User otis chodosh - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:13:17Z http://mathoverflow.net/feeds/user/1540 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129447/a-simple-and-good-reference-about-solitons/129477#129477 Answer by Otis Chodosh for A simple and good reference about solitons Otis Chodosh 2013-05-03T01:39:44Z 2013-05-03T01:39:44Z <p>The answers so far seem to be about "solitons" in general which just means a "self similar solution to some PDE." Ricci solitons meet this criteria, but in case you'd like more <em>Ricci soliton</em> focused materials, the following might be of some use:</p> <p>There is a recent survey of Cao which contains quite a lot of information, and references: <a href="http://arxiv.org/pdf/0908.2006v1.pdf" rel="nofollow">http://arxiv.org/pdf/0908.2006v1.pdf</a></p> <p>You may enjoy Hamilton's "Formations of Singularities in the Ricci Flow" <a href="http://www.ams.org/mathscinet-getitem?mr=1375255" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=1375255</a></p> <p>The work of Feldman--Ilmanen--Knopf where they construct rotationally symmetric expanding/shrinking Kahler-Ricci solitons on line bundles over projective space contains quite a bit of nice intuition and explanation about the solitons. You can find the paper here" <a href="ftp://134.76.12.4/pub/EMIS/journals/NYJM/JDG/p/2003/65-2-1.pdf" rel="nofollow">ftp://134.76.12.4/pub/EMIS/journals/NYJM/JDG/p/2003/65-2-1.pdf</a></p> <p>Finally, I'll mention an interesting (quite recent) result of Brendle, proving that the "Bryant Soliton" (constructed by Bryant in the note: <a href="http://www.math.duke.edu/~bryant/3DRotSymRicciSolitons.pdf" rel="nofollow">http://www.math.duke.edu/~bryant/3DRotSymRicciSolitons.pdf</a>) which is a rotationally symmetric steady soliton on $\mathbb{R}^3$, is the unique non-flat, 3d, $\kappa$-noncollapsed steady Ricci soliton. You can see this paper here: <a href="http://arxiv.org/pdf/1202.1264.pdf" rel="nofollow">http://arxiv.org/pdf/1202.1264.pdf</a> </p> http://mathoverflow.net/questions/96343/the-left-hand-side-of-the-ricci-flow-equation-at-the-initial-value/96363#96363 Answer by Otis Chodosh for the left hand side of the Ricci flow equation at the initial value Otis Chodosh 2012-05-08T19:16:36Z 2013-02-19T16:46:03Z <p>Misha's comment could be a bit misleading. In particular, it is not true that the Ricci flow should exist on a slightly bigger interval $(-\epsilon,T)$ with $g(0) = g_0$. One way to see this is by thinking about a theorem of Bando (see <a href="http://www.springerlink.com/content/v0764574t4764138/" rel="nofollow">http://www.springerlink.com/content/v0764574t4764138/</a> if you have access) which says that if $(M,g_t)$ is a solution for Ricci Flow on the interval $[0,T)$, then $g_t$ is real analytic with respect to the normal coordinate charts on $M$ for $t>0$. In particular, if $g_0$ was not real analytic, we cannot extend the flow backwards for any $\epsilon>0$ (the comment was only in charts, but by compacntess if we can do it in charts, we can do it on the whole manifold for some small $\epsilon>0$) because then Bando's theorem would imply that $g$ were real analytic.</p> <hr> <p>The correct statement is just like for the heat equation. We say that $f$ is a solution to the heat equation $$\frac{\partial f}{\partial t} = \Delta f$$ on $[0,T)$ if the above equation is satisfied for $t>0$ and $\lim_{t\searrow 0} f= f_0$. In particular, there is no "meaning" of the equation at $t=0$, only for $t>0$. Do not get confused by trying to apply ODE intuition to the PDE. Parabolic equations are not like ODE's in the sense that you can just "go in the direction of $\Delta f$".</p> <hr> <p>So, for completeness, here is what it means to be a solution to RF on the interval $[0,T)$ with initial data $g_0$:</p> <p>The metric $g_t$ is smooth for $t\in (0,T)$ and for such $t$, $g_t$ satisfies $$\frac{\partial g_t}{\partial t} = -2Ric_{g_t}.$$ You can think of this either in local coordinate charts, as Misha does, or just as a coordinate free equation for the symmetric 2-tensor $\frac{\partial}{\partial t} g_t$.</p> <p>Furthermore, we require that $g$ is continuous up to $t=0$ (because here we're only interested in solving RF with smooth initial data, if we wanted to start with rough data, we'd require a limit in some sense) and $$g_{t=0} = g_0$$ at each point in $M$. </p> <hr> <p>If you're confused, you should read up on the heat equation first. Its <em>exactly</em> the same. In particular, after reading about the heat equation, you should read about the De Turk trick, which transforms the RF into a strongly parabolic equation (i.e. heat-type equation) by fixing the diffeomorphism gauge. A quick google suggests the following chapter <a href="http://www.springerlink.com/content/0t673151r72133r7/" rel="nofollow">http://www.springerlink.com/content/0t673151r72133r7/</a> as a possible reference. Any book on Ricci Flow should have a good description of this.</p> http://mathoverflow.net/questions/122308/when-a-riemannian-manifold-is-of-hessian-typ/122321#122321 Answer by Otis Chodosh for When a Riemannian manifold is of Hessian Typ Otis Chodosh 2013-02-19T16:42:10Z 2013-02-19T16:42:10Z <p>I might be misunderstanding the question, but I believe that if there are functions $h$ and $k$ so that $$\text{Hess}_h = kg,$$ then $(M,g)$ must be (at least locally) a warped product $(a,b) \times_f N^{n-1}$. This follows from integrating along flowlines of $\nabla h$, to compare the induced metrics on different level sets of $h$. </p> <p>I don't know if this is was the first proof of this result, but the result I've stated above is proven (and discussed a bit more than I have here) in Cheeger-Colding's paper "Lower Bounds on Ricci Curvature and Almost Rigidity of Warped Products" on p 192-194 in <a href="https://www.math.sunysb.edu/~pedro/683s10/References/Cheeger-Colding%20-%20Ricci%20curvature%20and%20almost%20rigidity%20of%20warped%20products.pdf" rel="nofollow">this copy</a> of the paper.</p> http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120066#120066 Answer by Otis Chodosh for Geometric picture of scalar curvature Otis Chodosh 2013-01-28T00:06:51Z 2013-01-29T22:15:07Z <p>I'll add a few things which I've found helpful to get intuition to what Renato's already written. </p> <p>Scalar curvature has the very simple geometric interpretation as the volume defect of small balls (as Wikipedia says) $$Vol(B_r(p)) = Vol^{\mathbb{R}^n}(B_r^{\mathbb{R}^n}(0)) \left( 1- \frac{R_p}{6(n+1)} r^2 + O (r^2) \right)$$ as $r\to 0$. </p> <p>It's an instructive exercise to try to prove this. Hint: see problem 3 &amp; 4 in <a href="http://www.msri.org/c/document_library/get_file?uuid=1ec12439-821c-42f5-865f-130b81a7be7a&amp;groupId=14404" rel="nofollow">this problem set</a> [LINK SEEMS TO BE BROKEN -- if anyone wants a copy they can send me an email] from a <a href="http://www.msri.org/web/msri/scientific/workshops/summer-graduate-workshops/show/-/event/Wm9228" rel="nofollow">MSRI summer school</a>. I think the problems are due to Justin Corvino.</p> <p>As Renato has already discussed a bit, the question of "how flexible" scalar curvature is a great place to try to get some intuition. You might see:</p> <p><a href="http://www.ihes.fr/~gromov/PDF/1%5B77%5D.pdf" rel="nofollow">These notes</a> of Gromov, see p 87. The whole thing is great for all sorts of curvature intuition.