10. Let f_a:S^1\to S^1, a\in[0,1] be a strictly increasing family of homeomorphisms on the unit circle, \rho(a) be the rotation number of f_a. Poincare observed that \rho(a)=p/q if and only if f_a admits some periodic points of period q. In this case f_a^q admits fixed points.

Note that a\mapsto \rho(a) is continuous, and non-decreasing. However, \rho may not be strictly increasing. In fact, if \rho(a_0)=p/q and f^q\neq Id, then \rho is locked at p/q for a closed interval I_{p/q}\ni a_0. More precisely, if f^q(x) > x for some x, then \rho(a)=p/q on [a_0-\epsilon,a_0] for some \epsilon > 0; if f^q(x)  0; while a_0\in \text{Int}(I_{p/q}) if both happen.

Also oberve that if r=\rho(a)\notin \mathbb{Q}, then I_r is a singelton. So assuming f_a is not unipotent for each a\in[0,1], the function a\mapsto \rho(a) is a Devil’s staircase: it is constant on closed intervals I_{p/q}, whose union \bigcup I_{p/q} is dense in I.

9. Let X:M\to TM be a vector field on M, \phi_t:M\to M be the flow induced by X on M. That is, \frac{d}{dt}\phi_t(x)=X(\phi_t(x)). Then we take a curve s\mapsto x_s\in M, and consider the solutions \phi_t(x_s). There are two ways to take derivative:

(1) \displaystyle \frac{d}{dt}\phi_t(x_s)=X(\phi_t(x_s)).

(2) \displaystyle \frac{d}{ds}\phi_t(x_s)=D\phi_t(\frac{d}{ds}x_s)), which induces the tangent flow D\phi_t:TM\to TM of \phi_t:M\to M.

Combine these two derivatives together:

\displaystyle \frac{d}{dt}D_x\phi_t(x_s')=\frac{d}{dt}\frac{d}{ds}\phi_t(x_s) =\frac{d}{ds}\frac{d}{dt}\phi_t(x_s)=\frac{d}{ds}X(\phi_t(x_s)) =D_{\phi_t(x)}X\circ D_x\phi_t(x_s').

This gives rise to an equation \displaystyle \frac{d}{dt}D_x\phi_t=D_{\phi_t(x)}X\circ D_x\phi_t.


Formally, one can consider the differential equation along a solution x(t):
\displaystyle \frac{d}{dt}D(t)=D_{\phi_t(x)}X\circ D(t), D(0)=Id. Then D(t) is called the linear Poincare map along x(t). Suppose x(T)=x(0). Then D(T) determines if the periodic orbit is hyperbolic or elliptic. Note that the path D(t), 0\le t\le T contains more information than the above characterization.

8. Frink’s proof a metricization theorem. Suppose a topological space X is endowed with a function \rho(x,y) satisfying (a) \rho(x,y)=0 iff x=y; (b) \rho(y,x)=\rho(x,y); (c) \rho(x,y)\le 2\max\{\rho(x,z),\rho(z,y)\}.
Then there is a metric d on X induced by \rho:

d(x,y)=\inf\{\sum_{i=1}^n \rho(x_i,x_{i-1}): x_0=x,x_n=y,n\ge 1\}.

It is easy to see that d defines a metric on X. Moreover, it is proved that \rho(x,y)/4 \le d(x,y)\le \rho(x,y).

More generally, the third condition can be replaced by (d) there exists a positive function \phi such that if \rho(x,y)<\phi(\epsilon) and \rho(y,z)<\phi(\epsilon), then \rho(x,z)0. Define a sequence r_n inductively by r_1=1/2, r_{n+1}=\phi(r_n), n\ge 1. Clearly r_n\le 2^{-n}\to 0. Then we discretize the function \rho: let \hat \rho(x,y)=2^{-n} if \rho(x,y)\in[r_n,r_{n-1}). Then we check that the new function \hat \rho satisfies (c) and induces a metric d. Clearly \hat \rho induces the same topology on X as \rho.



7. Given a embedding of a convex sphere e: (S^2,g)\to \mathbb{R}^3, let \nu:e(S^2)\to S^2 be the Gauss map, that is, for each point x\in e(S^2), \nu(x) is the unit outer normal vector of e(S^2) at x. Clearly \nu is a diffeomorphism. The Gauss curvature K_g can be viewed as a function on the standard S^2 through \nu, and satisfies
\displaystyle \int_{S^2} \frac{x}{K_g(x)}d\mu=0. (*)

Minkowski problem. Given a smooth positive function K on S^2 satisfying (*).
Is there a Riemannian metric g on S^2 such that K_g=K?

This 2D version was also answered in the same paper of Nirenberg in 1953. For the high dimensional case see Pogorelov in 1969 and Cheng–Yau in 1976.

