User michele triestino - MathOverflow

Answer by Michele Triestino for Orbits of Thompson's group

2013-04-12T17:41:34Z

If you mean Thompson's $F$, please, specify the action on the circle.

If you mean Thompson's group $[F,F]$, the answer is yes, as it acts on the interval.

I think you are talking about Thompson's group $T$, and its dynamics on the circle was well described by Ghys and Sergiescu in

Sur un groupe remarquable de difféomorphismes du cercle.

The answer to your question is however rather trivial : the group contains every dyadic rotation, which act minimally on the circle. So every finite orbite must be dense, which is impossible.

Answer by Michele Triestino for Numerical calculation of Arnold tongue

2012-12-14T14:07:32Z

There is a good way to compute rotation number of a circle homeomorphism (this was the way Poincaré thinked of it): you calculate the rotation number buy its continued fraction in a direct way.

You start from a point $x$ and $f(x)$: this gives you a decomposition of the circle into points that are on the right side of $x$ (in $]x,f(x)[$) and points which are on its left side (in $]f(x),x[$). You look at $f^2(x)$ and you write $R$ if it is on the right side of $x$, $L$ otherwise. Iteranting $f$ you find a sequence of $R$'s and $L$'s. If you get $LLLLR$, for example, you record 4 (this is the number of $L$'s) and you approximate the rotation number of $f$ by $1/4$.

Renormalizing $f$, you iterate this process finding $\rho=[0,a_1,a_2,\ldots,a_k]$.

I won't be more precise here.

Every detail is very well explained in de Melo & van Strien's One-Dimensional Dynamics, section I.1.

You can find a paper by Bruin (Numerical determination of the continued fraction expansion of the rotation number) in which he compares different methods on Arnold tongues.

EDIT[update]: Recently, I wrote for myself some sage lines implementing the algorithm I described you. This is my second version, now working for rational numbers too : I was originally interested only in irrational rotation numbers (comments are welcome to improve it!).

L=8 #length for cf-expansion, depending on your computer, 8 or 9 suggested for a try run
A=100000 #maximum size of single element of the sequence

def partfrac(x):
    return x-floor(x)

##### computing rational approximations given continued fraction expansion
# input b=a continued fraction expansion
# input l=L length of computed expansion
def rational_approximation(b,l):
    p=[0,1]
    q=[1,b[1]]
    for i in range(1,l+1):
        p.append(b[i+1]*p[i]+p[i-1])
        q.append(b[i+1]*q[i]+q[i-1])
    return simplify(p[l+1]/q[l+1])


#computing rotation number of a given circle map f
def rotation(f):
    a=[0]
    orbit=[]
    orbit.append(partfrac(f(0)))
    if orbit[0]==0 :
        print 'map with a fixed point'
        return 0

    def shift(x):  #set f(0) as the origin + 1
        if partfrac(x)>orbit[0]:
            return partfrac(x)-1
        return partfrac(x)

    def first_return(p,pre_p,y):
        x=shift(f(y))
        while x<pre_p or x>p:
            x=shift(f(x))
        return x

    a.append(1)
    x=orbit[0]

    if shift(f(orbit[0]))==0:
        print 'map with periodic point of order 2'
        return 1/2

    if shift(f(orbit[0]))<0:
        while shift(f(x))<0:
            a[1]=a[1]+1
            x = shift(f(x))
            if a[1]>A:
                print 'approximatively 0'
                return 0
            if shift(f(x))==0:
                print 'periodic point'
                a[1]=a[1]+1
                return 1/a[1]
        orbit.append(shift(x))
        z = shift(f(x))
        a.append(0)
        while z>0:
            y = z
            z = first_return(shift(orbit[0]),shift(orbit[1]),z)
            a[2]=a[2]+1
            if a[2]>A:
                print 'approximatively rational'
                return 1/a[1]
            if z==0:
                print 'periodic point'
                a[2]=a[2]+1
                return rational_approximation(a,1)
        orbit.append(y)

    if shift(f(orbit[0]))>0:
        def shift(y):  #set f(0) as the origin
            if partfrac(y)>=orbit[0]:
                return partfrac(y)-1
                return partfrac(y)
        orbit.append(orbit[0]-1)
        a.append(0)
        while shift(f(x))>0:
            a[2] = a[2] + 1
            x = shift(f(x))
            if a[2]>A:
                print 'approximatively rational'
                return 1/a[1]
            if shift(f(x))==0:
                print 'periodic point'
                a[2]=a[2]+1
                return rational_approximation(a,1)
        orbit.append(shift(x))    
        z = shift(f(x))

