User thomas richard - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:38:07Z http://mathoverflow.net/feeds/user/8887 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130081/converse-to-milnors-theorem-on-manifolds-with-nonnegative-ricci-curvature Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature. Thomas Richard 2013-05-08T14:26:39Z 2013-05-09T00:23:28Z <p>Disclaimer : I suspect the question I am about to ask is really hard, but I just want to know the status of such questions.</p> <p>Thanks to Milnor, we know that the $\pi_1$ any compact manifold with nonnegative Ricci curvature has polynomial growth. </p> <p>I want to know if anything is known about the opposite direction : which manifolds whose $\pi_1$ has polynomial growth admit metric with nonnegative Ricci curvature ?</p> <p>At least if you allow non compact manifolds, the answer cannot be 'all of them' (because of Whitehead's three manifold), and I feel like there are compact counter examples but I am not able to cook one. </p> http://mathoverflow.net/questions/129100/reduction-of-antisymmetric-complex-matrices Reduction of antisymmetric complex matrices Thomas Richard 2013-04-29T14:31:51Z 2013-04-30T13:45:57Z <p>Let $E=\mathfrak{so}(n,\mathbb{C})$ be the Lie algebra of antisymmetric complex matrices. We consider the action of the complex orthogonal group $SO(n,\mathbb{C})$ on $E$ by conjugation. Is there a nice description of the orbits ? Something similar to the fact the antisymmetric matrices are orthogonaly similar to matrices with $2\times 2$ blocks on the diagonal whose characteristic polynomial is $X^2+\lambda^2$ ?</p> <p>That seems to be classic, but I wasn't able to locate results about this...</p> <p>Thanks !</p> http://mathoverflow.net/questions/126935/fixing-a-proof-of-the-systolic-inequality-for-higher-genus-surfaces Fixing a proof of the systolic inequality for higher genus surfaces Thomas Richard 2013-04-09T05:28:26Z 2013-04-09T11:00:46Z <p>I'm currently learning some stuffs about systolic inequalities. While reading the relevant sections (p329 to 340) in Berger's Panoramic View of Riemannian Geometry, I noticed a gap in one of the proofs (starting at page 331). The goal is to prove :</p> <blockquote> <p>For any non simply connected compact surface $(M,g)$, $Area(M,g)\geq Sys(M,g)^2/2$.</p> </blockquote> <p>Berger considers a periodic geodesic $c$ realizing the systole $L=Sys(M,g)$, pick a point $m$ on $c$ and claims that $Vol B(m,L/2)\geq L^2/2$, which is enough to show the result.</p> <p>However in the proof of the claim, he invokes the following fact : for <code>$r&lt;L/2$</code>, $B(m,r)$ is a topological disk. He says that this comes from the "very definition of the systole". But this claim is false, it is enough to consider a torus with a long thin finger glued at some point to see it (fig 7.26 on the page where the claim is made shows exactly this). In the paper "Systolic and inter-systolic inequalities", Gromov, facing the same kind of situation, just says "chop the fingers", while this is intuitively convincing I don't see a way to make it rigorous.</p> <p>My question is the following : is this fixable ?</p> <p>Thinking a little bit about it, it seems enough for the rest of the argument to show that all but one of the connected component of $M\backslash B(m,r)$ are disks, and that the systole $c$ doesn't meet the components which are disks. But I am not able to prove this at the moment.</p> <p>I'm aware that another proof is available, through estimating the homological systole. But I like the proof on the next page of Berger that the systolic ration grows at least like the square root of the genus, which uses the same argument.</p> <p>I should also say that I don't have access the article of Hebda to which Berger refers.</p> http://mathoverflow.net/questions/119827/algebraic-characterization-of-the-curvature-operator-of-symmetric-spaces Algebraic characterization of the curvature operator of symmetric spaces Thomas Richard 2013-01-25T11:40:03Z 2013-01-26T14:10:02Z <p>My question is the following :</p> <blockquote> <p>Given an algebraic curvature operator $R\in S^2_B(\Lambda^2\mathbb{R}^n)$, is there an a simple criterion to know if this curvature operator can occur as the curvature operator of symmetric space ?</p> </blockquote> <p>I would be almost equally happy if someone can point to a way to know if $R$ can be the curvature operator of some Riemannian manifold with reduced holonomy $G\subset SO(n,\mathbb{R})$.