User vamsi - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-20T08:22:02Zhttp://mathoverflow.net/feeds/user/3709http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/130888/vector-bundles-on-stein-manifoldsVector bundles on Stein ManifoldsVamsi2013-05-17T01:24:33Z2013-05-17T01:24:33Z
<p>This might be standard if true (if so, I shall be grateful if provided with a reference). Given a smooth map from a Stein manifold X to Gr(k,n) (the Grassmannian of k planes in C^n), is there a holomorphic map in its homotopy class?</p>
http://mathoverflow.net/questions/126419/density-of-smooth-functions-in-sobolev-spaces-on-manifoldsDensity of smooth functions in Sobolev spaces on manifoldsVamsi2013-04-03T18:01:08Z2013-04-19T12:52:12Z
<p>Hebbey defines the Sobolev space of functions on a Riemannian manifold (M,g) as the completion of smooth functions under the Sobolev norm. However, I have seen (elsewhere) that Sobolev spaces have been defined as the collection of locally integrable functions whose weak derivatives (w.r.t to the Levi-Civita connection) are in L^p. For compact manifolds, or for manifolds with bounded geometry and positive injectivity radius (like $\mathbb{R}^n$), I can see why these two definitions are equivalent. Are they equivalent for general noncompact manifolds?</p>
http://mathoverflow.net/questions/106282/alexandrov-bakelmann-pucci-maximum-principleAlexandrov-Bakelmann-Pucci maximum principleVamsi2012-09-03T22:33:10Z2012-09-04T21:25:09Z
<p>The ABP maximum principle states (roughly) that, if $a^{ij} \partial _i \partial _j u \geq f$, over a domain $\Omega$ in $\mathbb{R}^n$ (where $a^{ij} \geq C Id >0$), then (assuming sufficient regularity of the coefficients), $\sup _{\Omega} u \leq \sup _{\partial \Omega} u + C (\int _{\Omega} \vert f \vert^n )^{1/n}$. Its proof seems to be quite unconventional (the final inequality is proved by an inequality of the measures of certain sets constructed using the normal mapping). Is there an intuitive explanation of the proof? (Perhaps using some probabilistic ideas or something?) I mean, usually, in order to prove maximum principles, the key idea is to use that at a local max the second derivative is negative-definite. But, this one seems to use some strange ideas..</p>
http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-coverAn example of a complex manifold without a finite open coverVamsi2011-03-20T17:46:11Z2012-07-09T10:01:04Z
<p>Are there non-compact complex manifolds that
a) Don't embed in C^n (holomorphically)
and
b) Cannot be covered by a finite number of coordinate open sets?
If b) can be satisfied, then I think so can a) be by taking a product with a compact complex manifold. If one takes a Riemann surface of infinite genus, one does not have a "good" finite open cover, but I allow non-contractible open covers as well. Apologies in advance for this elementary question.</p>
http://mathoverflow.net/questions/92319/holder-estimates-for-the-complex-monge-ampere-equationHölder estimates for the Complex Monge-Ampere equationVamsi2012-03-26T22:58:00Z2012-06-26T13:46:56Z
<p>If on a bounded smooth, pseudoconvex domain in $\mathbb{C}^n$, $\mathrm{det} ( \mathrm{Hess}(u)) = f$ ($f>0$, $\mathrm{Hess}(u)>0$, $u=0$ on the boundary), if $f \in C^{k, \alpha}$, is $u \in C^{k+2, \alpha}$ ? (I mean, is there an apriori estimate on $u$ with the Hölder exponents of $f$ and $u$ being the same (equal to $\alpha$?)</p>
http://mathoverflow.net/questions/13615/atiyah-bott-localisation-applied-to-euler-characteristicAtiyah Bott localisation applied to Euler characteristicVamsi2010-02-01T02:26:05Z2012-04-28T15:46:45Z
<p>Suppose we have a torus action on a compact oriented manifold M. Assume the action has isolated fixed points. Why is it that the equivariant Euler class of the normal bundle at the fixed point (i.e. the tangent space at that point) is (upto a sign) the product of the weights of the action of the lie algebra of the Torus on the tangent space at that point? (Maybe this is obvious but I don't see it).</p>
