Label transport and topology [frap-0014]
✍️source

    A sequence of puzzle moves may return pieces to their positions while changing their orientations. The first question is what local return map records that change; the cycle criterion computes it. A related question asks how labels move between neighboring simplices. The projectivity construction of Izmestiev and Joswig makes that question precise, and the complete unfolding records the topology of one such transport system [izmestiev2002branched, Sections 2.1--2.3.3].

    1. Perspectivities along facet paths

      Let \(K\) be a pure \(d\)-dimensional simplicial complex. If neighboring facets \(\sigma \) and \(\tau \) share a ridge, \(\sigma \) has one vertex \(v(\sigma ,\tau )\) not in \(\tau \). The perspectivity

      \[h_{\sigma ,\tau }:V(\sigma )\longrightarrow V(\tau )\]

      swaps that exceptional vertex with \(v(\tau ,\sigma )\) and fixes the common ridge vertices. For a facet path \(\gamma =(\sigma _0,\ldots ,\sigma _n)\), its projectivity is the composite

      \[\langle \gamma \rangle =h_{\sigma _0,\sigma _1}\cdots h_{\sigma _{n-1},\sigma _n}.\]

      A closed facet path based at \(\sigma _0\) gives a permutation of \(V(\sigma _0)\). The set of these permutations is the projectivity group \(\Pi (K,\sigma _0)\leq \operatorname {Sym}(V(\sigma _0))\); changing the base facet conjugates the group when the dual graph is connected. The inverse path supplies the inverse permutation, so this is transport with a genuine composition law [izmestiev2002branched, Section 2.1].

      2. Coloring as a trivial-transport test

        A proper coloring of the vertices of \(K\) with \(d+1\) colors gives every facet a consistently labeled vertex set. Each perspectivity preserves those labels, so every projectivity is the identity. Thus proper coloring is a sufficient condition for \(\Pi (K)=1\); a nontrivial projectivity group is an obstruction to such a global labeling [izmestiev2002branched, Section 2.1].

        This is a useful comparison with a framed puzzle. A local frame assigns labels in one fiber, and a move transports them to another fiber. The framed transport and its total-orientation character measure the return action in the puzzle model. Projectivities do the analogous bookkeeping for simplex vertices, but the two constructions have different state spaces and move laws. No identification between their groups is being assumed.

        3. Complete unfolding as a transport cover

          Fix a base facet and a labeling of its vertices by \(0,\ldots ,d\). A facet reached by a path inherits a labeling by transporting along that path. Different paths can give different admissible labelings, precisely when their projectivity differs. The complete unfolding keeps these admissibly labeled facets as separate sheets and glues neighboring sheets only when their labels agree on the common ridge. It carries a canonical map back to \(K\); the base projectivity group acts by permuting these sheets. The complete unfolding itself has trivial projectivity group [izmestiev2002branched, Section 2.2, Proposition 2.2.2].

          The construction is therefore a topological record of transport failure: a loop in the base may close geometrically while changing the sheet label. The complete unfolding removes that ambiguity by remembering the transported labeling. This is analogous to a covering-space monodromy calculation; the branch locus is additional local-combinatorial data of the unfolding and is not inferred here from the puzzle model.

          4. The tetrahedral boundary example

            For the boundary complex \(\partial \Delta ^3\), the projectivity group is \(S_3\). Its complete unfolding is glued from 24 triangles. Identifying opposite sides of the triangulated hexagon gives a torus \(T^2\); the canonical map to the tetrahedral boundary, topologically \(S^2\), is a six-fold branched covering. There are four branch points on \(S^2\); each has three preimages, each of branching index two [izmestiev2002branched, Section 2.3.3, Figure 3].

            The example separates three pieces of data that are easy to conflate: the base triangulation, the permutation group generated by facet-path transport, and the topology of the unfolded surface. The torus is not a claim about a cube solver. It is a concrete model showing how local label transport can assemble a global branched cover.

            A comparison between these constructions must name the fiber, the local transport map, and the allowed paths before relating topology to puzzle orientation. The projectivity example supplies one transport vocabulary; the framed cube account supplies a different finite-state calculation. A bridge between them would require an explicit map between those data and a proof that it preserves composition, return maps, and the claimed topology.

