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 Jun15 comment Volume comparison theoremsOne way to find a formula is to write the unit $n$-sphere as a warped product over $[0,\pi]$ with unit $(n-1)$ sphere as a fiber, and $\sin r$ as the warping function. Then use Fubini theorem for warped product and induction. Is this homework? Jun15 comment Negative curvature in the middle of $R^{3}$I think the paper mathnet.or.kr/mathnet/thesis_file/… in [Bull. Korean Math. Soc. 49 (2012), No. 3, pp. 581–588] does exactly what you want. Jun15 comment Negative curvature in the middle of $R^{3}$What do you mean by the "standard Euclidean metric outside"? Do you mean that after removing your $N$ there is an isometry to the standard $\mathbb R^3$ minus a compact set? BTW, in dimension 2 any smooh nonpositive function on $\mathbb R^2$ can be realized as the scalar (or equaivalently sectional) curvature of a complete Riemannian metric. One simply solves the Jacobi equation $f_{xx}+Kf=0$ where $K$ is the sectional curvature, and then the metric is $dx^2+f^2dy^2$. See page 217 of [Kazdan-Warner, "Curvature Functions for Open 2-Manifolds", Annals of Math. 99, No. 2, (1974), pp. 203-219]. Jun14 comment Do there exist exotic 4-tori?As for curvature restrictions no exotic 4-torus has a metric of nonnegative Ricci curvature (this is due to Cheeger-Gromoll and is a corollary of their splitting theorem). Jun14 comment Do there exist exotic 4-tori?My understanding is that this is open, and in fact no exotic closed aspherical 4-manifolds are known. Jun11 comment Negative sectionnal curvature and constant curvature By "random groups with expanders" I meant the groups constructed by adding relations to a free group randomly according to edges of a expander. Sloppy language... Jun11 comment Negative sectionnal curvature and constant curvature In fact, these hyperbolic groups aren't fundamental groups of complete nonpositivaly curved (Riemannian) manifolds. Jun11 comment Negative sectionnal curvature and constant curvature Misha, there are torsion-free hyperbolic groups that aren't the fundamental group of a manifold in Class 1. Namely, Gromov's random groups with expanders, see Naor-Silberman's theorem 1.1 in arxiv.org/abs/1005.4084. Jun11 comment Explanations for mathematicians, about the falsifiability (or not) of string theoryVoted to closed since this is not a math research question. Specifically, I do not see how falsifability of the string theory affects mathematics. It it quite clear that the string theory has already had a huge impact on (some areas of) math regardless of whether it describes physical reality. Jun9 accepted Negative sectionnal curvature and constant curvature Jun8 comment 3 manifolds with diffeomorphic unit tangent bundlesIn the case when the 3-manifolds are lens spaces some partial classification is due to Sadeeb Ottenburger, see hss.ulb.uni-bonn.de/2009/1820/1820.htm, and arxiv.org/find/math/1/au:+Ottenburger_S/0/1/0/all/…. I am not sure whether there is a clean simple statement in this case. If memeory serves, he can handle all fundamental groups $\mathbb Z_r$ where $r$ is coprime to $6$. The proofs are surgery theoretic. Jun8 revised Negative sectionnal curvature and constant curvature deleted 4 characters in body Jun8 answered Negative sectionnal curvature and constant curvature May27 comment Universal covering of compact surfacesLee Mosher: this depends on how you define a compact surface of genus $g$. If the definition is "connected sum of $g$ tori" no classification is needed. If the definition is "compact oriented 2-manifold of Euler characteristic $2-2g$, then it is needed. I think, the polygon argument is easy, while the classification of surfaces is less so. May26 comment Universal covering of compact surfacesVoted to close because the answer is immediate from the classification of surfaces (as Daniele explains). May21 comment Closed geodesic loops around points in compact manifoldsAnton: the argument in my first comment surely works. The argument in the second comment does not give complete detail, but I think the point is that the gradient flow of the energy functional gives a deformation retraction of the manifold of broken geodesics loops onto a small neighborhood of a constant loop at $p$, and that small neighborhood is contractible. May20 comment Closed geodesic loops around points in compact manifoldsIn fact, one needs very little from the Morse theory because there is no critical points (e.g. the index theorem is not needed); basically, replace the loop space by the finite dimensional manifold of piecewise geodesic loops, and note that the energy functional has no critical points, so the finite dimensional manifold is contractible. The rest is elementary algebraic topology. What could be easier? May20 comment Closed geodesic loops around points in compact manifoldsSuppose $p$ is totally convex, so there is no geodesic loop based at $p$. By the main theorem of the Morse theory the loop space $\Omega_p M$ has a homotopy type of a CW complex with cells correponding to geodesic loops at $p$, so $\Omega_p M$ is