Metric entropy 1

Some basic thing, including the definition of metric entropy, is introduced in my early blog.

Among the other thing, there is something we need to focus on:

  1. Definition of metric entropy, and more general, topological entropy.

2. Spanning set and separating set describe of entropy.

3.amernov theorem:

$latex h_{\mu}(T)=\frac{1}{n}h_{\mu}(T^n)$.

Now we state the result of Margulis and Ruelle:

Let $latex M$ be a compact Riemannian manifold, $latex f: M\to M$ is a diffeomorphism and $latex \mu$ is a $latex f$-invariant measure.

Entropy is always bounded above by the sum of positive exponents;i.e.,

$h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).$

Where $latex dimE_i(x)$ is the multiplicity of $latex \lambda_i(x)$ and $latex a^{+}=max(a,0)$.

Pesin shows the inequality is in fact an equality if $latex f\in C^2$ and $latex m$ is equivalent to the Riemannian measure on $latex M$. So this is also sometimes known as Pesin’s formula.

F.Ledrappier and L.S.Young generate the result of Pesin.

One of their main results is:

$latex f: M\to M$ is a $latex C^2$ diffeomorphism, where $latex M$ is a compact Riemanian manifold, f is compatible with the Lesbegue measure on $latex M$, and

$latex h_m({f,\mu})=\int_{M}\lambda_idim(V_i)dm$

If and only if on the canonically defined quotation manifold $M/W_{\mu}$, i.e. the manifold mod unstable manifold $W_{\mu}$, the induced conditional measure $latex m_{\xi}$ is absolute continuous.

Remark: according to my understanding, equality just means in some sense we have the inverse estimate:

$latex h_{m}(f,\mu)\geq \int_{M}\lambda_idim(V_i)dm.$

This result may just mean near the fix point of $latex f$,i.e. the place charge the topology of the foliation, we have the inverse estimate. Such an inverse estimate will lead to control of the singularity of the push forward measure $latex m_{\xi}$ on the quation manifold.  So $latex m_{\xi} $ have good regularity. But this idea is not complete to solve the problem.

Now we begin to get a geometric explain and which will lead a rigorous proof of the inequality:

$latex h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).$

At first we could observe that the long time average $latex \lim_{n\to \infty}\frac{1}{n}log||Df^n||$ of $latex Df$ could be diagonal. Assume after diagonal the eigenvalue is

$latex \lambda_1\leq \lambda_2\leq \lambda_3\leq…\leq \lambda_{n-1}\leq \lambda_n$.

This eigenvalue could divide into 3 parts: <0,=0,>0.

This will lead to a direct sum decomposition of the tangent bundle $latex TM$:

$latex TM\simeq E_{u}\otimes E_s\otimes E_c.$

Where $E_u$ is the part corresponding to the eigenvalue>0, For this part we consider the more refinement decomposition:

$latex E_u=\otimes_{k=1}^rV_k$, $latex V_k$ is the eigenvector space of $latex \lambda_k$. The dimension of $V_k$ is $dim V_k$.

On the other hand, we have an equality of metric entropy:

$latex h_{m}(f)=\frac{1}{n}h_{m}(f^n)=\sup_{\alpha\in partition \ set}\frac{1}{n}h_m(f^n,\alpha)$.

For the later one, $latex \alpha$ is a measurable partition of $latex M$, then $latex \alpha$ could always be refine to a smaller partition $latex \beta$, and we have:

$latex h_{m}(f,\alpha)\leq h_{m}(f,\beta)$.

Now we arrive the central place of the proof:

every partition could be refined by a partition with the boundary of almost all cubes is parallel to the foliation. So  we focus ourselves on the portion $latex \beta$ and all boundaries of cubes in $latex \beta$ is parallel to the eigenvector.

Under this situation, we need only estimate the numbers of $latex \vee_{i=1}^nT^i\beta$. Estimating it is not very difficult. we need only observe the following two things:


$latex \lim{n\to \infty}$ exists a.e. in $latex M$. So this lead to the definition of foliation almost everywhere, and except a measurable zero set. In fact this set is the set of fix point of $latex M$ under $latex f$.


After a rescaling, every point which is not a fix point of $latex f$ could be understand as it is far away from fix points. Then the foliation could be understand as  a product space locally. The flow with the direction which the eigenvalue is less than 1 cold not change $latex \vee_{i=1}^nT^i\beta$. The direction with eigenvalue equal to 1 is just transition and just change the number of $latex \vee_{i=1}^nT^i\beta$ with polynomial growth. But the central thing is the direction with eigenvalue large than one and will make $latex \vee_{i=1}^nT^i\beta$ change with viscosity $latex e^{\lambda_i}$. and we product it and get :

$latex h_{m}(f,\mu)\leq \int_{M}\lambda_i dim(V_i)dm$.

In fact the proof only need $latex f$ to be $latex C^1$