most recent 30 from http://mathoverflow.net 2013-05-18T20:40:24Z http://mathoverflow.net/feeds/user/4696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130454/stein-manifolds-definiton/130510#130510 Mohan Ramachandran 2013-05-13T17:56:03Z 2013-05-13T17:56:03Z

<p>Compact complex manifolds are holomorphically convex.Stein's original definition had in addition to holomorphic convexity separation of points by holomorphic functions and local coordinates given by holomorphic functions .Stein himself called these holomorphically complete manifolds .He introduced it to study the solution of Cousin problems on complex manifolds . Remmert showed that a noncompact connected complex manifold is holomorphically convex iff it admits a proper holomorphic mapping to a Stein space . You can find some of these discussed in the book of Grauert and Remmert Theory of Stein spaces . From Stein's definition one can see that Stein manifolds admit strictly plurisubharmonic exhaustion functions .The converse is a deep theorem of Grauert.</p>

http://mathoverflow.net/questions/129507/hyperbolic-riemann-surface/129851#129851 Mohan Ramachandran 2013-05-06T15:18:04Z 2013-05-07T01:42:51Z

<p>If a domain on a Riemann surface has a component of the frontier with at least two distinct points then it is potential theoretic hyperbolic. This result goes back to Osgood. Existence of green 's function is connected to existence of barrier at some frontier point. The existence of barriers is local. The solution of the local problem can be found in the book of Ahlfors-Sario titled Riemann Surfaces .</p>

http://mathoverflow.net/questions/129822/greens-function-hyperbolic-riemann-surface/129838#129838 Mohan Ramachandran 2013-05-06T13:46:59Z 2013-05-06T13:46:59Z

<p>Potential theoretic hyperbolic implies poincare hyperbolic. This is just part of proof of uniformisation theorem. The other direction is false since any compact Riemann surface of genus at least two is poincare hyperbolic. </p>

http://mathoverflow.net/questions/119444/classification-of-non-compact-riemannian-manifold-with-ric-n-1-and-first-eige/119503#119503 Mohan Ramachandran 2013-01-21T19:13:20Z 2013-01-21T19:13:20Z

<p>Peter Li and Jiaping Wang have a series of papers on the structure of complete manifolds with<br> positive spectrum .Peter Li also has a survey article on this topic .You can find it on his home page .</p>

http://mathoverflow.net/questions/119198/when-is-a-smooth-projective-variety-a-fibration/119212#119212 Mohan Ramachandran 2013-01-17T20:57:10Z 2013-01-17T20:57:10Z

<p>By a theorem of Gromov and Schoen if the fundamental group of X is a proper amalgamated product or HNN extension then X maps surjectively to a curve .</p>

http://mathoverflow.net/questions/119113/essential-uniqueness-of-the-real-analytic-structure-on-mathbb-r/119201#119201 Mohan Ramachandran 2013-01-17T17:48:31Z 2013-01-17T17:48:31Z

<p>For Misha's comment and question 3 see the paper of David Minda Regular Analytic arcs and curves Colloq.Math 38(1977) no 1 73-82 .Regarding the vanishing theorem for real analytic manifolds this was proved by Henri Cartan Bulletin de la S.M.F tome 85 (1957) 77-99. Cartan assumed his manifolds were real analytically embedded in euclidean space . One just needs to know that the Complexification is Stein .This is due to Grauert's solution for the Levi problem. In case of one dimensional real analytic manifolds the complexification is Stein by the Runge type theorem of Behnke and Stein . You can also use the Behnke-Stein theorem to show that real analytic one manifolds have complete real analytic metrics and follow Milnor's proof. All proofs I know go through Complexifications .Existence of Complexification for real analytic manifold is due to Whitney and Bruhat</p>

http://mathoverflow.net/questions/117020/how-many-nonparabolic-ends-guarantee-a-nonconstant-harmonic-function-on-riemannia/117150#117150 Mohan Ramachandran 2012-12-24T15:55:10Z 2012-12-24T15:55:10Z

<p>Any open Riemannian manifold has enough harmonic function to separate points and give local coordinates .This a consequence of a Runge type theorem of Lax and Malgrange independently.In fact Greene and Wu showed that one can harmonically embed the open manifold in Euclidean space. The paper of Greene and Wu is in Annales Institut Fourier vol 25 (1975) 215-235</p>

http://mathoverflow.net/questions/87869/a-corollary-to-stone-weierstrass-theorem/87922#87922 Mohan Ramachandran 2012-02-08T19:02:24Z 2012-02-08T19:02:24Z