            References

              Reference. Branched Coverings, Triangulations, and 3-Manifolds [izmestiev2002branched]

              @misc{izmestiev2002branched,
 title = {Branched Coverings, Triangulations, and 3-Manifolds},
 author = {Izmestiev, Ivan and Joswig, Michael},
 year = {2002},
 archiveprefix = {arXiv},
 eprint = {math/0108202},
 url = {https://arxiv.org/abs/math/0108202v2}
}

              Context

                Further connections [frap-0007]

                  The total-orientation character can expose a constraint, but what survives when we change the local frame or the dimension? Framed orientations and transfer and dimension develop the first connections: wreath product records positions and local transport, while abelianized characters detect constraints when the allowed generators preserve them [marchetti2025rubik, Section 3.1]. Finite-index transfer organizes related calculations through coset representatives [mayer2015artin, Sections 3.1--3.3]. The dimension calculation explains how a local abelianization can lose this detecting power without deciding reachability.

                  The following outlooks ask different questions about transport and functions on a move group. Each needs its own construction; neither is a prerequisite for continuing to the layered cubes.

                  Label transport and topology [frap-0014]
                  ✍️source

                    A sequence of puzzle moves may return pieces to their positions while changing their orientations. The first question is what local return map records that change; the cycle criterion computes it. A related question asks how labels move between neighboring simplices. The projectivity construction of Izmestiev and Joswig makes that question precise, and the complete unfolding records the topology of one such transport system [izmestiev2002branched, Sections 2.1--2.3.3].

                    Perspectivities along facet paths

                      Let \(K\) be a pure \(d\)-dimensional simplicial complex. If neighboring facets \(\sigma \) and \(\tau \) share a ridge, \(\sigma \) has one vertex \(v(\sigma ,\tau )\) not in \(\tau \). The perspectivity

                      \[h_{\sigma ,\tau }:V(\sigma )\longrightarrow V(\tau )\]

                      swaps that exceptional vertex with \(v(\tau ,\sigma )\) and fixes the common ridge vertices. For a facet path \(\gamma =(\sigma _0,\ldots ,\sigma _n)\), its projectivity is the composite

                      \[\langle \gamma \rangle =h_{\sigma _0,\sigma _1}\cdots h_{\sigma _{n-1},\sigma _n}.\]

                      A closed facet path based at \(\sigma _0\) gives a permutation of \(V(\sigma _0)\). The set of these permutations is the projectivity group \(\Pi (K,\sigma _0)\leq \operatorname {Sym}(V(\sigma _0))\); changing the base facet conjugates the group when the dual graph is connected. The inverse path supplies the inverse permutation, so this is transport with a genuine composition law [izmestiev2002branched, Section 2.1].

                      Coloring as a trivial-transport test

                        A proper coloring of the vertices of \(K\) with \(d+1\) colors gives every facet a consistently labeled vertex set. Each perspectivity preserves those labels, so every projectivity is the identity. Thus proper coloring is a sufficient condition for \(\Pi (K)=1\); a nontrivial projectivity group is an obstruction to such a global labeling [izmestiev2002branched, Section 2.1].

                        This is a useful comparison with a framed puzzle. A local frame assigns labels in one fiber, and a move transports them to another fiber. The framed transport and its total-orientation character measure the return action in the puzzle model. Projectivities do the analogous bookkeeping for simplex vertices, but the two constructions have different state spaces and move laws. No identification between their groups is being assumed.

                        Complete unfolding as a transport cover

                          Fix a base facet and a labeling of its vertices by \(0,\ldots ,d\). A facet reached by a path inherits a labeling by transporting along that path. Different paths can give different admissible labelings, precisely when their projectivity differs. The complete unfolding keeps these admissibly labeled facets as separate sheets and glues neighboring sheets only when their labels agree on the common ridge. It carries a canonical map back to \(K\); the base projectivity group acts by permuting these sheets. The complete unfolding itself has trivial projectivity group [izmestiev2002branched, Section 2.2, Proposition 2.2.2].

                          The construction is therefore a topological record of transport failure: a loop in the base may close geometrically while changing the sheet label. The complete unfolding removes that ambiguity by remembering the transported labeling. This is analogous to a covering-space monodromy calculation; the branch locus is additional local-combinatorial data of the unfolding and is not inferred here from the puzzle model.

                          The tetrahedral boundary example

                            For the boundary complex \(\partial \Delta ^3\), the projectivity group is \(S_3\). Its complete unfolding is glued from 24 triangles. Identifying opposite sides of the triangulated hexagon gives a torus \(T^2\); the canonical map to the tetrahedral boundary, topologically \(S^2\), is a six-fold branched covering. There are four branch points on \(S^2\); each has three preimages, each of branching index two [izmestiev2002branched, Section 2.3.3, Figure 3].

