analytic K-homology sends (M,E,f) to f_*[D_E] [7, Theorems 6.1–6.2]; passage to finite-subcomplex colimits gives the convention used for B\Ginj. In particular the compact torus cycle used here defines a class without any compactness assumption on B\Ginj.

Let L be the line bundle on B\Ginj classified by c. The naturality of the index pairing follows at a finite stage from pullback of vector bundles and associativity of the Kasparov product:

The same identity holds for the virtual bundle [L]-[1]. Applying the twisted spin Dirac index theorem [3], in the dimension-two form of [8, Theorem 1], gives

The torus tangent bundle is trivial, its \widehat A class is one, and its untwisted complex Dirac index is zero. Thus the pushed-forward K-homology class has infinite order and is nonzero after tensoring with \Q.

Under Fourier transform \Cr(\Z^2)\cong C(\mathbb T^2), the assembly image of the torus spin Dirac class is the rank-zero Bott generator, up to orientation. This is the full-torus case of Emerson–Hudson’s assembly formula for a K-oriented subtorus [12, Theorem 1.3]: the dual subtorus is a point, whose shriek class is the compact Bott class.

For completeness we verify the required subgroup naturality at this particular cycle. Put A=\Cr(\Z^2), B=\Cr(\Ginj), and write j:A\to B for the isometric inclusion. If P\to T^2 is the principal \Z^2-bundle classified by the torus map, its Miščenko bundle is E_A=P\times_{\Z^2}A. The bundle obtained after the subgroup map is canonically

The last identification sends ([p,a],b) to [p,j(a)b]; the balancing relations and fiber inner products agree because j sends each subgroup unitary to the same group unitary. Extension of coefficients therefore commutes with the index Kasparov product. In explicit notation,

These indices are the assembly classes by the Miščenko comparison [26, Corollary 1.4 and diagram (1.6)]; see also [28, Proposition 3.7 and Theorem 4.1]. The comparison applies to the compact torus cycle, and its finite-subcomplex colimit handles the target B\Ginj. Consequently

by Proposition 8.8 and Lemma 8.9. This is integral vanishing, whereas the index pairing above proves infinite order of the source class.□