User marco mazzucchelli - MathOverflow

Generators of local homology groups of an isolated critical point

2012-03-23T19:34:56Z

This is a basic Morse theory question:

Let $M$ be a smooth manifold, and $f:M\to\mathbb{R}$ a smooth function with an isolated critical point $x$. Set $c:=f(x)$. The local homology of $f$ is the relative homology group $ C_{\star} (f,x):=H_{\star} ( \{f < c\}\cup\{x\},\{f < c\})$.

Assume that this group is non-trivial in some degree $d>0$. Let $\mathcal{G}$ be the family of (continuous) singular simplexes $\sigma:\Delta^d\to \{ f < c\}\cup\{x\}$ such that $f \circ \sigma (z) < c $ for all $z \in \partial \Delta^d$. Now, let $\mathcal{G}'\subset C_d(f,x)$ be the set of homology classes represented by the elements of $\mathcal{G}$.

Is $\mathcal{G}'$ a set of generators for $C_d(f,x)$?

The answer is clearly YES when $x$ is a non-degenerate critical point, or when it is a local maximum. With a tiny bit of help from Morse theory, I can also say that it is YES when $d$ is the sum of the Morse index and of the nullity of $x$. Finally, if $d$ is the minimal degree at which the local homology is non-zero, then the relative Hurewicz theorem implies that the local homology $C_{d} (f,x)$ is the same as the local homotopy $\pi_{d} ( \{f < c\}\cup\{x\},\{f < c\})$, and the answer is again YES.

How about the remaining cases? I was hoping that there would be a simple argument from algebraic topology that encompasses all the cases?

Piecewise constant functions with zero average

2011-12-02T16:24:39Z

Consider $0=t_0\leq t_1\leq...\leq t_n=1$, $f_0,...,f_{n-1}\in\mathbb{Z}$ and $F:[0,1]\to\mathbb{R}$ be such that

1) $F\equiv f_i$ on the interval $(t_i,t_{t+1})$, for all $i=0,...,n-1$,

2) $\displaystyle \int_0^1 F(t) dt=\sum_{i=0}^{n-1}(t_{i+1}-t_i)f_i=0$.

Does there exist an arbitrarily large prime number $p$ and a positive integer $k=k(p)$ such that $q:=p^k$ satisfies

$\displaystyle \sum_{i=1}^{q-1} F\left(\frac{i}{q}\right)=0$ ?

I know that the answer is YES when all the $t_j$'s are rational number: if $t_j=\frac{p_j}{q_j}$, then it suffices to choose $q\equiv 1$ mod $\mathrm{lcm}(q_1,...,q_{n-1})$.

Any idea for the general case?

Local homology of degenerate critical points

2010-10-05T11:27:42Z

Given a smooth function $f:M\rightarrow \mathbb R$ on a manifold, its local homology at a critical point $x$ is the group $$ C_\star(x) := H_\star ( M_{ < c} \cup { x } , M_{ < c} ) ,$$ where $H_\star$ denotes singular homology (with any coefficient group), $c=f(x)$, and $M_{ < c}$ is the space of those points $x\in M$ such that $f(x) < c$.

If $x$ is a non-degenerate critical point, then $C_\star(x)$ is completely determined by the Morse index of $f$ at $x$: the group $C_j(x)$ is equal to the coefficient group of the homology for $j=\mathrm{ind} (x)$, and is trivial for other values of $j$.

If $x$ is degenerate, the knowledge of $\mathrm{ind}(x)$ and $\mathrm{nul}(x)$ (this latter being the nullity of $f$ at $x$) is not enough to determine $C_\star (x)$. It is easy to build examples of functions on $\mathbb R^2$ having a critical point $x$ with local homology $C_1(x)= G\oplus ...\oplus G$ ($k$ times, where $k>1$ and $G$ is the coefficient group) and $C_j(x)=0$ for $j\neq 1$. For instance, consider the function $f:\mathbb R^2\rightarrow\mathbb R$ given by $$ f(x,y)=(y-2x^2)(y-x^2)(y+x^2)(y+2x^2). $$ Here, the origin is a critical point whose local homology (say, with $\mathbb Z_2$ coefficients) should be $C_1(0)=\mathbb Z_2\oplus\mathbb Z_2\oplus \mathbb Z_2$ and $C_j(0)=0$ for $j\neq 1$.

Does anybody know examples of functions having critical points whose local homology is nonzero in more then one degree?

If the answer to the previous question is yes (as I would expect), is it true that given $(n_1,d_1), ..., (n_r,d_r)$ there exists a function $f:M\rightarrow\mathbb R$ with a critical point $x$ whose local homology is given by $C_{d_j}(x)=G^{ \oplus n_j }$ and $C_d(x)=0$ for $d\neq d_1,...,d_r$?

Special Hamiltonian diffeomorphisms

2010-10-04T22:03:01Z

Is there any obstruction that prevents a Hamiltonian diffeomorphism on some symplectic manifold to be realized as the time-one map of the Hamiltonian flow of an autonomous Hamiltonian?

In the same spirit, is there any obstruction that prevents a Hamiltonian diffeomorphism on $T^*M$ (with the canonical symplectic structure) to be realized as the time-one map of the Hamiltonian flow of a Tonelli Hamiltonian (i.e. a Hamiltonian which is fiberwise convex and superlinear)?

Answer by Marco Mazzucchelli for When is the time evolution of a Hamiltonian system described by the geodesic flow on a Riemannian manifold?

2010-10-02T19:40:21Z

I wish to add an $\epsilon$ to the previous answer. Assume that your symplectic manifold is $T^*M$, where $M$ is a closed manifold, and consider a so called ``Tonelli Hamiltonian´´ $H:T^*M\rightarrow\mathbb R$, which is simply a function which is fiberwise (differentiably) convex and superlinear. Then consider the value $$ c(H)=\min_{u\in C^\infty(M;\mathbb R)} \max_{q\in M} H(q,du_q). $$

This special number in the literature is called ``Mañé critical value´´. Now, if you fix an energy value $h > C(H)$, you can easily build a new Hamiltonian $G$ such that

1) $H^{-1}(h)=G^{-1}(h')$ for some $h'\in\mathbb R$

2) $G(q,\lambda p)=\lambda^2 G(q,p)$ for each $\lambda>0$

3) $G$ is fiberwise convex

Now, the Legendre dual of the Hamiltonian $G$ will be a Finsler metric (not Riemannian in general).

The interesting thing here is that the critical value $c(H)$ can be defined in many other equivalent ways (in terms of Lagrangian action of closed loops, in terms of minimizing invariant measures, etc.). You can find more on this in the book by Contreras and Iturriaga: http://www.cimat.mx/~gonzalo/libro/lagrangians.pdf

Comment by Marco Mazzucchelli

2011-12-04T04:41:05Z

Thanks a lot GH!!! In Hamiltonian dynamics, this theorem should implies the following: if the average Maslov index of an orbit $\gamma$ is zero, then for infinitely many prime numbers $p$ the $p$-iteration of the orbit gamma has the same Maslov index as $\gamma$. Maybe one may get something useful out of this...

Comment by Marco Mazzucchelli

2011-12-02T22:25:35Z

@Greg It's not known to me. I would be happy with that already.

Comment by Marco Mazzucchelli

2011-12-02T17:10:48Z

Yes, all the $f_j$'s must be integer.

Comment by Marco Mazzucchelli

2011-12-02T16:32:43Z

Sorry, that was a typo.