                            The example separates three pieces of data that are easy to conflate: the base triangulation, the permutation group generated by facet-path transport, and the topology of the unfolded surface. The torus is not a claim about a cube solver. It is a concrete model showing how local label transport can assemble a global branched cover.

                            A comparison between these constructions must name the fiber, the local transport map, and the allowed paths before relating topology to puzzle orientation. The projectivity example supplies one transport vocabulary; the framed cube account supplies a different finite-state calculation. A bridge between them would require an explicit map between those data and a proof that it preserves composition, return maps, and the claimed topology.

                            Scrambling, spectra and search coordinates [frap-0015]

                              Random moves raise a different question from solving: when, and how quickly, does a scramble lose information about its starting state? A random walk depends on its allowed moves and their probabilities. We use the fixed-frame three-layer move group from the complete cube account, and write \(G\) for its reachable state group. A walk with step law \(\mu \) has distribution

                              \[\Pr [X_k=g]=\mu ^{*k}(g)\]

                              from the solved state, where convolution records the successive moves.

                              A parity calculation before mixing claims

                                Let \(\varepsilon :G\to \{+1,-1\}\) be the sign of the corner permutation. Each face quarter-turn or inverse is a four-cycle on corners, so \(\varepsilon =-1\); each face half-turn is a product of two transpositions, so \(\varepsilon =+1\). If the walk chooses uniformly from the twelve quarter-turns and inverse quarter-turns, then

                                \[\varepsilon (X_k)=(-1)^k\]

                                with probability one. Its support is therefore contained in one parity coset at every fixed time, while the uniform distribution on \(G\) gives equal mass to the two cosets. The total-variation distance from uniform is at least \(1/2\) for every \(k\). This walk cannot converge to uniform: the obstruction is periodicity, not a slow estimate of a mixing time.

                                The usual eighteen-move HTM law gives a different exact observable. Six half-turns have sign \(+1\) and twelve quarter-turns or inverses have sign \(-1\), hence one step has

                                \[\mathbb E[\varepsilon (X_1)]=\frac 6{18}-\frac {12}{18}=-\frac 13.\]

                                Independence gives \(\mathbb E[\varepsilon (X_k)]=(-1/3)^k\). Since uniform parity has expectation zero, the parity marginal alone gives the lower bound

                                \[\lVert \mathcal L(X_k)-\mathcal U_G\rVert _{\mathrm {TV}}\geq \frac 12\left (\frac 13\right )^k.\]

                                This bound decays and does not prove convergence by itself. A lazy walk, which includes the identity with positive probability, removes this specific periodicity; a convergence statement still needs a generated reachable component and an aperiodicity argument.

                                Averaging and finite-group spectra

                                  For a function \(f:G\to \mathbb C\), the one-step averaging operator is

                                  \[(T_\mu f)(g)=\sum _{s\in G}\mu (s)f(gs).\]

                                  Constants are fixed by \(T_\mu \). For a finite-dimensional unitary representation \(\rho \), its matrix coefficients

                                  \[f_{u,v}(g)=\langle u,\rho (g)v\rangle \]

                                  form invariant blocks for this operator: averaging a coefficient replaces \(\rho (g)\) by multiplication by the Fourier block

                                  \[\widehat \mu (\rho )=\sum _{s\in G}\mu (s)\rho (s).\]

                                  Thus the nonconstant spectrum describes which observables retain memory. If the step support generates the reachable group and every nontrivial block contracts on the relevant quotient, then these blocks decay and the walk converges to uniform on that component. The hypotheses matter: the quarter-turn walk above fails the aperiodicity test, so no spectral slogan can erase its parity obstruction.

                                  The same calculation explains why a prescribed scramble length is not a universal guarantee of randomness. A distance bound answers how far a state can be from solved; a distributional claim asks how the whole law projects onto every persistent observable. These are different certificates.

                                  Fourier coordinates for learned search

                                    Pan and Kondor use irreducible-representation matrix coefficients as a Fourier basis for learned value functions on permutation puzzles [pan2021fourier, Sections 4--4.1]. Their experiments compare learned scores that choose moves with conventional heuristic and value-based search; the paper explicitly discusses the distinction from an admissible lower bound [pan2021fourier, Section 6]. A learned score can rank promising moves while overestimating or underestimating the remaining distance. It is therefore evidence about search guidance, not a shortest-path certificate.

                                    For the cube, a safe lower bound can still be obtained from the exact projected distances in a pattern database as in computer solving and efficiency. A Fourier score may be combined with that bound as a tie-breaker or move-ordering device, but replacing the bound with the score changes the guarantee. The representation coordinates explain how a search function is organized; they do not, without an error theorem, make it admissible.