    for i in range(1,L):
        a.append(0)
        if shift(orbit[i+1])<shift(orbit[i]):
            while z>0:
                y = z
                z = first_return(shift(orbit[i]),shift(orbit[i+1]),z)
                a[i+2]=a[i+2]+1
                if a[i+2]>A:
                    print 'approximatively rational'
                    return rational_approximation(a,i)
                if z==0:
                    print 'periodic point'
                    a[i+2]=a[i+2]+1
                    return rational_approximation(a,i+1)
        if shift(orbit[i+1])>shift(orbit[i]):
             while z<0:
                 y = z
                 z = first_return(shift(orbit[i+1]),shift(orbit[i]),z)
                 a[i+2]=a[i+2]+1
                 if a[i+2]>A:
                     print 'approximatively rational'
                     return rational_approximation(a,i)
                if z==0:
                     print 'periodic point'
                     a[i+2]=a[i+2]+1
                     return rational_approximation(a,i+1)
         orbit.append(y)

    print a
    return rational_approximation(a,L)

Counting $(n,k)$-forests of binary trees

2010-07-13T15:21:52Z

Given $k,n\in\mathbf{N}$ with $n\ge k$, define the set $\mathcal{F}(k,n)$ of $(k,n)$-forests of binary rooted trees (where a $(k,n)$-forest is a collection of $k$ rooted trees, which have a totality of $n$ leaves).

My aim is to count the cardinality of $\mathcal{F}(k,n)$.

For example it is well known that $|\mathcal{F}(1,n+1)|$ is given by the $n$-th Catalan number $c_n=\frac{1}{n+1}{{2n}\choose{n}}$.

Using generating functions (and something like Maple), we have an explicit way to obtain any $f_{k,n}:=|\mathcal{F}(k,n)|$: if $C(z)$ denotes the generating function of $c_n$, then the $n-k$-th coefficient of $C(z)^k$ is exactly $f_{k,n}$.

Isn't there any explicit formula for such numbers?

Is there an asymptotic estimate and is it a good one?

Answer by Michele Triestino for Good probability measues on $S^1$ reprented by a kernel

2011-03-22T08:45:47Z

Perhaps this is not exactly what you are expecting, but it is anyway interesting.

In topological dynamics, when you have a compact group $\Gamma$ acting on the circle with Haar measure $dg$, you define a $\Gamma$-invariant probability measure on $S^1$ by the formula $$\mu(A):=\int_\Gamma Leb(gA)dg$$ where $A$ is any Borel subset of $S^1$ and $Leb$ denotes the Lebesgue measure on the circle.

This is an easy exercise, but if you want to find more, one possible reference is Navas' survey on Groups of circle diffeomorphisms

Answer by Michele Triestino for How about this book Topological Methods in Group Theory

2011-01-17T12:22:04Z

I can just review the first half of the book. In order, the short introduction of general topology is too short, a good reference would have been preferable. For the following, the theory of CW complexes requires a tough notation and technicalities in proofs and definitions. Furthermore, proofs are not so easy to read, since Geoghegan uses frequently expressions like "using Theorem 3.14.5 we get that..." so you are obliged to jump back many many times.

Though, I've found this book pretty good. The spirit is always geometric and you can find a good amount of interesting results of the last three decades.

Answer by Michele Triestino for Is Thompson's Group F amenable?

2011-01-17T11:58:53Z

Shavgulidze's back.

A new "proof" is now on the arxiv.

The first part contains what was correct in the previous preprint. In the last 5 pages he tries to give a suitable proof for the group $F$. The point is always the same: he looks for a good density on the space of partitions of the interval. The style is always the same: a few pages full of maths, with no explication.

Answer by Michele Triestino for Most memorable titles

2010-10-31T20:40:35Z

A Midsummer Knot's Dream, by Allison Henrich, Noël MacNaughton, Sneha Narayan, Oliver Pechenik, Robert Silversmith, Jennifer Townsend

It is quite funny to read

Answer by Michele Triestino for Which group does not satisfy the Tits alternative?

2010-09-14T16:22:18Z

There are also the Burnside's groups $B(m,n)$ for $n\ge 665$ odd: they are of exponential growth and have the law $x^n=1$ so that they cannot contain any free subgroup on two generators. The fact that they are not solvable follows by the theorem of Rosenblatt:

"A f.g. solvable group is of exponential growth if and only if it contains a free sub-semigroup on two generators."