</p> <p><a href="http://www.math.uchicago.edu/~shmuel/QuantCourse%20/Guthsystmet.pdf" rel="nofollow">This article</a> by Larry Guth, specifically ch 3. He discusses the "Geroch Conjecture" which asks if the $n$-dimensional torus admits a metric of positive scalar curvature (in two dimensions, it clearly does not, by Gauss-Bonnet). He gives references to the Schoen-Yau proof of the positive mass theorem and the Geroch Conjecture, both of which are great things to look at. </p> <p>As for "difference between Ricci and scalar curvature" one might consider the questions of which 3-manifolds admit metrics positive scalar/Ricci curvature. By <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214436922" rel="nofollow">Hamilton - Three-manifolds of positive Ricci curvature</a>, only compact manifolds which are covered by $S^3$ can admit metrics of $Ric > 0$. On the other hand, the work of Perelman shows that the $3$-manifolds with positive scalar curvature are exactly connect sums of $S^3,S^3/\Gamma, S^2\times S^1$. See <a href="http://arxiv.org/pdf/0907.2444v2.pdf" rel="nofollow">this paper</a> of Fernando Marques. </p> <p>See also <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183553962" rel="nofollow">the Yamabe problem</a>. The Yamabe problem is whether or not you can conformally change a metric so that it has constant scalar curvature. The answer is yes, by the combined efforts of Yamabe (whose original solution had an error), Trudinger (who found the error and initiated the path towards the eventual resolution), Aubin (who solved the problem for $n\geq6$ and not locally conformally flat) and finally Schoen (who dealt with the other cases with an amazing solution using ideas from general relativity). </p> <hr> <p>Some more:</p> <p>In addition to the Geroch conjecture work by Schoen-Yau (see also the works of Gromov-Lawson <a href="http://www.ams.org/mathscinet-getitem?mr=569070" rel="nofollow">here</a> and <a href="http://www.ams.org/mathscinet-getitem?mr=577131" rel="nofollow">here</a>), there have been a huge number of work on the question of which manifolds admit positive scalar curvature. See <a href="http://users.ictp.it/~pub_off/lectures/lns009/Stolz/Stolz.pdf" rel="nofollow">this survey</a> paper of Stolz. In addition, you might read the part about his conjectures on positive Ricci curvature. In spite of the fact that $Ric>0$ seems like a much stronger condition than $R>0$. However, there are no known examples of simply connected manifolds admitting positive scalar curvature but not positive Ricci curvature! In order to understand Stolz's conjectures you'll probably want to read about the spin-theoretic obstruction to positive scalar curvature. </p> <p>In particular, due to Schoen-Yau/Gromov-Lawson if two manifolds admits metrics of positive scalar curvature, then their connect sum does as well. As far as I know, the corresponding result is not known for positive Ricci.</p> <p>You also asked about positive curvature operator vs positive Ricci. For works on positive curvature operator, you should of course start with Hamilton's <a href="http://intlpress.com/JDG/archive/1986/24-2-153.pdf" rel="nofollow">paper</a> "4-manifolds with positive curvature operator." and then the later works of <a href="http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_32.pdf" rel="nofollow">Bohm-Wilking</a> and then <a href="http://math.stanford.edu/~brendle/icm2010.pdf" rel="nofollow">Brendle-Schoen</a> to see that positive curvature operator (and other conditions, e.g. positive isotropic curvature, $1/4$-pinched sectional curvature) is very restrictive. See Stolz's paper (section 4.6) for some examples of $Ric>0$, and you can find some examples admitting $Ric>0$ but not positive curvature operator. </p> <hr> <p>Of course, this answer is quite skewed towards the "global comparison geometry" point of view. One can also ask about analytic/metric consequences of curvature bounds. It turns out here, you probably need at least Ricci lower bounds, scalar curvature bounds are not really enough, as far as I know. See <a href="http://www.math.ucsb.edu/~wei/paper/06survey.pdf" rel="nofollow">this survey</a> (which also discusses some of the things above) or <a href="http://library.msri.org/books/Book30/files/colding.pdf" rel="nofollow">this one</a>.</p> <hr> <p>Finally, you might be interested in the work of <a href="http://math.berkeley.edu/~lott/LottVillani.pdf" rel="nofollow">Lott-Villani</a> and <a href="http://www-wt.iam.uni-bonn.de/~sturm/papers/paper47.pdf" rel="nofollow">Sturm</a>, who (building on work of many people, who I won't try to cite but you can read their papers for this) show that Ricci lower bounds have an amazing interpretation in the "metric-measure" sense. They show that Ricci lower bounds can be detected by "convexity of the entropy functional" as smooth measures "move along geodesics." Furthermore, this property is preserved under measured <a href="http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence" rel="nofollow">Gromov-Hausdorff convergence</a> (which is a very weak! On the other hand, of course one could try to to show that scalar curvature lower bounds are preserved under Gromov-Hausdorff convergence, but as far as I know this is an open problem. </p> http://mathoverflow.net/questions/117460/intuition-for-mean-curvature/117477#117477 Answer by Otis Chodosh for Intuition for mean curvature. Otis Chodosh 2012-12-29T05:20:24Z 2012-12-29T05:28:12Z <p>Perhaps a more geometric way of viewing mean curvature (at least for a hypersurface) is as follows: </p> <p>If $\varphi: \Sigma^n\hookrightarrow (M^{n+1},g)$ is an embedded (oriented) hypersurface, then we can naturally extend a family of embeddings by choosing a unit normal $\nu$ and flowing by unit speed, i.e.</p> <p>$$\varphi_t(x) := \exp_x(t \nu(x))$$</p> <p>For small enough $t \in (-\epsilon,\epsilon)$, this still gives an embedding $\varphi_t: \Sigma\to M$. Now, one could pull back the metric $g$ by $\varphi$ to obtain a metric on $\Sigma$. This metric gives a (Riemannian) volume form on $\Sigma$, $d\mu_t$. Then, </p> <p>$$\frac{\partial}{\partial t} d\mu_t = nH d\mu_t$$</p> <p>(I think this matches the normalization in @Joseph O'Rourke's answer, but of course the sign changes with the choice direction of the normal)</p> <p>This follows from the first variation formula. </p> <p>In other words, the mean curvature measures how the volume of the submanifold (locally) changes under flowing the surface in the direction of the unit normal.</p> http://mathoverflow.net/questions/114318/fixed-point-arguments-in-pde/114346#114346 Answer by Otis Chodosh for fixed point arguments in PDE Otis Chodosh 2012-11-24T16:44:14Z 2012-11-24T16:44:14Z <p>The <a href="http://en.wikipedia.org/wiki/Minimal_surface_equation" rel="nofollow">graphical minimal surface equation</a> is a great example of a PDE where Leray-Schauder fixed point theory is applied:</p> <p>$$\left(\delta_{ij} - \frac{D_i u D_j u}{1+|Du|^2}\right)D_{ij}u = 0$$</p> <p>This represents the condition that $graph(u)$ is a minimal surface, or in other words is a critical point for the area functional.</p> <p>For surfaces in $\mathbb{R}^3$, existence for the a slightly more general problem (<a href="http://en.wikipedia.org/wiki/Plateau%27s_problem" rel="nofollow">Plateau's problem</a>) was established by Douglass-Rado in the 1930's, using beautiful conformal methods. However, the higher dimensional problem required the introduction of the Leray-Schauder fixed point theorem. One version of this says</p> <blockquote> <p>For $\mathscr{B}$ a Banach space, and $T : \mathscr{B} \times [0,1] \to \mathscr{B}$ continuous, compact and so that $T(x,0) = 0$ for all $x \in \mathscr{B}$. Suppose also that there is some $M > 0$ so that if $x = T(x,\sigma)$ for $(x,\sigma) \in \mathscr{B} \times[0,1]$ then $$\Vert x \Vert_\mathscr{B} \leq M.