http://mathoverflow.net/questions/91981/monge-ampere-equations-concavityMonge Ampere equations (concavity) Vamsi2012-03-23T07:09:23Z2012-04-05T09:15:43Z
<p>The Monge-Ampere (whether real or complex, whether in a domain or on a manifold) equations usually studied are of the type $F(u, \nabla u, \mathrm{Hess}(u)) = 0$ where $F$ is a concave function of $\mathrm{Hess}(u)$. This is done for the purpose of $C^{2, \alpha}$ estimates. My question is: Has any work been done in the non-concave setting? (I know of Harvey and Lawson's work that gives us viscosity solutions, but I wish for more regular solutions) Especially with regard to the $C^{2,\alpha}$ estimate?</p>
http://mathoverflow.net/questions/91125/bedford-taylor-theoryBedford-Taylor theory Vamsi2012-03-13T21:39:53Z2012-03-14T18:29:17Z
<p>The Dirichlet problem for the Complex Monge-Ampere equation on a bounded pseudoconvex domain in $\mathbb{C}^n$ was studied in Bedford-Taylor's seminal paper wherein they defined $(dd^{c} u)^n$ for locally bounded plurisubharmonic $u$. But, they don't seem to use it in their Perron-type method, instead using a convex-measure-theoretic construction claiming that the upper envelope is not well-behaved. Now that we know more about psh functions, have people studied the Dirichlet problem without using the measure-theoretic construction of Goffman and Serrin? (and reproved Bedford-Taylor's results)</p>
http://mathoverflow.net/questions/90700/where-is-number-theory-used-in-the-rest-of-mathematics/91147#91147Answer by Vamsi for Where is number theory used in the rest of mathematics?Vamsi2012-03-14T02:37:57Z2012-03-14T02:37:57Z<p>If Arakelov geometry counts as number theory, then, <a href="http://arxiv.org/pdf/math/0401029v1.pdf" rel="nofollow">http://arxiv.org/pdf/math/0401029v1.pdf</a> demonstrates the computation of the Analytic torsion (a purely analytic object involving the product of determinants of laplacians) using the Arithmetic Riemann-Roch theorem.</p>
http://mathoverflow.net/questions/69922/atiyah-bott-yang-mills-connectionsAtiyah-Bott Yang-Mills connectionsVamsi2011-07-10T03:01:50Z2011-12-24T00:55:35Z
<p>In Atiyah-Bott's paper on Yang-Mills equations on Riemann surfaces, a special case of what they do is to prove that Unitary Yang-Mills connections over a R.S $M$ are in bijective correspondence with unitary representations of a central extension of the fundamental group. To this end, given a representation, they construct a unitary bundle as follows (and therein lies my confusion):
Take a degree one $U(1)$ bundle $Q$. Now, pull this bundle back to the universal cover $\tilde{M}$. The new bundle is a $U(1) \times \pi_1(M)$ bundle over $M$. Now apparently, one can "lift the representation of the central extension and thus form a unitary bundle" (with a connection). Why can one do this?</p>
http://mathoverflow.net/questions/76303/a-non-kahler-metric-on-projectivised-vector-bundlesA (non-Kahler) metric on projectivised vector bundlesVamsi2011-09-25T00:55:00Z2011-12-16T00:52:40Z
<p>Given a hermitian holomorphic vector bundle (E, h) on a complex manifold-with-a-metric (X,g), then consider the following (natural) construction of a metric on the total space of $\mathbb{P}(E)$ : Firstly, the metric $h$ induces a Fubini-study metric on the fibres of $\mathbb{P}(E)$ (just take local orthonormal frames to get a smooth fibre bundle, use the usual Fubini-study metric and then pullback to the holomorphic bundle). Secondly, the Chern connection of $h$ induces a splitting $T\mathbb{P}(E) = T\mathbb{C}\mathbb{P}^{r} \oplus TX$. Now put the direct sum metric. My question is: This seems like a very natural construction. Has this been studied before (in the sense of curvature properties etc)? If so, I'd be most grateful if a reference is pointed out. </p>
http://mathoverflow.net/questions/83286/mathematical-software-for-chern-weil-theoryMathematical software for Chern-Weil theoryVamsi2011-12-12T21:11:26Z2011-12-12T21:38:44Z
<p>I need to compute curvature forms and Chern-Weil forms for a given metric (in local coordinates) on a vector bundle. Is there are software package that does this? If I manage to compute the derivatives involved is there atleast a package that does the algebra part of it? </p>
