User vitali kapovitch - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:06:07Z http://mathoverflow.net/feeds/user/18050 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103396/geodesics-in-lens-spaces/103403#103403 Answer by Vitali Kapovitch for geodesics in Lens spaces Vitali Kapovitch 2012-07-28T19:37:32Z 2012-07-28T20:05:56Z <p>this is true for all spherical spaceforms. given two non-antipodal points on the sphere, the geodesic circle passing through them is unique. therefore given any point $p\in\mathbb S^n$ with $n>1$ at most finitely many geodesic circles through $p$ can project to closed geodesics of length less than $2\pi$ (or $\pi$ if the group contains $-Id$). On the other hand there will obviously be some shorter closed geodesics in the quotient too.</p> http://mathoverflow.net/questions/102517/how-can-i-tell-whether-a-manifold-is-homogeneous/103016#103016 Answer by Vitali Kapovitch for How can I tell whether a manifold is homogeneous? Vitali Kapovitch 2012-07-24T16:04:16Z 2012-07-24T16:04:16Z <p>as mentioned in various comments there are many geometric conditions of various type that imply homogeneity (or local homogeneity). Constant curvature, parallel curvature tensor, almost flat metric, quarter-pinched curvature, nonnegative bisectional curvature and so on. But in purely topological terms the best you can hope for is necessary conditions of the kind mentioned in the question you linked. Necessary and sufficient conditions are pretty much impossible because topological recognition problems are <strong>hard</strong>.</p> <p>For example, if you look at compact 2-connected homogeneous spaces (or even biquotients) then there are only finitely many of them in every dimension since nonabelian simple Lie groups have only finitely many irreducible representations in every dimension. so in principle the recognition problem for such manifolds ought to be straightforward: just check that that $\pi_1(M)=\pi_2(M)=0$ and compare $M$ to a finite list. However, that last step is actually quite tricky. Surgery theory sort of tells us how to do it but the method is hardly practical.</p> <p>To give a specific example from my own experience. When classifying biquotients with singly generated rational cohomology rings I and Wolfgang Ziller had to deal with one specific biquotient $M^{11}=G_2//SU(3)$ given by a representation $\rho:SU(3)\to G_2\times G_2$ where the representation on the left has index 2 and on the right has index 3. We were able to show that $M^{11}$ is almost diffeomorphic to the unit tangent bundle $T^1S^6$ which is the homogeneous space $SO(7)/SO(5)$. this means that they differ at most by a connected sum with an exotic sphere. but we couldn't decide if they are actually diffeomorphic. to do that one needs to compute the Eells-Kuiper invariant of $M$ which would couldn't do as that requires writing $M$ as a boundary.</p> http://mathoverflow.net/questions/102161/proving-the-existence-of-good-covers/102190#102190 Answer by Vitali Kapovitch for Proving the existence of good covers Vitali Kapovitch 2012-07-13T23:10:15Z 2012-07-14T14:10:18Z <p>you don't really need a whole lot of Riemannian geometry to prove this. Embed the manifold into $\mathbb R^n$ by Whitney and look at very small charts around points given by orthogonal projections onto the tangent spaces. the transition maps will be arbitrary close to identity in $C^2$. that means that a small round disk in one chart will remain strictly convex in nearby charts (because if $f(x)=|x|^2$ and $\phi$ is a transition map such that $\phi-Id$ has small first and second derivatives then $f\circ \phi$ is still strictly convex and hence has convex sublevel sets). This is is all you need to conclude that all intersections are contractible. I guess since the above argument doesn't use any Riemannian geometry notions it should qualify as an answer to the second question?</p> <p>Incidentally, does a good open cover always exist if a manifold is only topological?</p> http://mathoverflow.net/questions/101776/altitudes-of-a-triangle/101904#101904 Answer by Vitali Kapovitch for Altitudes of a triangle Vitali Kapovitch 2012-07-11T01:22:58Z 2012-07-11T01:55:33Z <p>I think this (and other similar facts) can be derived uniformly using elementary analyticity arguments. First you prove it for the round unit sphere. by rescaling this implies that it's true for the round sphere of any radius. now look at the cosine law in the simply connected space form of constant curvature $k$. since this formula is analytic in $k$, the "size" of the potential failure of the altitudes to intersect at the same point (measured in any reasonable way) will also be analytic in $k$ and since it's constantly zero for $k>0$ it must be constantly zero for all $k$. It should not be hard to make the above into a rigorous argument.</p> http://mathoverflow.net/questions/98990/geodesic-circles-on-riemannian-manifolds/99002#99002 Answer by Vitali Kapovitch for Geodesic circles on riemannian manifolds Vitali Kapovitch 2012-06-07T01:49:14Z 2012-06-07T01:49:14Z <p>I'm not sure but I believe you are asking if there always exists a closed geodesic such that it gives a distance preserving embedding of $S^1$ with respect to the length metric on the circle and the ambient metric on the target manifold. If this is the question then the answer is no.</p> <p>Balacheff, Croke and Katz in <a href="http://www.springerlink.com/content/ax605170431104q2/" rel="nofollow">"A Zoll counterexample to a geodesic length conjecture"</a> have an example of a metric on $S^2$ where the length of the shortest closed geodesic is strictly bigger than twice the diameter of $S^2$. This of course gives a counterexample to the above question too.</p> http://mathoverflow.net/questions/98609/algorithmically-unsolvable-problems-in-topology/98614#98614 Answer by Vitali Kapovitch for Algorithmically unsolvable problems in topology Vitali Kapovitch 2012-06-01T23:16:05Z 2012-06-01T23:46:58Z <p>It seems pretty clear to me that if you fix $n$ and look at finite simply connected n-dimensional simplicial complexes then the (rational) homotopy equivalence problem is decidable. It's pretty clear that construction of (rational) Postnikov towers is algorithmic. Comparing two Postnikov towers is a sequence of obstruction problems, each decidable. And you don't need to compare full Postnikov towers, it's enough to compare up to height $n$ (the rest are determined automatically). </p> http://mathoverflow.net/questions/98107/metric-deformations-from-non-negative-to-positive-curvature/98128#98128 Answer by Vitali Kapovitch for Metric Deformations from Non-Negative to Positive Curvature Vitali Kapovitch 2012-05-27T16:06:13Z 2012-05-27T16:06:13Z <p>As Benoît Kloeckner points out this is false for non simply connected manifolds with $RP^2\times RP^2$ being a counterexample (by Synge's theorem). For simply connected manifolds this is a well known open problem. </p> <p>BTW, Sha-Yang examples are only known not to admit nonnegative sectional curvature for connected sums of things like $S^n\times S^m$ for very large number of summands (by Gromov's betti number estimate). For, say, connected sum of 3 copies of $S^n\times S^m$ with $n,m>1$ nothing is known.