User bo - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:33:02Z http://mathoverflow.net/feeds/user/12904 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/134119/holomorphic-separation-and-the-existence-of-strictly-plursisubharmonic-functions Holomorphic separation and the existence of strictly plursisubharmonic functions Bo 2013-06-19T05:26:04Z 2013-06-19T05:26:04Z <p>Recall that a complex manifold is Stein if it is holomorphically convex and separable. If we assume holomorphically convex alone, then there is Cartan-Remmert reduction to say how far it is from being Stein. If we assume holomorphic separation alone, I learnt from Th 9.9(b) in p41 of Demailly's lecture notes <a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/estimations_l2.pdf" rel="nofollow">Here</a> that:</p> <p>"On a complex manifold X, if $O(X)$ locally separate points (i.e. for any $x \in X$, there is a nbhd of $U_x$ such that for any $y \in {U_x}-x$, there exists $f \in O(X)$ with $f(x) \neq f(y)$), then there exists a smooth nonnegative strictly plursisubharmonic function" </p> <p>The proof looks like first construct global PSH functions which are locally strictly PSH on small nbhd, then adding them together after multiplying some constants. I don't understand why that sum converges. Since the proof looks very straightforward and I may miss sth simple here. I would appreciate if anyone can help. Thanks! </p> http://mathoverflow.net/questions/117391/how-to-understand-the-diffeomorphism-in-the-cheeger-gromov-compactness How to understand the diffeomorphism in the Cheeger-Gromov compactness Bo 2012-12-28T07:02:11Z 2012-12-29T03:51:13Z <p>Recall that $(M_{k}, g_k,O_k)$ of complete pointed Riemannian manifolds converges smoothly in the sense of Cheeger-Gromov to $(M_{\infty}, g_{\infty}, O_{\infty})$ if there exists an exhaustion of open sets $U_k$ of $M_{\infty}$ containing $O_{\infty}$, and a sequence of diffeomorphism $\phi_k$ from $U_k$ to $V_k=\phi_k(U_k) \subset M_k$ with $\phi_k (O_{\infty})=O_k$, such that the pull back $(U_k, \phi_k^{\ast} (g_k))$ converges to $(M_{\infty}, g_{\infty})$ uniformly on compact sets in $M_{\infty}$. </p> <p>These days I am interested in how wild the diffeomorphism $\phi_k$ can be. Consider a sequence of rotationaly invariant metrics on $\mathbb{R}^n$ with uniformly bounded geometry and injective radius at the origin has a uniform lower bound (which can be ensured by for example, choosing a uniform bounded positive cone angle at infinity for the sequence metrics), fix the sequence of points to be the origin, by the compactness theorem we expect a limit metric which is still rotationaly invariant. However, I am not sure if the limit is necessarily rotationaly invariant. If so does it mean that in this special case, we can choose the sequence of diffeomorphism to be a sequence of rotations around the origin?</p> <p>I understand that a way to think about this symmetric case is to go over the general construction on diffeomorphism and the limit metric in the proof of Compactness Theorem. It is a little involved and I am still on the way to understand it. Any suggestions or help will be appreciated.