http://mathoverflow.net/questions/80989/a-specific-degeneration-of-a-rank-2-bundleA specific degeneration of a rank 2 bundleVamsi2011-11-15T16:19:08Z2011-11-16T11:58:06Z
<p>I wish to know if there is a rank 2 vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{P}^1$ such that $\mathbb{P}(E)$ when restricted to $\mathbb{P}^1 \times [0:1]$ is the $n$th Hirzebruch surface and when restricted to $\mathbb{P}^1 \times [x:y]$ is the $(n-2)$th Hirzebruch surface.</p>
http://mathoverflow.net/questions/79891/a-vector-bundle-with-a-given-jumping-lineA vector bundle with a given jumping lineVamsi2011-11-03T03:15:19Z2011-11-04T00:50:24Z
<p>I'd like to know if there exists a holomorphic rank 2 sub-bundle of $T\mathbb{P}^3$ which, when restricted to a given line is $\mathcal{O}(-a)\oplus \mathcal{O}(a)$, but is trivial when restricted to all other lines lying in a plane containing this line (i.e. this line is a jumping line of order $a$).</p>
<p>EDIT: From Angelo's answer, we see that there are no subbundles of the tangent bundle satisfying this property. A related question: Is there a vector bundle that has a given jumping line? (which is not a subbundle of the tangent bundle of course).</p>
http://mathoverflow.net/questions/78460/learning-arakelov-geometryLearning Arakelov geometryVamsi2011-10-18T14:29:10Z2011-10-19T00:26:42Z
<p>I have a complex analytic background (Griffiths and Harris, Huybrechts, Demailley etc). Also, I understand some PDE. I want to learn Arakelov geometry (atleast till the point I can "apply" computations of Bott-Chern forms and Analytic torsion to producing theorems of interest in Arakelov geometry). I know almost nothing of schemes or of number theory. I don't how much of these is needed to learn this stuff. I'd be grateful if any good references/suggestions are pointed out.</p>
http://mathoverflow.net/questions/78093/regarding-discrete-eigenvaluesRegarding Discrete EigenvaluesVamsi2011-10-14T02:29:36Z2011-10-14T16:47:30Z
<p>For many eigenvalue problems for differential operators (for example the quantum harmonic oscillator (HO)), unless we impose some behaviour at infinity, the eigenvalues will not be discrete.
But, suppose we consider the laplacian (of the usual round (Fubini-Study) metric) on $\mathbb{C}\mathbb{P}^1$. If we restrict ourselves to $\mathbb{C}$ and forget about infinity, then we get the eigenvalue problem : $\Delta f = \frac{1}{(1+r^2)^2} \lambda f$ (where $r$ is the distance from the origin in $\mathbb{C}$). Does this have the same eigenvalues as the original laplacian on the sphere? I mean, does forgetting the "boundary" conditions at infinity add new eigenvalues? In general, when is one allowed to "forget" what happens at infinity?</p>
http://mathoverflow.net/questions/75069/holomorphic-h-principle-for-compact-manifoldsHolomorphic h-principle for compact manifoldsVamsi2011-09-10T06:29:43Z2011-09-10T10:11:00Z
<p>The Oka principle for Stein manifolds says (roughly) that the only obstructions for "things" are topological obstructions (for instance every smooth complex vector bundle admits a holomorphic structure, etc). Is there a similar principle (atleast in some cases) for compact complex manifolds? Or atleast some version of a h-principle for compact manifolds?</p>
http://mathoverflow.net/questions/73439/when-is-a-given-matrix-of-two-forms-a-curvature-formWhen is a given matrix of two forms a curvature form?Vamsi2011-08-22T21:14:58Z2011-08-23T00:08:15Z
<p>Let's assume we are working over $\mathbb{R}^n$ (but feel free to change to domain to answer the question). I wish to know if the equation $F = dA + A \wedge A$ can be solved for a matrix of $1$-forms $A$, given a (smooth) matrix of two forms $F$ which satisfies the condition $dF =B \wedge F - F \wedge B$ for smooth matrix of one-forms $B$ (i.e. the Bianchi identity is satisfied). Notice that this is true for line-bundles (in fact over any convex open set). </p>
http://mathoverflow.net/questions/69904/branched-covers-of-compact-riemann-surfacesBranched covers of compact Riemann surfacesVamsi2011-07-09T19:47:55Z2011-07-10T09:40:32Z