$$ Then, there is a fixed point for $T(\cdot, 1)$. </p> </blockquote> <p>One of the reasons that I find this theorem very neat is because in some sense it treats a priori estimates as just that! In other words, you <em>never</em> need to show a solution to some relaxed equation holds or something like that, you just have to show that if there is a solution, then you have some estimates (and of course compactness). </p> <p>To use this, we define the operator $\hat T : C^{1,\beta}(\overline \Omega) \times [0,1] \to C^{2,\beta'}(\overline\Omega)$ as the solution operator to the linear PDE, solving for some $v$ satisfying $$\left(\delta_{ij} - \frac{D_i u D_j u}{1+|Du|^2}\right)D_{ij}v = 0 \text{ in \Omega}$$ $$v = \sigma \varphi \text{ on \partial\Omega}$$ Linear existence theory shows that this map is well defined and if we then compose it with the map $C^{2,\beta'} \hookrightarrow C^{1,\beta}$, we have a map $T: C^{1,\beta} (\overline\Omega) \times [0,1] \to C^{1,\beta} (\overline\Omega)$, which is compact because the inclusion $C^2\to C^{1,\beta}$ is.</p> <p>Furthermore, if $\sigma =0$, the $0$ solution clearly works.</p> <p>Thus, to prove that Leray-Schauder applies, one must show that the a priori estimate holds. This is a bit delicate, so I won't go into the details, but only remark that one needs to assume some geometric conditions on the boundary (mean convexity). If you're interested, you can find the details in Gilbarg-Trudinger, starting with 11.3 but sort of jumping around for the various bounds. </p> http://mathoverflow.net/questions/114167/ricci-flow-descending-from-an-universal-cover/114211#114211 Answer by Otis Chodosh for Ricci flow descending from an universal cover Otis Chodosh 2012-11-23T05:29:47Z 2012-11-23T05:29:47Z <p>I was going to post this as a comment, but it got too long. I'm not exactly sure how to answer the question as written, but the following might be enlightening: </p> <p>A crucial piece of information for your Ricci flow is that you start with bounded curvature, i.e. $$\sup_M |Riem|_{g(0)} &lt; \infty.$$ Then, according to <a href="http://www.intlpress.com/JDG/archive/1989/30-1-223.pdf" rel="nofollow">Shi (Deforming the metric on complete Riemannian manifold, JDG 1989)</a> there <em>exists</em> a Ricci flow $g(t)$ with bounded curvature, $g(t)$, defined for $t \in [0,T)$ where for $t \in [0,T)$ $$\sup_M |Riem|_{g(t)} &lt; \infty.$$ Shi makes no claims about uniqueness.</p> <p>Later, it was proved by <a href="http://arxiv.org/abs/math/0505447" rel="nofollow">Chen-Zhu (Uniqueness of the Ricci Flow on Complete Noncompact Manifolds)</a> that if $g_1(t), g_2(t)$ are solutions to Ricci flow for $t \in [0,T)$ and they <em>both</em> have bounded curvature in $[0,T)$, <em>and</em> $g_1(0) = g_2(0)$, then $g_1(t) = g_2(t)$ </p> <p>In particular, a corollary of Chen-Zhu's result is:</p> <blockquote> <p>If $\phi\in Isom(M,g(0))$ is an isometry of $(M,g(0))$ where $g(0)$ has bounded curvature, then <em>the</em> Shi solution to Ricci flow with initial data $g(0)$ still has $\phi \in Isom(M,g(t))$.</p> <p>A "fancy" way of saying this is $Isom(M,g(0)) \subseteq Isom(M,g(t))$.</p> </blockquote> <p>To prove this, if $\phi \in Isom(M,g(0))$, let $g(t)$ be the Shi solution to Ricci flow starting at $g(0)$. Clearly $\phi^*g(t)$ is still a solution to Ricci flow, but $\phi^*g(0) = g(0)$ by assumption that it is an isometry of $g(0)$. So $\phi^*g(t)$ is a solution to Ricci flow (obviously having bounded curvature) with initial data $g(0)$, so it must be $g(t)$!</p> <p>I assume that you were originally referring to <a href="http://arxiv.org/pdf/0906.4920v1.pdf" rel="nofollow">Kotschwar (Backwards Uniqueness of Ricci Flow)</a>. This result says the opposite of Chen-Zhu, and is not really what you want here. In particular, it says that if $g_1(t), g_2(t)$ are both solutions to Ricci flow on $t\in[0,T]$ with uniformly bounded curvature ($T&lt;\infty$), if $g_1(T) = g_2(T)$ then $g_1(t) = g_2(t)$ for all $t \in [0,T]$. In particular, this complements the Chen-Zhu corollary:</p> <blockquote> <p>Under the assumptions of uniformly bounded curvature, $Isom(M,g(t)) \subseteq Isom(M,g(0))$.</p> </blockquote> <p>Finally, I'll remark on the assumption of bounded curvature. In <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1246888488" rel="nofollow">Chen (Strong Uniqueness of the Ricci flow, JDG 2009)</a>, it is shown that one does not always need to assume bounded curvature in dimension $3$. Instead, Chen shows:</p> <blockquote> <p>If $(M,g(0))$ is complete, has bounded, nonnegative sectional curvature, $0 \leq Rm \leq K$, then any two $g_1(t),g_2(t)$ Ricci flows (which are complete for all $t\in[0,T)$) starting from $g(0)$ must agree.</p> </blockquote> <hr> <p>None of this exactly answers your question as it stands, but I don't think that being simply connected makes things any easier (but I could be wrong). On the other hand, as long as your initial manifold has bounded curvature, by Shi-Chen-Zhu, there is a unique (short time) solution to Ricci flow with bounded curvature. </p> http://mathoverflow.net/questions/56813/applications-of-the-notion-of-of-gromov-hausdorff-distance Applications of the notion of of Gromov-Hausdorff distance Otis Chodosh 2011-02-27T12:08:15Z 2012-10-19T14:58:20Z <p>I am looking for applications of the notion of <a href="http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence" rel="nofollow">Gromov-Hausdorff convergence</a> to prove theorems that a priori have nothing to do with it. Examples that I am aware of (thanks to wikipedia and google):</p> <ul> <li><p><a href="http://en.wikipedia.org/wiki/Gromov%27s_theorem_on_groups_of_polynomial_growth" rel="nofollow">Gromov's theorem</a> </p></li> <li><p>The <a href="http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence" rel="nofollow">wikipedia</a> page links to a <a href="http://www.springerlink.com/content/5t5qtxfyrjmp7rv7/" rel="nofollow">paper</a> that uses GH convergence to prove a stability result in cosmology.</p></li> </ul> <p>What are more examples? Ideally they would be along the lines of Gromov's theorem, or proofs of geometric facts, but I'm interested to hear about anything.</p> <p>As a subquestion, are there interesting applications of <a href="http://en.wikipedia.org/wiki/Gromov%27s_compactness_theorem_%28geometry%29" rel="nofollow">Gromov's compactness theorem</a> to prove results about manifolds with bounded Ricci which have nothing to do with GH convergence? </p> http://mathoverflow.net/questions/103287/a-existence-problem-of-p-q-metric/103326#103326 Answer by Otis Chodosh for A Existence Problem of (p,q) metric Otis Chodosh 2012-07-27T16:26:07Z 2012-07-27T16:26:07Z <p>I believe that you're a bit mistaken about the final claim. The correct statement should be that:</p> <blockquote> <p>There exists a "time-orientable lorentz" (i.e. a lorentzian metric with a nowhere vanishing timelike vector field) metric if and only if there exists a nowhere vanishing vector field (which happens if and only if the euler characteristic is zero).</p> </blockquote> <p>The proof of this is easy:</p> <p>(the only if direction is trivial) Suppose there is a nowhere vanishing vector field $X$. Pick any Riemannian metric $g$ and let $\omega$ be the dual $1$-form with $X$ with respect to $g$. Then, defining $$\tilde g = - 2\omega\cdot\omega + g,$$ this is a "time orientable lorentz metric" as desired.</p> <p>The same proof shows that </p> <blockquote> <p>There is a $(p,q)$-metric with $p$ linearly independent non-vanishing vector fields $X_i$ with $\tilde g(X_i,X_i) &lt; 0$ if and only if there are $p$ linearly independent nonvanishing vector fields on the manifold. </p> </blockquote> <p>This is something which can be detected by Euler classes, I think.</p> <p>I agree that this second conclusion is a bit unsatisfying, because it is natural to restrict to time orientable Lorentz metrics for physical reasons, but here it is not clear that it is a natural restriction. </p> http://mathoverflow.net/questions/102258/when-is-the-infimum-of-an-arbitrary-family-of-measurable-functions-also-measurabl/102259#102259 Answer by Otis Chodosh for When is the infimum of an arbitrary family of measurable functions also measurable? Otis Chodosh 2012-07-14T23:08:05Z 2012-07-14T23:08:05Z <p>Here is a set of of measurable functions with cardinality the continuum whose infimum is not (Borel) measurable:</p> <p>Let $S\subset [0,1]$ be a non-measurable set. For $t\in[0,1]$ let $f_t(x)$ be the function defined as follows:</p> <p>If $t \in S$ then $f_t(x) = 2$ for $t \not = x$ and $f_t(t) = 1$. If $t \not \in S$, then let $f_t(x) \equiv 2$. Then $f(x) := \inf_{t\in[0,1]} f_t(x)$ is $2$ on $S^c$ and $1$ on $S$, so is certainly not measurable.