# Metric entropy 2

I am reading the article “ENTROPY THEORY OF GEODESIC FLOWS”.

Now we focus on the upper semi-continuouty of the metric entropy map. The object we investigate is $(X,T,\mu)$, where $\mu$ is a $T-$invariant measure.

The insight to make us interested to this kind of problem is a part of variational problem, something about the existence of certain object which combine a certain moduli space to make some quantity attain critical value(maximum or minimum). The most simple example maybe Isoperimetric inequality and Dirichlet principle of Laplace. Any way, to establish such a existence result a classical approach is to proof the upper semi-continuouty and bounded for associate energy of the problem. In our case the semi-continuouty will be some thin about the regularity of the entropy map:

$E:M(X,T)\to h_{\mu}.$

We define the entropy at infinity:

$sup_{(\mu_n)}limsup_{\mu_n\to 0}h_{\mu_n}(T)$

Where $(u_n)_{n=1}^{\infty}$ varies in all sequences of measure coverage to $0$ in the sense for all $A\subset M$, $A$ measurable then $\lim_{n\to \infty} \mu_{n}(A)=0$.

Compact case

we say some thing about the compact case, In this case we have finite partition with smaller and smaller cubes, this could be understand as a sequences of smaller and smaller scales. A example to explain the differences is $\mathbb N^{\mathbb N},\sigma$, shift map on countable alphabet.

Because of this thing, there is a good sympolotic model, i.e.  h-expension, and it generalization  asymptotically  h-expension equipped on a compact metric space $X$ have been proved to be that the corresponding entropy map is upper semi-continous.

In particular $C^{\infty}$ diffeomorphisms on compact manifold is asymptotically h-expensive.

Natural problem but I do not understand very well:

Why it is natural to assume the measure to be probability measure in the non-compact space?

Non-compact case

$(X,d)$ metric space

$T:X\longrightarrow X$ is a continuous map.

$d_n(x,y)=\sup_{0\leq k\leq n-1}d(T^kx,T^ky)$, then $d_{n}$ is still a metric.

Easy to see $\frac{1}{n}h_{\mu}(T^n)=h_{\mu}(T)$. This identity could be proved by the cretition of entropy by $\delta$-seperate set and $\delta$-cover set.

Kapok theorem:

$X$ compact, for every ergodic measure $\mu$ the following formula hold:

$h_{\mu}(T)=\lim_{\epsilon \to 0}limsup_{n\to \infty}\frac{1}{n}logN_{\mu}(n,\epsilon,\delta)$.

Where $h_{\mu}(T)$ is the measure theoretic entropy of $\mu$.

Riquelme proved the same formula hold for Lipchitz maps on topological manifold.

Let $M_e(X,T)$ defined the moduli space of $T$-invariant portability measure.

Let $M_(X,T)$ defined the moduli space of ergodic $T$-invariant probability measure.

Simplified entropy formula:

$(X,d,T)$ satisfied simplified entropy formula if $\forall \epsilon >0$ surfaced small and $\forall \delta\in (0,1)$, $\mu\in M _e(X,T)$.

$h_{\mu}(T)=\limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta))$.

Simplified entropy inequality:

If $\epsilon>0$ suffciently small, $\mu \in M_{e}(X,T)$, $\delta\in (0,1)$.

$h_{\mu}(T)\leq \limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta))$.

Weak entropy dense:

$M_e(X,T)$ is weak entropy dense in $M(X,T)$. $\forall \lambda>0$, $\forall \mu\in M(X,T)$, $\exists \mu_n\in M_e(X,T)$, satisfied:

1. $\mu_n\to \mu$ weakly.
2. $h_{\mu_n}(T)>h_{\mu}(T)-\lambda$, $\forall \lambda>0$.

# Metric entropy 1

Some basic thing, include 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:

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

Now we state the result of Margulis and Ruelle:

Let $M$ be a compact riemannian manifold, $f:M\to M$ is a diffeomorphism and $\mu$ is a $f$-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 $dimE_i(x)$ is the multiplicity of $\lambda_i(x)$ and $a^{+}=max(a,0)$.

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

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

One of their main result is:

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

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

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

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

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

This result maybe just mean near the fix point of $f$,i.e. the place charge the topology of the foliation, we have the inverse estimate. Such a inverse estimate will lead a control of the singularity of the push forward measure $m_{\xi}$ on the quation manifold.  So $m_{\xi}$ 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:

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

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

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

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

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

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

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

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

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

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

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

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

Now we arrive the central place of the proof:

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

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

1.

$\lim{n\to \infty}$ exists a.e. in $M$. 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 $M$ under $f$.

2.

After a rescaling, every point which is not a fix point of $f$ 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 $\vee_{i=1}^nT^i\beta$. The direction with eigenvalue equal to 1 is just transition and just change the number of $\vee_{i=1}^nT^i\beta$ with polynomial growth. But the central thing is the direction with eigenvalue large than one and will make $\vee_{i=1}^nT^i\beta$ change with viscosity $e^{\lambda_i}$. and we product it and get :

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

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