<p>Let $S$ be a compact R.S of genus $\geq 2$. In the paper "Stable and unitary vector bundles on compact Riemann surfaces" (by Narasimhan and Seshadri), they claim that there is a branched covering map from the upper half plane to $S$ which is ramified at exactly one point (with index $N$) (i.e. $S$ is the quotient of $\mathbb{H}$ by the group $\langle A_i, B_i, C \vert \Pi [A_i, B_i] = C, [A_i,C] = [B_i, C] = I, C^N=I \rangle$). I am not able to access the reference (Grothendieck) pointed out in that paper. Is there any other reference (or any easy proof of this?) ? Also, does a similar fact hold for non-compact Riemann surfaces (compact minus a finite collection of points).</p>
http://mathoverflow.net/questions/69276/quillen-metric-definitionQuillen metric definitionVamsi2011-07-01T16:59:00Z2011-07-01T16:59:00Z
<p>I am confused as to why a regularised determinant is used in the definition of Quillen's metric on the determinant line bundle defined over the space of $\bar{\partial}$ operators on a (hermitian) vector bundle over a compact Riemann surface. I mean, why not use the metric induced by the vector bundle just by itself? Is the determinant introduced so as to define a smooth metric? For the record I am referring to Quillen's paper "On the determinants of Cauchy-Riemann operators" over a Riemann surface.</p>
http://mathoverflow.net/questions/5499/which-mathematicians-have-influenced-you-the-most/65804#65804Answer by Vamsi for Which mathematicians have influenced you the most?Vamsi2011-05-23T23:16:32Z2011-05-23T23:16:32Z<p>Simon Donaldson. His proofs involve (to quote wikipedia) a creative use of analysis. I loved his proof of the theorem of Narasimhan and Seshadri.</p>
http://mathoverflow.net/questions/64545/mehta-seshadri-and-parabolic-bundlesMehta-Seshadri and Parabolic bundlesVamsi2011-05-10T21:42:25Z2011-05-12T12:43:55Z
<p>In the original paper of Mehta-Seshadri, it seems like they treat the case of zero parabolic degree (i.e. they prove that zero parabolic degree stable parabolic bundles correspond to irreps of the fundamental group of the Riemann surface minus some points). But, as in the Narasimhan-Seshadri theorem, is the nonzero degree case obtained by considering representations of a central extension of the fundamental group? I'd be grateful if someone may point to a reference. </p>
http://mathoverflow.net/questions/59479/partial-bar-partial-lemma-for-contractible-domains$\partial \bar{\partial}$ lemma for contractible domainsVamsi2011-03-24T20:58:08Z2011-03-25T18:28:30Z
<p>Is every (p,p) ($p\geq1$) closed form in a contractible open set of $\mathbb{C}^n$, $\partial \bar{\partial}$ exact? We know that every d-exact (p,p) form on a compact Kahler manifold is $\partial \bar{\partial}$ exact (by the Hodge theorem), but unfortunately, that can't be applied here...</p>
http://mathoverflow.net/questions/48997/griffiths-and-harris-referenceGriffiths and Harris referenceVamsi2010-12-10T23:52:57Z2011-01-15T23:06:57Z
<p>Trying to read the section on Poincare duality from Griffiths and Harris is a nightmare. I want to know if there is a place where Poincare duality and intersection theory are done cleanly and rigorously in the order that GH do them (usually, one proves Poincare duality for singular cohomology and then defines the intersection pairing by cup product and proves that (using the Thom isomorphism) that indeed the intersection pairing counts the number of points taken with sign (and convert everything to forms using De Rham's theorem). GH on the other hand define intersection pairing and then proceed further. However, one has to wave their hands at relativistic speeds to make some things work here).</p>
http://mathoverflow.net/questions/35988/why-were-matrix-determinants-once-such-a-big-deal/50136#50136Answer by Vamsi for Why were matrix determinants once such a big deal?Vamsi2010-12-22T06:57:54Z2010-12-22T06:57:54Z<p>1) The Chern-Weil theory of characteristic classes is built upon determinants of functions of curvature forms of vector bundles.