</p> <p>I suspect this might be linked to the fact that $\Lambda^2\mathbb{R}^n=\mathfrak{so}(n,\mathbb{R})$ and how $R$ acts on $\mathfrak{g}\subset\mathfrak{so}(n,\mathbb{R})$. So far, the only link I found is that the image of $R$ has to be contained in $\mathfrak{g}$. I have no idea if this is sufficient. And I wonder if it is possible to write a condition NOT along the line of "There is a Lie subalgebra $\mathfrak{g}\subset\mathfrak{so}(n,\mathbb{R})$ such that..."</p> http://mathoverflow.net/questions/119443/how-to-generate-a-random-weyl-curvature-operator How to generate a random (Weyl) curvature operator ? Thomas Richard 2013-01-21T04:28:44Z 2013-01-21T04:37:01Z <p>Given a dimension $n$, the space of curvature operators is the space $S^2_B(\Lambda^2\mathbb{R}^n)$ of symmetric endomorphisms $R$ of $\Lambda^2\mathbb{R}^n$ which satisfy the first Bianchi identity :</p> <blockquote> <p>Given $x,y,z,t\in\mathbb{R}^n$, $\langle R(x\wedge y),z\wedge t\rangle+\langle R(y\wedge z),x\wedge t\rangle+\langle R(z\wedge x),y\wedge t\rangle=0$</p> </blockquote> <p>This the space where the curvature tensor of a Riemannian manifold lives.</p> <p>It is naturally equipped with the following action of $O(n,\mathbb{R})$ :</p> <blockquote> <p>$\langle g.R(x\wedge y),z\wedge t\rangle =\langle R(gx\wedge gy),gz\wedge gt\rangle$</p> </blockquote> <p>which splits into irreducible components :</p> <blockquote> <p>$S^2_B(\Lambda^2\mathbb{R}^n)=\mathbb{R}Id_{\Lambda^2\mathbb{R^n}}\oplus S^2_0\mathbb{R}^n\wedge id_{\mathbb{R}^n} \oplus \mathcal{W}$</p> </blockquote> <p>where $\wedge$ is the Kulkarni-Nomizu product and $\mathcal{W}$ is the space of Weyl tensors, that is tensors in $S^2_B(\Lambda^2\mathbb{R}^n)$ whose traces are all zero.</p> <p>In my research on Ricci flow, I am investigating Hamilton's ODE, which is ODE on $S^2_B(\Lambda^2\mathbb{R}^n)$. I am currently doing some numerical exploration, and haven't come with a good way of getting a "random" initial condition. I would like this choice to be invariant under the action of $O(n,\mathbb{R})$. This implies that we can treat each component of the previous decomposition separately. My question then splits in two subquestions :</p> <blockquote> <ul> <li><strong>Q1:</strong> To treat the first two parts of the decomposition, I just need to generate a random symmetric matrix on $\mathbb{R}^n$ in an $O(n,\mathbb{R})$ invariant way. I believe this is classic "random matrix theory", but I am unfortunately totally ignorant about this field. What would be an efficient algorithm to solve this problem ?</li> <li><strong>Q2:</strong> How to treat the Weyl part ? Can we design an efficient algorithm for generating a Weyl tensor ?</li> </ul> </blockquote> <p>PS: I know that the first question must be already treated somewhere, but while googling "random symmetric matrix" gave me interesting information, I wasn't able to recover an algorithm from that.</p> <p>PS2: If that helps, the software I'm using is FreeMat, an open source clone of matlab.</p> http://mathoverflow.net/questions/117058/rigorous-solution-to-ricci-flow-on-dumbbell-s3/117067#117067 Answer by Thomas Richard for Rigorous solution to Ricci Flow on dumbbell $S^3$ Thomas Richard 2012-12-23T03:27:50Z 2012-12-23T03:27:50Z <p>The neckpinch solution on a dumbell has been constructed by Angenant and Knopff. A whole chapter of the book "The Ricci Flow: An Introduction" by Chow and Knopff is devoted to the construction of such a solution. </p> <p>For the "weak solution" side, it has not been settled yet. Angenant and Knopff has some result in the rotationnaly symmetric case in a preprint that was put on the arXiv something like two years ago. </p> http://mathoverflow.net/questions/114747/dependence-of-the-blow-up-time-of-existence-of-an-ode-with-respect-to-initial-con Dependence of the blow-up time of existence of an ODE with respect to initial condition. Thomas Richard 2012-11-28T09:56:19Z 2012-11-28T10:43:47Z <p>Let $V$ be a smooth vector field on $\mathbb{R}^n$. Assume that the maximal solution to the Cauchy problem $x'=V(x), x(0)=x_0$ exist only for $t\in [0,T)$, where $T$ is finite, denote this time by $T(x_0)$. Is $T$ continuous with respect to $x_0$ ? Is it $C^1$ ?</p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/110445#110445 Answer by Thomas Richard for Slick ways to make annoying verifications Thomas Richard 2012-10-23T16:38:46Z 2012-10-23T19:26:25Z <p>A really elementary one :</p> <p>A linear map between two vector spaces of the same finite dimension is an isomorphism if and only if its kernel is zero.</p> <p>As an application, I like the proof of the existence of Lagrange interpolation polynomials.