User jeremy kahn - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:36:56Z http://mathoverflow.net/feeds/user/8252 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95701/the-deep-significance-of-the-question-of-the-mandelbrot-sets-local-connectedness/109036#109036 Answer by Jeremy Kahn for The deep significance of the question of the Mandelbrot set's local connectedness? Jeremy Kahn 2012-10-07T03:02:07Z 2012-10-07T03:02:07Z <p>If a connected compact $K \subset C$ is locally connected then the Riemann map $h\colon C \setminus \Delta \to C \setminus K$ extends continuously to $\partial \Delta$. For each $z \in K$, the boundary of the convex hull of $h^{-1}(\{z\})$ is the union of a set $\Lambda_z$ of chords; the union of these $\Lambda_z$ over all $z \in K$ is a closed set $\Lambda_K$ of disjoint chords; it is called a <em>lamination</em> of $\Delta$. We can reconstruct the convex hulls of each $h^{-1}(\{z\})$ from $\Lambda_K$, and when we collapse every convex hull to a point, we obtain a topological model for $K$. </p> <p>In the case where $K$ is the Mandelbrot set $M$, the lamination $\Lambda_M$ can be described combinatorially, so MLC would mean that we know the topology of $M$. </p> <p>There is a second answer which is more subtle and more important. For each $c \in M$, the filled Julia set $K_c$ of $z \mapsto z^2 + c$ is compact and connected; if it is locally connected the resulting lamination $\Lambda_c \equiv \Lambda_{K_c}$ is, in the right sense, invariant under $z \mapsto z^2$ on $\partial \Delta$. Even if $K_c$ is not locally connected, there is a way of defining what the lamination <em>would</em> be if $K_c$ were locally connected. Every invariant lamination appears as $\Lambda_c$ for <em>some</em> c, and MLC is equivalent to the statement there is a <em>unique</em> $c$ with a given lamination. We think of $\Lambda_c$ as describing the <em>combinatorics</em> of $K_c$, and we think of this uniqueness conjecture as "combinatorial rigidity"---two maps of the form $z \mapsto z^2 + c$ are conformally conjugate (and hence equal) if they are "combinatorially equivalent". </p> <p>(Actually, if $z \mapsto z^2 + c$ has an attracting periodic cycle, then the set of combinatorial equivalent parameters form an open subset of $C$, so the statement of combinatorial rigidity must be suitably modified in that case. It is known that structural stability is open and dense in the family of maps $z \mapsto z^2 + c$, so combinatorial rigidity implies that every $z \mapsto z^2 + c$ in this open and dense set must have an attracting periodic cycle; this is the implication that Eremenko alluded to in his answer. )</p> <p>In this sense MLC is closely analogous to Thurston's Ending Lamination Conjecture (proven by Brock, Canary, and Minsky), which says, broadly speaking, that a finitely generated Kleinian group is determined by the topology of its quotient and the ending laminations of its ends, which are also, when viewed appropriately, invariant laminations of the disk. </p> <p>There is a third answer which is more historical and empirical. We can prove MLC and combinatorial rigidity "pointwise" (or "laminationwise") by proving that for a given invariant lamination $\Lambda$, it appears as the lamination $\Lambda_c$ for a single $c$. This has been done in great many cases, first by Jean-Christophe Yoccoz, and then by Mikhail Lyubich, the author of this post, Genadi Levin, and Mitsuhira Shishikura. To prove this combinatorial rigidity for a given $c$ seems to require a detailed understanding of the geometry of the associated dynamical system, and this almost always leads to further results. So proving MLC would most