</p> http://mathoverflow.net/questions/97860/basic-question-about-rectifiability/97911#97911 Answer by Otis Chodosh for basic question about rectifiability Otis Chodosh 2012-05-25T04:48:07Z 2012-05-25T17:06:03Z <p>I think that the answer to the question is basically that one usually considers rectifiable sets in the sense you give and further specifies that they are measurable. </p> <p>On the other hand, I do not think you technically need measurability for a.e. tangent planes. Recall that your definition of rectifiability is equivalent to demanding that up to a set of measure zero, the set is contained in a countable union of $C^1$-manifolds. One way to get at the tangent plane of a rectifiable set is to say that a point $x \in S$ has a tangent plane $T_xM_i$ when $x\in M_i$ as long as for $x\in M_j$ for some other $j$ has $T_xM_j= T_xM_i$. Then one would have to check that this is defined fo $H^m$-a.e. for $x\in S$, but this follows from a sort of transversality argument, intuitively because $M_i$ and $M_j$ cannot intersect very often if they do not have the same tangent planes at their intersections. So, this argument should go through without reference to the measurability of $S$ at all.</p> <p>This said, as far as I know, people work with rectifiable sets which are also measurable, so it is just an issue of symantics whether or not it is included in the definition. </p> <p>Rectifiable (measurable) sets can still be very bad though. For example, if you take circles inside $B(0,1)\subset \mathbb{R}^2$, which are centered at the rational points and have radii which are square summable, then the union of these is a $1$-rectifiable set, because it is obviously a countable union of (smooth) $1$-manifolds. However it is dense in the ball! On the other hand, it has tangent planes almost everywhere! </p> <p>So, rectifiability does not really avoid "square filling things" in this sense. </p> <p>If it helps, an example of a non-rectifiable set is as follows. Take a triangle. Remove a regular hexagon inscribed in the triangle (side length 1/3 that of the triangle). This gives three triangles. Continue doing this and then take the intersection of all of these. This is a purely $1$-rectifiable set, roughly because if you project on to an axis perpendicular to a base of the triangle, then you get a cantor set, of $H^1$-measure zero and it is not hard to see that a $1$-rectifiable set must have at most one direction with this property (this set has three).</p> <hr> <p>This is in response to your comment on your answer:</p> <p>You ask why your original definition and this one are the same. Let me list three equivalent definitions of $k$-rectifiability:</p> <p>(1) Up to a set of measure zero $S \subset \cup_{i=1}^\infty M_i$ where the $M_i$ are $C^1$ $k$-dimensional submanifolds.</p> <p>(2) Up to a set of measure zero $S \subset \cup_{i=1}^\infty F_i(\mathbb{R}^k)$ for $F:\mathbb{R}^k\to \mathbb{R}^N$ (where $\mathbb{R}^N$ is the ambient space). </p> <p>(3) Up to a set of measure zero $S\subset \cup_{i=1}^\infty \text{image}(F_i)$ where the $F_i :T_i \to \mathbb{R}^N$ is $C^1$ and $T_i$ is a $k$-plane in $\mathbb{R}^N$ where if we denote $\Pi_{T_i}$ by the projection $\mathbb{R}^N\to T_i$, we have that $\Pi_{T_i} \circ F_i = Id$. </p> <p>The equivalence of (1) and (2) is in Leon Simon's book, I believe. It uses a.e. differentiability of Lipschitz functions and a Sard type theorem. To understand (3), notice that the bit about the projection just says that the images of $F_i$ are just graphs over the $T_i$. Graphs are $C^1$ submanifolds, so this (3) => (1). On the other hand any $C^1$ manifold is locally the graph over its tangent plane, so we can slice up a decomposition as in (1) into countably many such graphs, implying (3). </p> http://mathoverflow.net/questions/95316/physical-interpretation-of-robin-boundary-conditions Physical Interpretation of Robin Boundary Conditions Otis Chodosh 2012-04-27T01:46:58Z 2012-04-30T18:43:44Z <p>In a (bounded) domain $\Omega \subset \mathbb{R}^n$, if we're studying the Laplace equation or heat equation or such PDE's we can impose the Dirichlet <code>$u|_{\partial\Omega} \equiv 0$</code>, Neumann <code>$D_{\nu} u|_{\partial\Omega}\equiv 0$</code> or Robin (for $\alpha \in \mathbb{R}$) <code>$(D_{\nu} u + \alpha u)|_{\partial \Omega} \equiv 0$</code>.</p> <p>I know that, for example for the heat equation, Dirichlet eigenvalues correspond physically to the boundary being in contact with a (large) heat bath at $T=0$. Or, in the Laplace equation, if we're intersted in the modes supported by $\Omega$ (as a drum), Dirichlet boundary conditions can be thought of keeping the boundary from moving.</p> <p>Neumann boundary conditions, for the heat flow, correspond to a perfectly insulated boundary. For the Laplace equation and drum modes, I think this corresponds to allowing the boundary to flap up and down, but not move otherwise.</p> <hr> <p>My question is: what sort of physical interpretations are there for the Robin boundary conditions? <a href="http://en.wikipedia.org/wiki/Robin_boundary_conditions" rel="nofollow">Wikipedia</a> says that they are related to electromagnetic problems, but gives no details. I'd be happy with answers that are not necessarily physics-related, for example, if there was somewhere that Robin boundary conditions naturally arise in a mathematical context, I'd be interested to know about that as well.</p> http://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds What are "good" examples of spin manifolds? Otis Chodosh 2011-11-04T19:15:16Z 2011-11-06T18:38:05Z <p>I'm trying to get a grasp on what it means for a manifold to be spin. My question is, roughly:</p> <blockquote> <p>What are some "good" (in the sense of illustrating the concept) examples of manifolds which are spin (or not spin) (and why)?</p> </blockquote> <hr> <p>For comparison, I'd consider the cylinder and the mobius strip to be "good" examples of orientable (or not) bundles.</p> <hr> <p>I've read the answers to <a href="http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors" rel="nofollow">http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors</a> which are helpful, but I'd like specific examples (non-examples) to think about. </p> http://mathoverflow.net/questions/71533/wasserstein-geometry-of-measures-on-manifolds-related-to-the-generalized-legendre Wasserstein geometry of measures on manifolds related to the generalized Legendre transform and $d^2/2$-convexity Otis Chodosh 2011-07-29T00:11:14Z 2011-07-29T16:06:29Z <p>Let $(M,g)$ be a fixed closed Riemannian manifold, normalized to have volume 1. We'll write $d_M(x,y)$ for the (geodesic) distance between two points $x,y\in M$. I'm interested in the following class of functions $\varphi: M\to \mathbb{R}$.</p> <blockquote> <p>Call $\varphi$, <em>$d^2/2$-convex</em> when there is $\psi:M\to \mathbb{R}$ such that $$\varphi(x) = - \inf_{y\in M} \left[ \frac 12 d_M(x,y)^2 + \psi(y) \right]$$</p> <p>In this case, we say that $\varphi$ is the <em>generalized Legendre transform</em> of $\psi$, which we write $\varphi = \psi^c$</p> </blockquote> <p>In particular, one can show that if a $d^2/2$-convex function is bounded, Lipschitz (and thus differentiable $vol_M$ a.e.) with gradient bounded by the diameter of $M$. Also, $(\varphi^c)^c=\varphi$ if and only if $\varphi$ is $d^2/2$ convex.