</p> http://mathoverflow.net/questions/107458/manifold-whose-universal-covering-is-a-sphere-but-which-is-not-a-space-form Manifold whose universal covering is a sphere but which is not a space form? Thomas Richard 2012-09-18T11:54:16Z 2012-09-18T21:05:21Z <p>Let $M^n$ be a smooth manifold whose universal cover is homeomorphic $\mathbb{S}^n$, are there examples where $M^n$ is not homeomorphic to a space form ? </p> <p>The answer may vary if you replace homeomorphic by diffeomorphic, and I'm also interested in this question under this restriction.</p> <p>I came to this question while reading surveys about sphere theorems, where the non simply-connected case is harder to get (while you already know that the universal cover is a sphere, you need more work to show it is actually a space-form).</p> http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti/106792#106792 Answer by Thomas Richard for Coordinate-free derivation of the Einstein's field equation from the Hilbert action. Thomas Richard 2012-09-10T09:04:36Z 2012-09-10T09:04:36Z <p>This can be found in Besse "Einstein Manifolds", in chapter 4.</p> <p>The idea is to use Koszul formula for the Levi-Civitta connection to compute the derivative of the curvature with respect to the metric. Bianchi identities also help.</p> http://mathoverflow.net/questions/106719/nearly-constant-curvature-implies-nearly-isometric-to-a-space-form/106728#106728 Answer by Thomas Richard for Nearly constant curvature implies "nearly isometric" to a space form? Thomas Richard 2012-09-09T13:25:27Z 2012-09-10T08:34:15Z <p>This question is two-sided, and I'm not sure what you mean by "local diffeomorphism" so I'll treat both aspects. There is a local and a global version :</p> <h2>Local version :</h2> <blockquote> <p><strong>Q1 :</strong> Given a point $x$ in a Riemannian manifold $(M,g)$, can we find a constant curvature metric on a neighborhood of $x$ which is close to $g$ ?</p> </blockquote> <p>First remark : since we want a local statement, zero curvature is as good as constant curvature here. </p> <p>The answer to <strong>Q1</strong> is "yes", <em>without any restriction on the curvature</em>. This can be seen using normal coordinates centered at $x$ : in these coordinates, the metric at $x$ is euclidean and its distortion from being euclidean as one moves away from $x$ can be controlled (using that the curvature is bounded in a neighborhood of $x$, and that the injectivity radius at $x$ is positive).</p> <p><em>Edit :</em> The above statement is too complicated. As Anton's say in his comment to Agol's answer, use the exponential map ! The pull back of $g$ by the exponential map defines a riemannian metric on some neighborhood of the origin in $T_xM$ which is equal to $g_x$ at $x$, by continuity, this pullback stay close to the euclidean metric $g_x$ on $T_xM$ and this does the job.</p> <p>We can refine the question then :</p> <blockquote> <p><strong>Q1' :</strong> What can be said about the size of the neighborhood we obtained ?</p> </blockquote> <p>In this case we need to impose geometric restrictions on $(M,g)$. For instance, Cheeger and Anderson proved the following :</p> <blockquote> <p>For $n\in\mathbb{N}$, $k\in\mathbb{R}$, $i>0$ and $\varepsilon>0$, one can find $\delta>0$ such that in a $n$-manifold of Ricci curvature greater than $k$ and injectivity radius greater than $i$, any ball of radius $\delta$ admits a flat metric which is $\varepsilon$-close to $g$ in $C^0$-norm.</p> </blockquote> <p>See "$C^\alpha$-compactness for manifolds with Ricci curvature and injectivity radius bounded below."</p> <p>The proof uses more elaborate machinery than just normal coordinates : harmonic coordinates are used. If you stick to normal coordinates you can obtain a similar result but with stronger geometric assumptions.</p> <h2>Global version :</h2> <blockquote> <p><strong>Q2 :</strong> Under which condition does a manifold with almost $k$ curvature admit a $C^0$-close metric of constant curvature $k$ ?</p> </blockquote> <p>If you consider large spheres, they have almost zero cuvrvature but don't admit any flat metric, so you need to put some restrictions on the side of the manifolds.</p> <p>An example of theorem you can get is the following :</p> <blockquote> <p>For any $n\in\mathbb{N}$, $k\in\mathbb{R}$, $V>0$, $D>0$ and $\varepsilon>0$, there is a $\delta>0$ such that any $n$-manifold $(M,g)$ of diameter less than $D$, volume more than $V$, and sectional curvature between $k-\delta$ and $k+\delta$ admits a metric af constant sectional curvature $k$ which is $\varepsilon$-close to $g$ in $C^0$-norm.