User thomas rot - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:52:37Z http://mathoverflow.net/feeds/user/12156 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118962/is-a-manifold-with-flat-ends-of-bounded-geometry Is a manifold with flat ends of bounded geometry? Thomas Rot 2013-01-15T10:58:42Z 2013-01-15T23:44:13Z <p>A Riemannian manifold $(M,g)$ is said to have flat ends if the curvature tensor of $g$ vanishes outside a compact set $K$. I was wondering if such manifolds are of bounded geometry. Recall that a manifold is of bounded geometry if</p> <ol> <li>The curvature tensor and all its covariant derivatives are uniformly bounded.</li> <li>The injectivity radius has a uniform positive lower bound.</li> </ol> <p>It is obvious that a manifold with flat ends satisfies the first property, but it is not clear to me that a manifold with flat ends satisfies the second property. Two simple counterexamples come to mind. </p> <ol> <li>The manifold $M=\mathbb{R}^n-{0}$ has flat ends, but has no uniform lower bound on the injectivity radius.</li> <li>A cylinder $\mathbb{R}\times S_r^1$ of radius $r$ has injectivity radius $\pi r$. Take a countable union of such cylinders with decreasing radius $$M=\coprod_{n=1}^\infty \mathbb{R}\times S_{\frac 1n}^1.$$ This manifold is flat, but does not admit a uniform positive lower bound on the injectivity radius.</li> </ol> <p>Counterexample one might be excluded by assuming that $g$ is complete, and counterexample 2 might be excluded by demanding that $M$ is connected. Therefore my question is:</p> <blockquote> Is any connected and complete Riemannian manifold $(M,g)$ with flat ends of bounded geometry? </blockquote> http://mathoverflow.net/questions/112581/induced-maps-in-morse-homology Induced maps in Morse Homology Thomas Rot 2012-11-16T14:47:58Z 2012-12-09T12:26:37Z <p>Let $M,N$ be closed manifolds. Given a differentiable map $f:M\rightarrow N$, I am interested in computing $f_k:H_k(M)\rightarrow H_k(N)$, in Morse Homology. This problems seems difficult, and the only reference I have found is Schwarz' Morse Homology. His strategy is to factor $f$ as follows $$M\rightarrow M\times N\rightarrow \mathbb{R}^n\times N\rightarrow N.$$ where the first map is the graph of $f$, and the second map is an embedding of $M$ into some large $\mathbb{R}^n$, and the third map is a projection to $N$. The first two maps are embeddings of submanifolds, and it is not hard to see what the induced maps must be. Something similar happens with the projection map. </p> <p>This seems difficult to compute, because we need to construct an embedding $M\rightarrow\mathbb{R}^n$. I believe that the second and third step can be a simplified a bit, by choosing a suitable function $a$ on $M$ (with one minimum) and a function $b$ on $N$, and constructing an explicit map $C^k(M\times N,a\oplus b)\rightarrow C^k(N,b)$, which descends to homology. </p> <p>Has this approach been studied somewhere? Is there any literature on these functioral properties that I missed? Is anything known (and written down) for manifolds with boundary?</p> http://mathoverflow.net/questions/115671/is-the-set-of-all-smoothed-closed-simple-curves-on-mathbbr2-a-manifold/115696#115696 Answer by Thomas Rot for Is the set of all smoothed closed simple curves on $\mathbb{R}^2$ a manifold? Thomas Rot 2012-12-07T10:09:15Z 2012-12-07T10:09:15Z <p>Another nice reference for this kind of stuff is Klingenberg: Riemannian geometry. If one imposes less smoothness on the loops (closed curves), and assume the loops are $H^1$ (in Sobolev sense), the space of loops on a Riemannian manifold is a Hilbert manifold. To get the charts, one uses the exponential map on the base, to map ($H^1$) vector fields along a loop (which is morally a tangent vector to this curve) to a loop nearby. There are some issues with differentiability of the loops which define the charts, which can be overcome by approximations of smooth loops. The details are in the above mentioned reference.</p> <p>In this approach loops are allowed to self-intersect. It might be possible to study also a loop space of non-self intersecting loops, but I do not know what this space actually looks like, and if it is nicely embedded in the above mentioned space.