You can find details on paragraphs VII.C.27/28 of Pierre de la Harpe's book "Topics in Geometric Group Theory" (Chicago Lectures in Mathematics, 2000)

Burnside's Lemma and Geometry

2010-09-20T18:32:26Z

I think one of the most interesting results in Elementary Group Theory is the so-called "Burnside's Lemma", counting the numbers of orbits of a (finite) group action.

I wonder if there is any (interesting) application in Elementary Geometry (I mean Euclidean, hyperbolic or elliptic geometry).

Searching on Google, I've found the article "Applying Burnside’s lemma to a one-dimensional Escher problem" by T. Pisanski, but it sounds to me rather a combinatorial result.

Examples of non-diffeomorphic smooth manifolds with diffeomorphic tangent bundle

2010-07-13T12:06:58Z

Given a smooth manifold $M$, we define the differentiable structure on $TM$ in the usual way.

I would like to know examples of two smooth manifolds which are non-diffeomorphic, but their tangent bundles are.

Which is the smallest dimension in which one can find such examples?

What if I ask the same question for $k$ pairwise non-diffeomorphic manifolds?

Can we have $k=\infty$?

Comment by Michele Triestino

2013-04-17T08:29:12Z

Classical Gottschalk-Hedlund's Theorem (see Theorem 2.9.4 in Katok-Hasselblatt) tells you that if you have a recurrent point then your cocycle is a coboundary: $\phi_n(x)=u(T^n(x))-u(x)$, with $u$ continuous function on $X$.

Comment by Michele Triestino

2013-03-09T12:19:25Z

You can also give a look on Vincent Borelli's page math.univ-lyon1.fr/~borrelli/Enseignement.html But every material is in French!

Comment by Michele Triestino

2013-01-30T12:24:30Z

When considering the one-dimensional torus, you have lots of interesting examples. The firsts coming to my mind are Fuchsian groups (of first kind) or the action of Thompson's group $G$. In the differentiable world there are many beautiful results and conjectures, as you can read in a paper by Deroin, Klepstyn and Navas (arxiv.org/pdf/0806.1974.pdf)

Comment by Michele Triestino

2012-08-30T07:25:09Z

Maybe I would add the word "locally"... and from a mathematical point of view, I would say that $\mathbb{R}$ is not only an algebraic construction (that would maybe be true for algebraic numbers), but you are strongly considering it's topology.

Comment by Michele Triestino

2012-02-02T17:39:53Z

@Marty Thank you for your comment, that's why I wrote "so-called"

Comment by Michele Triestino

2011-03-23T08:14:44Z

I think you ask $\pi_n(f)=0$ for any $n$, isn't it?

Comment by Michele Triestino

2010-09-27T10:47:41Z

Gromov gave an example of a f.g. infinite torsion group acting on a space of nonpositive curvature in §4.5C of Mikhail Gromov, Hyperbolic groups. Essays in group theory, 75--263, Math. Sci. Res. Inst. Publ., 8, Springer, New York, 1987

Comment by Michele Triestino

2010-09-26T17:56:46Z

Evidently, I was thinking of a real vector space.. Are you interested in complex Hilbert spaces?

Comment by Michele Triestino

2010-09-26T17:50:33Z

Just a few questions: Why do you put so strict hypothesis (i.e. separability and Hilbert)? What about the finite-dimesional case (dim>1)? What about Banach spaces? And if we remove completeness?

Comment by Michele Triestino

2010-09-25T16:09:13Z

This is not the more appropriate place where to post such a question. Please, post it in math.stackexchange.com

Comment by Michele Triestino

2010-09-24T23:41:43Z

That's really astonishing me.. I was thinking of a negative answer, since the exponential map $\mathrm{Vect}(S^1)\longrightarrow\mathrm{Diff}(S^1)$ is not locally surjective (see J. Milnor "Remarks on infinite-dimensional Lie groups"). So the key point is exactly the conjugation you have shown!

Comment by Michele Triestino

2010-09-24T19:17:17Z

and yet another silly question: does the same conclusion of my previous comment hold for discrete spaces (I'm thinking of graphs, trees actually)?

Comment by Michele Triestino

2010-09-24T18:54:30Z

Hi Pablo! Just to see if I've got something.. If X is compact there are no horofunctions, are there?

Comment by Michele Triestino

2010-09-24T18:41:23Z

Maybe you are interested in this en.wikipedia.org/wiki/Extra_special_group

Comment by Michele Triestino

2010-09-24T07:48:17Z

thank you! you are right