Łukasz Grabowski
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Registered User
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I'm a research associate at Oxford University
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Mar 12 |
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Algebraicity of the “outer” boundary of the Mandelbrot set I clearly wrote that $\lambda$ and $\mu$ are assumed to be algebraic numbers |
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Mar 11 |
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Algebraicity of the “outer” boundary of the Mandelbrot set (the reason I asked my question for $M$ and not H's butterfly is I'm fairly sure this hasn't been studied for the H's butterfly, and I hoped perhaps it has been studied for $M$) |
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Mar 11 |
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Algebraicity of the “outer” boundary of the Mandelbrot set @Will, if you look at the Hofstadter's butterfly en.wikipedia.org/wiki/… and identify the vertical axis there with the interval [-i,i] and take $\lambda = iq$, where $q\in Q$ then $t_{\lambda,1}\in \bar{Q}$ (it's a trivial observation.) So it seems sensible to ask whether for the butterfly the function e.g. $x\mapsto t_{ix,1}$ maps $\bar{Q}$ to $\bar{Q}$. One definition of H's butterfly is as the parameters for which the orbit of 1 is bounded in a certain dyn. system defined by a linear recursion with non-const. coeff., so it's not too far away from $M$. |
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Mar 11 |
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Algebraicity of the “outer” boundary of the Mandelbrot set added 5 characters in body |
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Mar 11 |
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Algebraicity of the “outer” boundary of the Mandelbrot set @Gerald, I don't know how to do this example. My initial intuition, however, is that there should be some relatively simple description of $t_{\la,\mu}$: if we first look at very large $t$, where the orbit of $0$ is unbounded, then we go to smaller and smaller $t$, suddenly the orbit becomes bounded. I thought this critical parameter should have some relatively easy relation to the polynomial which we iterate (e.g. being a critical point of some function, etc.) Then again, one could phrase a similarly sounding naive intuition about why should the whole Mandelbrot set be a "simple object"... |
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Mar 11 |
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Algebraicity of the “outer” boundary of the Mandelbrot set Thanks for you reply - however if $M$ is the unit disk, then certainly $t$ would be algebraic - the point in question would be in the intersection of two (real) curves $x^2+y^2 =1$ and a line $y=ax+b$, where $a$ and $b$ are algebraic numbers depending on $\mu$ and $\lambda$. So as it stands for the moment, I still think my question makes sense. |
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Mar 11 |
asked | Algebraicity of the “outer” boundary of the Mandelbrot set |
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Feb 8 |
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pointwise ergodic theorem and mean sojourn time added question 2 |
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Feb 8 |
revised |
pointwise ergodic theorem and mean sojourn time typos |
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Feb 8 |
asked | pointwise ergodic theorem and mean sojourn time |
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Dec 16 |
awarded | ● Yearling |
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Dec 9 |
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Are amenable groups topologizable? I essentially stopped thinking about this, because I don't know how to proceed, but I thought I'd share one idea which at one point I thought was hopeful: G acts on a certain metric space X by isometries - namely fix a mean m on G and define X to be the set of subsets of G up to sets of mean 0. The metric on X is d(A,B) = m(A-B \cup B-A). There are various topologies on Isom(X) but I failed to prove any of them gives a Hausdorff non-discrete topology on G. |
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Dec 3 |
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Are amenable groups topologizable? @Simone: AFAI understand, Bohr compactif'n B(G) is in particular compact, so by Peter-Weyl thm if G embeds in B(G) then G is res. linear, and so by Malcev thm if it is fin. generated then it is res. finite. So to produce example of an amenable group G which doesn't embed into B(G) take a fin. generated simple amenable group, or easier take a fin.gen. solvable non-res. finite group (e.g. BS(1,n)). Taking a simple group is more convincing though, because the image in B(G) is trivial, whereas if the image in B(G) is infinite one could still hope for inducing a (Hausdorff) topology on G somehow... |
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Nov 28 |
awarded | ● Nice Question |
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Nov 28 |
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Are amenable groups topologizable? @Ben: you can use profinite topology, e.g. fix a prime number $p$; then integer $n$ is "near" to $0$ if large power of $p$ divides $n$. Other way to topologize integers is - take the action of integers $Z$ on the circle such that the generator $t$ of $Z$ acts by irrational rotation, and define $t^k$ to be "close" to the identity element iff $t^k$ is a rotation by a "small" angle. This is special case of "find an infinite cyclic subggroup of a compact group and induce the topology" |
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Nov 27 |
revised |
Are amenable groups topologizable? added 38 characters in body; edited body |
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Nov 27 |
asked | Are amenable groups topologizable? |
