By Ian F. Putnam
The writer develops a homology conception for Smale areas, which come with the fundamentals units for an Axiom A diffeomorphism. it really is according to materials. the 1st is a more robust model of Bowen's outcome that each such procedure is a twin of a shift of finite style lower than a finite-to-one issue map. the second one is Krieger's size workforce invariant for shifts of finite style. He proves a Lefschetz formulation which relates the variety of periodic issues of the procedure for a given interval to track information from the motion of the dynamics at the homology teams. The life of this type of thought used to be proposed by way of Bowen within the Seventies
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Extra info for A homology theory for Smale spaces
The map π may be written as the composition of three maps. The ﬁrst from Y s (y, Y /2) to Y s (y , Y ) sends z to [y , z]. The second from Y s (y , Y ) to X s (x ) is simply π. The third is the map from X s (x , X /2) to X s (x, X ) sends z to [x, z]. Each is deﬁned on an open set containing y, y and x , respectively and is an open map. The conclusion is that there exists some > 0 such that π(Y s (y, )) = U is an open set in X s (x) containing x. It then follows that X s (x) = ∪l≥0 ϕ−l (U ) = ∪l≥0 π(ψ −l (Y s (y, ))) ⊂ π(Y s (y)).
YL ) under ρL, . Consider now the factor map ρL, : ΣL,0 (π) → YL (πs ). 3. That is, let M ≥ 0 and consider M +1 points in ΣL,0 (π) which all have the same image under ρL, . Using our earlier notation, this set is written as (ΣL,0 (π))M (ρL, ). Each element of such an M + 1-tuple has the form (y0 , . . , yL , z0 ) and the condition that they have the same image under ρL, simply means that the y0 , . . , yL entries of each one are all the same. 32 2. DYNAMICS Thus, we could list them as (y0 , .
YL , z0 , . . , zM ) as above, we prove the ﬁrst statement only. Let 0 ≤ m ≤ M be ﬁxed. It follows from the hypothesis and our choice of K0 that [yl1 , yl2 ] is deﬁned and πs ([yl1 , yl2 ]) = [πs (yl1 ), πs (yl2 )] = πs (yl1 ). As πs is s-bijective, it follows that [yl1 , yl2 ] = yl1 . It also follows from our choice of that t(e0 (yl1 , zm )) = t(e0 (yl2 , zm )) and so we may compute, for k ≤ 0, ek (yl2 , zm ) = [e(yl1 , zm ), e(yl2 , zm )]k = ek ([yl1 , yl2 ], [zm , zm ]) = ek (yl1 , zm ).