most recent 30 from http://mathoverflow.net 2013-06-20T10:53:05Z http://mathoverflow.net/feeds/user/1988 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39885/what-spaces-have-well-known-horofunctions/39894#39894 valeri 2010-09-24T19:54:40Z 2010-09-24T19:54:40Z

<p>some remarks, not an answer. It seems, from [J. Cheeger and D. Gromoll, The splitting theorem for manifolds of nonnegative Ricci curvature, J. Diff. Geom. 6 (1971) 119-128.] it follows that horofunctions in Nil, Sol and \tilde Sl(2,R) are super-harmonic? (Ricci curvature is non-positive?) I know that the spaces of horofunction, Busseman functions coincide with Gromov ideal boundary (spaces of rays) when the sectional curvature is non-positive. (For non-negative curvature these ideal boundaries might be different. For the Heisenberg group Heis^{2n+1} with left-invariant metric the Gromov ideal boundary is the sphere S^{2n-1} with Carnot-Caratheodory metric, and geodesics equations are known - but not horofunctions :( )</p>

http://mathoverflow.net/questions/36496/abc-conjecture-follows-from-prime-factorization-is-np valeri 2010-08-23T21:04:27Z 2010-08-23T21:04:27Z

<p>Could somebody comment on the conjecture stated in the title:</p> <p>"abc-conjecture ( <a href="http://en.wikipedia.org/wiki/Abc_conjecture" rel="nofollow">http://en.wikipedia.org/wiki/Abc_conjecture</a> ) follows from the (prime-factorization is NP), see <a href="http://en.wikipedia.org/wiki/Integer_factorization" rel="nofollow">http://en.wikipedia.org/wiki/Integer_factorization</a> or <a href="http://en.wikipedia.org/wiki/Shor" rel="nofollow">http://en.wikipedia.org/wiki/Shor</a>'s_algorithm )" </p> <p>Or may be help with examples of similar claims, relating statements from Computer Science and Number theory. </p>

http://mathoverflow.net/questions/22050/is-there-a-common-name-for-the-complement-of-a-metric-space-in-its-completion/22063#22063 valeri 2010-04-21T15:11:02Z 2010-04-21T15:11:02Z

<p>Corona? Ideal boundary?</p>

http://mathoverflow.net/questions/18753/does-the-baker-campbell-hausdorff-formula-hold-for-vector-fields-on-a-compact-m/18792#18792 valeri 2010-03-19T20:28:02Z 2010-03-19T20:28:02Z

<p>Does it make any sense - if we consider vector fields as derivations of the algebra of smooth functions on M? Then each \phi_X is an automorphism of this algebra and there is an asymptotic formula for its (i.e., \phi_X) action. Product of two such asymptotic representations give as. formula for superposition \phi_X \circ \phi_Y - which leads to BCH for Z. (smth like in universal enveloping algebra of vector fields)... These asymptotic formulas not always converge thou </p>

http://mathoverflow.net/questions/16850/curvature-and-parallel-transport/17028#17028 valeri 2010-03-03T23:54:38Z 2010-03-03T23:54:38Z

<p>I believe, this is called Ambrose-Singer theorem. For the proof - you may introduce some coordinates (s,t) in the exp-image of plane spanned by X and Y, and define V(s,t) to be parallel along, say s-coordinate lines. Then compute how the derivative of V in t-direction changes along s-coordinate lines: it is D_s D_t V = R(X,Y)V since D_t D_s V \equiv 9, and s, t coordinate vectors commute - then integral of D_t V (s,t) = \int R(X,Y)V - D_t V(0,t) which is your parallel transport ... it may be something about this in doCarmo Riemannian geom, or Milnor Morse theory ...</p>

http://mathoverflow.net/questions/1554/point-singularity-of-a-riemannian-manifold-with-bounded-curvature/6338#6338 valeri 2009-11-21T00:54:59Z 2009-11-21T00:54:59Z

<p>What if you try a family of triangles, parallel to some two-direction s.t. their union contains singularity? (Like tetraedr for n=3)? Then their geometry (angles, sides, etc) are controlled from "outside" the singularity, so they all have uniformly bounded cirvature - including that one which contains singularity. Let the size then goes to zero. Does it mean that the tangent plane is defined at singularity and is R^n and so on ...? </p>

http://mathoverflow.net/questions/115486/convexity-of-hypersurfaces valeri 2012-12-05T21:59:55Z 2012-12-05T21:59:55Z

Dear Yong - glad to help!

http://mathoverflow.net/questions/115486/convexity-of-hypersurfaces valeri 2012-12-05T14:35:57Z 2012-12-05T14:35:57Z

In fact, as Mrowka mentioned in the ref, the answer is still &quot;yes&quot; for embedded hypersurface.

http://mathoverflow.net/questions/115486/convexity-of-hypersurfaces valeri 2012-12-05T14:34:50Z 2012-12-05T14:34:50Z

