User zhengyu hu - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:37:14Z http://mathoverflow.net/feeds/user/18119 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92876/on-vanishing-orders-of-an-ideal-via-the-restriction On vanishing orders of an ideal via the restriction Zhengyu Hu 2012-04-02T06:17:24Z 2012-04-03T05:26:26Z <p>Let $Y$ be a submanifold of a complex manifold $X$, and $a$ be an ideal on $X$ which does not vanish along the entire $Y$. Consider a point $\xi$ on $Y$, there are the vanishing order $ord_{\xi}a$ and the vanishing order $ord_{\xi}a\cdot O_{Y}$. Is there any relation between the two numbers?</p> http://mathoverflow.net/questions/91907/on-the-multiplicities-of-an-ideal-on-a-smooth-variety On the multiplicities of an ideal on a smooth variety Zhengyu Hu 2012-03-22T12:22:41Z 2012-03-22T14:28:42Z <p>Let $X$ be a smooth variety, $\xi$ be a point of $X$ and $\mathfrak{a}$ be an ideal sheaf. If we define $mult_{\xi} \mathfrak{a}$ to be the largest integer $p$ such that <code>$\mathfrak{a} \cdot \mathcal{O}_{X,\xi}\subseteq \mathfrak{m}_{\xi}^{p}$</code>, where <code>$\mathfrak{m}_{\xi}$</code> is the maximal ideal of <code>$\mathcal{O}_{X,\xi}$</code>, we have the map $\xi \mapsto mult_{\xi}\mathfrak{a}$.</p> <p>Is this map upper-semicontinuous?</p> http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring On the comparison of linear topologies on a local ring Zhengyu Hu 2011-12-01T13:06:16Z 2011-12-25T10:22:38Z <p>Let $R$ be a local ring, $a_{\lambda}$ be a decreasing net of ideals, indexed by a directed set, such that each $a_{\lambda}$ is contained in the nilradical ideal and $\bigcap a_{\lambda}=(0)$. Then for any integer $k$, there exists an index $\lambda$ such that $a_{\lambda}\subseteq m^{k}$. That is, the linear topology defined by $a_{\lambda}$ is finer than the topology defined by $m^{k}$.</p> <p>Could anyone give a proof of this statement or a counter example? Feel free to add some more assumptions...</p> http://mathoverflow.net/questions/82860/uniform-approximation-of-nakayama-zariski-decomposition Uniform approximation of Nakayama-Zariski decomposition Zhengyu Hu 2011-12-07T11:31:52Z 2011-12-07T12:35:23Z <p>Let $X$ be a complex smooth projective manifold, $L$ be a pseudo-effective $\mathbb{Q}$-divisor on $X$, and $A$ be an ample divisor. Is it true that, for any $\varepsilon>0$, there exists $\delta$ such that $\sigma_{E}(\|L\|)-\mathrm{ord}_{E}(\|L+\delta A\|)&lt;\varepsilon$ for any prime divisor $E$ on $X$?</p> <p>Moreover, let $\pi:Y \longrightarrow X$ be a birational projective morphism from a smooth manifold $Y$. Does the inequality $\sigma_{E} (\|f^{\ast}L\|)-ord_{E}(\|f^{\ast}L+\delta A\|)&lt;\varepsilon (1+ ord_{E}(K_{Y/X})$ hold?</p> <p>Notations: $ord_{E}(\|L\|)=\inf \frac{1}{m}ord_{E} Fix|mL|$, $\sigma_{E}(\|L\|)=\lim_{\delta \rightarrow 0} ord_{E}(\|L+\delta A\|)$.</p> http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle/82454#82454 Answer by Zhengyu Hu for 'Ampleness' of a big line bundle Zhengyu Hu 2011-12-02T13:41:26Z 2011-12-02T13:46:36Z <p>In general the answer is no, even $F$ is a line bundle itself. It is easy to see that a globally generated line bundle is nef, and if $F$ is not nef, and the segment between $F$ and $M$ does not intersect with the ample cone in $N^{1}(X)$, then $F \otimes M^{n}$ is numerically propotional to a divisor lies in the interior of the segment, thus is not nef.</p> http://mathoverflow.net/questions/82271/on-the-completion-of-a-complete-local-ring On the Completion of a complete local ring Zhengyu Hu 2011-11-30T13:37:05Z 2011-12-01T15:39:27Z <p>Let $(R,\mathfrak{m})$ be a complete local ring, $a_{\lambda}$ be a decreasing net of ideals in $R$, indexed by a directed set. Consider the completion under $a_{\lambda}$-topology $A=\underleftarrow{\lim} R/\mathfrak{a}_{\lambda}$. Is $A$ still complete under the $\mathfrak{m}$-topology?