We further continue the discussion of topological entropy. Here, we discuss various results that bound above and below the entropy of a given map.

**1. The topological entropy of Lipschitz maps **

Many of the dynamical systems of interest are actually given by compact manifolds and smooth maps . These are always Lipschitz with respect to appropriate metrics. Indeed, choose a Riemannian metric on and let denote the induced norm on the tangent spaces. Then is a Lipschitz constant for with respect to the metric on induced by the Riemannian metric. In this case, the entropy is always finite. We shall prove this in a more general context.

Let be a compact metric space, and for , let denote the the number of -balls necessary to cover (which is always finite). Then we call

the **ball dimension** of . For instance, an -cube has ball dimension . It follows more generally that a Riemannian -manifold has ball dimension .

The reason we shall use this concept below is that gives an upper bound for a minimal -spanning set of the space . (Recall that an -spanning set means that every point is -close to it.) In fact, if the -balls cover , then the centers of these form an -spanning set.

Theorem 1Let be a compact metric space with finite ball dimension . Suppose is a Lipschitz continuous mapping with constant , i.e. for all . Then the entropy of is finite and

The idea of this proof is simple. We can evidently assume (if this is not already the case, take very close to 1 and let ). Then we need to consider the metric and consider minimal -spanning sets with respect to this metric. However, .

Now consider , the minimal cover of by balls (with respect to the original metric ). Then any point of is close to one of the centers of these balls (with metric ). The previous remark implies that any point of is -close **with the metric **. In particular, the minimal -spanning set of has cardinality at most . Call this , as before; then

Now, by definition, is of the order . Plugging this into the last expression gives that

Letting , we get on the left side the topological entropy. This establishes the bound.

Given the remarks made earlier about the ball dimension of a manifold and Lipschitzness of smooth maps, it follows that:

Corollary 2Let be a compact Riemannian manifold of dimension , a smooth map. Then

**2. The topological entropy of maps of the circle **

We now consider a special (but important) low-dimensional case, namely that where the ground space is the unit circle . It is well-known from elementary algebraic topology that the homotopy classes of maps are in bijection with the integers, i.e. that the fundamental group of is the integers. In this way, we can associate to each continuous an invariant, called its **degree**, which turns multiplicativity to composition. For instance, the map has degree .

It is possible, as we now show, to connect this with topological entropy.

Theorem 3Let be continuous and of degree . Then .

The idea of the proof is simple–it’s basically to reverse the Lipschitz argument above. We shall use the following mini-lemma : Let be continuous and suppose we can decompose into arcs such that the image of each arc has length at most . Then has degree . The reason is as follows. Lift to a map such that . Then on the pull-back of each arc in the interval , can jump by at most 1 because is contained in a semicircle.

Now, let’s prove the theorem. Suppose . Pick between the two numbers. Then for large, the smallest -spanning subset of with respect to has cardinality at most . Call this set . These points are contained in small arcs that cover and such that has length at most . This is because we can take to be the -neighborhood of **in the metric **. It now follows by the previous mini-lemma that the degree of is at most . But the degree is multiplicative and must therefore be . This contradiction establishes the result.

I don’t think I’m going to actually prove it, but I’ll mention another (significantly deeper) version of this idea. First, recall that if is an oriented -manifold and continuous, then the top singular cohomology group is one-dimensional, and the associated map

is called the **degree** of . In the smooth case, which is what we care about, we can construct this (in view of the isomorphism between de Rham and singular cohomology) by taking an -form whose integral is nonzero and considering

this is the degree.

The following is the theorem that I will state without proof:

Theorem 4Let be a smooth map of an oriented compact, -dimensional manifold . Then we have .

For a proof, I refer you to Chapter 8 of Katok-Hassselblatt’s *Introduction to the Modern Theory of Dynamical Systems,* which is my primary source for this material.

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