                                    The displayed parity bounds and finite-operator framework are the results established here. They do not claim a full-cube mixing time, a universal scramble length, or a reproduced machine-learning benchmark. Those would require a declared move law, reachable component, statistical or spectral analysis, and independently checked computational evidence.

                                    Which moves survive a quotient? [frap-0019]

                                      A quotient of observations is useful only when the moves still have a well-defined action on the observations. Let \(X\) be a declared sticker-state set, let \(G\) act on \(X\), and let the surjection \(q:X\twoheadrightarrow Y\) identify the observations we want to retain (equivalently, take \(Y=q(X)\)). The move \(g\in G\) preserves the fibers of \(q\) when

                                      \[q(x)=q(x')\Longrightarrow q(gx)=q(gx')\qquad (x,x'\in X).\]

                                      Exactly in this case there is a unique induced map \(\bar g:Y\to Y\) with

                                      \[q\circ g=\bar g\circ q.\]

                                      If \(g\) is a bijection and \(X\) is finite, then \(\bar g\) is surjective and therefore bijective. In an infinite setting, require the inverse move to preserve fibers as well. Thus, for the finite cube action below, the fiber-preserving bijections form a subgroup; a quotient does not automatically inherit the whole puzzle action.

                                      The antipodal sticker quotient

                                        For the ordinary cube, use the surface-cell and frame convention in how the cube models connect and the complete three-layer cube. Represent a sticker by a pair \((p,n)\), where \(p\in \{-1,0,1\}^3\) satisfies \(\lVert p\rVert _\infty =1\) and \(n=\operatorname {sgn}(p_i)e_i\) for each nonzero coordinate \(p_i\). Thus corners, edges and face centers contribute \(8\cdot 3\), \(12\cdot 2\) and \(6\cdot 1\) pairs, respectively, for 54 stickers. The antipodal identification is the involution

                                        \[a(p,n)=(-p,-n),\qquad q(p,n)=\{(p,n),a(p,n)\}.\]

                                        Its sticker fibres pair opposite stickers. This is a quotient of the labelled sticker action; it is not the same as forgetting equal colors or quotienting by whole-cube rotations.

                                        Why one face turn fails

                                          The positive-\(z\) face turn uses

                                          \[u(x,y,z)=(y,-x,z)\]

                                          on every sticker whose cubie has \(z=1\), rotating both its position and its normal, and fixes the other cubies. Take a top sticker and its antipode, for example

                                          \[x=((1,1,1),e_z),\qquad a(x)=((-1,-1,-1),-e_z).\]

                                          The turn sends \(x\) to the sticker at \(((1,-1,1),e_z)\), while it fixes \(a(x)\). These outputs are not an antipodal pair, so \(q(ux)\ne q(u\,a(x))\). The single \(U\) move therefore does not descend to the antipodal quotient.

                                          A synchronized opposite-layer move

                                            The clockwise turn of the negative-\(z\) face, viewed from that face, is the inverse global rotation on \(z=-1\). Write it as \(d\); its inverse \(d^{-1}\) uses the same global rotation as \(u\). The composite \(u d^{-1}\) rotates both opposite layers by the same linear map and fixes the middle cubies. It commutes with the antipodal involution:

                                            \[a\,(u d^{-1})=(u d^{-1})\,a.\]

                                            Consequently it preserves every antipodal fiber and induces a bijection on \(Y\). A direct check of all 54 stickers gives the same result: \(U\) fails on the displayed pair, while the synchronized \(U D'\) move maps every antipodal pair to an antipodal pair. The surviving action is therefore an action of the fiber-preserving subgroup, not of the six ordinary face turns individually.

                                            What the cover does not provide

                                              A covering or quotient of the underlying shape supplies a map of shape data. It does not by itself supply a map between sticker-state spaces, a chosen fiber relation, or an equivariance proof for each legal move. Those data are separate: the explicit cube action records which stickers a face turn moves, while the quotient records which states are identified. A move descends only after the displayed fiber test has been checked.

                                              This is the precise limit of the comparison with the complete unfolding in label transport and topology. Projectivity and branched unfolding describe transport of simplex labels; the antipodal example describes descent of a particular sticker action. They pose analogous questions about information lost by identification, but neither construction is being identified with the other. A general lifting or descent classification is outside this note.

                                              The practical rule is short: state the state space, the quotient map and the move action before saying that a move survives identification. The antipodal calculation supplies both a negative and a positive finite example, while preserving the distinction between a shape cover and a compatible puzzle action.