</p> <p>Also, there are plenty of easier examples of manifolds of positive Ricci curvature other than those of Sha and Yang. For example homogeneous spaces and more generally biquotients of compact Lie groups with finite fundamental groups. <strong>All of them</strong> have positive Ricci curvature and nonnegative sectional curvature but almost none are known to admit positive sectional curvature.</p> <p>Moreover, there are lots of such examples with quasi-positive curvature ( where sectional curvature is positive on a dense open set of points in $M$) and the question is still open for such manifolds. See for example this paper by Wilking <a href="http://www.springerlink.com/content/b5yf3gdp0v8kruqq/" rel="nofollow">"Manifolds with positive sectional curvature almost everywhere."</a></p> <p>Among many other examples Wilking constructs such metrics on $S^2\times S^3$. This shows that even in the simply connected case either the Hopf conjecture (that a product of positively curved manifolds can not be positively curved) is false or the deformation conjecture for quasi-positively curvaed manifolds is false.</p> http://mathoverflow.net/questions/96776/existence-of-a-large-leaf-in-a-foliation-of-the-ball/98015#98015 Answer by Vitali Kapovitch for Existence of a large leaf in a foliation of the ball Vitali Kapovitch 2012-05-26T05:09:59Z 2012-05-26T05:09:59Z <p>I think one might be able to prove this without using the heavy duty machinery from the geometric measure theory that Anton mentioned by modifying the proof of the main result of the following paper of Gromov <a href="http://www.springerlink.com/content/1lnx26ecfnxkxtva/" rel="nofollow">"Isoperimetry of Waists and Concentration of Maps"</a>.</p> <p>He proves a stronger version of your statement but for the sphere instead a disk: for any continuous map $f: \mathbb S^n\to\mathbb R^{k}$ there is a fiber $F$ such that $vol(U_\epsilon(F))\ge vol (U_\epsilon(\mathbb S^{n-k}))$ for any $\epsilon>0$. </p> <p>This of course immediately gives <em>some</em> lower bound on the volume of the maximal leaf in your case because the double of a disk is bilipschitz to a sphere but I think it's likely that one can get a sharp bound too.</p> <p>There is a very <a href="http://arxiv.org/abs/0911.3972" rel="nofollow">readable explanation of Gromov's proof by one of his students posted on the arxiv</a>. That's what I looked at as reading Gromov's papers can be tough.</p> <p>The construction is roughly as follows. He looks at convex polyhedral partitions of $\mathbb S^n$ into $2^N$ (with $N\to\infty$) subsets of equal volume. Given a map $f: \mathbb S^n\to\mathbb R^{k}$ he defines a section of a certain vector bundle over the space of partitions and checks that this bundle has a nonzero top Stiefel-Whitney class so that this section must have a zero. by construction a zero of the section gives a fiber of $f$ passing through the center of mass of every convex set in the partition. The argument is very similar to (and is in fact a generalization of) the proof of the Borsuk-Ulam theorem.</p> <p>Furthermore one can make sure that the convex sets are $\delta(N)$ close to being k-dimensional with $\delta(N)\to 0$ as $N\to\infty$. One then passes to the limit to get a "partition" into convex sets of dimension $\le k$ and the fiber in question is the one that passes through the center of mass of every convex set with respect to the limit of the normalized volume measure. The key geometric part is to prove that for every $k$-dimensional convex set $C$ in the limit and the normalized limit measure $\mu$ on it one has that $$\mu(B_\epsilon(p)\cap C)\ge \frac {vol (U_\epsilon(\mathbb S^{n-k}))}{vol (\mathbb S^n)}$$ where $p$ is the center of mass of $C$. Since the elements of the partitions along the sequence had equal volume this yields the result.</p> <p>The proof of the inequality in the displayed formula is a convexity argument similar to the proof of the Bishop-Gromov volume comparison. I haven't checked the details but I think the whole construction can be adapted to a disk instead of a sphere by using convex partitions of the disk obtained by coning off the elements of the partitions of the sphere to give a sharp bound for the disk. This may even be discussed somewhere in Gromov's paper.</p> http://mathoverflow.net/questions/97830/nontrivial-pi-2diffm/97953#97953 Answer by Vitali Kapovitch for nontrivial $\pi_2(Diff(M))$ Vitali Kapovitch 2012-05-25T15:48:22Z 2012-05-25T16:02:23Z <p>Regarding the question about nontrivial $\pi_k(Diff(M))$ for $k>2$ there are plenty of such examples too. For example Farrell and Hsiang in <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=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;review_format=html&amp;s4=Hsiang&amp;s5=diffeomorphisms&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq&amp;r=3&amp;mx-pid=520509" rel="nofollow">"On the rational homotopy groups of the diffeomorphism groups of discs, spheres and aspherical manifolds."</a> computed the rational homotopy groups of $Diff(S^n)$ in the stability range ($i\lt n/6-7$ ) which in particular gives that in the stability range <code>$\pi_{4i-1}(Diff(S^n))\otimes \mathbb Q\ne 0$</code>. They have some other computations there too. Those are hard results however and if you are not interested in computing the homotopy groups exactly and only want to prove that they are not trivial then much more elementary considerations are sufficient. </p> <p>For example it's well known that for any odd $n$ the space $Aut(S^{n})$ (the identity component of space of self homotopy equivalences of $S^n$) is rationally equivalent to $S^{n}$ and moreover the obvious map $SO(n+1)\to Aut(S^n)$ is an epimorphism on $\pi_n \otimes \mathbb Q$. Since the map $SO(n+1)\to Aut(S^n)$ factors through $SO(n+1)\to Diff(S^n)\to Aut(S^n)$ it follows that $SO(n+1)\to Diff(S^n)$ is not zero on $\pi_n\otimes \mathbb Q$. This gives you lots of examples with nontrivial odd $\pi_i(Diff(M))$. If you want even ones too then one can use the same trick as in Allen Hatcher's example. </p> <p>Pick an element in $\pi_n(SO(n+1))\otimes \mathbb Q$ which maps to the generator of $\pi_n(S^n)\otimes\mathbb Q$ under the evaluation map and let <code>$\alpha: S^{n-1}\to \Omega SO(n+1)$</code> be the corresponding spheroid in $\pi_{n-1}(\Omega SO(n+1))\cong \pi_n(SO(n+1))$ . This gives you a map $\Phi: S^{n-1}\times S^n\times S^1\to S^n\times S^1$ given by $\Phi(x,y,t)=(\alpha(x)(t)(y),t)$. By construction, this is an (n-1)-spheroid in $Diff(S^n\times S^1)$. And it's clearly nontrivial because of the action of $\Phi$ on the cohomology. Therefore, $\pi_{n-1}(Diff(S^1\times S^n))\otimes \mathbb Q\ne 0$ for any odd $n>1$.