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What is the multidegree of a curve $C subset mathbbP^n times mathbbP^m$?

1

$begingroup$

What is the multidegree of a curve $C hookrightarrow mathbbP^n times mathbbP^m$?

I'm reading Notes on stable maps and quantum cohomology by W. Fulton and R. Pandharipande, and on page 14, there is the sentence

Let $H$ be the Hilbert scheme of genus $g$ curves in $mathbbP(W) times mathbbP^r$ of multidegree $(e,d)$.

I know what the Hilbert scheme is, but afaik one needs to specify a Hilbert polynomial $P$. Fixing the (arithmetic) genus defines the constant term, because $g = (-1)^d(P(0) - 1)$, where $d$ is the dimension. The degree of $P$ is given by saying that we consider curves, i.e. $deg(P) = 1$. So the leading coefficient is still missing.

Of course to define the Hilbet scheme (or even the Hilbert polynomial $P$) we also need to specify a very ample sheaf, or a closed embedding $mathbbP^n times mathbbP^m to mathbbP^N$ for some $N > 0$. I actually have two guesses here, but I don't know which is it:

1. simply take the Segre embedding. Here I have no clue what the multidegree could mean.


2. take the embedding defined by the very ample invertible sheaf $mathcalO_mathbbP^n(e) boxtimes mathcalO_mathbbP^m(d)$, and maybe fix $1$ as the leading coefficient for $P$?

algebraic-geometry projective-space projective-schemes moduli-space

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asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319

$endgroup$

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  Let $Hsubset mathbbP^n$ be a general hyperplane, then $e=Ccdot Htimes mathbbP^m$. Reversing the roles of $n,m$ you get $d$.
  $endgroup$
  – Mohan
  Apr 8 at 13:34
  

  
  
  
  


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  $begingroup$
  What is $C cdot H times mathbbP^m$? The number of intersections? And do you have any clue how to fix the Hilbert polynomial of such a curve, depending on $(e, d)$?
  $endgroup$
  – red_trumpet
  Apr 8 at 14:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hilbert polynomial will depend on the ample class.
  $endgroup$
  – Mohan
  Apr 8 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, but what is an obvious ample class? You suggest $H times mathbbP^m$ as a divisor?
  $endgroup$
  – red_trumpet
  Apr 8 at 16:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $HtimesmathbbP^m$ is not ample.
  $endgroup$
  – Mohan
  Apr 8 at 17:26
  

  
  
  
  

add a comment | 

1

$begingroup$

What is the multidegree of a curve $C hookrightarrow mathbbP^n times mathbbP^m$?

I'm reading Notes on stable maps and quantum cohomology by W. Fulton and R. Pandharipande, and on page 14, there is the sentence

Let $H$ be the Hilbert scheme of genus $g$ curves in $mathbbP(W) times mathbbP^r$ of multidegree $(e,d)$.

I know what the Hilbert scheme is, but afaik one needs to specify a Hilbert polynomial $P$. Fixing the (arithmetic) genus defines the constant term, because $g = (-1)^d(P(0) - 1)$, where $d$ is the dimension. The degree of $P$ is given by saying that we consider curves, i.e. $deg(P) = 1$. So the leading coefficient is still missing.

Of course to define the Hilbet scheme (or even the Hilbert polynomial $P$) we also need to specify a very ample sheaf, or a closed embedding $mathbbP^n times mathbbP^m to mathbbP^N$ for some $N > 0$. I actually have two guesses here, but I don't know which is it:

1. simply take the Segre embedding. Here I have no clue what the multidegree could mean.


2. take the embedding defined by the very ample invertible sheaf $mathcalO_mathbbP^n(e) boxtimes mathcalO_mathbbP^m(d)$, and maybe fix $1$ as the leading coefficient for $P$?

algebraic-geometry projective-space projective-schemes moduli-space

share|cite|improve this question

asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $Hsubset mathbbP^n$ be a general hyperplane, then $e=Ccdot Htimes mathbbP^m$. Reversing the roles of $n,m$ you get $d$.
  $endgroup$
  – Mohan
  Apr 8 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $C cdot H times mathbbP^m$? The number of intersections? And do you have any clue how to fix the Hilbert polynomial of such a curve, depending on $(e, d)$?
  $endgroup$
  – red_trumpet
  Apr 8 at 14:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hilbert polynomial will depend on the ample class.
  $endgroup$
  – Mohan
  Apr 8 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, but what is an obvious ample class? You suggest $H times mathbbP^m$ as a divisor?
  $endgroup$
  – red_trumpet
  Apr 8 at 16:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $HtimesmathbbP^m$ is not ample.
  $endgroup$
  – Mohan
  Apr 8 at 17:26
  

  
  
  
  

add a comment | 

$begingroup$

What is the multidegree of a curve $C hookrightarrow mathbbP^n times mathbbP^m$?

I'm reading Notes on stable maps and quantum cohomology by W. Fulton and R. Pandharipande, and on page 14, there is the sentence

Let $H$ be the Hilbert scheme of genus $g$ curves in $mathbbP(W) times mathbbP^r$ of multidegree $(e,d)$.

I know what the Hilbert scheme is, but afaik one needs to specify a Hilbert polynomial $P$. Fixing the (arithmetic) genus defines the constant term, because $g = (-1)^d(P(0) - 1)$, where $d$ is the dimension. The degree of $P$ is given by saying that we consider curves, i.e. $deg(P) = 1$. So the leading coefficient is still missing.

Of course to define the Hilbet scheme (or even the Hilbert polynomial $P$) we also need to specify a very ample sheaf, or a closed embedding $mathbbP^n times mathbbP^m to mathbbP^N$ for some $N > 0$. I actually have two guesses here, but I don't know which is it:

1. simply take the Segre embedding. Here I have no clue what the multidegree could mean.


2. take the embedding defined by the very ample invertible sheaf $mathcalO_mathbbP^n(e) boxtimes mathcalO_mathbbP^m(d)$, and maybe fix $1$ as the leading coefficient for $P$?

algebraic-geometry projective-space projective-schemes moduli-space

share|cite|improve this question

asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319

$endgroup$

share|cite|improve this question

asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319

share|cite|improve this question

asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319

asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319

asked Apr 8 at 12:24

red_trumpetred_trumpet

1,061319