We define a broad class of piecewise smooth plane homeomorphisms which have properties similar to the properties of Lozi maps, including the existence of a hyperbolic attractor. We call those maps Lozi-like. For those maps one can apply our previous results on kneading theory for Lozi maps. We show a strong numerical evidence that there exist Lozi-like maps that have kneading sequences different than those of Lozi maps. dimension for unimodal maps, there is a big difference. The one-dimensional analogue of the Lozi family is the family of tent maps. There we have one parameter and one kneading sequence. For the Lozi family we have two parameters, but infinitely many kneading sequences. Thus, by using a concrete formula (1), we immensely restrict the possible sets of kneading sequences. It makes sense to conjecture that in a generic nparameter subfamily of Lozi-like maps, n kneading sequences determine all other kneading sequences, at least locally. In fact, in our example at the end of this paper, we see that for the Lozi family two kneading sequences may determine the parameter values, and thus all kneading sequences.As we mentioned, we wrote [5] thinking about possible generalizations. Therefore for the Lozi-like maps (under suitable assumptions) all results of that paper hold, and the proofs are the same, subject only to obvious modification of terminology. There are only two exceptions, where in [5] we used the results of [2]. For those exceptions we provide new, general proofs in the section about symbolic dynamics.The paper is organized as follows. In Section 2 we provide the definitions and prove the basic properties of Lozi-like maps. In Section 3 we prove the existence of an attractor. In Section 4 we give two proofs about symbolic dynamics, that are different than in [5]. Finally, in Section 5 we present an example of a three-parameter family of Lozi-like maps. We also show that it is essentially larger than the Lozi family. This last result uses a computer in a not completely rigorous way, so strictly speaking it is not a proof, but a strong numerical evidence.
DefinitionsA cone K in R 2 is a set given by a unit vector v and a number ∈ (0, 1) bywhere || · || denotes the usual Euclidean norm. The straight line {tv : t ∈ R} is the axis of the cone. Two cones will be called disjoint if their intersection consists only of the vector 0. Clearly, the image under an invertible linear transformation of R 2 of a cone is a cone, although the image of the axis is not necessarily the axis of the image.Lemma 2.1. Let K u and K s be disjoint cones. Then there is an invertible linear transformation T : R 2 → R 2 such that the x-axis is the axis of T (K u ) and the y-axis is the axis of T (K s ).Proof. We will define T as the composition of three linear transformations. Choose two of the four components of R 2 (K u ∪ K s ), such that they are not images of each other under the central symmetry with respect to the origin. Then choose one vector in each of those components. There is an invertible linear transformation T...