</p> </blockquote> <p>The proof relies on Cheeger-Gromov compactness theorem for sequences of Riemannian manifolds. A (really) sketchy goes like that : we argue by contradiction, you take a sequence $\delta_i$ going to $0$, and you assume you can find a sequence of manifolds $(M_i,g_i)$ satisfying the hypothesis of the theorem with $\delta=\delta_i$ and not satisfying the conclusion of the theorem. Then up to a subsequence, the sequence has a limit which is a manifold of constant curvature $k$, by the very definition of Cheeger-Gromov convergence, this imply the for some $i$ large enough, $M_i$ admit a constant curvature $k$ metric $\varepsilon$-close to $g_i$, a contradiction.</p> <p>The lower bound on the volume is necessary (at least in the $k=0$ case) the so called "infranilmanifolds" admit metrics of curvature as close as wanted to $0$ with diameter bounded above but no flat metric.</p> <p>For the $k=1$ case, the bounds on the diameter is unnecessary because of Myers theorem.</p> <p>For the $k=-1$ case, I don't know if the hypothesis can be weakened.</p> http://mathoverflow.net/questions/106609/uniqueness-of-lipschitz-function-satisfying-differential-equation/106610#106610 Answer by Thomas Richard for Uniqueness of Lipschitz function satisfying differential equation Thomas Richard 2012-09-07T15:18:14Z 2012-09-07T15:37:48Z <p>I suppose you want uniqueness assuming some initial condition. </p> <p>Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.</p> <p>Maybe with stronger assumptions on $F$ you can do something assuming only that $X$ is Lipschitz but I'm not sure. Without the Lipschitz condition you don't have uniqueness assuming that the differential equation is satisfied a.e., Cantor stair is a counterexample.</p> <p>Edit : The uniqueness statement you want is true when $F$ is Lipschitz. In this case, with your assumptions, $X$ has a right derivative which is continuous. From that it's not too complicated to show that $X$ has a derivative everywhere which is continuous. And then the classical Cauchy-Lipschitz applies.</p> http://mathoverflow.net/questions/96554/how-do-you-call-a-map-which-sends-convergent-sequences-to-pre-compact-ones How do you call a map which sends convergent sequences to pre-compact ones ? Thomas Richard 2012-05-10T08:53:42Z 2012-05-10T08:53:42Z <p>In my work I encountered a map $f$ between two metric spaces $X$ and $Y$ that was not continuous (at least I couldn't prove it was), but I was able to prove that convergent sequences $(x_n)$ in $X$ were sent by $f$ to sequences lying in a compact set in $Y$ (in particular, any subsequence of $f(x_n)$ had a convergent subsequence).</p> <p>Do these kind of maps have a name ?</p> <p>Thanks.</p> http://mathoverflow.net/questions/358/examples-of-great-mathematical-writing/95007#95007 Answer by Thomas Richard for Examples of great mathematical writing Thomas Richard 2012-04-24T09:37:59Z 2012-04-24T09:37:59Z <p>"A panoramic view of Riemannian Geometry" by M. Berger is an example of excellent mathematical writing to my taste. This book is great to learn what are the questions of interest in the field, and what are the main results. Although you will not find detailed proofs of the results, the main ideas are often explained in an intuitive way. </p> http://mathoverflow.net/questions/89866/is-it-overkill-to-invoke-kirszbraun-theorem-to-prove-the-following-fact Is it overkill to invoke Kirszbraun theorem to prove the following fact ? Thomas Richard 2012-02-29T13:25:38Z 2012-02-29T13:25:38Z <p>Given a small enough convex triangle $(abc)$ in a (smooth or Alexandrov) surface $(X,d)$ of curvature greater than $-1$, let $(\overline{abc})$ be its comparison triangle in $\mathbb{H}^2$. Then there exist a $1$-Lipschitz map $\varphi$ from the interior of $(abc)$ to the interior of $(\overline{abc})$.</p> <p>Using Kirszbraun theorem (for Alexandrov spaces, as stated by Lang and Schroeder or Alexander, Kapovitch and Petrunin) this is simple. Just identify the sides of $(abc)$ and $(\overline{abc})$ in the usual way, this map is $1$-Lipschitz because of the curvature bound and we can apply Kirszbraun theorem to extend it to the interior of the triangle.</p> <p>I was wondering if one could explicitly build $\varphi$ in this specific case.</p> http://mathoverflow.net/questions/86807/smoothability-of-compact-alexandrov-surfaces-with-curvature-bounded-from-below Smoothability of compact Alexandrov surfaces with curvature bounded from below. Thomas Richard 2012-01-27T10:01:55Z 2012-01-29T03:16:01Z <p>Let $(X,d)$ be compact metric space of curvature greater than $-1$ (in the sense of comparison triangles), assume that its Hausdorff dimension is $2$. Then a result of Perelman says that $X$ is a 2-dimensional manifold.