likely mean having a thorough understanding of the geometry and dynamics of every map $z \mapsto z^2 + c$. </p> http://mathoverflow.net/questions/58815/polynomial-roots-and-convexity/108684#108684 Answer by Jeremy Kahn for Polynomial roots and convexity Jeremy Kahn 2012-10-03T02:46:05Z 2012-10-05T16:21:31Z <p>We can characterize those $P$ for which $\mathrm{Hull}(P) = \mathrm{Conv}(P)$. </p> <p>First suppose that the roots of $P$ do not lie on a line. We prove that $\mathrm{Hull}(P) = \mathrm{Conv}(P)$ if and only if there is an antiderivative $Q$ of $P$ for which $Q = A^2B$, and the roots of $B$ lie in the convex hull of $A$. The "if" is immediate and left to the reader. </p> <p>Note that a root $\beta$ of $P$ can lie on the boundary of $\mathrm{Conv}(\Pi_\omega)$ only if $\beta$ is a root of $\Pi_\omega$, or all the roots of $\Pi_\omega$ lie on a line. This latter case of course is impossible if the roots of $P$ do not lie on a line. </p> <p>Let $\beta$ be a root of $P$. We claim that for every neighborhood $N$ of $\Pi(\beta)$ there is a neighborhood $M$ of $\beta$ such that $\mathrm{Conv}(\Pi_y) \supset M$ for every $y \in \mathbb C \setminus N$. This certainly holds for any given $M$ when $y$ lies outside a large compact subset of $\mathbb C$, so we can think of $y$ ranging over a compact set. For each $y \neq \Pi(\beta)$, there is a ball of positive radius around $\beta$ in $\mathrm{Conv}(\Pi_y)$, and the size of the maximal such ball varies continuously, so the claim follows. </p> <p>Now, suppose that $\beta$ and $\gamma$ are adjacent vertices (extreme points) of $\mathrm{Conv}(P)$. Suppose $\gamma$ is <em>not</em> a root of $\Pi_{\Pi(\beta)}$. Then $\gamma$ lies in the interior of $\mathrm{Conv}(\Pi_{\Pi(\beta)})$, and we can find $\gamma'$ in the interior of $\mathrm{Conv}(\Pi_{\Pi(\beta)})$ so that $\overline{\gamma' \beta} \cap \mathrm{Conv}(P) = \beta$. Then $\overline{\gamma' \beta} \subset \mathrm{Conv}(\Pi_y)$ for all $y$ in a suitable neighborhood $N$ of $\Pi(\beta)$. On the other hand, $M \subset \mathrm{Conv}(\Pi_y)$ for a suitable neighborhood $M$ of $\beta$ and all $y \notin N$. Therefore $\overline{\gamma' \beta} \cap M \subset \mathrm{Hull}(P)$, and hence $\mathrm{Conv}(P) \subsetneq \mathrm{Hull}(P)$.</p> <p>So if $\mathrm{Conv}(P) = \mathrm{Hull}(P)$, then $\Pi(\beta) = \Pi(\gamma)$ for all adjacent vertices $\beta, \gamma$ of $\mathrm{Conv}(P)$. Letting $Q$ be $\Pi(\beta)$ for any extreme point $\beta$ of $\mathrm{Conv}(P)$, we see that every extreme point of $\mathrm{Conv}(P)$ is a root of $Q$, and hence a double root of $Q$. Moveover, if $\mathrm{Conv}(P) = \mathrm{Hull}(P)$, then $\mathrm{Conv}(P) \supseteq \mathrm{Conv}(Q)$, so $Q = A^2B$, where the roots of $B$ lie in the convex hull of the roots of $A$, the extreme points of $\mathrm{Conv}(P)$.</p> <p>Now suppose that the roots of $P$ lie on a line. We can assume that $P$ is real (with positive leading term) and all of the roots of $P$ are real as well. Let $\beta$ and $\gamma$ be the least and greatest root of $P$, respectively. Then, by a similar analysis, $\mathrm{Hull}(P) = \mathrm{Conv}(P)$ if and only if</p> <ol> <li>$\Pi_y$ has all real roots for some value of $y$,</li> <li>The roots of $\Pi_{\Pi(\beta)}$ have real part at least $\beta$, and</li> <li>The roots of $\Pi_{\Pi(\gamma)}$ have real part at most $\gamma$. </li> </ol> <p>You can easily verify that 1, 2, and 3 hold when $P$ has degree 3. I would be tempted to conjecture that 2 and 3 <em>always</em> hold (when the roots of $P$ are real).