K-theory for Cuntz-Krieger algebras arising from real quadratic maps

We compute the $K$-groups for the Cuntz-Krieger algebras $\mathcal{O}_{A_{\mathcal{K}(f_{\mu})}}$, where $A_{\mathcal{K}(f_{\mu})}$ is the Markov transition matrix arising from the \textit{kneading sequence }$\mathcal{K} (f_{\mu})$ of the one-parameter family of real quadratic maps $f_{\mu}$.

kneading theory [14], J. Guckenheimer [5] has classified up to topological conjugacy, a certain class of maps which includes the quadratic family. The idea of kneading theory is to encode information about the orbits of a map in terms of infinite sequences of symbols and to exploit the natural order of the interval to establish topological properties of the map. In the following, The itinerary of f µ (c) is called the kneading sequence of f µ and will be denoted by K(f µ ).
be the alphabet set. The sequences on N are ordered lexicographically. However, this ordering is not reflected by the mapping x → ε(x), because the map f µ reverses orientation on [c, 1]. To take this into account, for a sequence ε = (ε n ) ∞ n=0 of the symbols −1, 0, +1, another sequence θ = (θ n ) ∞ n=0 is defined by θ n = n i=0 ε i . If ε = ε(x) is the itinerary of a point x ∈ I then θ = θ(x) is called the invariant coordinate of x. The fundamental observation of J. Milnor and W. Thurston [14] is the monotonicity of the invariant coordinates: x < y ⇒ θ(x) ≤ θ(y).
Let us now consider only those kneading sequences that are periodic, i.e.
will then determine a Markov partition of I into n−1 line intervals {I 1 , I 2 , ..., , whose definitions will be given in the proof of Theorem 1. Thus, we will have a Markov transition matrix A K(fµ) defined by It is easy to see that this matrix A K(fµ) is not a permutation matrix and no row or column of A K(fµ) is zero. Thus, for each one of these matrices and following the work of J. Cuntz and W. Krieger [2], one can construct the Cuntz-Krieger algebra O A K(fµ) . In [3], J. Cuntz proved that for a r × r matrix A that satisfies a certain condition (I) (see [2]), which is readily verified by the matrices A K(fµ) . In We can now state and prove the following. and be the point on the unit interval [0, 1] represented by the sequence z i for i = 1, 2.... We have σ(z i ) = z i+1 for i = 1, ..., n − 1 and σ(z n ) = z 1 . Denote by ω the n × n matrix representing the shift map σ. Let C 0 be the vector space spanned by the formal basis {z ′ 1 , . . . , z ′ n }. Now, let ρ be the permutation of the set {1, . . . , n}, which allows us to order the points z ′ 1 , . . . , z ′ n on the unit interval . . , n and let π denote the permutation matrix which takes the formal basis {z ′ 1 , . . . , z ′ n } to the formal basis {x 1 , . . . , x n }. We will denote by C 1 the n − 1 dimensional vector space spanned by the formal basis {x i+1 − x i : i = 1, . . . , n − 1}. Set Thus, we can define the Markov transition matrix A K(fµ) as above. Let ϕ denote the incidence matrix that takes the formal basis {x 1 , . . . , x n } of C 0 to the formal basis {x 2 − x 1 , . . . , x n − x n−1 } of C 1 . Put η := ϕπ. As in [7] and [8], we obtain an endomorphism α of C 1 , that makes the following diagram commutative.
We have α = ηωη T (ηη T ) −1 . Remark that if we neglect the negative signs on the matrix α then we will obtain precisely the Markov transition matrix A K(fµ) . In fact, consider the (n − 1) × (n − 1) matrix β := 1 n L 0 0 −1 n R where 1 n L and 1 n R are the identity matrices of rank n L and n R respectively, with n L (n R ) being the number of intervals I i of the Markov partition placed on the left (right) hand side of the turning point of f µ . Therefore, we have The matrix γ K(fµ) makes the diagram commutative. Finally, set θ K(fµ) := γ K(fµ) ω. Then, the following diagram is also commutative. Now, notice that the transpose of η has the following where Y is an invertible (over Z) n× n integer matrix given by i is the inclusion C 1 ֒→ C 0 given by and X is an invertible (over Z) (n − 1) × (n − 1) integer matrix obtained from the (n − 1) × n matrix η T by removing the n-th row of η T . Thus, from the commutative diagram we will have the following commutative diagram with short exact rows where the map p is represented by the 1 × n matrix 0 . . . 0 1 and i.e., A ′ is similar to A T K(fµ) over Z and θ ′ is similar to θ T K(fµ) over Z. Hence, for example by [10], we obtain respectively Now, from the last diagram we have, for example by [9], Therefore, Next, we will compute Z n /(1 − θ K(fµ) )Z n . From the previous discussions and notations, the n × n matrix θ K(fµ) is explicitly given by Notice that this matrix θ K(fµ) completely describes the dynamics of f µ .
Finally, using row and column elementary operations over Z, we can find invertible (over Z) matrices U 1 and U 2 with integer entries such that Thus, we obtain where R = −1, L = +1, C = 0. Thus, we can construct the 5 × 5 Markov transition matrix A K(fµ) and the matrices θ K(fµ) , ω, ϕ, and π.  Remark 1. In the statement of Theorem 1 the case a = 0 may occur. This happens when we have a star product factorizable kneading sequence [4]. In this case the correspondent Markov transition matrix is reducible.
Remark 2. In [6] the authors have constructed a class of C * -algebras from the β-expansions of real numbers. In fact, by considering a semiconjugacy from the real quadratic map to the tent map [14], we can also obtain Theorem 1 using [6] and the λ-expansions of real numbers introduced in [4].
Remark 3. In [12] (see also [11]) and [13] the BF-groups are explicitly calculated with respect to other kind of maps on the interval.