</p> http://mathoverflow.net/questions/114201/reference-for-cohomology-theories/114221#114221 Answer by Thomas Rot for reference for (co)homology theories Thomas Rot 2012-11-23T08:20:18Z 2012-11-23T09:32:40Z <p>I really like the book of Bott and Tu for the De Rham theory. Hatchers book - freely available on his site - contains nice treatments of singular and cellular (co) homologies.</p> <p>About your comment. What relations between the theories are you looking for? The Eilenberg Steenrod axioms - <a href="http://www.encyclopediaofmath.org/index.php/Steenrod-Eilenberg_axioms" rel="nofollow">http://www.encyclopediaofmath.org/index.php/Steenrod-Eilenberg_axioms</a> - show that the singular, cellular, and de Rham theories are the same (you have to be a bit careful with the coefficients of course), on spaces where they are all defined. I believe the book of Bredon discusses this a bit, but I don't have it with me here (there is a very short passage in Hatcher). Group cohomology can be seen to be the cohomology of a certain space associated to the group. I does not really matter in which theory we compute these, because they will give the same results. I'm not familiar with the Daubault and sheaf cohomology, so I don't have anything relevant to add to this.</p> http://mathoverflow.net/questions/105370/when-does-a-hypersurface-have-contact-type/105374#105374 Answer by Thomas Rot for When does a hypersurface have contact-type? Thomas Rot 2012-08-24T08:56:22Z 2012-08-24T10:08:40Z <p>Not all hypersurfaces in $\mathbb{R}^{2n}$ are of contact type.</p> <p>Weinstein, in the paper: "On the hypothesis of Rabinowitz' periodic orbit theorems", where he defined the concept of contact type, gives a criterion. If $H^1(\Sigma)=0$, and $\Sigma$ is of contact type, then the characteristic line bundle comes with a distinguished orientation, determined by those vectors $\xi$ such that $\alpha(\xi)>0$ for all contact forms $\alpha$. This is independent of the contact form. For periodic orbits this induces a positivity criterion. In the same paper he also constructs an hypersurface in $\mathbb{R}^4$ which is not of contact type, by showing that the criterion is violated.</p> <p>Many hypersurfaces are of contact type, as you remarked. Another nice example are <em>mechanical hypersurfaces</em>. These are hypersurfaces arising as level sets from hamiltonians $H=T+V$, where $T$ is the kinetic energy term $T=\frac{1}{2}\vert p\vert^2$, and $V$ is a potential depending only on $q$. This works in general cotangent bundles.</p> http://mathoverflow.net/questions/101395/which-differential-equations-allow-for-a-variational-formulation Which differential equations allow for a variational formulation? Thomas Rot 2012-07-05T11:58:20Z 2012-07-06T10:16:02Z <p>Many ODE's and PDE's arising in nature have a variational formulation. An example of what I mean is the following. Classical motions are solutions $q(t)$ to Lagrange's equation $$\frac{d}{dt}\frac{\partial L(q,\dot q)}{\partial\dot q}=\frac{\partial L(q,\dot q)}{\partial q},$$ and these are critical points of the <em>functional</em> $$I(q)=\int L(q,\dot q)dt.$$ Of course one needs to be precise with what considers a solution to both equations. This amounts to specifying regularity and a domain of the functional. This example is an ODE, but many PDE examples are possible as well (for example electromagnetism, or more exotic physical theories). Once one knows a variational description of the problem, many more methods are available to solve the problem. </p> <p>Now I do not expect that any PDE or ODE can be viewed (even formally) as a critical point of a suitable action functional. This is because this whole set up reminds me of De Rham cohomology: "which one-forms (the differential equations) are exact (that is, the $d$ of a functional)?". The last sentence is not correct, but the analogy maybe is? Anyway, my question is:</p> <blockquote> <p>Are there any criteria to determine if a given differential equation admits a variational formulation?