</p> <hr> <p>I'm interested in these because of their relation to the theory of optimal transport. I'll briefly describe it here, but in theory one should not need to know anything about optimal transport to answer my questions (although it may help as it seems intimately related). I'm intersted in the 2-Wasserstein metric on $\mathcal{P}(M)$, the space of probability measures on $M$. Lying a bit, I'll say that this is defined to be, for $\mu,\nu\in \mathcal{P}(M)$ probability measures $$d^W(\mu,\nu)^2 : = \inf_{F:M\to M, F_*\mu = \nu} \int_M d(x,F(x))^2 dvol_M$$ Here, $F_*\mu = \nu$ means that for all Borel sets $A$, $\mu(F^{-1}(A)) = \nu(A)$. (In fact this is only really true when $\mu$ and $\nu$ are absolutely continuous wrt the volume measure, and it needs a bit of generalization to be really true. Anyways, what is relevant to the beginning of the question, is that for any measure $\nu \in \mathcal{P}(M)$, there is a <em>unique</em> (up to constants) $d^2/2$-convex function $\varphi$ such that $\nu = exp(\nabla \varphi)_* vol_M$, and it turns out this is the unique minimizer in the above definition of distance with $\mu = vol_M$. That is $$d^W(vol_M,\nu) = \int_M |\nabla \varphi|_g^2 d vol_M$$</p> <p>Furthermore, the geodesic from $vol_M$ to $\nu$ in $\mathcal{P}(M)$ (in the metric space sense) is given by $t\mapsto \exp(t\nabla \varphi)_* vol_M$.</p> <p>Now in this <a href="http://arxiv.org/abs/0901.1815" rel="nofollow">paper</a>, Sturm (not sure if he was the first to do this) shows that this gives a continuous involution on $\mathcal{P}(M)$, by taking the Legendre transform of the $d^2/2$-convex $\varphi$ (uniquely) associated to $\nu$ and setting $\nu^c := \exp(\nabla \varphi^c)_* vol_M$. </p> <hr> <p>I'm interested in various geometric properties related to this involution, which can be rephrased in elementary terms as follows (I've included my geometric interpretations in quotations):</p> <ol> <li><p>Is it true that for $t\in[0,1]$, $(t\varphi)^c = t (\varphi^c)$? "geodesics from $vol_M$ are mapped to other geodesics from $vol_M$" (I think I can prove this using some weird scaling arguments, but I'd like an elementary proof just from the definition, which should probably exist if it is true)</p></li> <li><p>What is the relation (if any) between $$\int_M |\nabla \varphi|_g^2 dvol_M$$ and $$\int_M |\nabla \varphi^c|_g^2 dvol_M$$ "how close are $d^W(vol_M,\nu)$ and $d^W(vol_M,\nu^c)$?"</p></li> <li><p>For any $d^2/2$-convex function $\varphi$, let $M$ be the supremum of $m$ such that $m\varphi$ is $d^2/2$-convex. One can show that $M\varphi$ is then $d^2/2$-convex (the set of $d^2/2$-convex functions is closed in $H^1$, see the Sturm paper). What does $\exp(M\nabla \varphi)_* vol_M$ look like? Is it in general totally singular, etc? "what do the endpoints of geodesics starting from $vol_M$ look like?"</p></li> <li><p>How should I think of this map geometrically, i.e. on the level of measures?</p></li> </ol> <p>I'm most interested in (2), followed by (3), but have included (4) just in case someone has any insight - as far as I can tell this is not very well understood and probably does not have a good answer right now. </p> http://mathoverflow.net/questions/70917/entropy-of-a-measure/70951#70951 Answer by Otis Chodosh for Entropy of a measure Otis Chodosh 2011-07-22T02:32:01Z 2011-07-22T02:32:01Z <p>In the vein of what Tapio suggested, one place to look for some ideas is <a href="http://math.berkeley.edu/~lott/LottVillani.pdf" rel="nofollow">the Lott-Villani paper</a> on optimal transport where they discuss various entropies (section 3.2 on p 923 on the version I linked)</p> <p>In particular they discuss the Shannon entropy as discussed by Tapio. A side note - it is also called the Boltzmann H-functional. (See also definition 3.28 for another possible direction to try)</p> <hr> <p>To answer your comment to Tapio's post:</p> <p>If you use $m$ as the counting measure, then I believe that Tapio's definition agrees with the "limiting version" of what you wrote in some sense. If $m$ is the counting measure, then any probability measure $\mu$ is absolutely continuous wrt $m$. Then, its clear that $d\mu/dm (x) = \mu(x)$, so plugging this into the Shannon entropy formula from Tapio's post, we get $$\tag{E} Ent(\mu|m) = \sum_{x\in \mathbb{N}} \mu(x) \log\mu(x)$$</p> <p>I believe that the signt discrepancy is because Tapio's formula is usually called the Boltzman H-functional, whereas Shannon entropy is usually referred to as the negative of what Tapio wrote there but I am not exactly sure on these semantics. </p> <p>Suppose that $\mu(n) = \frac {1}{2^n}$. Then $$Ent(\mu|m) = \sum_{n=1}^\infty \frac{n \log 2}{2^n} \in (0,\infty)$$</p> <p>Perhaps I have misunderstood your question/comment however? The entropy defined above does satisfy the first condition you give, i.e. that it takes its minimum on dirac measures (by jensen's inequality $Ent \geq 0$ and clearly $Ent(\delta_x|m) = 0$. I am a bit confused as to what a finitely additive translation probability measure on the natural numbers is? Does "finitely additive" mean a weaker condition than just "probability measure"? Can you give an example of one?</p> http://mathoverflow.net/questions/32690/planar-sets-where-any-line-through-the-center-of-mass-divides-the-set-into-two-re Planar sets where any line through the center of mass divides the set into two regions of equal area. Otis Chodosh 2010-07-20T20:28:10Z 2011-06-20T15:11:35Z <p>This question is influenced by the following riddle:</p> <blockquote> <p>You are given a rectangular set in the plane with a rectangular hole cut out (in any orientation). How do you cut the region into two sets of equal area?</p> </blockquote> <p><em>SPOILER ALERT!!</em> - The answer is that you can cut through the center of both rectangles, and because any line through the center of a rectangle divides it into two pieces of equal area, this cut works. </p> <hr> <p>I have been wondering about the following question - What sort of conditions on a set guarantee that it has this property, that any line through the center of mass divides it into two regions of equal area?</p> <p>The only thing that I have been able to think of is $\pi$ rotational symmetry around the center of mass. For example the rectangle has this symmetry. This symmetry means that in fact the two regions cut by any line through the center of mass are congruent, and not just equal area. </p> <p>Thus, my question is:</p> <blockquote> <p>Suppose that we have a planar (measurable) set $A \subset \mathbb{R}^2$ (with positive measure). If there is a point $a\in \mathbb{R}^2$ such that: for any line $\ell \subset \mathbb{R}^2$ through $a$, denoting the regions of $A$ on either side of the line $B$ and $C$ then we have $|B| = |C|$ (Lebesgue measure), then is it necessarily true that (1) $a$ is the center of mass of $A$ and (2) that $A$ has $\pi$ rotational symmetry around $A$ in the a.e. sense, i.e. if $\tilde A$ is $A$ rotated by $\pi$ around $a$ then the symmetric difference between $A$ and $\tilde A$ has measure zero, i.e. $|A \ \Delta \ \tilde A| = 0$.</p> </blockquote> <p>I feel like (1) should be true, but I'm not so sure about (2). If the answer to (2) is no, then what sort of sufficient conditions are there? I'm mostly just curious about the answer, so by all means feel free to strengthen the assumptions on $A$, like requiring it to be a region bounded by a smooth boundary, etc. Thanks!</p> http://mathoverflow.net/questions/37111/what-is-the-smallest-c-algebra-containing-the-standard-pseudodifferential-o What is the smallest $C^*$-algebra containing the "standard" pseudodifferential operators? Otis Chodosh 2010-08-30T02:41:28Z 2011-04-20T20:46:13Z <p>Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying $|\partial^\alpha_x \partial^\beta_\xi a(x,\xi)| \leq C_{\alpha,\beta} (1+|\xi|)^{-|\beta|}$ ) a $C^*$-algebra? </p> <p>In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?