User zhengyu hu - MathOverflow

On vanishing orders of an ideal via the restriction

2012-04-02T06:17:24Z

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?

On the multiplicities of an ideal on a smooth variety

2012-03-22T12:22:41Z

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 $\mathfrak{a} \cdot \mathcal{O}_{X,\xi}\subseteq \mathfrak{m}_{\xi}^{p}$ , where $\mathfrak{m}_{\xi}$ is the maximal ideal of $\mathcal{O}_{X,\xi}$ , we have the map $\xi \mapsto mult_{\xi}\mathfrak{a}$.

Is this map upper-semicontinuous?

On the comparison of linear topologies on a local ring

2011-12-01T13:06:16Z

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 $m^{k}$.

Could anyone give a proof of this statement or a counter example? Feel free to add some more assumptions...

Uniform approximation of Nakayama-Zariski decomposition

2011-12-07T11:31:52Z

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 $\sigma_{E}(\|L\|)-\mathrm{ord}_{E}(\|L+\delta A\|)<\varepsilon$ for any prime divisor $E$ on $X$?

Moreover, let $\pi:Y \longrightarrow X$ be a birational projective morphism from a smooth manifold $Y$. Does the inequality $\sigma_{E} (\|f^{\ast}L\|)-ord_{E}(\|f^{\ast}L+\delta A\|)<\varepsilon (1+ ord_{E}(K_{Y/X})$ hold?

Notations: $ord_{E}(\|L\|)=\inf \frac{1}{m}ord_{E} Fix|mL|$, $\sigma_{E}(\|L\|)=\lim_{\delta \rightarrow 0} ord_{E}(\|L+\delta A\|)$.

Answer by Zhengyu Hu for 'Ampleness' of a big line bundle

2011-12-02T13:41:26Z

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.

On the Completion of a complete local ring

2011-11-30T13:37:05Z

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?

On the completion of two topologies on a local ring

2011-12-01T08:49:42Z

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 $\widehat{a}=\bigcap \widehat{a_{\lambda}}$?

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 $0\longrightarrow \widehat{a} \longrightarrow \widehat{R}\longrightarrow (\underleftarrow{\lim}R/a_{\lambda})^{\wedge}$.

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...

On subadditivity of multiplier ideals

2011-11-28T13:17:30Z

If $\varphi$ and $\psi$ are two plurisubharmonic weights on an algebraic manifold $X$, then we have $\mathcal{J}(\varphi+\psi)\subseteq \mathcal{J}(\varphi)\mathcal{J}(\psi)$.

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 $\mathcal{J}(\varphi+\psi)\subseteq \mathcal{J}(\varphi)\mathcal{J}(\psi)$?

Maybe it's a simple question, but I'm not familar with the analytic settings...

How to prove the existence of divisorial Zariski decomposition?

2011-10-04T10:35:55Z

Let $L$ be a pseudo-effective divisor, we may define its numerical fixed part $N_{\sigma}(L)$ . How to prove it is a divisor? I know there is a proof in Nakayama's book, but I can't find this book.

Is [mD] very ample if D is ample?

2011-09-28T13:37:37Z

Let D be an ample R-divisor, is the round down [mD] very ample for any sufficiently divisible number m?

I think it's true. But I do not know how to arrange an argument.

Comment by Zhengyu Hu

2012-04-04T01:39:28Z

You are right, there are no further relations.

Comment by Zhengyu Hu

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 $a+p \subseteq m^{k}+p$.

Comment by Zhengyu Hu

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...

Comment by Zhengyu Hu

2012-04-02T11:56:25Z

I guess the equality should be expected in any case...

Comment by Zhengyu Hu

2011-12-07T12:36:40Z

I add some explanations for the notations... if I don't explain clearly enough, let me know about it...

Comment by Zhengyu Hu

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...

Comment by Zhengyu Hu

2011-12-02T15:29:56Z

I don't understand the notations and the Chevalley theorem... could you give a reference?

Comment by Zhengyu Hu

2011-12-02T15:25:59Z

I think "Thus the image of $a_{\lambda}$ is equal to the image of the intersection of all $a_{\mu}$, which is zero" is not a clear argument since in general this does not hold..

Comment by Zhengyu Hu

2011-12-02T15:10:02Z

Oh... I think your comment is a good counter example. It seems that works for any $M$.

Comment by Zhengyu Hu

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.

Comment by Zhengyu Hu

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...

Comment by Zhengyu Hu

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...)

Comment by Zhengyu Hu

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 $a=0$...

Comment by Zhengyu Hu

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...

Comment by Zhengyu Hu

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...