6. Weyl problem. Let g be a Riemannian metric on the 2-sphere S^2 with positive Gauss curvature. Does there exist an isometric embedding of (S^2,g) into \mathbb{R}^3?

H. Lewy proved in 1938 the existence under the assumption that the metric g is analytic, using his results on analytic Monge-Ampere equations. Nirenberg proved the existence in 1953 using his results on strong apriori estimates for nonlinear elliptic PDE in 2D. Aleksandrov obtained a generalized solution in 1948 as a limit of polyhedra, and Pogorelov proved the regularity of this generalized solution. In 1969 Pogorelov posed and solved Weyl’s problem for embedding into a three-dimensional Riemannian manifold.

5. Let (M^d,g) be a d-dimensional Riemannian manifold, R_g be the sectional curvature tensor induced by the Levi–Civita connection. Clearly R_g is determined by g? To what extend does R_g determine g? See here.

Answered here by Misha Kapovich:

(1) if d\ge 3 and M has nowhere constant sectional curvature, then a diffeomorphism preserving sectional curvature is necessarily an isometry.

Explain: specifying the sectional curvature of a metric is generally a very overdetermined problem in higher dimensions.

(2) if d=2, then there are counter-examples. Weinstein’s argument shows that every Riemannian surface admits different metrics of the same Gauss curvature (using flow orthogonal to the gradient of the curvature function).


4. Let f be a diffeomorphism on a manifold M^d, \mathcal{W}^k be an f-invariant expanding foliation: there exists \lambda>1 such that |Df(v)|\ge \lambda |v| for any v\in E:=T\mathcal{W}. Then the leaf volume \text{vol}(W(x,r)) grows polynomially.

Let \Lambda=\max\{\text{Jac}(Df|E_{y}):y\in M\}. Then for any r\ge 1, pick n=\lceil\log_\lambda r\rceil . Or equally, \lambda^{n-1} < r\le \lambda^{n}. Then f^nW(f^{-n}x,1)\supset W(x,r),
\displaystyle \text{vol}W(x,r)\le \text{vol}(f^nW(f^{-n}x,1))
\displaystyle =\int_{W(f^{-n}x,1)}\text{Jac}(Df^n|E_{y})dy\le \Lambda^{n}\cdot \text{vol}(W(f^{-n}x,1))
\displaystyle \le C\Lambda^{n}\le C\Lambda^{\log_\lambda r+1}=C\Lambda\cdot r^{\log_\lambda \Lambda}.

3. Hilbert's fourth problem is to find all geometries whose axioms are closest to those of Euclidean geometry for which lines are geodesics. He also provided an example, a Finsler metric on a convex body. More precisely, let Q\subset\mathbb{R}^n be a bounded convex body. Let x\in Q^o. Then for each vector v\in T_xQ, draw the line through x in the direction of v. This line intersects \partial Q at two points, say x^{\pm} (forward and backward). Then the Finsler F(x,\cdot) is given by \displaystyle F(x,v):=\frac{1}{2}\left(\frac{1}{|xx^+|}+\frac{1}{|xx^-|}\right)\cdot|v|. Clearly this definition depends on the shape of Q and the direction of v. So it may not be Riemannian. Note that the Finsler F is reversible. Moreover, the geodesic distance between two points x,y\in Q^o is given by \displaystyle d(x,y)=\frac{1}{2}|\log[a,b,x,y]|, where a,b are the points of intersections of the line from x to y, \displaystyle [a,b,x,y]=\frac{|ax|\cdot |by|}{|ay|\cdot |bx|}.

2. Suspension. Let \alpha:\mathbb{Z}^d\to \text{Diff}(M) be a group action on a manifold M. Then we consider the induced action of \mathbb{Z}^d on the product manifold M\times \mathbb{R}^d, where \hat\alpha(n)(x,v)=(\alpha(n)(x),v-n). Consider the quotient space N=M\times \mathbb{R}^d/\hat\alpha, on which there is a natural \mathbb{R}^d action that comes from shift \beta(u):(x,v)\mapsto (x,v+u). Clearly the later commutes with the \mathbb{Z}^d action and desends to an action \beta on N.


1. Let \phi_t be a flow on X, f=\phi_1 be the time-1 map, \mathcal{M}(\phi) be the set of flow-invariant probability measures, \mathcal{M}(f) be the set of f-invariant probability measures. Clearly \mathcal{M}(\phi)\subset\mathcal{M}(f). On the other hand, for each \mu\in \mathcal{M}(f), the measure \displaystyle I(\mu):=\int_0^1 (\phi_t)_\ast \mu dt is flow invariant. Therefore,
\mathcal{M}(\phi)=\{I(\mu): \mu\in\mathcal{M}(f)\}.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Please log in using one of these methods to post your comment: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: