# Siu decomposition

hello. I have the following question: Let $T$ be a positive closed current of bidimension $(p,p)$ then one has the Siu decomposition $T=R+S$ where $R$ is a positive closed current such that its Lelong number is zero along analytic sets of dimension greater or equal $p$. Now my question is why this current $R$ is the bigger with this property? $R$ is called the residual part while $S$ is the singualr one and $S$ is also positive.

thank you

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I think you are talking about the result in Siu's paper "Analyticity of sets associated to Lelong numbers and extension of closed positive currents, Invent. Math. 27 (1974) 53–156." But I do not quite understand what you are asking. What do you mean by "$R$ is bigger with this property"? – Paul Jul 6 '11 at 12:30
He wrote not "is bigger" bit "is the bigger", so probably meant "is the biggest" (e.g. in Spanish "el" = the, "más grande" = bigger, but "el más grande" = the biggest). – Noam D. Elkies Jul 6 '11 at 15:15

In my opinion, the point is not that $R$ is the biggest in whatever sense you may give to this, but that the decomposition $T=R+S$ with $S=\sum \lambda_j [A_j]$ ($A_j$ being a $p$-dimensional analytic set), and $R$ having zero Lelong number along any $p$-dimensional analytic set.

The uniqueness is clear because for $x$ generic in $A_j$, $\nu(R+S,x)=\lambda_j \nu([A_j],x)=\lambda_j$, which determines thus uniquely $\lambda_j$, and therefore $S$ and $R$.

Now, if you really want to see that $R$ is the biggest current (in the sense of positivity of currents) such that $R$ has zero Lelong number along any $p$-dimensional analytic set, then you can proceed this way: assume that $T=S'+R'$ is another such decomposition. Then for $x$ generic in $A_j$, $\nu(T,A_j)=\nu(T,x)=\nu(S',x)=\nu(S',A_j)$. But it is a classical fact (see e.g Demailly, Complex analytic and differential Geometry, Proposition 8.16) that $S'-\nu(S', A_j) [A_j]$ is a closed positive current, so that $S' \geqslant S$, and therefore $R' \leqslant R$, which concludes.

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Indeed, the crucial property that Henri points out in his answer that if T is a closed positive current of bidimension $(p,p)$ and $A$ an irreducible analytic set of dimension $p$ then $T-\nu(T,A)[A]$ is again positive is exactly what you need in order show the existence part of the decomposition above (more precisely the convergence of the series in the weak topology). Thus, I really think that our answers are more or less equivalent! – diverietti Jul 7 '11 at 8:01

I am not sure if I understand correctly your question, but anyway I'll try to answer: the point is that Siu's decomposition is in fact unique, but let me explain better.

Let $X$ be a complex manifold and $T$ a closed positive current of bidimension $(p,p)$. Then, there is a unique decomposition of $T$ as a (possibly ﬁnite) weakly convergent series $$T=\sum_{j\ge 1}\lambda_j[A_j]+R,\quad\lambda_j\ge 0,$$ where $[A_j]$ is the current of integration over an irreducible $p$-dimensional analytic set $A_j\subset X$ and where $R$ is a closed positive current with the property that $\dim E_c(R)< p$ for every $c>0$. Here $E_c(R)$ is the set of point $x\in X$ such that the Lelong number $\nu(R,x)$ of $R$ at $x$ is greater than or equal to $c$.

To prove the uniqueness, assume that $T$ admits such a decomposition. Then, the $p$-dimensional components of $E_c(T)$ are $(A_j)_{\lambda_j\ge c}$, for $\nu(T,x)=\sum\lambda_j\nu([A_j],x)+\nu(R,x)$ is non zero only on $\bigcup A_j\cup\bigcup E_c(R)$, and is equal to $\lambda_j$ generically on $A_j$. In particular $A_j$ and $\lambda_j$ are unique.

Thus $R$ is neither the biggest nor the littlest... It is just unique!

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Hi Simone! You've been faster than I on this one ;-) But I think the OP's question makes sense. Indeed, you could write $T=(S+R/2)+R$, which would give a decomposition in the OP's sense. – Henri Jul 6 '11 at 22:13
Hi Henri! Did you mean $T=(S+R/2)+R/2$? Anyway, I don't think so, because $R/2$ is not of the form "integration on analytic set of dimension $p$". – diverietti Jul 6 '11 at 22:17
Yes I mean $R/2$ sorry! Of course $R/2$ is not of the form you say because it is the residual part, thus has zero Lelong number along any $p$-dim analytic set. It is just a matter of definition here, cf the OP question. – Henri Jul 6 '11 at 22:21
Right, a matter of definitions... That's why I tried to fix also notations in my answer! – diverietti Jul 6 '11 at 22:36

R is the biggest with such property since S is the sum of currents of integration on varieties of dimension p. Moreover, a dimension (p,p) current can not have positive Lelong number along varieties of dimension greater than p.

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There is still something to say so far, cf the third § of my answer. – Henri Jul 6 '11 at 22:10