</p> <h2>Claim</h2> <p>$(X,d)$ can be Gromov-Hausdorff approximated by a sequence of Riemannian surfaces $(M_i,g_i)$ such that $\int_{M_i}|K_{g_i}|dv_{g_i}$ is bounded.</p> <p>To see this :</p> <ol> <li><p>A version of the Gauss-Bonnet Theorem holds (<a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jmsj/1225113599%20%22Machigashira%22" rel="nofollow">Machigashira, The Gaussian curvature of an Alexandrov surface</a>), which implies that $(X,d)$ is an Alexandrov surface with bounded <em>integral</em> curvature.</p></li> <li><p>Any such surface can be approximated by smooth Riemannian surfaces with bounded integral curvature. See Reshetnyak, Geometry IV, Encyclopaedia of Mathematical Sciences. </p></li> </ol> <h2>Question</h2> <p>Is any compact Alexandrov surface of curvature greater than $-1$ approximated by a sequence of smooth compact Riemannian surfaces with curvature bounded from below (by -1, or something else if this helps) ?</p> <p>Maybe this is classic but I didn't found explicit results of this kind.</p> <p>Thanks.</p> http://mathoverflow.net/questions/86906/complexity-of-matching-red-and-blue-points-in-the-plane Complexity of matching red and blue points in the plane. Thomas Richard 2012-01-28T16:23:46Z 2012-01-28T21:29:41Z <p>I'm just asking because I'm curious. I was seeking references on the following problem, that a friend exposed to me last holidays :</p> <h2>Problem</h2> <p>Given $n$ red points and $n$ blue points in the plane in general position (no 3 of them are aligned), find a pairing of the red points with the blue points such that the segments it draws are all disjoint.</p> <p>This problem is always solvable, and admits several proof. A proof I know goes like this :</p> <p>Start with an arbitrary pairing, and look for intersections of the segments it defines, if there are none you're done. If you found one, do the following operation :</p> <pre><code>r r r r \ / | | X =&gt; | | / \ | | b b b b </code></pre> <p>(uncross the crossing you have found), you may create new crossings with this operation. If you repeat this operation, you cannot cycle, because the triangle inequality shows that the sum of the length of the segments is strictly decreasing. So you will eventually get stuck at a configuration with no crossings.</p> <h2>Questions</h2> <ol> <li>What is the complexity of the algorithm described in the proof ?</li> <li>What is the best known algorithm to solve this problem ?</li> </ol> <p>I wouldn't be surprised to learn that this problem is a classic in computational geometry, however googling didn't give me references. Since some computational geometers are active on MO, I thought I could get interesting answers here.</p> http://mathoverflow.net/questions/82710/regularity-of-solutions-to-a-linear-degenerate-parabolic-pde Regularity of solutions to a linear degenerate parabolic pde Thomas Richard 2011-12-05T15:56:18Z 2011-12-09T16:57:40Z <p>I've encountered the following problem which is causing me some trouble :</p> <p>Let $(M^2,g)$ be a smooth compact Riemannian surface (with constant curvature for instance) and consider a smooth function $u:M\times [0,1) \to \mathbb{R}$ which is a solution of the following PDE : $\partial_t u(x,t)=A(x,t)\Delta u(x,t)$ where $A$ is a function which is smooth on $M\times [0,1)$ and satisfies : $C_1|1-t|^\alpha \leq A(x,t)\leq C_2$.</p> <p>It is not hard to see that $u$ is uniformly bounded with the maximum principle, a simple computation also shows that $\int_M |\nabla u(x,t)|^2dv_g$ decreases with $t$, which gives $L^p$ convergence of $u(.,t)$ as $t$ goes to $1$.</p> <p>My question is : does $u$ have a continuous extension to $M\times[0,1]$ ?</p> <p>Any insight or reference on this kind of problem is welcome.</p> <p>Thanks.</p> <p>EDIT :</p> <p>Some further observations :</p> <ul> <li><p>For the application I have in mind, I would be happy with just a sequence $t_i$ going to $1$ such that $u(.,t_i)$ uniformly converges.</p></li> <li><p>The statement in the previous item is true if we replace $M$ by an interval, thanks the Rellich-Kondrachov compactness theorem, in fact, with $M$ of dimension 2,we are exactly at the critical exponent given by Rellich-Kondrachov Theorem.</p></li> <li><p>The statement is true if $A$ depends only on $t$ (not on $x$), just put $t'=\int_0^tA(\tau)d\tau$ and you get that $u(x,t')$ satisfies the usual heat equation.</p></li> </ul> <p>Thanks again.</p> http://mathoverflow.net/questions/81995/any-compactness-theorem-for-manifolds-which-has-ricci-lower-bound/81999#81999 Answer by Thomas Richard for any compactness theorem for manifolds which has Ricci lower bound? Thomas Richard 2011-11-27T11:20:48Z 2011-11-27T11:20:48Z <p>A well-known theorem of Gromov says that if $(M_i,g_i)$ is a sequence of manifolds with $Ric\geq -kg$ and fixed dimension, then it is a converging subsequence in the pointed Gromov-Hausdorff topology. References for this fact are Gromov's "Metric Structures for Riemannian and non-Riemannian spaces" and Burago-Burago-Ivanov's "A Course in Metric Geometry".</p> <p>A lot of investigation has been made on the possible limiting length space, see "On the structure of spaces with Ricci curvature bounded below." by Cheeger and Colding, and subsequent papers.</p> <p>On the other hand, one can have smoother convergence with stronger geometric requirements. For instance, one has $C^\alpha$ compactness when the injectivity radius is bounded from below, see "$C^\alpha$-compactness for manifolds with Ricci curvature and injectivity radius bounded below", by Anderson and Cheeger.</p> http://mathoverflow.net/questions/80452/diagonalizability-of-the-curvature-operator/80458#80458 Answer by Thomas Richard for diagonalizability of the curvature operator Thomas Richard 2011-11-09T07:19:56Z 2011-11-09T09:22:34Z <p>I am not aware of any systematic treatment of this question. However there are two standard examples.</p> <p>1) If $(M,g)$ is an hyper surface in a constant curvature manifold, then its curvature operator is given by $R=A\wedge A-kI$ where $I$ is the identity and $A$ is the second fundamental form. Then if $e_i$ diagonalise $A$, $e_i\wedge e_j$ diagonalise $R$.</p> <p>2) Another example is rotationnaly symmetric metrics on $\mathbb{R}\times S^n$, the proof can be found in Riemannian Geometry by Petersen.</p> http://mathoverflow.net/questions/74214/examples-where-its-useful-to-know-that-a-mathematical-object-belongs-to-some-fam/74240#74240 Answer by Thomas Richard for Examples where it's useful to know that a mathematical object belongs to some family of objects Thomas Richard 2011-09-01T08:11:33Z 2011-09-01T08:11:33Z <p>In differential geometry, the use of geometric flows seem to fit in this framework. One could see Hamilton-Perelman proof as an instance of this phenomenon, but maybe the lack of canonicity of Ricci Flow with surgery disqualifies it. </p> <p>However, Hamilton's theorem on 3-manifolds with $Ric>0$ and Brendle-Schoen differentiable sphere theorem would be interesting examples maybe. In these case, Ricci Flow (without surgery) creates a deformation between your initial metric (with $Ric>0$ or strictly quarter pinched) and metric of constant curvature $1$, since only quotients of the sphere bear such metrics, the initial manifold was a quotient of a sphere.</p> <p>I'm sure there are also interesting examples in harmonic map heat flow (starting with the work of Eels and Sampson) but I don't know them well enough.</p> http://mathoverflow.net/questions/68546/teaching-a-pedagogy-course/68590#68590 Answer by Thomas Richard for Teaching a pedagogy course Thomas Richard 2011-06-23T06:46:11Z 2011-06-23T06:46:11Z <p>I don't know if this qualify as "pedagogy" but something i've found useful when I began teaching was the acting lessons I had in high school. I don't know how it is in the US but here in france you can begin teaching as a PhD student with very few experience in talking in front of people. </p> <p>I find it quite important to know how to use your voice and your body when you give a lecture (or a math talk). Of course acting lessons seems to be a waste of time at this level but a few exercise on breathing and speaking loud enough could be of some use I guess.</p> http://mathoverflow.net/questions/67046/metrically-singular-alexandrov-space Metrically singular Alexandrov space. Thomas Richard 2011-06-06T15:39:18Z 2011-06-07T11:40:12Z <p>Perelman's stability theorem shows in particular that a finite dimensional compact Alexandrov space $(X,d)$ such that $X$ is not a topological manifold cannot be approximated in the Gromov-Hausdorf topology by Riemannian manifolds of the same dimension whose sectional curvature is bounded from below.</p> <p>My questions are : </p> <p>(1) Are there examples of compact Alexandrov spaces (say non-negatively curved) $(X,d)$ such that $X$ is a topological manifold but $(X,d)$ cannot be approximated by Riemannian manifolds of non-negative sectional curvature ?</p> <p>(2) Does it change something if we allow the manifolds $(M_n,g_n)$ in the sequence to have a lower bound on the sectional curvature which is only going to zero as $n$ goes to infinity ?</p> <p>Thanks.