</p> http://mathoverflow.net/questions/96592/why-does-the-asymptotic-cone-fill-the-holes/96593#96593 Answer by Vitali Kapovitch for Why does the asymptotic cone fill the holes? Vitali Kapovitch 2012-05-10T17:24:22Z 2012-05-10T17:24:22Z <p>when you take Gromov-Hausdorff limits you have to restrict to complete spaces if you want a well-defined notion because otherwise the limits are not unique. After all the Gromov-Hausdorff distance between $\mathbb R^2$ and <code>$\mathbb R^2\backslash \{pt\}$</code> is zero. so no such constructions can see infinitesimal holes. </p> http://mathoverflow.net/questions/93483/smooth-representatives-for-elements-of-pi-7-textexotic-s7/93506#93506 Answer by Vitali Kapovitch for Smooth representatives for elements of $\pi_7(\text{exotic$S^7$})$ Vitali Kapovitch 2012-04-08T15:57:14Z 2012-04-08T16:30:23Z <p>Duràn wrote down an explicit formula for such map in <a href="http://www.springerlink.com/content/v775g60776g6m813/" rel="nofollow">"Pointed Wiedersehen Metrics on Exotic Spheres and Diffeomorphisms of $S^6$"</a>. That is he wrote an explicit formula for an exotic diffeomorphism from $S^6$ to $S^6$ which is homotopic but not isotopic to the identity. This the produces an explicit homeomorphism from $S^7$ to an exotic sphere by glueing as described by Ryan in his answer.</p> <p>Geometric properties of that particular map were later studied by various people. For example, it's written down explicitly on page 1 in <a href="http://www.ime.usp.br/~eufrasio/matcont/pt/edicoes.php?idvolume=29" rel="nofollow">"Bootstrapping $Ad$-Equivariant Maps, Diffeomorphisms and Involutions"</a> by Duràn and Rigas and Sperança (this link is freely accessible unlike the first one).</p> http://mathoverflow.net/questions/93434/when-is-the-connected-sum-of-manifolds-orientation-independent/93504#93504 Answer by Vitali Kapovitch for When is the connected sum of manifolds orientation-independent? Vitali Kapovitch 2012-04-08T15:08:38Z 2012-04-08T15:22:50Z <p>In the oriented category it's not true that if <code>$M\#N$</code> is oriented diffeomorphic to <code>$M\#(-N)$</code> then $N$ admits an orientation reversing diffeomorphism. I suspect there are easier counterexamples but the following works. Take $M^{4k+3}=S^3\times \mathbb{CP}^{2k}$ with $k\ge 1$. Then there exists an exotic sphere $\Sigma^{4k+3}$ such that $\Sigma$ does not admit an orientation reversing diffeomorphism but <code>$M\#\Sigma$</code> is oriented diffeomorphic to $M$ (and hence also to <code>$M\#(-\Sigma)$</code>). Note that $M$ obviously admits an orientation reversing diffeomorphism because $S^3$ does.</p> <p>I read about this fact in a paper by Belegradek, Kwasik and Schultz <a href="http://arxiv.org/abs/0912.4874" rel="nofollow">"Codimension two souls and cancellation phenomena"</a>. Not sure if this is the earliest reference, perhaps Igor can clarify this - he visits MO regularly.</p> <p>More specifically they show that if $I(M)$ is the inertia group of $M$ (the group of oriented exotic $4k+3$-spheres $\Sigma$ such that the standard homeomorphis <code>$M\#\Sigma\to M$</code> is homotopic to a diffeomorphism) then $I(S^3\times \mathbb{CP}^{2k})\cap bP_{4(k+1)}$ has index 2 in $bP_{4(k+1)}$ where $bP_{4(k+1)}$ is the group of exotic $(4k+3)$-spheres bounding parallelizable manifolds. It's known that $bP_{4(k+1)}$ is cyclic or order exponentially growing in $k$ so any nontrivial element $\Sigma$ of $I(S^3\times \mathbb{CP}^{2k})\cap bP_{4(k+1)}$ of order different from 2 works.</p> http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms/92438#92438 Answer by Vitali Kapovitch for To which extent can one recover a manifold from its group of homeomorphisms Vitali Kapovitch 2012-03-28T03:58:01Z 2012-03-28T03:58:01Z <p>This is more of a longish comment rather but I'd like to point out that while Igor's reference in principle gives a complete answer, actually reading off any specific information about $M$ (such as its dimension) from some topological invariants of $Diff(M)$ is likely hard. </p> <p>Moreover, if one relaxes the categories somewhat then the answer to Misha's question can even be negative! Specifically, one can ask if the homotopy type of the monoid of self homotopy equivalences of $M$ determines $M$ up to homotopy type. I actually don't know the answer to this but if one relaxes the category even further and looks at <em>the rational homotopy type</em> then in contrast with the diffeomorphism case the answer is actually <strong>NO</strong>.</p> <p>Specifically, it's rather easy to compute that the rational homotopy type of the identity component $Aut(M)$ of the monoid of self homotopy equivalences of an equal rank biquotient of Lie groups $M=G//H$ that satisfies Halperin's conjecture (which says that in this case $H^\ast (M,\mathbb Q)$ has no negative degree derivations) is determined by rational homotopy and homology groups of $M$. In this case $Aut(M)$ is rationally equivalent to a product of finitely many odd dimensional spheres and one can write an explicit (if somewhat ugly) formula for the dimensions of the spheres that show up in terms of $\pi_\ast(M)\otimes \mathbb Q$ and $H_\ast(M,\mathbb Q)$.</p> <p>But there are plenty of examples of such biquotients in dimensions above 5 which have distinct rational types but the same rational homotopy and homology. For example, one can take $G//T$ where $G$ is a simply connected Lie group and $T\le G\times G$ is a torus of the same rank as rank $G$. All such biquotients satisfy Halperin's conjecture so the formula I mention above applies. It is then clear that rational homology and homotopy groups of $G//T$ are completely determined by $G$ but the rational type of $G//T$ can be different depending on the embedding $T\to G\times G$. There are infinitely many such examples already in dimension 6 of the form $(S^3\times S^3\times S^3)//T^3$. Still, in this case one can read off for example, the dimension of $M$ from the knowledge of the rational homotopy groups of $Aut(M)$ but I don't know how to get such formula for a general closed simply connected manifold $M$ (and I'm not even sure if it's possible).</p> http://mathoverflow.net/questions/90419/is-the-space-of-directions-an-inner-metric-space-for-inner-metric-space-of-curvat/90422#90422 Answer by Vitali Kapovitch for Is the space of directions an inner metric space for inner metric space of curvature $\ge k$? Vitali Kapovitch 2012-03-07T03:13:39Z 2012-03-07T03:13:39Z <p>I'm not familiar with that reference but a standard example for this is by Stephanie Halbeisen <a href="http://www.springerlink.com/content/wm7wxk00kt4yjygf/" rel="nofollow">"On tangent cones of Alexandrov spaces with curvature bounded below".