</p> <p>Added after posted:</p> <p>To make the point of "understand the diffeomorphism $\phi_k$" clear, let me continue on the rotationally symmetric example, Let us put one more restrition on the sequence $(M_k, g_k, O_k)=(R^n, g_k,O)$ that the scalar curvature $R(O)$ attain the maximum for each $g_k$, then after the taking limit we can get a metric $(M_\infty, g_\infty,O_\infty)$ which also have $R(O_\infty)$ as a maximum on $M_\infty$, But if we take $O_k$ to be a sequence points $P_k$ suitably close to the origin $O$, the limit will be a new metric $(N_\infty, h_\infty,P_\infty)$. It is reasonably to believe the new metric could be the same as $(M_\infty, g_\infty,O_\infty)$. However $R(P_\infty)$ might not be the maximum of scalar curvature on $N_\infty$ in general, So it is natural to find a maximum of scalar curvature on $N_\infty$ first, name it $Q_\infty \in N_\infty$, then check how far the pull back sequence $\Phi_k(Q_\infty) \in M_k$ is from $O$. It sounds to me that it is important to understand the behavior of $\Phi_k$. However I feel that it is a little wild even for this example.</p> http://mathoverflow.net/questions/109502/perelmans-example-on-nonuniqueness-of-tangent-cones-at-infinity Perelman's example on nonuniqueness of tangent cones at infinity Bo 2012-10-13T00:54:32Z 2012-10-13T04:37:30Z <p>Perelman has an example on manifolds with nonunique tangent cones at infinity. The paper is <a href="http://library.msri.org/books/Book30/contents.html" rel="nofollow">here</a>. It is a complete manifold with positive Ricci curvature, Euclidean volume growth, and quadratic curvature decay. The metric has the form $ds^2=dt^2+A(t)^2 dx^2+B^2 (t) dy^2+C^2 (t) dz^2$, with a particular choice of $A,B,C$ as functions of $t$ given in the paper linked above. While the $Ric>0$ is easier to verify, I have difficulty to understand why the tangent cones are not unique. My question is: what different cross sections (I mean the set ${r=1}$ on the tangent cones at infinity) they have if we pick different sequences $r_i \rightarrow +\infty$ in $(M,\frac{1}{r_i^2}g, p)$ to get the tangent cone. Since different tangent cones must have the same cone angle, those cross sections must have the same Hausdorff measure. In this sense it is harder for me to imagine why they could be different. Any help will be appreciated. </p> http://mathoverflow.net/questions/103361/cone-angle-at-infinity-for-product-of-cones cone angle at infinity for product of cones Bo 2012-07-28T03:14:08Z 2012-07-28T08:50:48Z <p>Let $A=\lim_{r \rightarrow +\infty} \frac{Vol(B(o,r))}{\omega_{n} r^{n}}$ for any Riemannian manifold $(\mathbb{M}^{n},g)$ with nonnegative Ricci curvature. Here $\omega_{n}$ is the volume of unit ball in $\mathbb{R}^n$. We call $A$ the cone angle at infinity or asymptotic volume ratio, and the manifold is cone-like if $A>0$. I got a direct question with this definition. Let $(\mathbb{M}^{n},g)$ be a metric product of two manifolds $(\mathbb{M}_1^{k},g_1,A_1)$ and $(\mathbb{M}_2^{n-k},g,A_2)$ both with nonnegative curvature and cone-like. Does the cone angle of the product manifold only depends on $k$, $n$, $A_1$ and $A_2$. It sounds like the new cone angle will be some average of $A_1$ and $A_2$, but I failed to get a clean formula even assuming product of two rotationally symmetric cones in $R^3$ (like surface of evolution). Is there something simple I miss? Thanks!</p> http://mathoverflow.net/questions/97969/convex-functions-from-surfaces-of-revolution convex functions from surfaces of revolution Bo 2012-05-25T18:10:16Z 2012-05-25T18:10:16Z <p>Li-Tam (Ann of math 1987) proved some estimates on nonnegative harmonic functions on manifolds with sec $\geq 0$ outside a compact set. I just want to get some feeling on how nontivial those estimates are. so I tried surface of revolution with positive curvature. In this simple case, those estimates are reduced the following question on convex functions from the graph of the surface. </p> <p>Assume that $f$ is a smooth convex function on $[0,+\infty]$ with $f(0)=f^{\prime}(0)=0$ and $\lim_{x \rightarrow +\infty} f(x)=\lim_{x \rightarrow +\infty} f^{\prime}(x)=+\infty$ ($f^{\prime}$ is stricty increasing), </p> <p>Then can we show:</p> <p>$$\lim_{x \rightarrow +\infty} \frac{\log[\int_{C}^{x} \frac{f^{\prime}(\tau)}{\tau} d\tau \int_{C}^{x} \tau f^{\prime} (\tau) d\tau]} {\log[f(x)]}=2$$ for any fixed number $C>0$?