</p> <hr> <p>If not, then is there any nice characterization by the $C^*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace? </p> <hr> <p><b> Edit: </b>Per Yemon Choi's comments below, the above question seems somewhat hopeless. As described <a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=MR&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=MR90d%3A47016&amp;s5=&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">here</a>, $\Psi^0(\mathbb{R})$ is a Fréchet $*$-algebra with a topology stronger than the operator topology. I assume that this is the topology given by the seminorms on symbols: $$\Vert a \Vert_{\alpha,\beta} = \sup_{x,\xi \in \mathbb{R}} (1+|\xi|)^{|\beta|} |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)|.$$</p> <p>So, in addition to the above question, I am adding the following question, to make it so that there might be an answer:</p> <p>Is there a reasonable description of the smallest $C^*$-algebra containing $\Psi^0$? </p> http://mathoverflow.net/questions/60704/do-manifolds-with-no-ricci-lower-bounds-for-any-metric-exist Do manifolds with no Ricci lower bounds for any metric exist? Otis Chodosh 2011-04-05T15:27:29Z 2011-04-05T16:12:54Z <p>Is there a smooth (noncompact) manifold $M$ such that for any Riemannian metric $g$ on $M$ there are $p_i$ and unit tangent vectors $v_i \in T_{p_i}M$ such that $Ricc(g)|_{p_i}(v_i,v_i) \leq -i$? This seems unlikely, but I'm not sure how to prove it.</p> <hr> <p>Alternatively, what is the simplest example of a fixed $(M,g)$ with no lower Ricci bounds in the sense above? It seems that some conformal change of the standard Euclidean metric could accomplish this, but I dont see a simple way to do this.</p> http://mathoverflow.net/questions/59944/relationship-between-spectrum-geometry-and-almost-isometry/60013#60013 Answer by Otis Chodosh for Relationship between spectrum geometry and almost-isometry Otis Chodosh 2011-03-29T19:22:50Z 2011-03-29T19:28:21Z <p>The above comments are still mostly valid with $\epsilon$-isometries:</p> <p>if $(M,g)$ and $(N,g)$ are Riemannian manifolds with diameters less than $D$, then</p> <p>$f:M\to N$, $x\mapsto n_0$ for some $n_0\in N$ has</p> <p>$$|d_N(f(x),f(x')) - d_M(x,x')| = d_M(x,x') \leq D$$</p> <p>and for all $y \in N$, because $$d_N(y,n_0) \leq D$$</p> <p>this $f$ is a $D$-isometry. So, just having a $\epsilon$-isometry for some large $\epsilon$ should not be seen as an unlikely occurrence (however, the (even weaker) notion of quasi-isometry that you used before <em>is</em> an interesting idea, but some sort noncompactness plays a big role in things, as should be obvious from all of these answers.</p> <hr> <p>On the other hand, $\epsilon$-isometries and the spectrum are certainly related in some ways, particularly with lower Ricci bounds:</p> <p>For example in the paper by Cheeger and Colding, "On the structure of spaces with Ricci curvature bounded below. III." (J. Differential Geom., 54(1):37–74, 2000.) they prove that for manifolds with appropriate Ricci lower bounds, under Gromov-Hausdorff convergence the spectrum and eigenfunctions converge in some sense. </p> <p>For example, their Theorem 7.11 says:</p> <blockquote> <blockquote> <p>For $M_1^n$, $M_2^n$, Riemannian manifolds satisfying $$Ric \geq -(n-1)$$ and $$diam(N_1^n) \leq d &lt; \infty$$ Then for all $N &lt; \infty$ and $\epsilon > 0$ there is a $\delta(n,d,\epsilon,N) > 0$ such that if $$d_{GH}(M_1^n,M_2^n) &lt; \delta$$ then for $j\leq N$, we have that $|\lambda_{j,1} - \lambda_{j,2}| &lt;\epsilon$, where $\lambda_{i,k}$ is the $i$-th eigenvalue on the $M_k$.</p> </blockquote> </blockquote> <p>You can get more information about this <a href="http://arxiv.org/abs/math/0612107" rel="nofollow">here</a>, which also links to some interesting looking work by Lott about how the same question with the laplacian on $p$-forms.</p> <hr> <p>Just to clarify how $d_{GH}$ is related to $\epsilon$-isometries, in case it is unclear. It is a theorem that the Gromov-Hausdorff distance is $&lt;\epsilon$ if there exists an $\epsilon/2$-isometry and vice versa. A good place to read about this is <a href="http://books.google.com/books?id=afnlx8sHmQIC&amp;printsec=frontcover&amp;dq=course+in+metric+geometry&amp;source=bl&amp;ots=JwC08A-Ze6&amp;sig=YdYs4tF4UPJXqZJiYUexjWyKHt8&amp;hl=en&amp;ei=WjCSTZ34MMKnhAeC4NiEDw&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=6&amp;ved=0CD4Q6AEwBQ#v=onepage&amp;q&amp;f=false" rel="nofollow">here</a></p> <hr> <p>Thus, this should give you some sort of condition on how close two compact manifolds can be in the Gromov-Hausdorff topology, if you know their spectrum. Rescale them so that $Ric \geq -(n-1)$, and then applying Theorem 7.11 you can get a lower bound on $\epsilon$. I'm not sure how explicit the $\delta$ is in their proof, however, if this is important to you.</p> http://mathoverflow.net/questions/52631/converse-to-bishop-gromov-inequality Converse to Bishop-Gromov Inequality Otis Chodosh 2011-01-20T15:16:04Z 2011-01-24T20:54:23Z <p>Is the converse to the <a href="http://en.wikipedia.org/wiki/Bishop%E2%80%93Gromov_inequality" rel="nofollow">Bishop-Gromov Inequality</a> true? </p> <p>In other words, if, for a complete $n$-dimensional Riemannian manifold $M$, there is $k \in \mathbb{R}$, such that defining $V_k(R)$ to be the size of a ball of radius $R$ in the standard space $S^n_k$, for any $p \in M$ and $R\geq 0$ we have that </p> <p>$$vol_M(B_R(p)) \leq V_k(R)$$</p> <p>then is it true that on $M$, $Ric \geq (n-1)k$?</p> http://mathoverflow.net/questions/52902/proving-theorems-by-using-functions-with-fixed-points/52937#52937 Answer by Otis Chodosh for Proving theorems by using functions with fixed points. Otis Chodosh 2011-01-23T10:28:37Z 2011-01-24T08:25:00Z <p>In addition to ODE existence theorems, there are also uses for PDE existence/uniqueness theorems. An example of that is constructing weak solutions to the linear Boltzmann equation. I think this example is interesting because it is more of a philosophy, not so much precise "fixed point theorem" that is used here. </p> <p>The linear Boltzmann equation is: </p> <p>$\partial_t f + v\cdot \nabla_x f = Kf -af + Q$</p> <p>where </p> <p>$Kf = \int k(t,x,v,v') f(t,x,v')dv'$</p> <p>By Duhamel's principle, we know that a strong solution would satisfy</p> <p>$f(t,x,v) = f_0(x-tv,v) + \int_0^t (Kf - af + Q)(s,x-(t-s)v,v)ds$.</p> <p>We basically use this as our definition of a weak solution. Thus, we can rephrase the search for a weak solution as looking for a fixed point to the operator </p> <p>$g \mapsto F[f,Q] + \tau g$</p> <p>where </p> <p>$F[f_0,Q] = f_0(x-vt,v) + \int_0^t Q(s,x-(t-s)v,v)ds$</p> <p>and</p> <p>$\tau g = \int_0^t (Kf - af)(s,x-(t-s)v,v)ds$.</p> <p>Notice that the series</p> <p>$\sum_{n\geq 0} \tau^n[F[f_0,Q]]$</p> <p>would be such a fixed point if we had appropriate convergence (just hit it with $\tau$ and see what happens), so basically, we've reduced the problem to bounding the operator $\tau$ in the appropriate space which we would like weak solutions to live. As I mentioned above, this doesn't really use any "fixed point theorems" but is clearly still a fixed point argument.</p> http://mathoverflow.net/questions/48951/how-to-show-the-cardinality-of-nonisometric-compact-metric-spaces-is-the-continuu How to show the cardinality of nonisometric compact metric spaces is the continuum Otis Chodosh 2010-12-10T18:12:30Z 2010-12-10T18:31:47Z <p>It is asserted in <em>A Course in Metric Geometry</em> by Burago, Burago, Ivanov that</p> <blockquote> <p>there can be no more than continuum of mutually nonisometric compact spaces</p> </blockquote> <p>How is this proven? </p> <p>Its clear that there must be at least a continuum of mutually nonisometric compact spaces, i.e. $([0,\alpha], d_{\mathbb{R}})$ for $\alpha>0$ are a family of nonisometric metric spaces, but I don't know enough set theory to have any ideas how to bound the cardinality from above. A first guess was that the fact that compact metric spaces are totally bounded should be useful?