</p> http://mathoverflow.net/questions/59279/geometry-of-whitehead-manifolds Geometry of Whitehead manifolds. Thomas Richard 2011-03-23T10:24:13Z 2011-03-29T01:53:18Z <p>I'm currently studying some problems about the Whitehead manifold $W$ (the open 3-manifold which is contractible but not homeomorphic to $\mathbb{R}^3$). Does there exists some survey paper on its properties ?</p> <p>I'm particularly interested in geometric results in the spirit of "Taming 3-manifolds using scalar curvature" by Chang, Weinberger and Yu (<a href="http://www.ams.org/mathscinet-getitem?mr=2721617" rel="nofollow">MathSciNet page</a>) which shows that $W$ admits no metrics of uniformly positive scalar curvature.</p> <p>Thanks.</p> http://mathoverflow.net/questions/48564/harmonic-coordinates-on-riemannian-manifolds Harmonic coordinates on Riemannian manifolds Thomas Richard 2010-12-07T15:11:54Z 2010-12-07T15:23:28Z <p>I'm trying to read the paper of Jost and Karcher on the existence of harmonic coordinates on a ball whose size only depend on the injectivity radius and a two sided bound on the curvature.</p> <p>Unfortunately, my german skills are quite low and make the reading really slow. Does there exist another place to find this proof (in English or French) ?</p> <p>Thanks</p> http://mathoverflow.net/questions/37595/roadmap-to-learning-about-ricci-flow/37599#37599 Answer by Thomas Richard for Roadmap to learning about Ricci Flow? Thomas Richard 2010-09-03T11:02:22Z 2010-09-03T11:02:22Z <p>To understand Perelman's proof of the Poincaré Conjecture, you need a solid background in Riemannian geometry. Many books can be used for an introduction to this field. There are two books I like on this subject : Riemannian Geometry, by Gallot, Hulin and Lafontaine and Riemannian Geometry by Petersen. </p> <p>After, you can try to learn about Ricci flow, a good starting point is Chow and Knopff's "The Ricci Flow : an Introduction". It covers the basics of Ricci flow including Hamilton's theorem that on a compact 3-manifold with $Ric>0$, the (normalized) flow will converge to constant curvature.</p> <p>Then, if you want to go into Perelman's work, there is the book "Ricci Flow and the Poincaré Conhecture" by Morgan and Tian. However you also have to understand Thurston's Geommetrization Conjecture, so you need a solid background in 3-manifold topology, I don't know the references for this part, maybe Thurston's lecture notes ? </p> <p>Another interesting road is to study the proof of the differentiable sphere theorem by Brendle and Schoen, a good reference is Brendle's "Ricci Flow and the Sphere Theorem".</p> <p>I Hope that was helpful.</p> http://mathoverflow.net/questions/37468/lipschitz-equivalence-of-riemannian-metrics/37477#37477 Answer by Thomas Richard for Lipschitz equivalence of Riemannian metrics Thomas Richard 2010-09-02T09:30:48Z 2010-09-02T09:30:48Z <p>As Dmitri says any two Riemannian metrics on a <strong>compact</strong> manifold are Lipschitz equivalent. The proof is quite simple. </p> <p>Consider $g$ and $h$ two metrics on $M$, Let $UM$ be the unit tangent bundle, since $M$ is compact, $UM$ is compact. Then you see that $f:UM\to \mathbb{R}$ defined by $f(x)=\frac{g(x,x)}{h(x,x)}$ is continuous and strictly positive. By compactness, it is bounded above and below by positive constants.</p> http://mathoverflow.net/questions/129975/technical-question-on-perimeter-of-level-sets Comment by Thomas Richard Thomas Richard 2013-05-07T15:13:54Z 2013-05-07T15:13:54Z Isn't $L_t$ equal to $\{u=t\}$ rather than $\{u\geq t\}$ ? http://mathoverflow.net/questions/129100/reduction-of-antisymmetric-complex-matrices/129219#129219 Comment by Thomas Richard Thomas Richard 2013-05-01T06:02:14Z 2013-05-01T06:02:14Z Thanks, I'll have a look at this ! http://mathoverflow.net/questions/129095/maximum-principle-for-a-non-uniformly-parabolic-operator Comment by Thomas Richard Thomas Richard 2013-04-29T14:23:31Z 2013-04-29T14:23:31Z Since your equation is uniformly parabolic for $t\in [0,T]$ ($T$ bounded), most of the arguments should carry on... http://mathoverflow.net/questions/128231/is-the-connected-sum-of-knots-an-isometry Comment by Thomas Richard Thomas Richard 2013-04-21T16:45:37Z 2013-04-21T16:45:37Z What is this distance between knots ? (pure curiosity) http://mathoverflow.net/questions/127946/isoperimetric-profile Comment by Thomas Richard Thomas Richard 2013-04-18T15:52:03Z 2013-04-18T15:52:03Z I'm not sure it works, but have you tried plugging into $|\partial B_r(p)|\geq h(|B_r(p)|/|M|)$ the expansion of $|\partial B_r(p)|$ and $|B_r(p)|$ in term of $r$ ? http://mathoverflow.net/questions/127934/only-finitely-many-fundamental-groups-in-mn-k-v-d Comment by Thomas Richard Thomas Richard 2013-04-18T15:39:06Z 2013-04-18T15:39:06Z No that's not it. Lemma 38 says that a presentation of the $\pi_1$ is given by $\langle\ \Gamma\|\ R\ \rangle$ where $\Gamma$ is a finite set of generators (the $\gamma_i$'s), and $R$ is a set of relations each of which is of the form $\gamma_i\gamma_j=\gamma_k$ for some triple $(i,j,k)$, but that doesn't mean that you have all the relations corresponding to each triple $(i,j,k)$, you only have some of them. http://mathoverflow.net/questions/127934/only-finitely-many-fundamental-groups-in-mn-k-v-d Comment by Thomas Richard Thomas Richard 2013-04-18T11:43:50Z 2013-04-18T11:43:50Z maybe I should have added that the $\gamma_i$'s are taken from a set of generators. http://mathoverflow.net/questions/127829/when-constant-scalar-curvature-implies-einstein/127832#127832 Comment by Thomas Richard Thomas Richard 2013-04-18T09:36:12Z 2013-04-18T09:36:12Z Just a caveat, for this metric on $\mathbb{S}^2\times\mathbb{S}^2$ not to be Einstein, the two $\mathbb{S}^2$ need to have different radius. And a remark about Agol's comment : what I like in it is that it in fact doesn't admit any Einstein metric, just because in dimension 3 Einstein is equivalent to constant sectional curvature. I wonder wether $\mathbb{S}^1\times\mathbb{S}^3$ enjoys the same property or not... http://mathoverflow.net/questions/127934/only-finitely-many-fundamental-groups-in-mn-k-v-d Comment by Thomas Richard Thomas Richard 2013-04-18T09:22:55Z 2013-04-18T09:22:55Z The answer is just at the top of page 277, Lemma 38. It shows that you can always find a presentation of the $\pi_1$ of a Riemannian manifold where all the relations are of the form $\gamma_i\gamma_j=\gamma_k$. Be careful, for a given group, this presentation might feel quite akward (I wonder how to present $\mathbb{Z}/7\mathbb{Z}$ in this way ?). What theorem 64 on the bottom on the page shows is that there is a bound on the number of generators needed for this presentation, not another one. http://mathoverflow.net/questions/126935/fixing-a-proof-of-the-systolic-inequality-for-higher-genus-surfaces/126959#126959 Comment by Thomas Richard Thomas Richard 2013-04-09T16:01:48Z 2013-04-09T16:01:48Z I like this one, thanks a lot ! http://mathoverflow.net/questions/126935/fixing-a-proof-of-the-systolic-inequality-for-higher-genus-surfaces Comment by Thomas Richard Thomas Richard 2013-04-09T16:01:23Z 2013-04-09T16:01:23Z (L is of course the systole...) http://mathoverflow.net/questions/126935/fixing-a-proof-of-the-systolic-inequality-for-higher-genus-surfaces Comment by Thomas Richard Thomas Richard 2013-04-09T16:01:05Z 2013-04-09T16:01:05Z @J. Martel : The estimate $Vol(B(m,sys/2))\geq L^2/2$ is true with $m$ a point on the systole, not for any $m$. Katz's answer and the comments that follow contain all the 'fixing' I needed. http://mathoverflow.net/questions/126935/fixing-a-proof-of-the-systolic-inequality-for-higher-genus-surfaces/126959#126959 Comment by Thomas Richard Thomas Richard 2013-04-09T14:12:13Z 2013-04-09T14:12:13Z Sorry I'm a bit slow... The fact that $S_+$ is not contractible implies that it has length greater than the systole, but I don't see how to derive a contradiction from this. http://mathoverflow.net/questions/126935/fixing-a-proof-of-the-systolic-inequality-for-higher-genus-surfaces/126959#126959 Comment by Thomas Richard Thomas Richard 2013-04-09T11:24:01Z 2013-04-09T11:24:01Z Thanks, but I'm not sure I understand. How do you show that $\gamma(-r)$ and $\gamma(r)$ are in the same connected component of <code>$S&#95;r=\{d(m,p)=r\}$</code> (call it $S_r'$) ? Also, the next step in Berger is to notice that each of the components of $S_r'$ linking $\gamma(-r)$ to $\gamma(r)$ has length greater than $2r$. But to do this, you need to show that the curve $\gamma$ with $\gamma([-r,r])$ replaced by one of the two component of $S_r'$ is not contractible, I am not sure how to do this without knowing in advance that $S_r'$ bounds a disk containing $m$. http://mathoverflow.net/questions/124473/ck-topology-of-metrics Comment by Thomas Richard Thomas Richard 2013-03-14T06:16:04Z 2013-03-14T06:16:04Z Another comment, in some sense these kind of convergence of metrics are often not the &quot;best&quot; ones from a geometric point of view. The problem is that they are not invariant under diffeomorphisms. You might be interested in what is called &quot;Cheeger-Gromov convergence&quot;, which is &quot;$C^k$ convergence of metrics up to diffeomorphisms&quot;.