</p> http://mathoverflow.net/questions/44326/most-memorable-titles/64316#64316 Answer by Jeremy Kahn for Most memorable titles Jeremy Kahn 2011-05-08T19:39:22Z 2011-05-08T19:39:22Z <p><a href="http://www.maths.ed.ac.uk/~aar/papers/exotic.pdf" rel="nofollow">On manifolds homeomorphic to the 7-sphere</a></p> <p>In which Milnor proves there is more than one. </p> http://mathoverflow.net/questions/47421/how-to-calculate-the-feigenbaum-constant-to-high-precision/63534#63534 Answer by Jeremy Kahn for How to calculate the Feigenbaum constant to high precision? Jeremy Kahn 2011-04-30T15:34:45Z 2011-04-30T15:34:45Z <p>One idea for computing the Feigenbaum constant is as follows: $\delta$ is the unique expanding eigenvalue for period-doubling renormalization at its fixed point $F$. So take any real-analytic 1-parameter family of univalent maps that is transverse to the stable manifold of renormalization (for example, the real quadratic family). Then the iterated renormalizations of this 1-parameter family will converge to the unstable manifold of renormalization. The unstable manifold is of course mapped to itself by renormalization and the derivative at the fixed point will be $\delta$. </p> <p>It should be possible to implement this numerically, by keeping track of the power series for a real-analytic family of real-analytic univalent map, and applying the renormalization by replacing $f_\lambda$ with $f_\lambda \circ f_\lambda$ (and suitably rescaling). Because renormalization acts as a contraction on these 1-parameter families, this procedure should be computationally stable, and provide a number of digits proportional to the number of times one renormalizes. </p> <p>One classical reference for the stable/unstable manifold picture for period doubling renormalization is Iterated Maps on the Interval as Dynamical Systems by Collet and Eckmann. </p> http://mathoverflow.net/questions/63178/tricks-of-the-trade Tricks of the Trade Jeremy Kahn 2011-04-27T14:36:41Z 2011-04-27T15:03:42Z <p>Can you name a mathematical theorem that is simple to state and relatively simple to prove, was essential to your research or to a work you found interesting and significant, has the potential to be applied in a wide variety of fields, and is not part of the curriculum of what "every mathematician should know"? </p> http://mathoverflow.net/questions/61034/optimal-countdown/61362#61362 Answer by Jeremy Kahn for Optimal Countdown Jeremy Kahn 2011-04-12T04:15:41Z 2011-04-12T04:15:41Z <p>We can prove that $\log N_k \sim k \log k$ as follows:</p> <p>If we want to combine a set of $k$ numbers using the four arithmetic operations, we can think of inputting the numbers (in any order) along with the operations into an RPN calculator. There are $k!$ ways of ordering the numbers, $C_{k-1} = \frac1{k}{2k-2 \choose k-1}$ ways of choosing places to insert the arithmetic operations (without running out of numbers on the stack) and $4^{k-1}$ ways of choosing which of the four operations we will insert at each place, for a grand total of $4^{k-1}\frac{(2k-2)!}{(k-1)!}$ ways of combining $k$ numbers with the four operations. If we are given $k$ numbers and we can work with any subset of them (as in the original formulation of $N_k$), then there are $$\sum_{i=1}^k {k \choose i} 4^{i-1}\frac{(2i-2)!}{(i-1)!} = \sum_{i=1}^k 4^{i-1} C_{i-1} \frac{k!}{(k-i)!