Mixing
Divisors split into a sections
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Suppose we have family $(pi:U longrightarrow X), p_i,f)$ be in $overlineM_g,n(P^r,d).$ I want to know what does the following statement mean:
the cartier divisors $f^astt_i$ split into sections $q_i,_j$
We have $f:Ulongrightarrow P^r$. I think it means that when we look at a fibre of $x$ which is a stable curve then zero scheme of$f^astt_i$ as a cartier divisor is $q_i,1(x) +q_i,2(x) +cdots +q_i,d(x)$.
$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
Am i right?
algebraic-geometry
asked 2 days ago
david
625
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Suppose we have family $(pi:U longrightarrow X), p_i,f)$ be in $overlineM_g,n(P^r,d).$ I want to know what does the following statement mean:
the cartier divisors $f^astt_i$ split into sections $q_i,_j$
We have $f:Ulongrightarrow P^r$. I think it means that when we look at a fibre of $x$ which is a stable curve then zero scheme of$f^astt_i$ as a cartier divisor is $q_i,1(x) +q_i,2(x) +cdots +q_i,d(x)$.
$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
Am i right?
algebraic-geometry
asked 2 days ago
david
625
If you're doing this kind of math you should learn LaTeX, and the code used in MathJax is mostly (not entirely) the same as what is done in LateX (albeit far less elaborate). Thus a thing like $$(pi:U longrightarrow X), p_i,f)$$ should be between just ONE pair of dollar signs or ONE pair of double dollar signs rather than several, and rather than entirely excluding some parts of it from MathJax, as you did. And something like $q_i,2$ should not be coded as q_i,_2 but as q_i,2. Note that they look different: $q_i,2$ versus $q_i,_2. qquad$
– Michael Hardy
2 days ago
$ldots,$and something like $P^r$ should also be between only ONE pair of dollar signs. $qquad$
– Michael Hardy
2 days ago
Can you elaborate a bit more? Are the $t_i$ the images of the marked points on your stable curve?
– Samir Canning
2 days ago
Sorry you are right.$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
– david
2 days ago
It means essentially that. If $picolon Cto S$ is a smooth separated finitely presented morphism of relative dimension $1$, then any section $s$ of $pi$ defines a relative effective Cartier divisor in $C$ over $S$. Now, if you have an arbitrary relative effective Cartier divisor on $C$ over $S$, you can ask whether it is a linear combination of sections of $pi$, in which case one might call it split. See e.g. Chapter 1 of Katz–Mazur for more on relative divisors.
– user501746
14 hours ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have family $(pi:U longrightarrow X), p_i,f)$ be in $overlineM_g,n(P^r,d).$ I want to know what does the following statement mean:
the cartier divisors $f^astt_i$ split into sections $q_i,_j$
We have $f:Ulongrightarrow P^r$. I think it means that when we look at a fibre of $x$ which is a stable curve then zero scheme of$f^astt_i$ as a cartier divisor is $q_i,1(x) +q_i,2(x) +cdots +q_i,d(x)$.
$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
Am i right?
algebraic-geometry
asked 2 days ago
david
625
Suppose we have family $(pi:U longrightarrow X), p_i,f)$ be in $overlineM_g,n(P^r,d).$ I want to know what does the following statement mean:
the cartier divisors $f^astt_i$ split into sections $q_i,_j$
We have $f:Ulongrightarrow P^r$. I think it means that when we look at a fibre of $x$ which is a stable curve then zero scheme of$f^astt_i$ as a cartier divisor is $q_i,1(x) +q_i,2(x) +cdots +q_i,d(x)$.
$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
Am i right?
algebraic-geometry
asked 2 days ago
david
625
edited 2 days ago
asked 2 days ago
david
625
asked 2 days ago
david
625
asked 2 days ago
david
625
625
If you're doing this kind of math you should learn LaTeX, and the code used in MathJax is mostly (not entirely) the same as what is done in LateX (albeit far less elaborate). Thus a thing like $$(pi:U longrightarrow X), p_i,f)$$ should be between just ONE pair of dollar signs or ONE pair of double dollar signs rather than several, and rather than entirely excluding some parts of it from MathJax, as you did. And something like $q_i,2$ should not be coded as q_i,_2 but as q_i,2. Note that they look different: $q_i,2$ versus $q_i,_2. qquad$
– Michael Hardy
2 days ago
$ldots,$and something like $P^r$ should also be between only ONE pair of dollar signs. $qquad$
– Michael Hardy
2 days ago
Can you elaborate a bit more? Are the $t_i$ the images of the marked points on your stable curve?
– Samir Canning
2 days ago
Sorry you are right.$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
– david
2 days ago
It means essentially that. If $picolon Cto S$ is a smooth separated finitely presented morphism of relative dimension $1$, then any section $s$ of $pi$ defines a relative effective Cartier divisor in $C$ over $S$. Now, if you have an arbitrary relative effective Cartier divisor on $C$ over $S$, you can ask whether it is a linear combination of sections of $pi$, in which case one might call it split. See e.g. Chapter 1 of Katz–Mazur for more on relative divisors.