                                              Backlinks

                                                Five encounters with a cube [frap-001C]

                                                  Put a solved three-layer cube on a table and turn its right face once. The move is familiar, but it raises concrete puzzles: which labelled pieces moved, why can a visually plausible arrangement be impossible, and how can another person undo the turn without watching the first hand? Begin with these experiences, then introduce only the distinctions needed to explain them.

                                                  One turn is larger than the visible face

                                                    The shaded strip is only the visible part of the layer: the same stickers cross the front, right, back, top, and bottom faces; viewed from above, the top-face part is the corresponding rightmost strip. A whole-cube rotation would change the viewing frame, whereas this face move keeps the six centres fixed and changes one layer. The layered-cube family makes this distinction for every \(N\); the complete three-layer cube spells it out for the six outer generators.

                                                    A sticker needs a home and a frame

                                                      A player sees colours; a proof must also track each sticker's cubie and transported local frame. How the cube models connect follows one sticker through incidence position, cubie fibre, and fixed-frame coordinate. A colour quotient forgets labels, while a larger cube adds positions, so the state space changes with the question.

                                                      A plausible state can still be impossible

                                                        Take a solved cube and twist one corner in place. Every sticker still has a cubie-sized home, and the three colours at that corner look locally consistent. But the corner orientations now have sum \(1\) modulo \(3\), whereas every face turn preserves sum \(0\). No sequence of legal turns can produce this state.

                                                        The complete three-layer cube derives the matching edge and permutation restrictions from the six generators. The calculation is useful at the table: it tells a solver when a scrambled-looking object was assembled illegally, and it tells a proof which state space is worth searching.

                                                        A human algorithm has to preserve these legal states while it borrows pieces from completed areas.

                                                        A formula often disturbs pieces that were already placed. The disturbance is acceptable only when the setup is undone and the completed pieces return. A working slot, a trigger, and an inverse setup make that promise visible. The complete layer-first method repeats this pattern through the cross, the first layer, the middle layer, and the last layer. Its completeness depends on the preservation and termination checks, not on the formula's familiarity.

                                                        Undoing a scramble checks the notation

                                                          The worked replay and projected lower bound fixes the viewing convention, records the ten tokens, and applies their inverse. Sticker replay returns to solved. A corner-only projected search finds no solution of depth at most two, so the evidence is the bounded statement

                                                          \[3\le d_{\rm HTM}\le 10.\]

                                                          The inverse word is an upper-bound witness; the projected search is a lower-bound witness. The computer account explains why these checks do not prove that ten is shortest. A state, a word, and a replay log are enough to test the notation before asking for a general solver.

                                                          Changing one ingredient changes the problem

                                                            Once the move, its frame, its local repair, and its replay are explicit, three variations separate the effects of enlarging the cube, changing the shape, and forgetting labels.

                                                            The layered-cube family develops the first variation; the incidence and facet construction in incidence, symmetry and moves develops the second; and which moves survive a quotient? gives the descent condition for the third. Framed orientations, label transport and topology, and scrambling, spectra and search then study the extra information each variation introduces or forgets. Their hypotheses stay attached to the concrete move that prompted the question.

                                                            A face turn can therefore be drawn, named, replayed, and checked before it is generalized. The resulting state maps and invariants support later formal certification or measured comparisons, while the examples here make clear which claim each later programme would have to establish.

                                                            How the notes are organized [frap-000C]

                                                              The developed mathematics follows one practical question: what data must a move carry so that we can later compose moves and test reachability? Incidence, symmetry and moves answers it in order. First, polytopes, flags and facet symmetry describe the shape and show their role in triangle and tetrahedron examples. Next, facet moves turn a local facet rotation into an action on incidence positions, with tetrahedron and cube calculations that make the construction concrete. Only then do positions, orientations and global constraints separate a piece's destination from transport in its local frame, while transfer and the effect of dimension connect that transport with a group homomorphism and compute the relevant local abelianizations.

                                                              The cube examples above connect the general mathematics with solving and play. The later accounts use those concrete states and moves as their starting point:

                                                              The transition between the abstract facet action and the fixed-frame cube coordinates is recorded in how the cube models connect. Read it before the all-\(N\) law if the distinction between incidence positions, cubie fibres and observations is new.