} \le 16^k k! \le (16k)^k$$ ways of choosing a subset and then arranging and combining the elements of the subset with the arithmetic operations. Hence $\log N_k \le k(\log k + \log 16)$. </p> <p>The lower bound is a little bit more interesting. Just by using addition and multiplication, we can prove that $N_{b+r-2} \ge b^r - 1$: We take as our $b + r - 2$ numbers $2, 3, \ldots b-1, 1, b, \ldots b^{r-1}$ (of course we are assuming that $b \ge 2$). Then we can write any positive integer $n &lt; b^r$ as $\sum_{i=0}^{r-1} a_i b^i$, with $0 \le a_i \le b-1$, and then, by collecting the terms with a given "digit" $a_i$, we can write $n$ as a sum of terms of the form $a(b^{i_{a1}} + \ldots + b^{i_{aj_a}})$, where each $a$, $0 \le a \le b-1$, appears at most once. Of course, we can throw out the term with $a=0$, and not write the 1 when $a=1$, so we can write our number with $2, 3, \ldots, b-1, 1, b, \ldots, b^{r-1}$. </p> <p>If we allow subtraction as well we can use Francois's idea (and the same set of numbers) to show that $N_{b+r-2} \ge ((2b - 1) ^ r - 1)/2$ when $b \ge 2, r \ge 1$. </p> <p>Even with only addition and multiplication, we obtain (roughly) $N_k \ge (\epsilon k)^{(1-\epsilon) k}$ for $k$ large given $\epsilon > 0$, and hence $\log N_k \ge (1-\epsilon) k \log k$ when $k$ is large given $\epsilon$. So $\log N_k \sim k \log k$.</p> <p>The next question to ask is whether $N_k^{1/k}/k$ has a limit, and if so, what is is. We have proven that $\limsup N_k^{1/k}/k \le 16$, but we have not even proven that $\liminf N_k^{1/k}/k > -\infty$. </p> http://mathoverflow.net/questions/61034/optimal-countdown/61095#61095 Answer by Jeremy Kahn for Optimal Countdown Jeremy Kahn 2011-04-08T20:13:02Z 2011-04-08T20:13:02Z <p>For convenience, we write $(a_1, \ldots, a_k) \le (b_1, \ldots, b_k)$ when $a_i \le b_i$ for each $i$. When $k = 3$, the solution $(2, 3, 10)$ is optimal among $(n_1, n_2, n_3) \le (10, 20, 100)$. For $k = 4$, you can make $1 \ldots 79$ with $(2, 3, 5, 33)$. This is optimal for $(n_1, n_2, n_3, n_4) \le (2, 4, 6, 50)$. These results were obtained with a perl script I wrote that you can find at www.math.sunysb.edu/~kahn/countdown. The last result, for example, was obtained by running </p> <p>countdown 2 4 6 50</p> <p>and took 7 minutes to run on my MacBook Pro. </p> http://mathoverflow.net/questions/95701/the-deep-significance-of-the-question-of-the-mandelbrot-sets-local-connectedness/95704#95704 Comment by Jeremy Kahn Jeremy Kahn 2012-12-28T05:03:39Z 2012-12-28T05:03:39Z If you draw the Mandelbrot set by coloring a pixel black when the center of the pixel lies in the set, it will appear to be disconnected, and the islands will appear to be islands. This is true even if you draw it at a high resolution. It's only when you color in the pixels where the escape time is his that you see, very clearly, that the Mandelbrot set is connected. http://mathoverflow.net/questions/108676/maximum-of-a-function-of-one-variable Comment by Jeremy Kahn Jeremy Kahn 2012-10-07T01:13:43Z 2012-10-07T01:13:43Z Alexandre, do you mean to say that $D$ is an ideal hyperbolic quadrilateral in the Poincare disk? (In other words, the arcs of $D$ are orthogonal to the unit circle). Otherwise, there is another parameter needed to define $D$, namely the angle that the arcs of $D$ make with the unit circle. http://mathoverflow.net/questions/61034/optimal-countdown/61362#61362 Comment by Jeremy Kahn Jeremy Kahn 2011-04-14T00:31:47Z 2011-04-14T00:31:47Z For example, with the numbers 1, 2, 3, and 4, in order, and the operation *, there are five ways that we can enter them into an RPN calculator: 1 2 * 3 * 4 *, 1 2 3 * * 4 *, 1 2 * 3 4 * *, 1 2 3 * 4 * *, and 1 2 3 4 * * *. In general, there must always be more numbers inputted into the calculator than there are operations applied: this condition is what gives us the Catalan numbers. http://mathoverflow.net/questions/61034/optimal-countdown/61095#61095 Comment by Jeremy Kahn Jeremy Kahn 2011-04-09T23:20:38Z 2011-04-09T23:20:38Z (2, 3, 14, 60) gives 86. This is optimal for $(n_1, n_2, n_3, n_4) \le (5, 10, 20, 100)$.