Independent roots do not supply coherent rational roots

Abstract

An explicit reduced abelian group contains a nonzero element with independently chosen roots of every positive degree.

Index positive degrees by j+1 for natural j. Let B be the direct sum of cyclic groups ZMod(2*(j+1)), and let e_j be the element with coefficient one in coordinate j and zero elsewhere. Put c_B=e_0. Let N be the additive subgroup generated by (j+1)*e_j-c_B for all j, and set G=B/N and c=c_B+N. Write b_j=e_j+N. These are explicit algebraic definitions, with no topology or continuity condition.

Theorem 1.1 (A divisible element in a reduced group).

$$((c\ne 0)\wedge (((\mathrm{nsmul}\left(2,c\right)=0)\wedge (((\forall n\in Nat,(0<n)⟹\exists b\in G,\mathrm{nsmul}\left(n,b\right)=c)\wedge (\forall f\in \mathrm{AddHom}\left(Q,G\right),f=0))))))$$

Proof. Machine-checked in Lean as D5/S3/ConceptDynamics/Coding/PrimitiveLifts/IndependentRootCoherenceSeparation.independent_roots_without_rational_coherence (✓ std3). ∎

Source. Repository-derived.

Commentary.

The concrete element c is nonzero, 2c=0, every positive natural n admits b in G with nb=c, and every additive homomorphism f:Q->G is zero. The roots may be chosen explicitly as b_(n-1). The theorem concerns all positive degrees, without a finite cutoff. The absence of any nonzero homomorphism from Q says that G is reduced: a nonzero divisible subgroup would admit a nonzero map from Q by successive compatible divisions.

The defining relation immediately gives (j+1)b_j=c, and the first cyclic summand gives 2c=0. To see that c survives, take QZ=Q/<1>, the quotient of the additive rational group by its integer subgroup. Send e_j to the class of 1/(2*(j+1)). This is a well-defined homomorphism on each cyclic summand because its order divides 2*(j+1), and the direct-sum universal property extends it to B. Every relation maps to 1/2-1/2=0. But c_B maps to the nonzero class of 1/2: no integer lies strictly between zero and one. Hence c is nonzero in G.

Let D be the direct sum of ZMod(j+1), and let beta:B->D reduce each coordinate modulo j+1. All defining relations lie in its kernel, so it descends to pi:G->D. The exact kernel of pi is the cyclic subgroup . Indeed, if beta(v)=0, choose an integer representative a_j for each coefficient of the finitely supported v. Its reduction is zero precisely when j+1 divides a_j. Thus a_j=(j+1)k_j for an integer k_j, and the image of the jth summand in G is k_jc. Summing over the finite support places v+N in . Conversely pi(c)=0, so every integer multiple of c lies in the kernel. This is the concrete kernel calculation; it does not assume a quotient splitting.

Now let f:Q->G be additive. For each rational q and coordinate j, write q=(j+1)(q/(j+1)). Additivity shows that the jth coordinate of pi(f(q)) is j+1 times an element of ZMod(j+1), hence zero. Thus f(q) lies in . Apply this conclusion at q/2: f(q/2)=kc for some integer k. Then f(q)=2f(q/2)=k(2c)=0. This proves that every such f is zero, while each equation nb=c separately has the explicit solution above.

For the single period-one return atlas, take the carrier G with deck translations and time map T(x)=x+c. A signed chart is A(t,z)=z+tc for integer t. Its seam law is A(t,z)=A(s,w) exactly when t=s+k and z+kc=w for an integer k; its normalized phase is the unique element of Fin 1. Thus the primitive base period is one, while T itself has order two. In the multiplicative ReturnAtlas API the deck group is Multiplicative G, with holonomy corresponding to c. The translation R_n(x)=x+b_(n-1) satisfies R_n^n=T for every positive n and commutes with every deck translation. These roots even commute pairwise because G is abelian. A coherent rational system of translations would instead be an additive map a:Q->G with a(1)=c, which the theorem excludes. Every map commuting with all deck translations is itself a translation, as follows by evaluating it at zero. Thus the same obstruction excludes a deck-equivariant rational action whose time-one map is T. Pairwise commutation supplies no denominator compatibility. This is a standard reduced torsion group construction used as an explicit sharpness example; no literature priority or physical continuity assertion is intended.

Theorem 1.2 (Every nontrivial root fiber is infinite).

$$\forall m\in Nat,(1<m)⟹\mathrm{Infinite}\left(\mathrm{Subtype}\left(\lambda x:G,\mathrm{nsmul}\left(m,x\right)=c\right)\right)$$

Proof. Machine-checked in Lean as D5/S3/ConceptDynamics/Coding/PrimitiveLifts/IndependentRootCoherenceSeparation.infinite_root_fiber (✓ std3). ∎

Source. Repository-derived.

Commentary.

In this exact quotient G and for every natural m>1, the subtype of all x with mx=c is infinite. An explicit injection from Nat sends n to u_n=(n+1)b_(m(n+1)-1). The defining relation gives mu_n=c. Under the same residual projection pi, its only possibly nonzero coordinate is i_n=m*(n+1)-1, with coefficient n+1 modulo m*(n+1). Since 0<n+1<m*(n+1), that coefficient is nonzero. The indices i_n are distinct, so evaluation at i_n distinguishes u_n from every u_k with k different from n. This proves infinitude at every nontrivial degree, rather than at just one degree. Degree one has the unique root c. No integer linear independence is asserted: all these roots satisfy the common equation m*u_n=c.

Consequently, finite-root-fiber compactness cannot be applied to these full root-choice sets: its finiteness hypothesis fails at every degree m>1 in the original counterexample. Infinite fibers alone do not exclude coherent roots. Here the separate proved facts that every additive map Q->G is zero and c is nonzero supply the actual obstruction to rational coherence. This coordinate argument is a repository derivation for the displayed group; it makes no claim of literature novelty or a new general compactness theorem.

References

• Truth anchor: D5/S3/ConceptDynamics/Coding/PrimitiveLifts/IndependentRootCoherenceSeparation.independent_roots_without_rational_coherence
• Truth anchor: D5/S3/ConceptDynamics/Coding/PrimitiveLifts/IndependentRootCoherenceSeparation.infinite_root_fiber