                                                              1. General Rubik puzzles: the incidence actions, framed orientations and transfer developed below.
                                                              2. The layered-cube family: piece types, layer turns and conventions as \(N\) varies.
                                                              3. The layered-cube law: all-\(N\) state coordinates and a constructive characterization of reachability.
                                                              4. The complete three-layer cube: explicit outer generators, impossible states, the standard count and a worked cleanup.
                                                              5. Human methods and computer search: recognizable cases, algebraic effects, usability, move costs and optimality. The complete layer-first method supplies one full human construction, and computer solving and efficiency supplies exact small search, admissible bounds and subgroup reduction.
                                                              6. From move sequences to play: notation, scrambling, recognition, replay and complete solves, including the worked replay and projected lower bound.

                                                              Further connections relate the developed orientation and transfer calculations to label transport and topology and scrambling and search functions. The quotient-compatible move account gives the criterion for an identification of observations to inherit a puzzle action. Together these results compare transport, randomization and information loss around the same concrete move.

                                                              Five encounters with a cube [frap-001C]

                                                                Put a solved three-layer cube on a table and turn its right face once. The move is familiar, but it raises concrete puzzles: which labelled pieces moved, why can a visually plausible arrangement be impossible, and how can another person undo the turn without watching the first hand? Begin with these experiences, then introduce only the distinctions needed to explain them.

                                                                One turn is larger than the visible face

                                                                  The shaded strip is only the visible part of the layer: the same stickers cross the front, right, back, top, and bottom faces; viewed from above, the top-face part is the corresponding rightmost strip. A whole-cube rotation would change the viewing frame, whereas this face move keeps the six centres fixed and changes one layer. The layered-cube family makes this distinction for every \(N\); the complete three-layer cube spells it out for the six outer generators.

                                                                  A sticker needs a home and a frame

                                                                    A player sees colours; a proof must also track each sticker's cubie and transported local frame. How the cube models connect follows one sticker through incidence position, cubie fibre, and fixed-frame coordinate. A colour quotient forgets labels, while a larger cube adds positions, so the state space changes with the question.

                                                                    A plausible state can still be impossible

                                                                      Take a solved cube and twist one corner in place. Every sticker still has a cubie-sized home, and the three colours at that corner look locally consistent. But the corner orientations now have sum \(1\) modulo \(3\), whereas every face turn preserves sum \(0\). No sequence of legal turns can produce this state.

                                                                      The complete three-layer cube derives the matching edge and permutation restrictions from the six generators. The calculation is useful at the table: it tells a solver when a scrambled-looking object was assembled illegally, and it tells a proof which state space is worth searching.

                                                                      A human algorithm has to preserve these legal states while it borrows pieces from completed areas.

                                                                      A formula often disturbs pieces that were already placed. The disturbance is acceptable only when the setup is undone and the completed pieces return. A working slot, a trigger, and an inverse setup make that promise visible. The complete layer-first method repeats this pattern through the cross, the first layer, the middle layer, and the last layer. Its completeness depends on the preservation and termination checks, not on the formula's familiarity.

                                                                      Undoing a scramble checks the notation

                                                                        The worked replay and projected lower bound fixes the viewing convention, records the ten tokens, and applies their inverse. Sticker replay returns to solved. A corner-only projected search finds no solution of depth at most two, so the evidence is the bounded statement

                                                                        \[3\le d_{\rm HTM}\le 10.\]

                                                                        The inverse word is an upper-bound witness; the projected search is a lower-bound witness. The computer account explains why these checks do not prove that ten is shortest. A state, a word, and a replay log are enough to test the notation before asking for a general solver.

                                                                        Changing one ingredient changes the problem

                                                                          Once the move, its frame, its local repair, and its replay are explicit, three variations separate the effects of enlarging the cube, changing the shape, and forgetting labels.

                                                                          The layered-cube family develops the first variation; the incidence and facet construction in incidence, symmetry and moves develops the second; and which moves survive a quotient? gives the descent condition for the third. Framed orientations, label transport and topology, and scrambling, spectra and search then study the extra information each variation introduces or forgets. Their hypotheses stay attached to the concrete move that prompted the question.

                                                                          A face turn can therefore be drawn, named, replayed, and checked before it is generalized. The resulting state maps and invariants support later formal certification or measured comparisons, while the examples here make clear which claim each later programme would have to establish.

                                                                          Which moves survive a quotient? [frap-0019]

                                                                            A quotient of observations is useful only when the moves still have a well-defined action on the observations. Let \(X\) be a declared sticker-state set, let \(G\) act on \(X\), and let the surjection \(q:X\twoheadrightarrow Y\) identify the observations we want to retain (equivalently, take \(Y=q(X)\)). The move \(g\in G\) preserves the fibers of \(q\) when

                                                                            \[q(x)=q(x')\Longrightarrow q(gx)=q(gx')\qquad (x,x'\in X).\]

                                                                            Exactly in this case there is a unique induced map \(\bar g:Y\to Y\) with

                                                                            \[q\circ g=\bar g\circ q.\]

                                                                            If \(g\) is a bijection and \(X\) is finite, then \(\bar g\) is surjective and therefore bijective. In an infinite setting, require the inverse move to preserve fibers as well. Thus, for the finite cube action below, the fiber-preserving bijections form a subgroup; a quotient does not automatically inherit the whole puzzle action.