<p>In your case we can find a holomorphic function on the plane that uniformly approximates the given continuous function .It is a consequence of the following .Suppose K is a compact measure zero subset of the plane whose complement in the plane is connected then every continuous function on K can be uniformly approximated by entire functions. Hartogs-Rosenthal says any continuous function on K can be uniformly approximated by functions holomorphic in a neighbourhood of K.By Runge's theorem, functions holomorphic in a neighbourhood of K can be uniformly approximated on K by entire functions. In your case since gamma is an arc its complement is connected .Since it is smooth it has measure zero . </p>

http://mathoverflow.net/questions/86333/the-levi-form-of-the-distance-squared-function-in-a-non-positively-curved-kaehle/86460#86460 Mohan Ramachandran 2012-01-23T16:18:23Z 2012-01-23T16:18:23Z

<p>By the Hessian comparison theorem the square of the distance function on X is strictly convex. On Kahler manifolds strictly convex functions are strictly plurisubharmonic .By Grauert's solution of the Levi problem X is Stein.See R E Greene and H H Wu Springer LNM 699 . There is an example of a complete simply connected negatively curved Hermitian manifold which is not Stein due to P Klembeck.So we need the Kahler assumption.</p>

http://mathoverflow.net/questions/80430/general-form-of-schwarz-reflection-principle/80837#80837 Mohan Ramachandran 2011-11-13T18:06:29Z 2011-11-13T18:06:29Z

<p>For reflection across analytic arcs see Caratheodory's book Conformal Representation pages 87-90 or his book Theory of Functions vol 2 pages 101-104 .</p>

http://mathoverflow.net/questions/59479/partial-bar-partial-lemma-for-contractible-domains/59586#59586 Mohan Ramachandran 2011-03-25T17:38:50Z 2011-03-25T17:45:42Z

<p>If in addition you assume that your domain is pseudoconvex then by a theorem of A.Aeppli what you want is true.The paper is titled :On the cohomology structure of Stein manifolds 1965 (Proc.Conf.Complex Analysis(Minneapolis Minn 1964) pages 58 to 70 Springer, Berlin. David's example indicates why one might need a hypothesis like pseudoconvexity.</p>

http://mathoverflow.net/questions/43614/is-every-nonnegatively-curved-plane-conformal-to-the-complex-plane/58679#58679 Mohan Ramachandran 2011-03-16T20:05:15Z 2011-03-16T20:43:02Z

<p>Cheng-Yau proved that: A complete Riemannian manifold with non-negative Ricci curvature and at most quadratic growth for volumes of balls as the radius goes to infinity is parabolic.</p> <p>EDIT (by Igor Belegradek). Various criteria for parabolicity are found in the <a href="http://www.ams.org/journals/bull/1999-36-02/S0273-0979-99-00776-4/home.html" rel="nofollow">survey</a> of Grigoryan. In particular, on page 177 it is mentioned that a a complete Riemannian manifold with at most quadratic volume growth is parabolic. For complete open nonnegatively curved surfaces the volume growth is at most quadratic by Bishop-Gromov. On the other hand, there are parabolic complete manifolds with arbitrary fast volume growth (see page 180 of the same survey). Finally, the very first proof of parabolicity of complete nonnegatively curved plane seems to be due to Blanc-Fiala (1941); the reference is in Huber's paper mentioned in my comment to Anton's answer.</p>

http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains/56669#56669 Mohan Ramachandran 2011-02-25T19:27:43Z 2011-02-25T20:14:42Z

<p>An alternate argument to the one given by BS in the smooth case is to note that by the implicit funcion theorem the hypersurface is locally two sided.If it is not globally two sided then one can construct a connected two sheeted covering of euclidean space which is a contradiction.Then we note that euclidean space in dimension atleast two has one end.This implies that the complement of the hypersurface has one nonrelatively compact component and the result follows.This proof works for any closed embedded locally two sided hpersurface.</p>

http://mathoverflow.net/questions/36050/embeddings-and-triangulations-of-real-analytic-varieties/36213#36213 Mohan Ramachandran 2010-08-20T18:47:34Z 2010-08-22T15:10:54Z

