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What is the definition of the "moving part" and "fixed part" of a linear system$|L|$?

I think the fixed part should be defined to be the greatest effective divisor $F$ such that $D-F\geq 0$ for every $D$ in the system, and the moving part is the linear system $|M|=|L|-F$.

Thus the fixed part is the codimension 1 part in the base locus.

$|M|$ may not be point free?(but I think for the curves, it is basepoint free)

If $|M|$ defines a rational map (morphism on some open subset) to $P^k$, what is its relation to the rational map defined by $|L|$?

When people say moving a divisor in the moving part, does it always mean using implicitly the Bertini theorem?

Is there any reference on the moving and fixed part of linear system?

  • "Is this already enough for $|M|$ to be base point free?" What is "this"? – RghtHndSd Apr 12 '14 at 13:35
  • I want to mean the base points do not contain a codim 1 suset. But I think it may not be right. –  Apr 12 '14 at 13:56
  • I believe the term "fixed part" is synonymous with "base locus". This is the collection of points that are common to all divisors in the linear system. This differs from your post in which you restrict to divisors rather than points. I can't think of any example where your definition of fixed part leads to something besides the empty set. – RghtHndSd Apr 12 '14 at 14:35
  • Oops, I was restricting my thoughts in the above comments to complete linear systems. – RghtHndSd Apr 12 '14 at 14:41
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    @rghthndsd: what you say above is not quite correct. "Fixed part" means "codimension 1 part of the base locus". –  Apr 13 '14 at 09:14
  • Yes, I wanted to mean that. –  Apr 13 '14 at 09:19

1 Answers1

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Some quick answers:

-Yes, the $\geq$ version of your definition is correct. (In the definition of moving part you have a typo; replace $D$ by $F$.)

-In general $|M|$ is not basepoint-free. Certainly that is true for curves, because there "fixed part" and "base locus" coincide. For surfaces $|M|$ might not be basepoint-free, but it is a theorem of Zariski that $|kM|$ is for some natural number $k$. For dimension 3 and more nothing like this is true: look up the definition of "movable linear system".

-Any rational map can be extended to codimension 1. In your situation, the rational map defined by $|L|$ will extend to the morphism defined by $|M|$.

-"Moving" just means choosing another effective divisor in the same linear system. No need to refer to Bertini's theorem.

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    Thanks for your answer, a small question: what is the "movable linear system" you mentioned? Is there reference for that? –  Apr 13 '14 at 09:37
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    Dear @mqx: that just means a linear system with empty fixed part. –  Apr 13 '14 at 09:38
  • What does it mean by "For dimension 3 and more nothing like this is true:look up the definition of "movable linear system"? –  Apr 13 '14 at 09:40
  • And can you explain a little bit in the following? Thanks! http://math.stackexchange.com/questions/484740/dimension-of-complete-linear-system-d-where-d-ae-with-e-base-point-fr?rq=1 –  Apr 13 '14 at 09:48
  • Dear mqx: all I can do is repeat what I said --- look up the definition of movable, and think about why it does not imply basepoint-free. –  Apr 13 '14 at 14:29
  • Oh I see,thanks! –  Apr 13 '14 at 15:09