                                                                            The antipodal sticker quotient

                                                                              For the ordinary cube, use the surface-cell and frame convention in how the cube models connect and the complete three-layer cube. Represent a sticker by a pair \((p,n)\), where \(p\in \{-1,0,1\}^3\) satisfies \(\lVert p\rVert _\infty =1\) and \(n=\operatorname {sgn}(p_i)e_i\) for each nonzero coordinate \(p_i\). Thus corners, edges and face centers contribute \(8\cdot 3\), \(12\cdot 2\) and \(6\cdot 1\) pairs, respectively, for 54 stickers. The antipodal identification is the involution

                                                                              \[a(p,n)=(-p,-n),\qquad q(p,n)=\{(p,n),a(p,n)\}.\]

                                                                              Its sticker fibres pair opposite stickers. This is a quotient of the labelled sticker action; it is not the same as forgetting equal colors or quotienting by whole-cube rotations.

                                                                              Why one face turn fails

                                                                                The positive-\(z\) face turn uses

                                                                                \[u(x,y,z)=(y,-x,z)\]

                                                                                on every sticker whose cubie has \(z=1\), rotating both its position and its normal, and fixes the other cubies. Take a top sticker and its antipode, for example

                                                                                \[x=((1,1,1),e_z),\qquad a(x)=((-1,-1,-1),-e_z).\]

                                                                                The turn sends \(x\) to the sticker at \(((1,-1,1),e_z)\), while it fixes \(a(x)\). These outputs are not an antipodal pair, so \(q(ux)\ne q(u\,a(x))\). The single \(U\) move therefore does not descend to the antipodal quotient.

                                                                                A synchronized opposite-layer move

                                                                                  The clockwise turn of the negative-\(z\) face, viewed from that face, is the inverse global rotation on \(z=-1\). Write it as \(d\); its inverse \(d^{-1}\) uses the same global rotation as \(u\). The composite \(u d^{-1}\) rotates both opposite layers by the same linear map and fixes the middle cubies. It commutes with the antipodal involution:

                                                                                  \[a\,(u d^{-1})=(u d^{-1})\,a.\]

                                                                                  Consequently it preserves every antipodal fiber and induces a bijection on \(Y\). A direct check of all 54 stickers gives the same result: \(U\) fails on the displayed pair, while the synchronized \(U D'\) move maps every antipodal pair to an antipodal pair. The surviving action is therefore an action of the fiber-preserving subgroup, not of the six ordinary face turns individually.

                                                                                  What the cover does not provide

                                                                                    A covering or quotient of the underlying shape supplies a map of shape data. It does not by itself supply a map between sticker-state spaces, a chosen fiber relation, or an equivariance proof for each legal move. Those data are separate: the explicit cube action records which stickers a face turn moves, while the quotient records which states are identified. A move descends only after the displayed fiber test has been checked.

                                                                                    This is the precise limit of the comparison with the complete unfolding in label transport and topology. Projectivity and branched unfolding describe transport of simplex labels; the antipodal example describes descent of a particular sticker action. They pose analogous questions about information lost by identification, but neither construction is being identified with the other. A general lifting or descent classification is outside this note.

                                                                                    The practical rule is short: state the state space, the quotient map and the move action before saying that a move survives identification. The antipodal calculation supplies both a negative and a positive finite example, while preserving the distinction between a shape cover and a compatible puzzle action.

                                                                                    Related

                                                                                      Definition. Framed transports [frap-000Q]

                                                                                        Let \(I\) be a finite set of positions, \(\Omega \) a nonempty set, and \(K\) a group acting faithfully on \(\Omega \). At each position \(i\), choose a set \(X_i\) of local labels and a bijective local frame \(c_i:\Omega \to X_i\). Keep the fibers distinct by using the tagged union

                                                                                        \[E=\coprod _{i\in I}X_i=\{(i,y):i\in I,\ y\in X_i\}.\]

                                                                                        A framed transport is a permutation of \(E\) given by a position permutation \(\pi \in \operatorname {Sym}(I)\) and elements \(a_i\in K\), with

                                                                                        \[g(i,c_i(x))=(\pi (i),c_{\pi (i)}(a_i x)).\]

                                                                                        Write \(g=(\pi ,a)\). Its local factors are indexed by the source position. Nonempty fibers determine \(\pi \) uniquely, and faithfulness determines each \(a_i\) uniquely. Conversely every such pair gives a permutation: in framed coordinates its inverse sends \((j,x)\) to \((\pi ^{-1}(j),a_{\pi ^{-1}(j)}^{-1}x)\).

