gpt 6 astra ultra attempt on Math-259

Conditional fiber collapse and periodic-data tests for the Anosov infranil conjecture

Definitions and mathematical statement

An Anosov diffeomorphism $f:M\to M$ of a closed smooth manifold has an invariant splitting $TM=E^s\oplus E^u$ and constants $C>0$, $0<\lambda<1$ with $\|Df^nv^s\|\le C\lambda^n\|v^s\|$ and $\|Df^{-n}v^u\|\le C\lambda^n\|v^u\|$ for $n\ge0$. An infranilmanifold is a compact quotient of a simply connected nilpotent Lie group by a torsion-free discrete affine group with finite holonomy. The conjecture asks for a homeomorphism $h$ and a hyperbolic infranil automorphism $A$, with no eigenvalue of its inducing differential on the unit circle, such that \[ h\circ f=A\circ h. \] The precise infranil automorphism convention follows the source; topological conjugacy, not differentiable conjugacy, is requested.

Formulation and concept references: [S1].

Short English statement

Is every uniformly hyperbolic diffeomorphism topologically equivalent to an algebraic hyperbolic automorphism on an infranilmanifold?

Research attempt

Attribution and scope.

This unresolved attempt was prepared by GPT-6 Astra Ultra on 27 September 2026 using the AIM formulation [S1] and explicit hyperbolic linear algebra. The desired equivalence is topological conjugacy. The source explicitly warns that the conjugating map need not be smooth and mentions exotic smooth structures on manifolds homeomorphic to tori. Those examples do not contradict the stated topological conclusion. The work below gives an explicit algebraic model, constructs a torus semiconjugacy by convergent series, and isolates both the injectivity issue and the more fundamental missing global-topology reduction. No classification of all Anosov manifolds is assumed.

A model with completely checkable dynamics.

Let $A\in\mathrm{GL}(d,\mathbb Z)$ have no eigenvalue of modulus one. It induces a diffeomorphism of $\mathbb T^d=\mathbb R^d/\mathbb Z^d$. The real generalized eigenspaces split as $E^s\oplus E^u$, according to eigenvalue modulus below or above one. Choose a number $\theta<1$ larger than all stable eigenvalue moduli and all unstable inverse moduli. Jordan-block powers are a polynomial in the iterate times an exponentially decaying factor. Absorbing that polynomial into a slightly larger exponential rate yields constants $C$ such that \[ \|A^j|_{E^s}\|\le C\theta^j, \qquad \|A^{-j}|_{E^u}\|\le C\theta^j\quad(j\ge0). \] The constant distributions descend to invariant subbundles of the tangent bundle, proving directly that the induced map is Anosov. The unstable directions are not a global expansion in every direction; the splitting is essential.

For a concrete surface model take \[ A=\begin{pmatrix}2&1\\1&1\end{pmatrix}. \] Its eigenvalues are $(3\pm\sqrt5)/2$, with product one and one on each side of one. This verifies all parts of the Anosov definition without relying on a numerical Lyapunov exponent along selected orbits.

Periodic-point data give a necessary conjugacy test.

For any positive integer $m$, a fixed point of $A^m$ corresponds to $x\in\mathbb R^d$ with $(A^m-I)x\in\mathbb Z^d$, modulo $\mathbb Z^d$. Since $A^m-I$ is invertible over $\mathbb R$, \[ \#\operatorname{Fix}(A^m)=|\det(A^m-I)|. \] This counts the finite cokernel of the integral matrix, equivalently the index of its image lattice. Hyperbolicity guarantees invertibility for every $m$, so no root-of-unity case is hidden in the formula. For the displayed cat map, let $t_m=\operatorname{tr}(A^m)$. Cayley--Hamilton gives \[ t_0=2,\quad t_1=3,\quad t_{m+1}=3t_m-t_{m-1}, \qquad \#\operatorname{Fix}(A^m)=t_m-2. \] The first counts are $1,5,16,45,121$.

A topological conjugacy bijects fixed points at every iterate and preserves least periods. Thus a proposed algebraic model must satisfy these checks. Passing finitely many checks is insufficient, and even all periodic counts would not construct a conjugacy. Conversely, differences in derivatives at periodic points need not obstruct a merely topological conjugacy; derivative multipliers are relevant to stronger smooth-conjugacy questions. This keeps the diagnostic aligned with the actual target.

An explicit semiconjugacy when the topology is already a torus.