</a></p> <p>The example is necessarily infinite dimensional as it's well known that this can not happen in finite dimensions. Also, it's worth noting that there is no canonical definition of a space of directions for infinite dimensional Alexandrov spaces. The definition that Halbeisen uses is a metric completion of equivalence classes of geodesic segments starting at $p$. But there are other natural definitions possible (say, by looking at ultralimit of pointed blow ups of $X$ at $p$). All these definitions agree for finite dimensional Alexandrov spaces but not for infinite dimensional ones.</p> http://mathoverflow.net/questions/90124/3rd-homotopy-group-of-a-compact-simple-lie-group/90154#90154 Answer by Vitali Kapovitch for 3rd homotopy group of a compact Simple Lie Group Vitali Kapovitch 2012-03-03T22:11:05Z 2012-03-03T23:20:14Z <p>This number is called the index of the map $\phi: SU(2)\to G$. It can be defined for any homomorphism $\phi:H\to G$ where $H$ is simple. Algebraically it can be computed as follows. Since $\mathfrak h$ is simple the restriction of the Killing form of $\mathfrak g$ to $\mathfrak h$ is a constant multiple of the Killing form of $\mathfrak h$. That constant is the index of $\phi$. In the specific case you are asking about for a simple root $\alpha$ the index can also be written as $\frac{(\alpha_{max},\alpha_{max})}{(\alpha,\alpha)}$ where $\alpha_{max}$ is the longest simple root of $\mathfrak g$. Note that from the classification of compact simple Lie groups this can only be equal to 1,2 or 3.</p> <p>See <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=1266842" rel="nofollow">Onishchik, "Topology of transitive transformation groups"</a>, §3.10 and §17.2 for details on this.</p> http://mathoverflow.net/questions/89542/a-followup-on-non-homogeneous-spaces/89550#89550 Answer by Vitali Kapovitch for A followup on non-homogeneous spaces. Vitali Kapovitch 2012-02-26T04:14:29Z 2012-02-26T04:35:04Z <p>Yes, <a href="http://www.springerlink.com/content/n068604337723788/" rel="nofollow">Eschenburg constructed an infinite family of simply connected 7-dimensional examples</a> and proved that many of them are not homotopy equivalent to homogeneous spaces. His examples are biquotients however. In fact the only known examples of closed positively curved manifolds are biquotients or spaces of cohomogeneity one and above dimension 13 only homogeneous examples (and space forms) are known.</p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89359#89359 Answer by Vitali Kapovitch for Example of a manifold which is not a homogeneous space of any Lie group Vitali Kapovitch 2012-02-24T03:03:49Z 2012-02-24T04:56:31Z <p>Apart from already mentioned non simply connected examples most simply connected manifolds are also not homogeneous. One easy criterion is that simply connected homogeneous spaces are rationally elliptic, i.e. they have finite dimensional total rational homotopy. That is because any connected Lie group is rationally homotopy equivalent to a product of odd dimensional spheres. so a homogeneous space is elliptic by a long exact homotopy sequence.</p> <p>Most simply connected manifolds are not rationally elliptic. For example the connected sum of more than two $CP^2$'s or $S^2\times S^2$'s. This is easily seen by looking at their minimal models. But even without computing minimal models it's known that an elliptic manifold $M^n$ has nonnegative Euler characteristic and has the total sum of its Betti numbers $\le 2^n$. So anything that violates either of these conditions such as the connected sum of several $S^3\times S^3$'s is definitely not rationally elliptic and hence can not be a homogeneous space or even a biquotient.</p> <p>As for higher genus surfaces it should not be hard to show that they can not be homogeneous spaces $G/H$ even if you don't assume that $G$ acts by isometries. If $G/H=S^2_g$ and the $G$ action is effective then for any proper normal $K\unlhd G$ which does not act transitively on $S^2_g$ we must have $K/(K\cap H)=S^1$. But then $G/K$ is also 1-dimensional and hence also a circle which is obviously impossible. This reduces the situation to the case of $G$ being simple which can also be easily ruled out for topological reasons.</p> http://mathoverflow.net/questions/88033/existing-proofs-of-rokhlins-theorem-for-pl-manifolds Existing proofs of Rokhlin's theorem for PL manifolds Vitali Kapovitch 2012-02-09T20:27:42Z 2012-02-10T01:18:43Z <p>I'm looking for a comprehensive reference to existing proofs of Rokhlin's theorem that a 4-dimensional closed spin PL manifold has signature divisible by 16. I'm specifically interested in direct proofs (if any such exist) which do not rely on the fact that $\pi_i(PL/O)=0$ for small $i$.</p> <p>The most commonly cited reference seems to be the book by Kirby <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=1001966" rel="nofollow">"The Topology of 4-manifolds"</a>. But the proof there is for smooth manifolds and I'm not sure why it works for PL manifolds although I've seen it claimed in various places that it does. The same is said about Rokhlin's original proof but I don't know why that's true either. I would also like to know if other proofs for PL manifolds exist. I'm particularly interested to know if there is a PL proof based on the Atiyah-Singer index theorem. </p> http://mathoverflow.net/questions/87343/is-the-lipschitz-unique-for-close-alexandrov-space-other-than-dimension-4/87353#87353 Answer by Vitali Kapovitch for Is the Lipschitz unique for close Alexandrov space other than dimension 4? Vitali Kapovitch 2012-02-02T18:16:41Z 2012-02-02T21:57:34Z <p>First, I want to point out that unlike for manifolds there is no notion of a geometric Lipschitz structure for general Alexandrov spaces as they are not locally modeled on a fixed space topologically. One can still ask whether you can have two Alexandrov spaces which are homeomorphic but not bi-Lipschitz homeomorphic but that is a much more metrically rigid question. This is an open problem for $n>4$. I believe your second question is open too mostly because there are so few examples of positively curved smooth manifolds. $\mathbb{CP}^2$ and $\mathbb S^4$ are all the known orientable ones in dimension 4 and as far as I know neither is known to admit an exotic lipschitz structure.