</p> <p>At first I thought it was just a simple calculus and $\liminf \geq 2$ by Cauchy-Schwarz, but I tried and failed to get the $\limsup$ part. I can only show that $\limsup \leq 2$ when $\lim_{x \rightarrow +\infty} \frac{f(x)}{xf^{\prime}(x)}$ exists. However, this assumption seems not true in general (although I don't have a counterexample yet). I will greatly appreciate if anyone suggests a solution. </p> http://mathoverflow.net/questions/96715/entire-functions-of-one-complex-variable-with-prescribed-value-and-order entire functions of one complex variable with prescribed value and order. Bo 2012-05-11T21:35:36Z 2012-05-12T01:17:03Z <p>In the complex plane, say $a_n \rightarrow \infty$ and $d_n$ and $A_n$ are arbitrary complex numbers. can we find an entire function with $f(a_n)=A_n$ with order $d_n$? (here "order" means $f(z)-A_n$ has a zero with order $d_n$) </p> <p>If without restrictions on the order, it is an exercise from Ahlfors 3rd edition P197 No.1. Following his hint, an answer is like $\sum_{n} g(z) \frac{A_n}{g^{\prime}(a_n)} \frac{e^{r_n(z-a_n)}}{(z-a_n)}$ for some suitable chosen $r_n$ where $g(z)$ is an entire functions with simple zeros at $a_n$. I am not sure how to do with the requirement on orders. It sounds like a standard result, I greatly appreciate if anyone with an idea or reference to this problem.</p> http://mathoverflow.net/questions/91885/the-largest-eigenvalue-of-the-matrix-a-with-a-iji-times-j-mod-p-for-p-is-a the largest eigenvalue of the matrix A with A_{ij}=(i \times j) mod p for p is a prime. Bo 2012-03-22T05:42:03Z 2012-03-26T20:45:45Z <p>For a prime p, consider the $(p-1) \times (p-1)$ matrix A with entry to be $A_{ij}=(i \times j) mod$ $p$. every row (column) is permutation of 1 to p-1, such a permutation is useful in one version of proof of Fermat's little theorem. Here the question is if the largest eigenvalue is always p(p-1)/2. also anything happens for the rank of it. except for p=2,3, the rank might always be p-2.</p> <p>the student taking a course (general intro to math) I am TAing asked me it. but it is embarrassing to say I don't know how to work it out. I know almost zero about primes and I believe this might be a standard result. I am grateful if anyone suggests a hint or any reference. </p> http://mathoverflow.net/questions/87488/4-dim-compact-positively-curved-manifolds-with-a-nontrivial-killing-vector-field 4-dim compact positively curved manifolds with a nontrivial Killing vector field. Bo 2012-02-03T22:55:26Z 2012-02-03T22:55:26Z <p>Kleiner-Hsiang (JDG 1989) proved such a manifold is homeomorphic to $S^4$ or $CP^{2}$. an interesting corrollary is that $S^2 \times S^2$ does not admit positively curved metric with countinuous symmetry. They asked in their paper if it is diffeomorphic. I don't know much on this area, only noticed one subsequent work Searle-Yang. Seems that there are preprints on arXiv attempting to do the problem in full generality. Not sure if it was fully settled. Anyone knows the precise status of this problem?</p> http://mathoverflow.net/questions/77275/example-for-busemann-function-is-not-an-exhaustion-when-ricci-ge-0/79204#79204 Answer by Bo for Example for Busemann function is not an exhaustion when Ricci $\ge 0$ Bo 2011-10-26T22:40:21Z 2011-10-26T22:40:21Z <p>Sorry it's not an answer. I just came across a similar question recently and found a reference. </p> <p>If the manifold has Ricci nonnegative and Euclidean volume growth, then the Busemann function is an exhaustion function. This is a result due to Zhongmin Shen.