contractible, i.e. $\pi_i(M)=0$ for $i>0$. Since $M$ is connected, it is contractible. No closed manifold is contractible by Poincare diality. I do not think there is a proof without Morse theory in disguise. May19 comment Volume growth of covers and growth of deck-transformation groupsThis is very helpful. I see now that I have had some silly confusions about growth. May19 comment Volume growth of covers and growth of deck-transformation groupsBy "equal" I mean that the growth functions have the same growth type, i.e. they dominate each other, where $g$ dominates $f$ if and only if $f(t)\le Ag(At+B)+B$ for all $t$ and some constants $A, B$. May19 asked Volume growth of covers and growth of deck-transformation groups May11 comment The center of a derived subgroup in an amenable groupJuschenko-Monod recently constructed a f.g. simple infinite amenable group $G$. Clearly it has trivial center and its derived subgroups all equal $G$. May9 comment Converse to Milnor’s theorem on manifolds with nonnegative Ricci curvature. The introduction of arxiv.org/abs/math/0109167 lists a number of examples (maybe all known ones?). I have not been following the subject for the last several years but I do not recall any new examples since that paper was written. May1 comment Does a *topological* manifold have an exhaustion by compact submanifolds with boundary?Misha, it wasn't me. Apr30 comment Purely parabolic Kleinian groupsYes, It does extend to the Karlsson-Noskov setting. Apr30 comment Purely parabolic Kleinian groupsI see. This should also extend to the setting of Karlsson-Noskov paper on isometry groups of spaces with contractive bordifications, such as e.g. visibility spaces. Apr30 comment Purely parabolic Kleinian groupsThe original question was actually a toy case of the same question for purely parabolic isometry groups of Hadamard manifolds. Neither argument extends to this setting, as far as I can see. Apr29 comment Purely parabolic Kleinian groupsThanks, Misha! I do not understand what you mean by the Zariski closure argument (what's is special about the Zarisky closure of a purely parabolic group?), but the other one I get: the group generated by high powers of parabolic elements with disjoint fixed points at infinity is geometrically finite (it visibly has a fundamental polyhedron with 4 faces), and hence it contains a hyperbolic element. Apr29 asked Purely parabolic Kleinian groups Apr24 comment Visual boundaries of universal covers of finite-volume nonpositively curved manifoldsSven, I forgot to mention that the lower curvature bound is never needed. Apr24 comment Visual boundaries of universal covers of finite-volume nonpositively curved manifolds(cont) The hyperbolic element is some word in the above hyperbolic elements. Doesn't this work? Apr24 comment Visual boundaries of universal covers of finite-volume nonpositively curved manifoldsI do not know enough of the locally symmetric case to see why the answer is yes (as Misha says), but if so, then it seems you can combine the above to conclude that the answer is always yes, i.e., use rank rigidity to decompose the universal cover as a product of locally symmetric or rank one factors. Fix a point at infinity $z$, pick a tangent vector $v$ in its direction, project it to factors, and approximate the projections by axis of hyperbolic elements, which commute and stabilize the product of axes, which is a flat. Arguing in that flat find a hyperbolic element with axis ending at $z$. Apr24 comment Visual boundaries of universal covers of finite-volume nonpositively curved manifoldsIf $M$ has rank one, then the answer is yes, see Theorem III.3.4 in Ballmann's DMV book. Apr19 answered What does a mathematician expect from mathematics education? Apr5 comment noncompact manifold with two ends splits?Removing two points from a 2-dimensional torus gives a manifold with two ends, but it is not the product of $\mathbb R$ and another manifold because that other manifold would have to be a circle for dimension reasons. The claim that the Busemann function of a line has no critical points is false. Mar18 comment Does a Riemannian manifold with bounded geometry admit an isometric proper embedding into Euclidean space with uniformly thick tubular neighborhood@Malte: completeness is implies by a lower injectivity radius bound (all geodesics extend by a definite amount). The graph of $sin(1/x)$ is a counterexample to your last sentence. Mar6 asked Hyperbolic groups with infinitely generated commutator subgroups Feb20 asked Non-tame 3-manifolds covered by the Euclidean space Feb15 awarded ● Nice Question Feb14 asked Linear groups that are nonlinear over the integers Feb12 answered Connected sum of topological manifolds Feb7 asked Hyperbolic 3-manifolds with no geometrically finite structure Jan12 accepted Is the volume functional contiunuous for compact manifolds with lower bounds on volume? Jan12 revised Is the volume functional contiunuous for compact manifolds with lower bounds on volume?added 32 characters in body Jan12 answered Is the volume functional contiunuous for compact manifolds with lower bounds on volume?