<br> *Edit: Actually, $\mathbb {RP}^4$ may be a candidate for this. If I understand things correctly there exists an exotic smooth $\mathbb{RP}^4$ by a result of <a href="http://www.jstor.org/stable/2006987" rel="nofollow"> Fintushel and Stern</a>. I'm actually somewhat confused by this because Fintushel and Stern only say that their example is homotopy equivalent to $\mathbb{RP}^4$ but don't mention homeomorphism. But I've seen people claim that Fintushel and Stern example produces a topological $\mathbb{RP}^4$. Perhaps someone in the area can clarify this. But assuming this is the case one can hope to show that their example is not bilipschitz to the standard $\mathbb{RP}^4$. I don't think this is known however. As far as I know the only tools for distinguishing lipschitz structures on 4-manifolds are due to Sullivan and Donaldson in <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=1032074" rel="nofollow">"Quasiconformal 4-manifolds"</a>. They are Lipschitz versions of Donaldson invariants and I think that theory only works for manifolds with $b_2>1$ which is not the case here.*</p> <p>Note that if you could construct two homeomorphic but not bilipschitz positively curved Alexandrov manifolds $X$ and $Y$ of dimension 4 (or more generally positively curved Alexandrov spaces of dimension 4) then you'd also get a counterexample to your first question given by spherical joins $X* X$ and $Y*Y$. This is because any bilipschizt homeomorphism between $X* X$ and $Y* Y$ would need to preserve the topological strata. So it would send $X$ to $Y$ and since they are convex in $X*X$ and $Y*Y$ respectively the resulting homeomorphism $X\to Y$ would be bilipschitz with respect to the original metrics on $X$ and $Y$.</p> http://mathoverflow.net/questions/70635/manifold-with-all-geodesics-of-morse-index-zero-but-no-negatively-curved-metric/86975#86975 Answer by Vitali Kapovitch for Manifold with all geodesics of Morse index zero but no negatively curved metric? Vitali Kapovitch 2012-01-29T19:28:35Z 2012-01-29T21:07:46Z <p>As mentioned by Rbega the question should be amended to ask whether it's true that a closed manifold $M$ without conjugate points admits a metric of <strong>non-positive</strong> (rather than negative) curvature (otherwise a torus is an obvious counterexample). In that form this is a well-known open problem. The exponential map at any point is a universal covering of $M$ and the geodesics in $\tilde M$ are unique. This does show that $M$ is aspherical but that is a long way from admitting a metric of nonpositive curvature.</p> <p>There are some partial results suggesting that fundamental groups of manifolds without conjugate points share some properties of fundamental groups of nonpositively curved manifolds. In particular, there is a <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=0847526" rel="nofollow">result of Croke and Shroeder</a> that if the metric is analytic then any abelian subgroup of $\pi_1(M)$ is embedded quasi-isometrically. By the following observation of Bruce Kleiner the analyticity condition can be removed: Croke and Schroeder show that even without assuming analyticity for any $\gamma\in\pi_1(M)$ its minimal displacement $d_\gamma$ satisfies $d_{\gamma^n}=nd_\gamma$ for any $n\ge1$. This then implies that $d_\gamma=\lim_{n\to\infty} d(\gamma^nx,x)/n$ for any $x\in\tilde M$. This in turn implies that the restriction of $d$ to an abelian subgroup $H \simeq \mathbb Z^n$ extends to a norm on $\mathbb R^n$. This implies that $H$ is quasi-isometrically embedded.</p> <p>This result implies for example that nonflat nilmanifolds can not admit metrics without conjugate points and more generally that every solvable subgroup of the fundamental group of a manifold without conjugate points is virtually abelian.</p> <p>But it's unlikely that any such manifold admits a metric of non-positive curvature. It is more probable that its fundamental group must satisfy some weaker condition such as <em>semi-hyperbolicity</em> but even that is completely unclear. The natural <em>bicombing</em> on $\tilde M$ given by geodesics need not satisfy the fellow traveler property (at least there is no clear reason where it should come from).</p> <p>So it might be worth trying to look for counterexamples and the first place I would look is among groups that are semi-hyperbolic but not $CAT(0)$. Specifically, any $CAT(0)$ group has the property that centralizers of non-torsion elements virtually split. This need not hold in a semi-hyperbolic group with the simplest example given by any nontrivial circle bundle over closed surfaces of genus $>1$. To be even more specific one can take the unit tangent bundle $T^1(S_g)$ to a hyperbolic surface. Note however that it's known that a closed homogenous manifold without conjugate points is flat so if there is a metric without conjugate points on $T^1(S_g)$ it can not be homogeneous. *Edit: Actually, this last remark is irrelevant as $T^1(S_g)$ can not admit any homogeneous metrics at all.*</p> http://mathoverflow.net/questions/86807/smoothability-of-compact-alexandrov-surfaces-with-curvature-bounded-from-below/86868#86868 Answer by Vitali Kapovitch for Smoothability of compact Alexandrov surfaces with curvature bounded from below. Vitali Kapovitch 2012-01-27T23:31:45Z 2012-01-29T03:16:01Z <p><em>Edit: Addressing Igor's comment I'd like to correct the references I gave. The correct reference for the exact argument I sketch should be the original book by Alexandrov <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=29518" rel="nofollow">"Intrinsic Geometry of Convex Surfaces"</a>(Chapter 7, section 6, Lemmas 1-3). In the later book by Alexandrov and Zalgaller a similar (but necessarily more complicated) argument is given for the case of surfaces with bounded integral curvature (Theorem 10, page 84). One can reconstruct the original proof in the easier case of curvature bounded below from that one but it requires some work.</em></p> <p>This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=216434" rel="nofollow">"Intrinsic geometry of surfaces"</a>).</p> <p>The general structure the proof is as follows. You take a very fine triangulation of $X$ and substitute the curved triangles by triangles with the same sides in the space form of constant curvature $-1$. This will give you a polyhedral surface of curvature $\ge -1$ (you only need to check that the cone angles at vertices are $\le 2\pi$ which is immediate from the definition of an Alexandrov space). Away from the cone points the metric will be smooth of constant curvature $-1$. The cone points can then be easily smoothed to get a smooth metric of $\sec\ge -1$. The hard part is to show that the resulting polyhedral surface is close to the original space $X$.</p> <p>A more interesting result is another theorem of Alexandrov ( <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=29518" rel="nofollow">"Intrinsic Geometry of Convex Surfaces"</a>) that locally any 2-dimensional Alexandrov space of $curv\ge -1$ is isometric to a level set of a convex function in $\mathbb H^3$. When the lower bound is 0 he proved an even sharper result that any 2-sphere of curvature $\ge 0$ is isometric to the boundary of a convex body in $\mathbb R^3$.