</p> http://mathoverflow.net/questions/82358/on-the-completion-of-two-topologies-on-a-local-ring On the completion of two topologies on a local ring Zhengyu Hu 2011-12-01T08:49:42Z 2011-12-01T08:56:45Z <p>Let $(R,m)$ be an (integral,excellent,regular...) local ring, and $(\widehat{R},\widehat{m})$ be its completion. Let ${a_{\lambda}}$ be a decreasing net of ideals, and $\widehat{a_{\lambda}}$ be the completion in $\widehat{R}$. If we denote $a=\bigcap a_{\lambda}$, such that $V(a)=\bigcup V(a_{\lambda})$ in $\mathrm{Spec}R$, do we have $\widehat{a}=\bigcap \widehat{a_{\lambda}}$?</p> <p>I have some ideas on this question. First consider an exact sequence $0\longrightarrow a \longrightarrow R\longrightarrow \underleftarrow{\lim}R/a_{\lambda}$, taking completion we have an exact sequence $0\longrightarrow \widehat{a} \longrightarrow \widehat{R}\longrightarrow (\underleftarrow{\lim}R/a_{\lambda})^{\wedge}$. </p> <p>Since $R/ a_{\lambda}\hookrightarrow \widehat{R}/\widehat{a_{\lambda}}$ is injective, we have an injection $\underleftarrow{\lim}R/a_{\lambda} \hookrightarrow \underleftarrow{\lim}\widehat{R}/\widehat{a_{\lambda}}$. Thus, if we can embedd $(\underleftarrow{\lim}R/a_{\lambda})^{\wedge}$ into $\underleftarrow{\lim}\widehat{R}/\widehat{a_{\lambda}}$, we have the conslusion. But I don't know under which assumptions this statement would be true...</p> http://mathoverflow.net/questions/82080/on-subadditivity-of-multiplier-ideals On subadditivity of multiplier ideals Zhengyu Hu 2011-11-28T13:17:30Z 2011-11-28T13:17:30Z <p>If $\varphi$ and $\psi$ are two plurisubharmonic weights on an algebraic manifold $X$, then we have $\mathcal{J}(\varphi+\psi)\subseteq \mathcal{J}(\varphi)\mathcal{J}(\psi)$. </p> <p>My question is, if we only allow algebraic functions in such ideals, that is, replace $\mathcal{J}(\varphi)$ by $\mathcal{J}(\varphi) \bigcap \mathcal{O}$, where $\mathcal{O}$ is the algebraic structure sheaf of $X$. Do we still have $\mathcal{J}(\varphi+\psi)\subseteq \mathcal{J}(\varphi)\mathcal{J}(\psi)$? </p> <p>Maybe it's a simple question, but I'm not familar with the analytic settings...</p> http://mathoverflow.net/questions/77122/how-to-prove-the-existence-of-divisorial-zariski-decomposition How to prove the existence of divisorial Zariski decomposition? Zhengyu Hu 2011-10-04T10:35:55Z 2011-10-04T19:26:18Z <p>Let <code>$L$</code> be a pseudo-effective divisor, we may define its numerical fixed part <code>$N_{\sigma}(L)$</code>. How to prove it is a divisor? I know there is a proof in Nakayama's book, but I can't find this book.</p> http://mathoverflow.net/questions/76640/is-md-very-ample-if-d-is-ample Is [mD] very ample if D is ample? Zhengyu Hu 2011-09-28T13:37:37Z 2011-10-01T14:15:37Z <p>Let D be an ample R-divisor, is the round down [mD] very ample for any sufficiently divisible number m?</p> <p>I think it's true. But I do not know how to arrange an argument.</p> http://mathoverflow.net/questions/92876/on-vanishing-orders-of-an-ideal-via-the-restriction Comment by Zhengyu Hu Zhengyu Hu 2012-04-04T01:39:28Z 2012-04-04T01:39:28Z You are right, there are no further relations. http://mathoverflow.net/questions/92876/on-vanishing-orders-of-an-ideal-via-the-restriction Comment by Zhengyu Hu Zhengyu Hu 2012-04-04T01:37:37Z 2012-04-04T01:37:37Z Yes, the order of $a$ is the largest integer $k$ such that $a\subseteq m^{k}$, while the order of $a\cdot O_{Z}$ is largest integer $k$ such that $a+p \subseteq m^{k}+p$. http://mathoverflow.net/questions/92876/on-vanishing-orders-of-an-ideal-via-the-restriction