Papers
Topics
Authors
Recent
Search
2000 character limit reached

The sectional curvature of the infinite dimensional manifold of Hölder equilibrium prababilities

Published 19 Nov 2018 in math.DS, cond-mat.stat-mech, math-ph, math.DG, math.MP, and math.PR | (1811.07748v9)

Abstract: Here we consider the discrete time dynamics described by a transformation T:M→MT:M \to M, where TT is the shift and M=1,2,...,d<sup>NM={1,2,...,d}<sup>\mathbb{N}. It is known that the infinite-dimensional manifold N\mathcal{N} of H\"older equilibrium probabilities is an analytical manifold and carries a natural Riemannian metric. Given a normalized H\"older potential AA denote by μA∈N\mu_A \in \mathcal{N} the associated equilibrium probability. The set of tangent vectors XX to the manifold N\mathcal{N} at the point μA\mu_A coincides with the kernel of the Ruelle operator for AA. The Riemannian norm ∣X∣=∣X∣<em>A|X|=|X|<em>A of the vector XX, which is tangent to N\mathcal{N} at the point μA\mu_A, is described via the asymptotic variance, that is, satisfies ∣X∣<sup>2  =</sup>⟨X,X⟩=lim⁡</em>n→∞1n∫(∑i=0<sup>n−1</sup>X∘T<sup>i</sup>)<sup>2</sup> dμA|X|<sup>2\,\,=</sup> \langle X, X \rangle =\lim</em>{n \to \infty} \frac{1}{n} \int (\sum_{i=0}<sup>{n-1}</sup> X\circ T<sup>i</sup> )<sup>2</sup> \,d \mu_A. Consider an orthonormal basis XiX_i, i∈Ni \in \mathbb{N}, for the tangent space at μA\mu_A. Given two unit tangent vectors XX and YY the curvature K(X,Y)K(X,Y) satisfies     K(X,Y)=14[ ∑i=1<sup>∞</sup>(∫X Y Xi dμA)<sup>2</sup>−∑i=1<sup>∞</sup>∫X<sup>2</sup>Xi dμA  ∫Y<sup>2</sup>Xi dμA ].\,\,\,\,K(X,Y) = \frac{1}{4}[\, \sum_{i=1}<sup>\infty</sup> ( \int X \,Y\, X_i \,d \mu_A)<sup>2</sup> - \sum_{i=1}<sup>\infty</sup> \int X<sup>2</sup> X_i \,d \mu_A\, \,\int Y<sup>2</sup> X_i \,d \mu_A \,]. When the equilibrium probabilities μA\mu_A is the set of invariant Markov probabilities on 0,1<sup>N⊂</sup>N{0,1}<sup>\mathbb{N}\subset</sup> \mathcal{N}, introducing an orthonormal basis a^y\hat{a}_y, indexed by finite words yy, we show explicit expressions for K(a^x,a^z)K(\hat{a}_x,\hat{a}_z), which is a finite sum. These values can be positive or negative depending on AA and the words xx and zz. Words x,zx,z with large length can eventually produce large negative curvature K(a^x,a^z)K(\hat{a}_x,\hat{a}_z). If x,zx, z do not begin with the same letter, then K(a^x,a^z)=0K(\hat{a}_x,\hat{a}_z)=0.

Citations (7)

Summary

No one has generated a summary of this paper yet.

Paper to Video (Beta)

No one has generated a video about this paper yet.

Whiteboard

No one has generated a whiteboard explanation for this paper yet.

Open Problems

We haven't generated a list of open problems mentioned in this paper yet.

Continue Learning

We haven't generated follow-up questions for this paper yet.