</p> http://mathoverflow.net/questions/45594/how-small-can-the-set-of-p-such-that-the-lp-norms-are-different-for-two-fixe How small can the set of $p$ such that the $L^p$ norms are different for two fixed functions? Otis Chodosh 2010-11-10T20:18:07Z 2010-11-11T09:28:00Z <p><strike>What does it tell you about two functions if their $L^p$ norms are the same for all $p\in[1,\infty]$? Certainly they could be related by composition with a diffeomorphism with Jacobian of norm 1, or even one could be a "pulled apart version of the other one" in the sense of $x^2\chi_{[0,2]}$ vs $x^2 \chi_{[0,1]} + (x-1)^2 \chi_{[2,3]}$. To try to ignore the second type of issue, I'll restrict to smooth functions, and ask the following precise question:</p> <p>Given $f,g\in C^\infty(\mathbb{R})$ such that $\Vert f \Vert_{L^p(\mathbb{R})} = \Vert g \Vert_{L^p(\mathbb{R})} &lt;\infty$ for all $p\in [1,\infty]$ is it necessarily true that $f(x) = g(\pm x + C)$ for some constant $C$ and for a choice of either $+$ or $-$ for all $x \in \mathbb{R}$? </strike></p> <p>As Qiaochu Yuan shows in his answer, smoothness doesn't solve the issue of "pulling apart" at all. Thus, I am interested in the following:</p> <p>What is the "smallest" (in whatever sense) but still nonempty subset of $S \subset [1,\infty]$ such that there is $f,g\in C^\infty(\mathbb{R})$ such that $S$ is the set of $p$ such that $\Vert f \Vert_{L^p} \neq \Vert g \Vert_{L^p}$?</p> http://mathoverflow.net/questions/38492/elementary-proof-of-bounds-on-discrete-derivative-applied-to-1ns Elementary proof of bounds on discrete derivative applied to $(1+n)^s$ Otis Chodosh 2010-09-12T17:26:00Z 2010-09-12T17:48:52Z <p>I would like to show that for $s \in \mathbb{R}$ and a nonnegative integer $k$ $$\triangle^k ((1+n)^s) \lesssim (1+|n|)^{s-k}$$</p> <p>where $\triangle$ is the discrete derivative, i.e. $\triangle^1 ((1+n)^s) = (2+n)^s - (1+n)^s$. </p> <hr> <p>This is easy when $s \in \mathbb{Z}$, and in the continuous analogue because $$\partial_x^k (1+x)^s = (s)(s-1)\cdots (s-k+1) (1+x)^{s-k}$$</p> <hr> <p>I think that you can use the generalized binomial theorem to prove this, but I was wondering if there was anything more straightforward, e.g. some kind of convexity argument to use the continuous case.</p> <hr> <p>Note: I wasn't sure about the tags, feel free to re-tag as appropriate. </p> http://mathoverflow.net/questions/36549/coefficients-from-stone-weierstrass-versus-fourier-transform/36582#36582 Answer by Otis Chodosh for Coefficients from Stone Weierstrass versus Fourier Transform Otis Chodosh 2010-08-24T18:43:25Z 2010-08-24T18:43:25Z <p>Re-reading your question, I think that I see what you are asking.</p> <p>Per @Andrea Ferretti's comments, you have to be careful to distinguish between ${e^{inx}}$ and $span \ {e^{inx}}$. You certainly are interested the latter. Sorry if my comments were sloppy and confusing above.</p> <p>So, I think that the it goes like this:</p> <p>From some corollary of Stone-Weirstrauss you can show that $span \ {e^{inx}}$ is dense in $C(\mathbb{S}^1)$ with the supremum norm. Because we know that $C(\mathbb{S}^1)\hookrightarrow L^2([0,1])$ has its image a dense subset of $L^2([0,1])$ and we know that if $f_n \to f$ in the supremum topology on $C(\mathbb{S}^1)$, then the images also converge in $L^2([0,1])$.</p> <p>Thus, by this reasoning, for $f\in L^2([0,1])$ we can find $f_n \in span \ {e^{inx}}$ such that $f_n = L^2([0,1])$. Lets write $$f_n = \sum_{k\in \mathbb{Z}} c_k^{(n)} e^{ikx}$$ where all but finitely many of the $c_k^{(n)}$ are zero (this is because in the span of infinitely many objects we only take a finite number of them to add together) </p> <p>Now, what I think you are asking is: what can we say about the coefficients $c_k^{(n)}$? The answer is that they converge to the $k$-th Fourier coefficient of $f$ as $n\to\infty$ because $$\hat f(k) = \langle f, e^{ikx} \rangle = \lim_{n\to\infty} \langle f_n ,e^{ikx}\rangle = \lim_{n\to\infty} c_k^{(n)}$$</p> <p>In fact if $c_k^{(n)}$ are arbitrary complex numbers, defining $f_n$ as above, we see that $$\Vert f - f_n \Vert_{L^2} = \sum_{k\in \mathbb{Z}} |\hat f(k) - c_k^{(n)}|^2$$ assuming convergence. Thus, if $(c_k^{(n)})_k \to (\hat f(k))_k$ as $n\to\infty$ in $\ell^2(\mathbb{Z})$ then $f_n\to f$ in $L^2$, which is a pretty weak condition.</p> http://mathoverflow.net/questions/35054/characterizing-the-harmonic-oscillator-creation-and-annihilation-operators-in-a-r Characterizing the harmonic oscillator creation and annihilation operators in a rotationally invariant way Otis Chodosh 2010-08-09T22:28:03Z 2010-08-17T22:48:56Z <p>I am interested in a characterization of the creation and annihilation operators that is in some sense invariant under $O(n)$ rotations of $\mathbb{R}^n$:</p> <h2>Background</h2> <p>The Harmonic Oscillator on $\mathbb{R}^n$ is the differential operator </p> <p><code>$$H := \sum_{k=1}^n \left[x_k^2-\frac{\partial^2}{\partial x_k^2}\right] = |x|^2 + \nabla.$$</code></p> <p>It is not hard to see that the $L^2(\mathbb{R}^n)$ eigenvalues are exactly <code>$\{n,n+2,n+4,\dots\}$</code>. Furthermore, the annihilation operator is an operator on Schwartz functions on $\mathbb{R}^n$ $$C_k := \frac {1}{\sqrt {2}}\left( x_k + \frac{\partial}{\partial x_k}\right)$$ and the creation operator as its adjoint (with the $L^2$ inner product) $$C_k^\dagger = \frac {1}{\sqrt {2}}\left( x_k -\frac{\partial}{\partial x_k}\right).$$ If we let $V_{n+2m}$ be the set of eigenfunctions with eigenvalue $n+2m$, we can show that $$C_k^\dagger V_{n+2m} \subset V_{n+2(m+1)}$$ $$C_k V_{n+2m} \subset V_{n+2(m-1)}$$ (where we define $V_r=0$ if $r$ is not an eigenvalue) and $V_n$ is spanned by $e^{-|x|^2/2}$. It turns out that $V_{n+2m}$ is isomorphic to the space of degree $m$ homogeneous polynomials in $n$ variables, which I'll denote $\mathcal{P}^n_m$, by the isomorphism $p \mapsto p(C^\dagger)$ i.e. $x_1x_2 \mapsto C_1^\dagger C_2^\dagger$, etc. All of this and more can be found <a href="http://math.mit.edu/~rbm/iml/Chapter4.pdf" rel="nofollow">here</a> starting on page 86 (with some slightly different notation than I've used here). </p> <h2>Motivation</h2> <p>One of the problems with this whole business is that even though $H$ and thus $V_{n+2m}$ are invariant under rotations of $\mathbb{R}^n$, the $C_k$ and $C_k^\dagger$ are not. We made an arbitrary choice of coordinates when we defined them. This leads to a non-cannonical choice of basis for the eigenspaces, and has been giving me problems in my research. My question is thus:</p> <h2>Question</h2> <p>Even though there is no cannonical choice of basis for $V_{n+2m}$, is there some characterization of the creation and anihilation operators that is invariant under $O(n)$ rotations of $\mathbb{R}^n$. That is, what is an $O(n)$-invariant characterization of the space of operators $$\text{span}_\mathbb{R}\{C_1^\dagger,C_2^\dagger,\dots, C^\dagger_n\}$$</p> <p>If this is not possible, then as an alternative answer, I am interested in insight into how rotations and the operators interact. </p> http://mathoverflow.net/questions/34811/discrete-hairy-ball-theorem/34821#34821 Answer by Otis Chodosh for Discrete Hairy Ball Theorem Otis Chodosh 2010-08-07T07:33:00Z 2010-08-07T07:33:00Z <p>By interpolating "legal configuration" if it were to exist, I think you can obtain a nowhere vanishing, continuous vector field on $S^2$. Suppose that for some $n$ there is a legal configuration on the $n\times n \times n$ cube. For each 1x1 face, put a vertex at the center. Then connect vertexes whose 1x1 face's touch. (basically something like a dual graph, but I don't know what the real terminology is). </p> <p>Doing this, you get $(n-1)^2$ squares on each face of the Rubik's cube, $n-1$ squares on each edge of the Rubik's cube, and one triangle for each vertex of the cube. Now, put the arrow from the 1x1 face at the associated vertex. For each square or triangle we can now linearly interpolate to get a vector field over the whole thing. These will patch back together to form a continuous vector field on $S^2$ because on the lines they are glued along, the value on each piece is linear interpolation between the same two vectors. Thus, basically it remains to check that given a "legal configuration" you cannot interpolate to a zero vector. </p> <p>The squares are not too bad, because the most that two of the vectors being interpolated can be off by is $90^\circ$, so you can't get a zero vector.