<p>If you just want a proper 1-1 real analytic map whose image is a real analytic variety then the result is theorem 2 page 593 of a paper of Tognoli and Tomassini in Ann.Scuola.Norm.Pisa (3) vol 21 yr 1967 pages 575-598. This means there is no control over the differential of the map.I am assuming that the real analytic space has finite dimension.If the dimension of the Zariski tangent spaces of a connected reduced real analytic space is bounded then it can be real analytically embedded in euclidean space, see paper by Aquistapace Broglia and Tognoli Ann.Scuola.Norm.Pisa (4) vol 6 yr 1979 p 415-426.</p>

http://mathoverflow.net/questions/34763/continuous-holomorphic-on-a-dense-open-holomorphic/34842#34842 Mohan Ramachandran 2010-08-07T15:51:54Z 2010-08-08T19:25:41Z

<p>If the curve C is rectifiable then the answer is yes.Under the assumption that C is rectifiable your question is known as Painleve's Theorem.It follows from the strong version of Cauchy's theorem which is stated as follows: If C is a simple closed rectifiable curve in the plane and f is holomorphic in the interior and continuous in the closed bdd region enclosed by C then the integral of f over C is zero.See for example the book by Behnke and Sommer page 119 (the book is in German).You can also find a proof of the strong form of Cauchy's theorem in the book titled:Elements of the topology of plane sets of points by M H A Newman,2nd edn page 187. If the jordan arc has positive area the answer to the question is no.See pages 122- 123 of the article in the amer math monthly vol 81 no 2 pages 115-137 year 1974.The paper is by Lawrence Zalcman who has other papers on this topic. </p>

http://mathoverflow.net/questions/28947/least-collaborative-mathematician/28970#28970 Mohan Ramachandran 2010-06-21T17:15:40Z 2010-06-21T17:50:07Z

<p>How about Marina Ratner.I believe she has had no collaborators</p>

http://mathoverflow.net/questions/21796/extension-of-strictly-plurisibharmonic-functions-on-a-kahler-manifold/21878#21878 Mohan Ramachandran 2010-04-19T19:44:30Z 2010-04-19T19:44:30Z

<p>If a complex manifold has a strictly plurisubharmonic function then it cannot contain positive dimensional compact analytic sets.This is clearly a necessary condition.So you might start considering your question on Stein manifolds.</p>

http://mathoverflow.net/questions/18454/fundamental-groups-of-noncompact-surfaces/18492#18492 Mohan Ramachandran 2010-03-17T13:57:53Z 2010-03-17T17:19:16Z

<p>If you assume the existence of smooth structure on a noncompact surface then it is easy to show the existence of a proper morse function with no local maximum.This shows that the surface is homotopic to a one dim CW complex. This is the smooth version of Igor's answer.</p> <p>EDIT BY ANDY PUTMAN: Mohan isn't registered and thus isn't able to comment, but he sent me an email with more details. The result is true in all dimensions : any noncompact smooth n-manifold is homotopy equivalent to an n-1 complex. The key is to construct a strictly subharmonic morse exhaustion function. The subharmonicity prevents the function from having local maxima. Details of this can be found in his paper "Elementary Construction of Exhausting Subsolutions of Ellitpic Operators", which was joint with Napier and was published in L’Enseignement Math´ematique, t. 50 (2004), p. 1–24.</p>

http://mathoverflow.net/questions/14278/haar-measure-on-a-quotient-references-for/14294#14294 Mohan Ramachandran 2010-02-05T16:57:23Z 2010-02-05T16:57:23Z

<p>You can find it in Federer Geometric Measure Theory pages 121-129.</p>

http://mathoverflow.net/questions/13414/kahler-manifold-which-is-not-algebraic/13416#13416 Mohan Ramachandran 2010-01-29T20:53:59Z 2010-01-29T20:53:59Z

<p>generic complex tori in complex dimension 2 or higher. MR</p>

http://mathoverflow.net/questions/4847/various-cartans-lemmata/9637#9637 Mohan Ramachandran 2009-12-23T20:20:49Z 2009-12-23T20:20:49Z

<p>The Cartan Lemma referred to here is due to Henri Cartan.It is used when one wants to prove that a holomorphic vector bundle over complex n space is trivial. </p>

http://mathoverflow.net/questions/129684/affine-varieties-as-stein-surfaces Mohan Ramachandran 2013-05-05T02:20:52Z 2013-05-05T02:20:52Z

@kaavek. Varieties can be singular.

http://mathoverflow.net/questions/129684/affine-varieties-as-stein-surfaces Mohan Ramachandran 2013-05-05T00:38:07Z 2013-05-05T00:38:07Z

Affine varieties over complex numbers are always Stein spaces.