Assume now that $f:\mathbb T^d\to\mathbb T^d$ is a diffeomorphism whose action on the fundamental group is a hyperbolic integral matrix $A$. A lift has the form \[ F(x)=Ax+\varphi(x), \] where $\varphi$ is continuous, bounded and $\mathbb Z^d$-periodic. The lift is a homeomorphism. Let $P_s,P_u$ be the linear projections onto the two invariant subspaces, and seek $H(x)=x+u(x)$ such that $H\circ F=A\circ H$. This requires \begin{equation}\label{wsref-0-61-32-35-35-3a-63-6f-68-6f-6d-6f-6c-6f-67-79} u(Fx)-Au(x)=-\varphi(x). \end{equation} The following uniformly convergent series solve it: \[ \begin{split} u_s(x)&=-\sum_{j=0}^{\infty} A_s^jP_s\varphi(F^{-j-1}x),\\ u_u(x)&=\sum_{j=0}^{\infty} A_u^{-j-1}P_u\varphi(F^jx), \qquad u=u_s+u_u. \end{split} \] The exponential bounds and boundedness of $\varphi$ prove convergence uniformly in $x$. Reindexing the stable series gives $u_s(Fx)=A_su_s(x)-P_s\varphi(x)$, and reindexing the unstable series gives the analogous identity. Their sum is (a255:cohomology). These sign checks matter: the stable part uses backward iterates, whereas the unstable part uses forward iterates.

The function $u$ is periodic. Indeed $F^j(x+z)=F^j(x)+A^jz$ for all integral $j$ and $z\in\mathbb Z^d$; $A^jz$ remains integral, including negative $j$, because $A\in\mathrm{GL}(d,\mathbb Z)$. Periodicity of $\varphi$ makes every summand periodic. Therefore $H$ descends to a continuous map $h:\mathbb T^d\to\mathbb T^d$ satisfying \[ h\circ f=A\circ h. \] The maps $x\mapsto x+t u(x)$ descend to a homotopy from the identity to $h$, so $h$ has degree one. It is surjective: if it omitted a point, its degree would factor through the punctured torus, whose top homology is zero. No claim that those homotopy maps are homeomorphisms is used. The conclusion so far is semiconjugacy, not conjugacy.

A quantitative sufficient condition for injectivity.

Suppose in addition that $f$ is expansive with constant $\delta>0$: if $d(f^n x,f^n y)<\delta$ for every $n\in\mathbb Z$, then $x=y$. Anosov diffeomorphisms have such constants by their local product structure, an established dynamical fact. If \[ 2\sup_x d(h(x),x)<\delta, \] then $h$ is injective. If $h(x)=h(y)$, semiconjugacy gives $h(f^n x)=h(f^n y)$ for every positive and negative iterate. The triangle inequality bounds $d(f^n x,f^n y)$ by the displayed left side, so expansivity forces equality. A continuous bijection of the compact torus to itself is a homeomorphism, completing conjugacy under this extra quantitative condition.

The series estimate supplies \[ \|u\|_\infty\le \left(\sum_{j\ge0}\|A_s^jP_s\|+ \sum_{j\ge0}\|A_u^{-j-1}P_u\|\right)\|\varphi\|_\infty. \] Thus sufficiently small lift displacement relative to an available expansivity constant gives a concrete criterion. This is not claimed to cover all toral Anosov diffeomorphisms or to reprove the stronger known rigidity theorems. For a large perturbation the bound may exceed $\delta$, and bounded distance of lifted orbits is not the same as the small distance required by expansivity on the torus.

Two obstructions to turning the calculation into a classification.

First, a degree-one surjection need not be one-to-one. On the circle, choose a continuous nondecreasing lift $g:\mathbb R\to\mathbb R$ with $g(t+1)=g(t)+1$, constant on a nontrivial interval and strictly increasing elsewhere in a period. It descends to a degree-one surjection that collapses an arc. Taking its product with identities gives the same countertest on a torus. This example is not asserted to be the semiconjugacy of an Anosov map; it shows that the degree argument alone has not proved injectivity. Hyperbolicity must enter again through an additional argument, such as the quantitative one above or a stronger global rigidity theorem.

Second, the entire series construction assumed the manifold was a torus and the induced matrix hyperbolic. A general Anosov splitting does not supply a global integral affine coordinate system. Even proving a particular manifold has the homotopy type of an infranilmanifold would still require the appropriate topological identification and dynamical conjugacy. The existence of contracting and expanding subbundles is differential information; it does not directly prove that the fundamental group is virtually nilpotent.

The statement also concerns diffeomorphisms, not time-one maps of Anosov flows. Such a time-one map has a central flow direction and fails the displayed two-subbundle hyperbolicity definition. Importing nontransitive flow examples without checking that direction would produce spurious counterexamples to the catalogue assertion.

Verification and outcome.

Exact matrix powers check the periodic-point formula and recurrence. The semiconjugacy is verified by reindexing absolutely uniformly convergent series, including its lattice equivariance. The injectivity step exposes its required smallness hypothesis rather than replacing boundedness by expansivity. The strongest result here is an explicit semiconjugacy construction and a sufficient conjugacy criterion within an already algebraically modeled topology. The first general gap is obtaining the infranil topological model from an arbitrary Anosov system; no construction or counterexample resolves it. The full classification problem remains UNRESOLVED.

Sources

• S1 Open Problems from the Workshop 'Emerging Applications of Measure Rigidity' (AIM, 2004), Section 2.1, Conjecture 3. https://aimath.org/WWN/measrigid/measrigid.pdf. Formulation source.