</p> <p>The triangles are a little trickier because things get twisted around, but if you try to write down the possible cases, you can see that there is basically only one type of "legal" corner configuration, and it doesn't interpolate to a zero vector. </p> <p>This seems to show that you don't even need to assume anything special about corner squares, you can allow them to point in the 8th illegal position, and there are still no "legal configurations."</p> http://mathoverflow.net/questions/130385/the-isoperimetric-problem-for-domains-constrained-to-lie-between-two-parallel-pla Comment by Otis Chodosh Otis Chodosh 2013-05-12T02:42:36Z 2013-05-12T02:42:36Z fyi your title is very misleading: in common mathematical terminology a &quot;minimal surface&quot; minimizes the surface area with no constraints on volume (actually &quot;minimal&quot; only means a critical point of area). You are discussing &quot;isoperimetric surfaces&quot; . Also, could you clarify exactly what your question is, it seems like you've found an answer, up to integrating an ODE.... http://mathoverflow.net/questions/127986/applications-of-pseudodifferential-operators-to-pde Comment by Otis Chodosh Otis Chodosh 2013-04-18T16:21:42Z 2013-04-18T16:21:42Z I'm not 100% sure, but I think you can do things like Schauder estimates and a lot of the other parts of &quot;elliptic theory&quot; with psiDO's. I've never read it, but I once glanced at Taylor's book &quot;Pseudodifferential Operators and Nonlinear PDE&quot;: <a href="http://www.unc.edu/math/Faculty/met/nonlin.html" rel="nofollow">unc.edu/math/Faculty/met/nonlin.html</a> which seems to have a bit of this. Anyways, it might also have some other topics you'd be interested in. http://mathoverflow.net/questions/127946/isoperimetric-profile Comment by Otis Chodosh Otis Chodosh 2013-04-18T16:15:28Z 2013-04-18T16:15:28Z @Thomas Richard, thats exactly what you do. I think the only issue that might be a tiny bit tricky is getting a good expansion of $r$ in terms of $|B_r(p)|$, but I think that this is not too hard.. http://mathoverflow.net/questions/120485/fourier-analytic-proofs Comment by Otis Chodosh Otis Chodosh 2013-02-01T04:26:52Z 2013-02-01T04:26:52Z Central limit theorem? http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120066#120066 Comment by Otis Chodosh Otis Chodosh 2013-01-29T22:16:46Z 2013-01-29T22:16:46Z Hi Renato, I have just filled out the registration, so I'll see you there! Also, I agree that the link is broken now. I think that it was working when I put it up, so perhaps MSRI is having some bug. If anyone is interested in the meantime I'd be happy to share the pdf of the worksheet with them. Just send me an email. http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120066#120066 Comment by Otis Chodosh Otis Chodosh 2013-01-28T03:15:08Z 2013-01-28T03:15:08Z @Deane Yang, Added some more! Feel free to add/modify what is there! http://mathoverflow.net/questions/118130/what-is-known-about-analogous-results-of-kazdan-and-warner-in-higher-dimensions Comment by Otis Chodosh Otis Chodosh 2013-01-05T19:11:05Z 2013-01-05T19:11:05Z *by my first sentence, I mean an identity between scalar curvature and the Gauss-Bonnet integrand (as @unknown comments) http://mathoverflow.net/questions/118130/what-is-known-about-analogous-results-of-kazdan-and-warner-in-higher-dimensions Comment by Otis Chodosh Otis Chodosh 2013-01-05T19:10:16Z 2013-01-05T19:10:16Z ...giving a contradiction. You may be interested in the following MO post: <a href="http://mathoverflow.net/questions/30035/some-questions-about-scalar-curvature" rel="nofollow" title="some questions about scalar curvature">mathoverflow.net/questions/30035/&hellip;</a>. http://mathoverflow.net/questions/118130/what-is-known-about-analogous-results-of-kazdan-and-warner-in-higher-dimensions Comment by Otis Chodosh Otis Chodosh 2013-01-05T19:09:41Z 2013-01-05T19:09:41Z @Ritwik, there is no possibility of such an identity holding. If $*e(TM)$ was some multiple of the scalar curvature $R$, this would imply that $\int R = \int *e(TM) dV = &lt;e(TM),[M]&gt;$, and this is a topological invariant of the (smooth structure) topology of $M$. (see <a href="http://en.wikipedia.org/wiki/Generalized_Gauss%E2%80%93Bonnet_theorem#Further_generalizations" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a>). However, one can show that in dimensions $\geq 3$, all manifolds admit a metric of negative scalar curvature, e.g. $S^3$. So, if the above identity held, we could evaluate $&lt;e(TM),[M]&gt;$ on the standard metric and this other one... http://mathoverflow.net/questions/118008/which-metric-spaces-have-this-superposition-property Comment by Otis Chodosh Otis Chodosh 2013-01-04T01:47:21Z 2013-01-04T01:47:21Z It seems that if you start with a Riemannian manifold with this property it must be highly symmetric. For example, it must be a symmetric space: take a unit speed geodesic through a point $p$ and consider the space $A = \gamma([-\epsilon,\epsilon))$ and $B = \gamma((-\epsilon,\epsilon])$. Then the associated isometry should be a inversion around $p$. I have no idea if this is sufficient, although it would be pretty cool if it was. Also, I'm not quite sure what to do if you modified your definition to demand that $A,B$ are closed. http://mathoverflow.net/questions/115644/parallel-orthogonal-complex-structures-on-complexified-tangent-bundle Comment by Otis Chodosh Otis Chodosh 2012-12-06T22:06:00Z 2012-12-06T22:06:00Z @Robert - I see, thanks! http://mathoverflow.net/questions/115644/parallel-orthogonal-complex-structures-on-complexified-tangent-bundle Comment by Otis Chodosh Otis Chodosh 2012-12-06T21:58:56Z 2012-12-06T21:58:56Z Have you thought about the &quot;multiplication by $i$ endomorphism&quot; $J \in End(T\mathbb{R}^3\otimes \mathbb{C})$? This clearly cannot be a &quot;non-complexified&quot; complex structure because $\mathbb{R}^3$ is odd dimensional. http://mathoverflow.net/questions/114995/the-mean-curvature-of-a-hypersurface Comment by Otis Chodosh Otis Chodosh 2012-11-30T16:20:58Z 2012-11-30T16:20:58Z The formula for graphical mean curvature is naturally in a divergence form. A hypersurfsce is locally graphical, so that gives what you want. Any book on minimal surfaces will give you the formula, the one I have in front of me is colding minicozzi eqn 1.6 on p 2. The 0 could replaced by mean curvature you're interested in nonminimal surfaces. http://mathoverflow.net/questions/114356/how-to-show-that-integral-of-borel-function-fx-y-over-y-is-borel-function-of-x Comment by Otis Chodosh Otis Chodosh 2012-11-24T18:14:15Z 2012-11-24T18:14:15Z This is a part of Fubini's theorem (assuming that $g \in L^1(\mathbb{R}^2)$) See big Rudin 8.8 (c). The main idea is to approximate by simple functions, where your statement is obvious. http://mathoverflow.net/questions/114167/ricci-flow-descending-from-an-universal-cover/114211#114211 Comment by Otis Chodosh Otis Chodosh 2012-11-23T19:40:57Z 2012-11-23T19:40:57Z Oh, sorry I missed the update!