                                                                                        This finite-position model isolates the coordinate mechanism of [marchetti2025rubik, Section 3.1, Definition 7 and equation (5)]. Which local label sets and local group describe a particular puzzle must still be justified for that puzzle. The requirement that a transport moves whole fibers is part of the model.

                                                                                        Under composition these transports form a wreath product \(K\wr \operatorname {Sym}(I)\). The normal subgroup \(K^I\) acts within fibers, and position permutations reorder its factors. In TauCeti.WreathProduct, a pair is stored using the destination-indexed function \(A_j=a_{\pi ^{-1}(j)}\). Its permutation action is TauCeti.WreathProduct.imprimitiveToPerm_apply; the nonempty-fiber and faithful-action hypotheses give its injectivity in TauCeti.WreathProduct.imprimitiveToPerm_injective. The bijections \(c_i\) identify that action on \(I\times \Omega \) with the action on \(E\) above.

                                                                                        Lemma. Orientation checked cycle by cycle [frap-000U]

                                                                                          Let \(g=(\pi ,a)\) be a framed transport on finitely many positions \(I\), with a faithful local group \(K\) on nonempty \(\Omega \) and bijective local frames. On a cycle

                                                                                          \[i_0\longmapsto i_1\longmapsto \cdots \longmapsto i_{r-1} \longmapsto i_0\]

                                                                                          of \(\pi \), the return transport in the first fiber has framed map \(a_{i_{r-1}}\cdots a_{i_1}a_{i_0}\). Its class in \(K_{\mathrm {ab}}\) is that cycle's contribution to \(\chi (g)=\prod _{i\in I}[a_i]\). Consequently, identity return transport on every position cycle implies \(\chi (g)=1\). This is the cycle criterion.

                                                                                          Proof. 

                                                                                            Apply the coordinate action \(r\) times. Each local map is applied after the preceding one, giving the stated product with the first map on the right. In the abelianization its factors may be reordered; multiplying over disjoint cycles uses each factor exactly once. A fixed position is a cycle of length one and must also be checked.

                                                                                            This is a sufficient test derived from the canonical character of [marchetti2025rubik, Section 3.1, p. 7]. A generator's finite order alone does not make its return map on each shorter position cycle the identity; the local transport on that cycle is the quantity to check.

                                                                                            Theorem. The abelianized total orientation [frap-000S]

                                                                                              Let \(I\) be finite, let \(K\) act faithfully on a nonempty \(\Omega \), and choose bijective frames \(c_i:\Omega \to X_i\). For a framed transport \(g=(\pi ,a)\), let \(K_{\mathrm {ab}}=K/[K,K]\) be its abelianization, where \([K,K]\) is generated by \(uvu^{-1}v^{-1}\), and write \([a]\) for the class of \(a\). The total-orientation character

                                                                                              \[\chi (g)=\prod _{i\in I}[a_i]\quad \hbox {in }K_{\mathrm {ab}}\]

                                                                                              is a homomorphism, independent of replacing the frames by \(c_i'=c_i t_i\) for \(t_i\in K\). This is the canonical character in [marchetti2025rubik, Section 3.1, p. 7]; the proof below gives both assertions directly in source-indexed coordinates.

                                                                                              Proof. 

                                                                                                For \(h=(\sigma ,b)\), the composition law and commutativity of the target give

                                                                                                \[\chi (hg)=\prod _i[b_{\pi (i)}][a_i] =\left (\prod _j[b_j]\right )\left (\prod _i[a_i]\right ) =\chi (h)\chi (g).\]

                                                                                                The middle equality reindexes using the bijection \(\pi \). In the new frames the local map is \(a_i'=t_{\pi (i)}^{-1}a_i t_i\). Its abelianized product contains each \([t_i]\) and its inverse exactly once, so these cancel and leave \(\chi (g)\).

                                                                                                Frame independence here concerns changes within the chosen group of frames. Reversing a common orientation convention can also change the identification of the target group. The target and its quotient map are Abelianization and Abelianization.of; the total product and its frame independence are the argument above.