</p> </blockquote> http://mathoverflow.net/questions/96863/how-to-explain-the-condition-c-in-critical-point-theory/96876#96876 Answer by Thomas Rot for How to explain the condition (C) in critical point theory? Thomas Rot 2012-05-14T07:34:00Z 2012-05-14T07:34:00Z <p>The condition forces things that look like critical points, to be critical points. Through various techniques, e.g. minimax methods, linking methods, morse theory, one obtains sequences that seem to converge to critical points. That is sequences $x_i$ that satisfy $||\nabla f(x_i)||\rightarrow 0$. and $||f(x_i)||$ is bounded. Let us call these Palais-Smale sequences. If a function/functional $f$ satisfies: all Palais-Smale sequences for $f$ have converging subsequences, we say $f$ satisfies the Palais-Smale condition.</p> <p>In general Palais-Smale sequences do not need to have a subsequence which converge to a limit. A very elementary and standard example on $\mathbb{R}$ is the following. Let $f(x)=\arctan(x)$, and the sequence $x_i=i$. It is easy to verify that the sequence is Palais-Smale, but does not have a converging subsequence.</p> <p>In this example we could get away with another property: properness (the preimage of every compact is compact). However, if $f:M\rightarrow\mathbb{R}$ is defined on a non-locally compact space (e.g. Banach/Hilbert spaces/manifolds), it can never be a proper map. The Palais-Smale condition is a condition which is satisfied by many interesting functionals, and ensures the convergence issues one would like.</p> http://mathoverflow.net/questions/88090/the-connected-components-of-the-free-loop-space The connected components of the free loop space Thomas Rot 2012-02-10T10:32:30Z 2012-02-10T17:58:33Z <p>I am trying to understand the topology (in terms of homology groups) of the free loop space $\Lambda M$ of nice spaces (Complete Riemannian connected finite dimensional manifolds $M$). I see the free loop space (of H^1 loops) as a Hilbert manifold, cf. Klingenbergs book. If the manifold $M$ has a non-trivial fundamental group, the free loop space has as many connected components as there are conjugacy classes in $\pi_1(M)$. How much do these components of $\Lambda M$ differ? Are these components all homotopy equivalent? For the circle the answer is yes, because all components of the free loop space are homotopy equivalent to the circle itself.</p> <p>The following question is related to my question</p> <p><a href="http://mathoverflow.net/questions/34927/are-the-path-components-of-a-loop-space-homotopy-equivalent" rel="nofollow">http://mathoverflow.net/questions/34927/are-the-path-components-of-a-loop-space-homotopy-equivalent</a></p> <p>However, I cannot seem to use the answer to this question directly, because I cannot concatenate two free loops, but maybe I am missing something obvious.</p> http://mathoverflow.net/questions/60044/groups-and-asymmetry/60047#60047 Answer by Thomas Rot for groups and asymmetry Thomas Rot 2011-03-30T09:58:53Z 2011-03-30T09:58:53Z <p>If you start with a very symmetric object (for example a sphere), you have a large symmetry group. If you break the symmetry (for example. you color two antipodal points of the above mentioned sphere), the symmetry group becomes smaller (in this case one is left with rotations on the axis through these points and some reflections). The amount that the group is reduced can be understood as some measure of asymmetry (of the sphere with marked points, with regards to the sphere without marked points). Or is this not something you are after?</p> http://mathoverflow.net/questions/51913/pde-two-dimensional-inhomogeneous/51939#51939 Answer by Thomas Rot for PDE - Two Dimensional Inhomogeneous?? Thomas Rot 2011-01-13T09:36:30Z 2011-01-14T19:57:44Z <p>I think you should look into the theory of fundamental solutions, or Greens functions. Greens functions are solutions to the equation</p> <p>$$G_{xx}+G_{yy}=\delta$$</p> <p>Which are explicitly know in this case (I think it is $\frac{\log(x^2+y^2)}{4\pi}$, but you have to look this up). Solutions of the original equation are then</p> <p>$$U=G\star f+U'$$</p> <p>With $\star$ the convolution and $U'$ a solution to the homogeneous Laplace equation $$U_{xx}'+U_{yy}'=0.