2) Feynman path integrals require determinants (but typically in infinite dimensions).</p>
http://mathoverflow.net/questions/50025/problems-where-we-cant-make-a-canonical-choice-solved-by-looking-at-all-choices/50104#50104Answer by Vamsi for Problems where we can't make a canonical choice, solved by looking at all choices at onceVamsi2010-12-21T23:07:43Z2010-12-21T23:07:43Z<p>Sard's theorem provides such an example. Given a random smooth map between two manifolds (lets say compact and of the same dimension), there is no canonical way of constructing a regular value. But, Sard's theorem looks at this entire set and proves it is of full measure (and hence there exists atleast one regular value). This result enables us to prove many such results like (the transversality theorem for instance).</p>
http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/49069#49069Answer by Vamsi for Proofs that require fundamentally new ways of thinkingVamsi2010-12-11T19:30:03Z2010-12-11T19:30:03Z<p>Proving that subgroups of free groups are free requires the knowledge of topology, a completely different field which a priori does not have anything to do with groups.</p>
http://mathoverflow.net/questions/46165/chern-classes-generating-cohomologyChern classes generating cohomologyVamsi2010-11-16T00:29:40Z2010-11-16T18:57:47Z
<p>The fact that Chern classes are Hodge classes (and are rational combinations of algebraic cycles) is a part of the proof of the "Gauss Bonnet theorem" (as given in Griffiths and Harris). So my question is why is the fundamental class of every algebraic variety a rational combination of them?</p>
http://mathoverflow.net/questions/40670/a-non-elliptic-pdeA non-elliptic PDEVamsi2010-09-30T20:18:31Z2010-09-30T20:18:31Z
<p>I wish to know if this PDE can be solved (for a real smooth function $\rho$) on a compact complex surface X :
$\bar{\partial}\partial \rho \wedge \bar{\partial}\partial \rho + \bar{\partial}\partial \rho \wedge \bar{\partial}\partial \sigma + \bar{\partial}\partial \sigma \wedge \bar{\partial}\partial \sigma = 0$ where $\sigma$ is a given smooth real function on X.
(Note that, one has solutions to this in $\mathbb{C}^2$ if $\sigma$ is real analytic). The problem is that the linearisation of this equation is not elliptic. The motivation for solving this is to produce a trivial line bundle with a hermitian metric so that its Chern character forms (not cohomology classes) are zero.</p>
http://mathoverflow.net/questions/39828/how-do-you-decide-whether-a-question-in-abstract-algebra-is-worth-studying/40551#40551Answer by Vamsi for How do you decide whether a question in abstract algebra is worth studying?Vamsi2010-09-29T23:15:00Z2010-09-29T23:15:00Z<p>This paper has a very nice introduction (it is on "pointless topology"). So apparently, one may come up with very random definitions for their own sake and hope someone "applies" them to more "concrete" problems.