http://mathoverflow.net/questions/129507/hyperbolic-riemann-surface Mohan Ramachandran 2013-05-03T17:23:02Z 2013-05-03T17:23:02Z

If you look at the tags it looks like theOP is interested in the existence of positive Green's function which is certainly true for many values of the radius of the disc that is removed.

http://mathoverflow.net/questions/126957/volume-of-complex-submanifolds Mohan Ramachandran 2013-04-09T20:20:12Z 2013-04-09T20:20:12Z

This is a consequence of the lower bound for volumes of intersection of analytic sets with ball of fixed radius with center on the analytic set.See for example page 190 of Chirka Complex Analytic Sets

http://mathoverflow.net/questions/126831/a-question-from-otto-forsters-book-on-riemann-surfaces Mohan Ramachandran 2013-04-08T18:34:10Z 2013-04-08T18:34:10Z

The argument is similar to the proof of Nakayama's lemma .Take everything on (1) to one side and multiply by the adjugate matrix. t

http://mathoverflow.net/questions/126597/injectivity-radius-of-the-completion-of-a-manifold Mohan Ramachandran 2013-04-05T21:53:39Z 2013-04-05T21:53:39Z

If the injectivity radius is strictly positive the metric is automatically complete.

http://mathoverflow.net/questions/126615/noncompact-manifold-with-two-ends-splits Mohan Ramachandran 2013-04-05T18:51:08Z 2013-04-05T18:51:08Z

@Agol: I believe you mean non-negative curvature.

http://mathoverflow.net/questions/126489/is-there-non-simple-connected-projective-varietyover-c-with-trivial-etale-funda Mohan Ramachandran 2013-04-04T16:29:50Z 2013-04-04T16:29:50Z

For the question exactly as stated in the body of the question the answer is yes since any finitely presented group is the fundamental group of a compact complex manifold.

http://mathoverflow.net/questions/112199/differential-equations-and-axiom-of-choice/112225#112225 Mohan Ramachandran 2013-01-23T19:19:38Z 2013-01-23T19:19:38Z

A similar proof can be found in the paper of Wolfgang Walter American Math Monthly vol 78 1971 pages 170-173 .

http://mathoverflow.net/questions/119113/essential-uniqueness-of-the-real-analytic-structure-on-mathbb-r Mohan Ramachandran 2013-01-19T19:18:47Z 2013-01-19T19:18:47Z

Yes.One needs a elliptic PDE with real analytic coefficients.One of the reasons SCV is complicated is that the PDE is overdermined elliptic system .

http://mathoverflow.net/questions/119113/essential-uniqueness-of-the-real-analytic-structure-on-mathbb-r Mohan Ramachandran 2013-01-17T18:31:43Z 2013-01-17T18:31:43Z

In fact Bochner showed real analytic embedding of compact real analytic manifolds with real analytic metrics in euclidean by using eigenfunctions of the laplacian.

http://mathoverflow.net/questions/119113/essential-uniqueness-of-the-real-analytic-structure-on-mathbb-r Mohan Ramachandran 2013-01-17T18:27:29Z 2013-01-17T18:27:29Z

The only other technique I know is to use existence of real analytic metrics and work with harmonic functions for these metrics .However all constructions I know of real analytic metrics are by complex analytic methods .

http://mathoverflow.net/questions/119113/essential-uniqueness-of-the-real-analytic-structure-on-mathbb-r Mohan Ramachandran 2013-01-17T17:54:11Z 2013-01-17T17:54:11Z

It is a very difficult open problem to prove theorems about real analytic manifolds without complexifying.

http://mathoverflow.net/questions/117020/how-many-nonparabolic-ends-guarantee-a-nonconstant-harmonic-function-on-riemannia/117150#117150 Mohan Ramachandran 2012-12-24T17:31:49Z 2012-12-24T17:31:49Z

Yes there is no control over growth of these functions.OP only asked about nonconstant harmonic functions .In the case of parabolic ends one can find a proper harmonic function along that end.This is a theorem of Mitsuru Nakai.

http://mathoverflow.net/questions/93330/why-is-the-fundamental-group-of-a-compact-riemann-surface-not-free Mohan Ramachandran 2012-04-09T15:56:33Z 2012-04-09T15:56:33Z

@Georges:Another argument If the fundamental group is free there is a continuous map from the surface to a finite one complex which induces an isomorphism of first cohomology. The cup product on first cohomology of the one complex is zero but this is not the case for the surface.