</p> <p>BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too. </p> http://mathoverflow.net/questions/85802/hopf-tori-in-s3/85811#85811 Answer by Vitali Kapovitch for Hopf Tori in $S^3$ Vitali Kapovitch 2012-01-16T14:03:22Z 2012-01-16T14:11:18Z <p>Your definition of the Clifford torus is off. The usual definition of the Clifford torus is the set $(z_1,z_2)\in\mathbb C^2$ in the unit sphere $|z_1|^2+|z_2|^2=1$ with $|z_1|^2=|z_2|^2=\frac 1 2$. This is a square torus isometric to $\mathbb R^2/\Gamma_c$ with $\Gamma_c$ generated by $(2\pi/\sqrt 2, 0), (0, 2\pi/\sqrt 2)$ (<strong>not $(2\pi, 0), (0, 2\pi)$</strong>) which is isometric to $\mathbb R^2/\Gamma_1$ by a $\pi/4$ rotation.</p> http://mathoverflow.net/questions/85694/behavior-of-sectional-curvature-under-metric-deformations/85756#85756 Answer by Vitali Kapovitch for Behavior of sectional curvature under metric deformations Vitali Kapovitch 2012-01-15T19:23:31Z 2012-01-15T22:22:10Z <p>Formula 2) is the correct one in general except it's the derivative of the sectional curvature i.e of $\frac{k_t(X,Y)}{|X\wedge Y|^2_t}$ (and not just of $k_t(X,Y)$) for an orthonormal frame $X,Y$ with respect to the original metric. This accounts for the last term in formula 2. For $k'(0)$ itself the correct formula is $$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R(X,Y)Y,X)$$</p> <p>As Deane said, this is a straightforward calculation. But since the OP seems to be struggling with it I'll supply some details. Let $\nabla^t$ be the Levi-Civita connection for $g_t$. Then $\nabla^t_XY=\nabla_XY+tS(X,Y)+O(t^2)$ where $S$ is a (2,1)-tensor and $\nabla=\nabla^0$. Let's first compute $k'(0)$ in terms of $S$. As usual let's work in normal coordinates around $p$ with $X,Y$ coordinate fields.</p> <p>We have $$\langle R^t(X,Y)Y,X\rangle_t=\langle R^t(X,Y)Y,X\rangle_0+t\cdot h(R(X,Y)Y,X)+O(t^2)=$$</p> <p>$$=\langle \nabla^t_X\nabla^t_YY,X\rangle_0-\langle \nabla^t_Y\nabla^t_XY,X\rangle_0+t\cdot h(R(X,Y)Y,X)+O(t^2)$$ Next we expand the first term</p> <p>$$\langle \nabla^t_X\nabla^t_YY,X\rangle_0=\langle \nabla^t_X(\nabla_YY+tS(Y,Y)),X\rangle_0+O(t^2)=$$</p> <p>$$\langle \nabla_X\nabla_YY,X\rangle_0+t\langle S(X,\nabla_YY)+\nabla_XS(Y,Y), X\rangle_0+O(t^2)=$$</p> <p>$$\langle \nabla_X\nabla_YY,X\rangle_0+t\langle \nabla_XS(Y,Y), X\rangle_0+O(t^2)$$</p> <p>where in the last equality we used that $\nabla_YY(p)=0$. After a similar computation for $\langle \nabla^t_Y\nabla^t_XY,X\rangle_0$ we get that</p> <p>$$k'(0)=\langle\nabla_XS(Y,Y)-\nabla_YS(X,Y), X\rangle_0+h(R(X,Y)Y,X)$$</p> <p>$$=\nabla_X S(Y,Y,X)-\nabla_YS(X,Y,X)+h(R(X,Y)Y,X)$$</p> <p>where we lowered the index and turned $S$ into a $(3,0)$-tensor $S(X,Y,Z)=\langle S(X,Y),Z\rangle_0$</p> <p>Lastly, recall that</p> <p>$\langle \nabla_XY,Z\rangle=\frac{1}{2}[X\langle Y,Z\rangle+Y\langle X,Z\rangle -Z\langle X, Y\rangle]$ for coordinate fileds. This easily gives</p> <p>$S(Y,Y,X)=\frac{1}{2}[Yh(X,Y)+Yh(X,Y)-Xh(Y,Y)]=\nabla_Yh(X,Y)-\frac{1}{2}\nabla_Xh(Y,Y)$ and $S(X,Y,X)=\frac{1}{2}\nabla_Yh(X,X)$ written invariantly as tensors.</p> <p>Altogether this gives $$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R(X,Y)Y,X)$$ as promised.</p> <p>Formulas 1) and 3) are not true in general but might be true in the specific circumstances where they are applied in the papers in question. </p> http://mathoverflow.net/questions/85539/conformally-flat/85542#85542 Answer by Vitali Kapovitch for Conformally-flat Vitali Kapovitch 2012-01-13T00:56:32Z 2012-01-13T00:56:32Z <p>I'm not quite sure what you mean by <em>always</em> non-positively curved. If you are asking if this metric is non-positively curved for any $f$ then this is false. If you are asking for conditions on $f$ ensuring that the resulting metric is non-positively curved then there is a general formula:</p> <p>Let $(M,g)$ be a Riemannian manifold and let $\tilde g=e^{2f}g$ be a new metric on $M$. Let $p\in M$ and let $u,v$ be orthonormal with respect to $g$ vectors in $T_pM$ and $\sigma$ the 2-plane spanned by them. </p> <p>Then <code>$e^{2f}\tilde{K}_\sigma =K_\sigma-[Hess_f(u,u)+Hess_f(v,v)+|\nabla f|^2-\langle \nabla f, u \rangle^2-\langle \nabla f, v \rangle^2]$</code>.</p> <p>This formula is in Besse btw (Theorem 1.159) but it's written slightly differently there.</p> <p>In the special case you are interested in $g$ is the canonical metric on $\mathbb R^n$ and hence $\tilde K$ is nonpositive iff $H_f(u,u)+H_f(v,v)+|\nabla f|^2-\langle \nabla f, u \rangle^2-\langle \nabla f, v \rangle^2\ge 0$ for any $p$ and any orthonormal $u$ and $v$ in $T_pM$. Note that for example it's always true if $f$ is convex.</p> http://mathoverflow.net/questions/85420/minimum-number-of-partial-views-to-cover-a-smooth-shape/85429#85429 Answer by Vitali Kapovitch for Minimum Number of Partial Views to Cover a Smooth Shape Vitali Kapovitch 2012-01-11T17:35:56Z 2012-01-11T17:35:56Z <p>This is very far from a full answer but for a centrally symmetric <strong>smooth</strong> compact convex body $S$ in $\mathbb R^n$ the "visibility" number is $n+1$. In particular, for a ball in $\mathbb R^3$ it's 4. The interesting part is to bound the minimal number from below.</p> <p>Suppose $k$ points $p_1,\ldots, p_k$ is enough. Let $U_i$ be the region of the boundary of $S$ visible from $p_i$. Observe that since $S$ is smooth and centrally symmetric, no $U_i$ contains any antipodal points. That means that their projections to $\mathbb{RP}^{n-1}$ are injective and we get a covering of $\mathbb{RP}^{n-1}$ by $k$ contractible sets. Therefore $k$ is bigger than or equal to the 1+(cup-length of $\mathbb{RP}^{n-1})=1+(n-1)=n$. (This is a very easy to prove and well-known fact about Lusternik-Schnirelmann category). This gives a lower bound $k\ge n$. But it can be improved to $n+1$. Suppose $n$ points are enough. If we remove one point then by above the projections of the sets $U_1,\ldots U_{n-1}$ don't cover $\mathbb{RP}^{n-1}$. Hence there is a pair of antipodal poits in $S$ not covered by any of $U_1,\ldots U_{n-1}$. That means that both of these points must be covered by $U_n$ which is impossible. Therefore $k\ge n+1$.</p> <p>Note that smoothness of $S$ is key in the above argument. For example a double of a round cone in $\mathbb R^n$ along the base is visible from just two points for any $n$. </p> <p>Lastly, let me echo Igor Rivin's remark that for a non-convex compact body the visibility number can be arbitrary high so I don't think anything meaningful can be said there.</p> <p><em>edit:</em> I see from Joseph O'Rourke's comment that this is a well-studied problem so I assume the above must be well-known.</p> http://mathoverflow.net/questions/84536/the-inertia-subgroup-of-theta-n-for-lie-groups/84548#84548 Answer by Vitali Kapovitch for The inertia subgroup of $\Theta_n$ for Lie groups Vitali Kapovitch 2011-12-29T20:40:51Z 2011-12-29T20:40:51Z <p>There is no natural Lie group structure on a connected sum of a Lie group and an exotic sphere. Where would it possibly come from?