<a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183550014" rel="nofollow">http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183550014</a></p>
http://mathoverflow.net/questions/126419/density-of-smooth-functions-in-sobolev-spaces-on-manifoldsComment by VamsiVamsi2013-04-04T06:21:46Z2013-04-04T06:21:46ZFor bounded domains in Euclidean space, this is true (Evans' book). It gets tricky when one wants to approximate by smooth functions that smooth upto the boundary.http://mathoverflow.net/questions/110633/monge-ampere-operatorComment by VamsiVamsi2012-10-26T15:48:21Z2012-10-26T15:48:21ZThis isn't a complete answer, but may help. For smooth u_k, u_k >v is an open set. For non-smooth u, this is not necessarily the case. It may have a boundary. Just outside the boundary max(u,v) = v and inside max(u,v) = u. So, whilst testing the current against test functions whose support "ends" at the boundary, it isn't obvious (to me) that dd^c(max(u,v)) = dd^c (u). http://mathoverflow.net/questions/91981/monge-ampere-equations-concavity/92043#92043Comment by VamsiVamsi2012-03-23T22:57:23Z2012-03-23T22:57:23ZI apologise for (possibly) misunderstanding your reply, but in my case I require the Monge-Ampere equation to be elliptic. The concavity of the equation (as a function of Hermitian matrices) is under question. http://mathoverflow.net/questions/91852/when-does-the-tangent-bundle-of-a-manifold-admit-a-flat-connectionComment by VamsiVamsi2012-03-21T23:26:31Z2012-03-21T23:26:31ZWhen n=3 and the manifold is orientable, it always does (orientable three manifolds are parallelisable)http://mathoverflow.net/questions/78460/learning-arakelov-geometry/78463#78463Comment by VamsiVamsi2011-10-18T19:41:27Z2011-10-18T19:41:27ZThanks for the answer. I also want to know if there are any applications of Analytic torsion outside Arakelov geometry. If not, I guess I would have to learn the scheme stuff....http://mathoverflow.net/questions/78093/regarding-discrete-eigenvalues/78107#78107Comment by VamsiVamsi2011-10-14T15:48:53Z2011-10-14T15:48:53ZThis is true, but, what if I don't demand that my eigenfunctions lie in L^2 (I just want them to be smooth functions satisfying the differential equation). I mean, in the example I gave, if one can prove that every eigenfunction has a limit as $r\rightarrow \infty$, then it extends to a function on the sphere and hence the eigenvalues will be the same.http://mathoverflow.net/questions/73439/when-is-a-given-matrix-of-two-forms-a-curvature-formComment by VamsiVamsi2011-08-22T22:03:18Z2011-08-22T22:03:18ZThe Bianchi identity guarantees this ($dtr(F^k) = ktr(dF F^{k-1})=ktr([B,F]F^{k-1})=0$)http://mathoverflow.net/questions/69904/branched-covers-of-compact-riemann-surfacesComment by VamsiVamsi2011-07-10T00:13:08Z2011-07-10T00:13:08ZIndeed it is a central extension of the fundamental group.http://mathoverflow.net/questions/69904/branched-covers-of-compact-riemann-surfaces/69906#69906Comment by VamsiVamsi2011-07-10T00:11:55Z2011-07-10T00:11:55ZThanks! I apologise for my question, but this won't work for a compact R.S minus a finite number of points would it? If it doesn't, then is there an analogue? Also, would the branching index of $S^{'}$ not be $N$ times the number of sheets in $S_1$?http://mathoverflow.net/questions/69276/quillen-metric-definitionComment by VamsiVamsi2011-07-03T11:20:21Z2011-07-03T11:20:21ZYes Paul, I mean the restriction of the $L^2$ metric.http://mathoverflow.net/questions/64545/mehta-seshadri-and-parabolic-bundles/64772#64772Comment by VamsiVamsi2011-05-13T15:46:19Z2011-05-13T15:46:19ZYeah, thanks. I realised that yesterday (Biquard gives us a projective representation of the fundamental group which then is a representation of an extension of the same).http://mathoverflow.net/questions/64545/mehta-seshadri-and-parabolic-bundlesComment by VamsiVamsi2011-05-10T23:14:25Z2011-05-10T23:14:25Z@Mattia: Thanks. Do you happen to have an online copy of the paper? I can't seem to find it at all?http://mathoverflow.net/questions/64545/mehta-seshadri-and-parabolic-bundlesComment by VamsiVamsi2011-05-10T23:12:11Z2011-05-10T23:12:11Z@Dmitri, I mean : Let $V$ be a stable bundle on a smooth curve $C$ and suppose the parabolic degree (=deg(E) + sum of weights of the flags) is not zero. Then, is it true that one can naturally associate a representation of the central extension of $\pi_1$ of $C$ minus some points to $V$?http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-coverComment by VamsiVamsi2011-04-08T17:27:05Z2011-04-08T17:27:05ZIf you have a good finite open cover, the betti numbers would be finite (I think) by a Mayer-Vietoris argument.http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover/59044#59044Comment by VamsiVamsi2011-03-21T20:02:04Z2011-03-21T20:02:04ZActually I don't want them to have infinitely many connected components. I want finitely many components. Thanks for the reference anyway.