Dear Yong, your first question was about hypersurface - as it is still in the title. The answer to that question - &quot;no&quot; Nadirashvili surface is minimal (H=0) inside the unit ball. It is immersed and open (complete non-compact without boundary). When you change to the boundary of a bounded domain - the answer is &quot;yes&quot;. To prove this you may consider the smallest ball which contains this domain - it touch your hypersurface at some point, therefore it has bigger curvature at this point than the boundary of the ball, which is strictly positive. Done.

http://mathoverflow.net/questions/115486/convexity-of-hypersurfaces valeri 2012-12-05T11:26:39Z 2012-12-05T11:26:39Z

no boundary - complete open surfaces, i.e., counter-examples to your conjecture, I think.

http://mathoverflow.net/questions/115486/convexity-of-hypersurfaces valeri 2012-12-05T10:32:26Z 2012-12-05T10:32:26Z

<a href="http://mathoverflow.net/questions/52851/the-geometry-of-nadirashvilis-complete-bounded-negative-curvature-surface" rel="nofollow" title="the geometry of nadirashvilis complete bounded negative curvature surface">mathoverflow.net/questions/52851/&hellip;</a>

http://mathoverflow.net/questions/39885/what-spaces-have-well-known-horofunctions/39894#39894 valeri 2010-09-25T08:21:57Z 2010-09-25T08:21:57Z

@Igor - &quot; haven't though enough of nonpositive Ricci curvature&quot; - me too :). My guess was that in Hadamard manifolds horofunctions are smooth, so Laplacian comparison with flat tangent space (where horofunctions are linear) via exponential map will do. Probably not ... I can not state smth right now.

http://mathoverflow.net/questions/39885/what-spaces-have-well-known-horofunctions/39894#39894 valeri 2010-09-24T21:28:23Z 2010-09-24T21:28:23Z

to Pablo: there are some papers by Kaplan (if I remember correctly) and by Marenich in Geom Dedicata 66:2 with calculations. To Igor: of course, Cheeger Gromoll for non-negative, but what I mean is that for non-positive their arguments will give super-harmonic, is this wrong?

http://mathoverflow.net/questions/22050/is-there-a-common-name-for-the-complement-of-a-metric-space-in-its-completion/22063#22063 valeri 2010-04-21T19:50:19Z 2010-04-21T19:50:19Z

yes, usually they are (I am not sure about all) compactifications.

http://mathoverflow.net/questions/22050/is-there-a-common-name-for-the-complement-of-a-metric-space-in-its-completion/22063#22063 valeri 2010-04-21T16:30:51Z 2010-04-21T16:30:51Z

I believe, &quot;ideal boundary&quot; pretty much common - see ideal boundaries of Hadamard manifolds (Tits boundary, Gromov ideal boundary etc, there are also ideal boundaries of horofunctions, Buseman functions, distance-like functions); or equivalent notions from the geometric group theory. Also there are some &quot;functional&quot; ideal boundaries usualy called corona (spaces) - Martin boundary, Furstenberg boundary, etc. Stone-Cech compactification adds the biggest, in some sense, corona space. I am not sure about references, may be start with wik <a href="http://en.wikipedia.org/wiki/Compactification_(mathematics" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a>)

http://mathoverflow.net/questions/16850/curvature-and-parallel-transport/17028#17028 valeri 2010-03-04T08:09:38Z 2010-03-04T08:09:38Z

The area enters because coordinate vectors may be not unit and not normal, but in the formula R(X,Y) they are - after you &quot;normalize&quot; coordinate vectors, say S and T - you have S\wedge T in the integral equals X\wedge Y \times area. After you prove this for small parallelogram - just note that both parallel operator - holonomy around small loop - and integral are additive - if you divide big area in two (many pieces) the formula for small pieces gives the same formula for big film.

http://mathoverflow.net/questions/1554/point-singularity-of-a-riemannian-manifold-with-bounded-curvature/6338#6338 valeri 2009-11-21T08:55:33Z 2009-11-21T08:55:33Z

What if we measure distances between vertices of the tetraedr in incomplete manifold M^n and then find points in complete - say R^n having the same mutual distances, and then continue this correspondence by linearity? Bounded curvature then implies that this is &quot;nice&quot; map (C{1,\alpha} or {2} - not sure) ?

http://mathoverflow.net/questions/1554/point-singularity-of-a-riemannian-manifold-with-bounded-curvature/6338#6338 valeri 2009-11-21T01:12:04Z 2009-11-21T01:12:04Z

I mean singularity is inside the triangle. No small loop just for n=2, where my suggestion would not work. For n&gt;2 (simply connected &quot;annulus&quot;) - we may compare - map the whole tetraedr to the similar one in complete manifold with bounded curvature. Then take the foliation from the image (pull back) - the distortion is under control -&gt; we have &quot;nice&quot; foliation of the completion of the target manifold. Does it work? (valeri)