</p> <p>See P400 Lemma 3.4 in Shen, Zhongmin, Complete manifolds with nonnegative Ricci curvature and large volume growth. Invent. Math. 125 (1996), no. 3, 393–404. </p> <p>But I did not check the details of this result. </p> http://mathoverflow.net/questions/79182/how-to-compute-kobayashi-distance-of-compact-kaehler-manifolds-with-postive-ricci How to compute Kobayashi distance of compact Kaehler manifolds with postive Ricci curvature? Bo 2011-10-26T18:52:56Z 2011-10-26T19:27:26Z <p>Recently I just learned the Kobayashi distance on complex manifolds and wants to get some feeling of how it looks like on exmaples of manifolds with positive Ricci curvature. I have a feeling that the Kobayashi distance on those manifolds should vanish since those are not very "hyperbolic".</p> <p>A simple example is extended complex plane. then its Kobayashi distance vanishes since the automorphism group can contract one point very close to the origin while keeping the origin fixed. I believe it is similarly true for all complex projective spaces with usual complex structure since automorphism group is known. Can we have more examples? Or is there a counterexample (an example of Kaehler manifold with positive Ricci curvature but not identically vanishing Kobayashi distance)?</p> http://mathoverflow.net/questions/117391/how-to-understand-the-diffeomorphism-in-the-cheeger-gromov-compactness Comment by Bo Bo 2012-12-28T22:05:31Z 2012-12-28T22:05:31Z Thanks Igor for suggesting equivariant version of convergence. Sorry for the vague question. Just add one more example to clarify. http://mathoverflow.net/questions/109502/perelmans-example-on-nonuniqueness-of-tangent-cones-at-infinity/109514#109514 Comment by Bo Bo 2012-10-14T03:12:12Z 2012-10-14T03:12:12Z Oh, Now I can see it. Thank you very much! http://mathoverflow.net/questions/103361/cone-angle-at-infinity-for-product-of-cones/103369#103369 Comment by Bo Bo 2012-07-29T03:30:21Z 2012-07-29T03:30:21Z Thanks for your patience! It is very helpful! http://mathoverflow.net/questions/101610/complete-or-open-kahler-manifold-and-simply-connected/101616#101616 Comment by Bo Bo 2012-07-09T07:37:54Z 2012-07-09T07:37:54Z In Yau's the open problem list, one asks &quot;Is any complete Kahler manifold with positive Ricci curvature is a Zariski open subset of some compact Kahler manifold?&quot; Any connection? http://mathoverflow.net/questions/91885/the-largest-eigenvalue-of-the-matrix-a-with-a-iji-times-j-mod-p-for-p-is-a/92131#92131 Comment by Bo Bo 2012-05-11T21:37:32Z 2012-05-11T21:37:32Z sorry have not been on for a while, surprised to see that you two developed the problem to a new level, I will try to learn it! Many thanks! http://mathoverflow.net/questions/91885/the-largest-eigenvalue-of-the-matrix-a-with-a-iji-times-j-mod-p-for-p-is-a Comment by Bo Bo 2012-03-22T17:49:00Z 2012-03-22T17:49:00Z Thank both of you for the help! still not sure if the rank is also always p-2 for $p \geq 5$ Maybe I will run some numeric test. http://mathoverflow.net/questions/87488/4-dim-compact-positively-curved-manifolds-with-a-nontrivial-killing-vector-field Comment by Bo Bo 2012-02-09T03:11:17Z 2012-02-09T03:11:17Z Thanks for pointing it out. I am just not sure if that paper is well acknowledged. http://mathoverflow.net/questions/79182/how-to-compute-kobayashi-distance-of-compact-kaehler-manifolds-with-postive-ricci/79185#79185 Comment by Bo Bo 2011-10-26T22:29:37Z 2011-10-26T22:29:37Z Thank you very much for the quick answer. What happens if we change &quot;compact&quot; to &quot;complete noncompact&quot; or change &quot;Ricci positive&quot; to &quot;Ricci nonnegative&quot;? Is there a similar argument? I know almost zero of algebraic geometry.