– user501746
14 hours ago
add a comment |Â
If you're doing this kind of math you should learn LaTeX, and the code used in MathJax is mostly (not entirely) the same as what is done in LateX (albeit far less elaborate). Thus a thing like $$(pi:U longrightarrow X), p_i,f)$$ should be between just ONE pair of dollar signs or ONE pair of double dollar signs rather than several, and rather than entirely excluding some parts of it from MathJax, as you did. And something like $q_i,2$ should not be coded as q_i,_2 but as q_i,2. Note that they look different: $q_i,2$ versus $q_i,_2. qquad$
– Michael Hardy
2 days ago
$ldots,$and something like $P^r$ should also be between only ONE pair of dollar signs. $qquad$
– Michael Hardy
2 days ago
Can you elaborate a bit more? Are the $t_i$ the images of the marked points on your stable curve?
– Samir Canning
2 days ago
Sorry you are right.$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
– david
2 days ago
It means essentially that. If $picolon Cto S$ is a smooth separated finitely presented morphism of relative dimension $1$, then any section $s$ of $pi$ defines a relative effective Cartier divisor in $C$ over $S$. Now, if you have an arbitrary relative effective Cartier divisor on $C$ over $S$, you can ask whether it is a linear combination of sections of $pi$, in which case one might call it split. See e.g. Chapter 1 of Katz–Mazur for more on relative divisors.
– user501746
14 hours ago
If you're doing this kind of math you should learn LaTeX, and the code used in MathJax is mostly (not entirely) the same as what is done in LateX (albeit far less elaborate). Thus a thing like $$(pi:U longrightarrow X), p_i,f)$$ should be between just ONE pair of dollar signs or ONE pair of double dollar signs rather than several, and rather than entirely excluding some parts of it from MathJax, as you did. And something like $q_i,2$ should not be coded as q_i,_2 but as q_i,2. Note that they look different: $q_i,2$ versus $q_i,_2. qquad$
– Michael Hardy
2 days ago
If you're doing this kind of math you should learn LaTeX, and the code used in MathJax is mostly (not entirely) the same as what is done in LateX (albeit far less elaborate). Thus a thing like $$(pi:U longrightarrow X), p_i,f)$$ should be between just ONE pair of dollar signs or ONE pair of double dollar signs rather than several, and rather than entirely excluding some parts of it from MathJax, as you did. And something like $q_i,2$ should not be coded as q_i,_2 but as q_i,2. Note that they look different: $q_i,2$ versus $q_i,_2. qquad$
– Michael Hardy
2 days ago
$ldots,$and something like $P^r$ should also be between only ONE pair of dollar signs. $qquad$
– Michael Hardy
2 days ago
$ldots,$and something like $P^r$ should also be between only ONE pair of dollar signs. $qquad$
– Michael Hardy
2 days ago
Can you elaborate a bit more? Are the $t_i$ the images of the marked points on your stable curve?
– Samir Canning
2 days ago
Can you elaborate a bit more? Are the $t_i$ the images of the marked points on your stable curve?
– Samir Canning
2 days ago
Sorry you are right.$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
– david
2 days ago
Sorry you are right.$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
– david
2 days ago
It means essentially that. If $picolon Cto S$ is a smooth separated finitely presented morphism of relative dimension $1$, then any section $s$ of $pi$ defines a relative effective Cartier divisor in $C$ over $S$. Now, if you have an arbitrary relative effective Cartier divisor on $C$ over $S$, you can ask whether it is a linear combination of sections of $pi$, in which case one might call it split. See e.g. Chapter 1 of Katz–Mazur for more on relative divisors.
– user501746
14 hours ago
It means essentially that. If $picolon Cto S$ is a smooth separated finitely presented morphism of relative dimension $1$, then any section $s$ of $pi$ defines a relative effective Cartier divisor in $C$ over $S$. Now, if you have an arbitrary relative effective Cartier divisor on $C$ over $S$, you can ask whether it is a linear combination of sections of $pi$, in which case one might call it split. See e.g. Chapter 1 of Katz–Mazur for more on relative divisors.
– user501746
14 hours ago
add a comment |Â
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If you're doing this kind of math you should learn LaTeX, and the code used in MathJax is mostly (not entirely) the same as what is done in LateX (albeit far less elaborate). Thus a thing like $$(pi:U longrightarrow X), p_i,f)$$ should be between just ONE pair of dollar signs or ONE pair of double dollar signs rather than several, and rather than entirely excluding some parts of it from MathJax, as you did. And something like $q_i,2$ should not be coded as q_i,_2 but as q_i,2. Note that they look different: $q_i,2$ versus $q_i,_2. qquad$
– Michael Hardy
2 days ago
$ldots,$and something like $P^r$ should also be between only ONE pair of dollar signs. $qquad$
– Michael Hardy
2 days ago
Can you elaborate a bit more? Are the $t_i$ the images of the marked points on your stable curve?
– Samir Canning
2 days ago
Sorry you are right.$t_0$,$t_1$,...,$t_r$ is basis for $H^0(P^r,O(1))$.
– david
2 days ago
It means essentially that. If $picolon Cto S$ is a smooth separated finitely presented morphism of relative dimension $1$, then any section $s$ of $pi$ defines a relative effective Cartier divisor in $C$ over $S$. Now, if you have an arbitrary relative effective Cartier divisor on $C$ over $S$, you can ask whether it is a linear combination of sections of $pi$, in which case one might call it split. See e.g. Chapter 1 of Katz–Mazur for more on relative divisors.
– user501746
14 hours ago