$$ Of course this requires some regularity of $f$. If $f$ is a compactly supported distribution this has a distributional solution. If one assumes a certain smoothness as well, this will induce smoothness of the solution.</p> <p>More on this theory can be found in the chapter on fundamental solutions of</p> <p>MR2680692 Duistermaat, J. J. ; Kolk, J. A. C. Distributions. Theory and applications. Translated from the Dutch by J. P. van Braam Houckgeest. Cornerstones. Birkhäuser Boston, Inc., Boston, MA, 2010. xvi+445 pp. ISBN: 978-0-8176-4672-1 </p> http://mathoverflow.net/questions/52032/examples-of-non-metrizable-spaces/52055#52055 Answer by Thomas Rot for Examples of non-metrizable spaces Thomas Rot 2011-01-14T09:25:49Z 2011-01-14T09:44:43Z <p>Function spaces are sometimes not metrizable. Let $X$ and $Y$ be topological spaces, and $C(X,Y)$ be the space of continuous maps from $X$ to $Y$, topologized in the compact open topology. Then $C(X,Y)$ need not be metrizable (it is if $X$ is a compact, and $Y$ is a metric space, it is though).</p> http://mathoverflow.net/questions/51753/conditional-composition-of-partial-functions-vs-relations/51758#51758 Answer by Thomas Rot for conditional composition of partial functions vs relations Thomas Rot 2011-01-11T13:31:30Z 2011-01-11T13:31:30Z <p>I have never heard of this, why do you need it?. I don't think it will be easier to define using relations. But here is my go at it. Let $f:A\rightarrow B$, $g:B\rightarrow C$, be partial functions, and $h:A\rightarrow C$, a function. We can define the composition of two partial functions by $$g\circ f={(a,c)\in A\times C\;|\; (a,b)\in f\quad (b,c)\in g \text{ for some } b\in B}.$$ Denote by $\pi_1:A\times C\rightarrow A$ the projection to the first coordinate (the domain). Then the partial composition can be defined by $$\text{condcmp}(a,b,c)=g\circ f \cup(c\cap(\pi^{-1}(A\setminus\pi_1(g\circ f)).$$ I doubt this can be simplified much more, and I don't think it is cleaner than the description you gave.</p> http://mathoverflow.net/questions/51747/on-pi-1f-omega-with-omega-convex/51756#51756 Answer by Thomas Rot for On $\pi_{1}(f(\Omega))$ with $\Omega$ convex Thomas Rot 2011-01-11T12:40:28Z 2011-01-11T12:40:28Z <p>In general it is not true. We have the following however. </p> <p>$\Omega$ is open. If $f$ is injective, $\Omega$ is homeomorphic to it's image $f(\Omega)$ via $f$ by Brouwers invariance of domain. The induced map $f_*:\pi_1(\Omega)\rightarrow\pi_1(f(\Omega))$ is hence an isomorphism. If $\Omega$ is convex, $\pi_1(\Omega)={1}$, hence $\pi_1(f(\Omega))={1}$.</p> <p>We don't need smoothness of $f$ at all, only continuity.</p> http://mathoverflow.net/questions/122266/detecting-non-transversality Comment by Thomas Rot Thomas Rot 2013-02-21T14:51:41Z 2013-02-21T14:51:41Z James, you can always perturb it so that the unstable connection disappears, with arbitrary small perturbations (of the function, or of the metric (since $f$ is assumed to be morse)). http://mathoverflow.net/questions/122266/detecting-non-transversality Comment by Thomas Rot Thomas Rot 2013-02-19T14:41:41Z 2013-02-19T14:41:41Z Disclaimer: this is not my field of research. There is a vast literature on numerical algorithms trying to detect heteroclinic connections. Some algorithms not only compute a candidate connection, but also give a proof that a connection exists within a certain error bound. This might be interesting to pursue. It is however a hard problem, because the connections between saddles (in two dimensions) are not stable (small perturbations of the function, or the metric, destroy them). I doubt that you can find general analytic algorithms. http://mathoverflow.net/questions/118962/is-a-manifold-with-flat-ends-of-bounded-geometry/119033#119033 Comment by Thomas Rot Thomas Rot 2013-01-17T12:40:03Z 2013-01-17T12:40:03Z Thank you for your very informative answer. http://mathoverflow.net/questions/118962/is-a-manifold-with-flat-ends-of-bounded-geometry Comment by Thomas Rot Thomas Rot 2013-01-15T17:18:55Z 2013-01-15T17:18:55Z @Misha: Thanks. I think I understand the idea in principle, but have to think a little more about the covers appearing in the classification. I would still be very much