Comment by Zhengyu Hu Zhengyu Hu 2012-04-03T05:26:16Z 2012-04-03T05:26:16Z I figure it out... In fact, the two vanishing orders has no relation.. $f=y+x^{n}$ would give an example... http://mathoverflow.net/questions/92876/on-vanishing-orders-of-an-ideal-via-the-restriction Comment by Zhengyu Hu Zhengyu Hu 2012-04-02T11:56:25Z 2012-04-02T11:56:25Z I guess the equality should be expected in any case... http://mathoverflow.net/questions/82860/uniform-approximation-of-nakayama-zariski-decomposition Comment by Zhengyu Hu Zhengyu Hu 2011-12-07T12:36:40Z 2011-12-07T12:36:40Z I add some explanations for the notations... if I don't explain clearly enough, let me know about it... http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring/82472#82472 Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T15:35:54Z 2011-12-02T15:35:54Z In fact, if we dot not require something on $a_{\lambda}$, this is clearly false. e.g. Take $f\in \widehat{R}$ be a transcendental function, then $(f)+m^{k}\bigcap R$ is not contained in $m^{k}$... So the crucial point is to see under which assumption the two topologies coincide... http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring/82472#82472 Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T15:29:56Z 2011-12-02T15:29:56Z I don't understand the notations and the Chevalley theorem... could you give a reference? http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring/82469#82469 Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T15:25:59Z 2011-12-02T15:25:59Z I think &quot;Thus the image of $a_{\lambda}$ is equal to the image of the intersection of all $a_{\mu}$, which is zero&quot; is not a clear argument since in general this does not hold.. http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle/82454#82454 Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T15:10:02Z 2011-12-02T15:10:02Z Oh... I think your comment is a good counter example. It seems that works for any $M$. http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle/82454#82454 Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T14:01:30Z 2011-12-02T14:01:30Z In fact, many line bundles satisfy the property that the segment does not intersect with the ample cone. Anti-ampleness is not required. http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T10:49:25Z 2011-12-02T10:49:25Z You can forget this condition... I just want to know in which cases $a_{\lambda}$ induces a finer topology. When $R$ is complete, this is by a Theorem of Chevalley (1946), and I am interested in the case $R$ is not complete... http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T10:43:25Z 2011-12-02T10:43:25Z You might see my comment before, this condition is just a translation of $V(a)=\bigcup V(a_{\lambda})$ in $\mathrm{Spec}R$. By passing $R$ to $R/a$, we have this condition (maybe under some more hypothesis...) http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T09:57:37Z 2011-12-02T09:57:37Z I just want to know in which canses the linear topology induced by $a_{\lambda}$ is finer than that induced by $a+m^{k}$, where $a=\bigcap a_{\lambda}$. In this question, I assume $a=0$... http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T09:55:59Z 2011-12-02T09:55:59Z noetherian, and you can add more assumptions if you would... auch like $R$ is excellent, or $a_{\lambda}$ satisfies some properties... http://mathoverflow.net/questions/82421/how-to-prove-weak-lower-semicontinuity-of-the-norm-in-a-banach-space Comment by Zhengyu Hu Zhengyu Hu 2011-12-02T02:10:00Z 2011-12-02T02:10:00Z I don't know the question... The norm is continuous of course... If you mean a function, try to write it as $\sup f_{lambda}$ with each $f_{\lambda}$ continuous...