</p> <p>Also, the implication you are trying to derive is wrong. Note that $S^3$ is a Lie group and hence so is a product of several $S^3$s. I'm not an expert on surgery theory but it's well-known that the <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=0275453" rel="nofollow">inertia group of a product of spheres is trivial</a> which means that the opposite of the statement you are after holds: the connected sum of a product of spheres with an exotic sphere is <em>never</em> orientably diffeomorphic to the original manifold. So for example the inertia group of $S^3\times S^3\times S^3$ is trivial while $\Theta_9$ has order 8.</p> http://mathoverflow.net/questions/20868/rational-homotopy-theory-of-a-punctured-manifold/84348#84348 Answer by Vitali Kapovitch for Rational homotopy theory of a punctured manifold Vitali Kapovitch 2011-12-27T01:17:24Z 2011-12-27T04:06:37Z <p>This is an old question but I hope the following is still of interest.</p> <p>If $M^n$ is a closed simply connected manifold then the inclusion <code>$M^n\backslash \{ pt\}\hookrightarrow M^n$</code> is $(n-1)$-connected which means that the induced map of minimal models is an isomorphism through dimension $(n-2)$. Next note that <code>$M^n\backslash \{ pt\}$</code> has zero homology in degrees above $n-2$. It's a general fact that given a minimal model up to dimension $k$ of a space whose cohomology vanishes in degrees above $k$ the rest of the minimal model is determined uniquely (and constructively) from the model up to degree $k$. This provides an easy recipe for computing the minimal model of <code>$M^n\backslash \{ pt\}$</code> which can be more explicitly described as follows.</p> <p>If $(\Lambda V, d)$ is a minimal model of $M^n$ then consider the following dga $(A,d)=(\Lambda V\oplus\Lambda \langle z\rangle/(z^2), d)$ with $\deg z=n-1, V\cdot z=0$ and $dz=[M]$ - the fundamental class of $M$. This is a model (non-minimal and even a non-free one!) of <code>$M^n\backslash \{ pt\}$</code>. In practice it's easier to directly compute the minimal model of $A$ by the general procedure outlined above.</p> <p>Here are a couple of examples.</p> <p>Let $M^4=\mathbb {CP}^2$. Its minimal model is $(\Lambda \langle x,y\rangle,d)$ with deg x=2, deg y=5, dx=0, dy=x^3. Up to degree $n-2=2$ this is simply given by $\Lambda\langle x\rangle$ with dx=0. Next we need another generator to make $H^4=0$ (which is currently generated by $[M]=x^2$) so we add $z$ of deg $3$ such that $dz=x^2$. Now the model $(\Lambda \langle x,z\rangle,d)$ with deg x=2, deg z=3, dx=0, dy=x^2 already has $H^i=0$ for $i\ge 4$ so we don't need to add anything else. The resulting model is easily recognized as the model of $\mathbb S^2$ which is of course not surprising since <code>$\mathbb{CP}^2\backslash\{pt\}$</code> is a Hopf disk bundle over $\mathbb{CP}^1$.</p> <p>A more interesting example: Let $M=\mathbb S^3\times\mathbb S^5$. Its minimal model is generated by $x,y$ with $\deg x=3, \deg y=5$ and $dx=0,dy=0$. Applying our recipe the model of <code>$\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$</code> will be the same through dimension 6. Next, we need to kill off cohomology in degree 8 which is currently generated by $[M]=xy$. So we need another generator $z$ of degree 7 with $dz=xy$. However, adding such generator introduces more cohomology in degrees 10 and 12 generated by $xz$ and $yz$. So we need two more generators $a$ and $b$ with $da=xz, db=yz$. However, adding those introduces yet more cohomology and we need to keep adding more generators. This will continue forever because <code>$\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$</code> is rationally hyperbolic.</p> <p>Lastly, let me mention that operations such as cell attachments (or in this case cell deletions) are usually easier handled by Quillen Lie algebra models which are better suited to work with cofibrations (while Sullivan models are better suited for fibrations). In this particular case it's especially easy. If $(\mathbb L_V,d)$ is a minimal Quillen Lie model of $M^n$ then the model of <code>$M^n\backslash \{ pt\}$</code> is obtained by simply removing a single generator from $V$ corresponding to the fundamental class of $M$.</p> http://mathoverflow.net/questions/50933/a-non-formal-space-with-vanishing-massey-products/83832#83832 Answer by Vitali Kapovitch for A non-formal space with vanishing Massey products? Vitali Kapovitch 2011-12-18T22:21:09Z 2011-12-20T23:10:40Z <p><em>Edit: Let me give it one more try -- I think I've fixed the example.</em> </p> <p>The following is the simplest simply connected example I know (note that André's example has a large fundamental group).</p> <p>Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,y,z, a_1,a_2,b_1,b_2, c_1,c_2,\ldots\rangle,d)$ with $\deg u = \deg v = \deg w = 3, \deg x = \deg y = \deg z =8, \deg a_i = \deg b_i =\deg c_i= 10$ and the following differentials: </p> <p> $du=dv=dw=0, dx=uvw, dy=dz=0, da_1=xu+yv, da_2=xu+zw, db_1=xv+yw,$ </p> $db_2=xv+zu, dc_1=xw+yu, dc_2=xw+zv$ </p> Keep adding generators to kill off all cohomology in degrees above 11. I believe this algebra finally has all Massey products equal to zero. For degree reasons the only possible nontrivial Massey products can be for two classes of degree 3 and one of degree 6, e.g. something of the form $\langle [u], [vw], [v]\rangle$. As far as I can tell all of them vanish. </p> <p>On the other hand this algebra is not formal. Indeed, suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V, d),0)$. Let $I=\ker \phi$. Since $\phi$ is a quasi-isomorphism, every closed element of $I$ is exact. Now, since $[y]$ and $[z]$ are independent cohomology classes of degree 8 we must have that $I^{\le 8}=\langle x+k_1y+k_2z\rangle$ for some $k_1,k_2\in\mathbb Q$. Then the class $[u(x + k_1y+k_2z)] = [ux] + k_1[u][y]+k_2[uz]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal. </p> <p>Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.</p> http://mathoverflow.net/questions/83585/does-the-following-condition-imply-the-homotopy-type-of-a-wedge-of-spheres/83587#83587 Answer by Vitali Kapovitch for Does the following condition imply the homotopy type of a wedge of spheres? Vitali Kapovitch 2011-12-16T02:23:16Z 2011-12-16T02:23:16Z <p>No, this is false. For example take $\mathbb{CP}^2\vee S^1\vee S^3$. It admits a cell decomposition and cohomology groups as you describe but clearly has a different homotopy type then a wedge of spheres because the cohomology ring structure is different.