interested in a more elementary argument, which does depend on the classification. http://mathoverflow.net/questions/118962/is-a-manifold-with-flat-ends-of-bounded-geometry Comment by Thomas Rot Thomas Rot 2013-01-15T13:28:47Z 2013-01-15T13:28:47Z @Thomas Richard, This paper: <a href="http://www.math.sciences.univ-nantes.fr/~carron/flat_end.pdf" rel="nofollow">math.sciences.univ-nantes.fr/~carron/flat_end.pdf</a> claims that the the number of ends is finite (I think they implicitly assume that $M$ is connected). http://mathoverflow.net/questions/115671/is-the-set-of-all-smoothed-closed-simple-curves-on-mathbbr2-a-manifold/115696#115696 Comment by Thomas Rot Thomas Rot 2012-12-07T13:05:07Z 2012-12-07T13:05:07Z @David C: Yes, Hilbert manifolds are nice for this purpose, because one has a gradient flow. http://mathoverflow.net/questions/96863/how-to-explain-the-condition-c-in-critical-point-theory/96876#96876 Comment by Thomas Rot Thomas Rot 2012-11-23T09:43:36Z 2012-11-23T09:43:36Z Thus if this condition does not hold, there is a sequence such that $|\grad f(x_i)|$-&gt;0 in S. The condition forces this sequence to have a limit, and the limit must be a critical point. http://mathoverflow.net/questions/112581/induced-maps-in-morse-homology/112591#112591 Comment by Thomas Rot Thomas Rot 2012-11-20T09:55:35Z 2012-11-20T09:55:35Z Thank you, that clarifies a lot! I misinterpreted the example the first time. http://mathoverflow.net/questions/112581/induced-maps-in-morse-homology/112784#112784 Comment by Thomas Rot Thomas Rot 2012-11-19T14:51:53Z 2012-11-19T14:51:53Z Thank you. I will order a copy of the book. http://mathoverflow.net/questions/112581/induced-maps-in-morse-homology/112591#112591 Comment by Thomas Rot Thomas Rot 2012-11-19T14:48:33Z 2012-11-19T14:48:33Z Thank you for your answer. However, I am not sure I understand your example... What is the problem exactly? I can assume without loss of generality that the metrics chosen make $F$ into a isometry right? This will make sure that isolated trajectories of the gradient flow on the left are mapped to isolated trajectories to the right. Or is the problem in the orientations? I believe you can fix this by fixing the orientations of the complexes. Or am I missing something obvious? http://mathoverflow.net/questions/113781/characterization-of-bounded-geometry-reference-request Comment by Thomas Rot Thomas Rot 2012-11-19T12:33:49Z 2012-11-19T12:33:49Z I have to check, don't have time now, if the statement you want is actually in there, but a lot of these type of theorems are proven in the PhD thesis of Jaap Eldering <a href="http://arxiv.org/pdf/1204.1310v2.pdf" rel="nofollow">arxiv.org/pdf/1204.1310v2.pdf</a> . See for example Lemma 2.6 http://mathoverflow.net/questions/107825/intuition-behind-thom-class/107892#107892 Comment by Thomas Rot Thomas Rot 2012-09-23T12:29:58Z 2012-09-23T12:29:58Z You can edit your original post, as not to clutter up the answers. http://mathoverflow.net/questions/105824/what-is-the-non-intuitive-part-in-sphere-eversion-turning-inside-out Comment by Thomas Rot Thomas Rot 2012-08-29T12:40:22Z 2012-08-29T12:40:22Z The 2 dimensional analog is actually impossible. And it was proven that it was possible, before people had an explicit construction on how to do it. http://mathoverflow.net/questions/104908/loop-space-de-rham-cohomology Comment by Thomas Rot Thomas Rot 2012-08-17T12:50:06Z 2012-08-17T12:50:06Z Of course the question I linked is asking fof singular homology. I do not know anything about the De-Rham complex on infinite dimensional spaces. http://mathoverflow.net/questions/104908/loop-space-de-rham-cohomology Comment by Thomas Rot Thomas Rot 2012-08-17T12:28:46Z 2012-08-17T12:28:46Z This might be of interest <a href="http://math.stackexchange.com/questions/48637/homology-of-the-loop-space" rel="nofollow" title="homology of the loop space">math.stackexchange.com/questions/48637/&hellip;</a> . Once you computed the homology of the based loop space, it is possible via the free loop space fibration $\Omega M\rightarrow \Lambda M\rightarrow M$ to write down spectral sequence which relates the homologies of $M,\Omega M$, and $\Lambda M$.