</p> http://mathoverflow.net/questions/83445/hyper-complex-and-quaternionic-kahler-geometry/83470#83470 Answer by Vitali Kapovitch for Hyper-Complex and quaternionic Kahler Geometry Vitali Kapovitch 2011-12-14T21:29:35Z 2011-12-14T22:16:38Z <p>By a recent result of <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=364613&amp;vfpref=html&amp;r=1&amp;mx-pid=2793859" rel="nofollow">Gauduchon, Moroianu and Semmelmann</a> if a positive quaternion Kahler manifold admits <strong>any</strong> almost complex structure then it must be the complex Grassmannians $Gr_2(\mathbb C^{n+2})$. The latter is not hypercomlex so there are no hypercomplex manifolds which are also positive Quaternion Kahler. Note that it's even easier if you assume that the hypercomplex structure is compatible with the quaternion Kahler structure which the above argument does not assume but I think the original question did assume. That implies that the twistor bundle is trivial which is well-known not to be possible for positive quaternion Kahler manifolds. I don't know what happens in the negative quaternion Kahler case but I suspect there are no examples there either at least among compact ones. So I would guess that any compact hypercomplex manifold with holonomy in $Sp(n)\cdot Sp(1)$ must be hyper-Kahler. </p> http://mathoverflow.net/questions/104556/what-can-be-the-cohomology-ring-of-a-manifold Comment by Vitali Kapovitch Vitali Kapovitch 2012-08-12T19:44:28Z 2012-08-12T19:44:28Z are you requiring the manifold to be closed? http://mathoverflow.net/questions/103228/find-an-action-of-mathbbz-2-on-mathbbcp1-which-is-compatible-with-the/103230#103230 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-27T00:37:15Z 2012-07-27T00:37:15Z @Zhaoting Wei even if you allow fixed points you won't get much. it's easy to see that the only order 2 diffeomorphisms of $\mathbb CP^1$ that commute with the action of $SL(2,\mathbb R)$ are the identity and the complex conjugation. http://mathoverflow.net/questions/102517/how-can-i-tell-whether-a-manifold-is-homogeneous Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-19T16:11:39Z 2012-07-19T16:11:39Z can you give an example of what you think qualifies as &quot;useful and practical&quot; when speaking of criteria? topological? geometric? there are numerous geometric conditions of various sorts (such as given in Lee Mosher's answer) that would imply this but I don't think that's what you are looking for. Purely topological conditions are basically impossible though. http://mathoverflow.net/questions/102161/proving-the-existence-of-good-covers/102225#102225 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-14T15:36:04Z 2012-07-14T15:36:04Z you are quite right and even Whitney is not needed for this. http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds/102181#102181 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-14T15:21:01Z 2012-07-14T15:21:01Z @Lee Mosher, Thanks again. you are absolutely right. when I was thinking about it yesterday it wasn't immediately clear to me how to slide the sphere off the pieces in the graph of contractible spaces (it's easy to do geometrically in the npc case which is what Agol was using in both of his arguments) but you are right - this is a purely topological statement once you know that both the edge and the vertex spaces are contractible. http://mathoverflow.net/questions/102161/proving-the-existence-of-good-covers/102225#102225 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-14T14:23:34Z 2012-07-14T14:23:34Z I think Mariano might object to using curvature explanations as being too Riemannian but the argument using convexity of the composite function (which was what I had in mind all along) doesn't formally use it although it of course amounts the same thing. http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds/102181#102181 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-13T23:32:47Z 2012-07-13T23:32:47Z @Lee Mosher thanks. but my question really was about that last <i>easy to deduce</i> step :=) http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds/102181#102181 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-13T22:42:33Z 2012-07-13T22:42:33Z nice! thanks again! http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds/102181#102181 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-13T21:36:31Z 2012-07-13T21:36:31Z Thanks! I understand the second argument (it's very nice and simple) but not the first. how do you put a npc metric on the universal cover? not all graph manifolds admit npc metrics. http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds/102181#102181 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-13T21:04:09Z 2012-07-13T21:04:09Z what is the simplest proof that a graph manifold with $\pi_1$-injective tori/Klein bottle is aspherical? I know this is classical (I think originally due to Whitehead?) but is there an elementary proof of this? http://mathoverflow.net/questions/102154/finite-subgroups-of-spin9/102167#102167 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-13T19:27:36Z 2012-07-13T19:27:36Z the fact that the action is free should surely simplify matters. after all there is a classification of spherical spaceforms by Wolf and one can presumably check which of the groups on his list are conjugate to subgroups of $Spin(9)$. This hardly sounds pleasant though. http://mathoverflow.net/questions/101776/altitudes-of-a-triangle/101904#101904 Comment by Vitali Kapovitch Vitali Kapovitch 2012-07-11T15:47:27Z 2012-07-11T15:47:27Z @Igor, Yes, I realize that this is not quite a proof yet in the hyperbolic case (I didn't try to make it precise anyway) but I don't see any issues making it work there too. you can first prove it for triangles where the altitudes actually do intersect in the space itself and no division by zero is needed and then (again using analyticity in, say Poincare model) extend it to the general case. http://mathoverflow.net/questions/98609/algorithmically-unsolvable-problems-in-topology/98614#98614 Comment by Vitali Kapovitch Vitali Kapovitch 2012-06-02T01:34:02Z 2012-06-02T01:34:02Z @Fernando Muro, yes, that's why I put rational in parentheses to include both cases. sorry of that was not clear. http://mathoverflow.net/questions/98609/algorithmically-unsolvable-problems-in-topology/98617#98617 Comment by Vitali Kapovitch Vitali Kapovitch 2012-06-01T23:55:03Z 2012-06-01T23:55:03Z @Joseph, note that Novikov's theorem does not assume simply connectednes and uses that fact very strongly. http://mathoverflow.net/questions/98166/is-the-singular-homology-of-a-real-algebraic-set-always-finitely-generated/98192#98192 Comment by Vitali Kapovitch Vitali Kapovitch 2012-05-30T00:54:42Z 2012-05-30T00:54:42Z @Dan Ramras the singular case follows from above because any algebraic set has finitely many smooth strata and you can choose $f_v$ that works for all of them. this easily yields a smooth gradient like vector field for $f_v$ tangent to $M$ on $M$ and without critical points outside a large ball.