Mixing
Structure of degree $n$ part of $Hom(mathbb P^1, mathbb P^1)$
$begingroup$
Consider the Hom scheme $X=Hom(mathbb P^1, mathbb P^1)$ over an algebraically closed field $k$. Using deformation theory, I think we can show locally at a point representing a degree $n$ morphism, $X$ is smooth of dimension $2n+1$.
So if we consider the subscheme $X_n$ of $X$ consist of degree $n$ morphisms, it has a natural action by $X_1 times X_1=Aut(mathbb P^1) times Aut(mathbb P^1)= PGL(2) times PGL(2)$ by composing morphism from left and right sides.
The question: is $X_n$ affine and connected? What is the quotient $X_n/PGL(2) times PGL(2)$ ?
algebraic-geometry
asked Jan 27 at 6:51
zzyzzy
2,6431420
$endgroup$
add a comment |
$begingroup$
Consider the Hom scheme $X=Hom(mathbb P^1, mathbb P^1)$ over an algebraically closed field $k$. Using deformation theory, I think we can show locally at a point representing a degree $n$ morphism, $X$ is smooth of dimension $2n+1$.
So if we consider the subscheme $X_n$ of $X$ consist of degree $n$ morphisms, it has a natural action by $X_1 times X_1=Aut(mathbb P^1) times Aut(mathbb P^1)= PGL(2) times PGL(2)$ by composing morphism from left and right sides.
The question: is $X_n$ affine and connected? What is the quotient $X_n/PGL(2) times PGL(2)$ ?
algebraic-geometry
asked Jan 27 at 6:51
zzyzzy
2,6431420
$endgroup$
add a comment |
$begingroup$
Consider the Hom scheme $X=Hom(mathbb P^1, mathbb P^1)$ over an algebraically closed field $k$. Using deformation theory, I think we can show locally at a point representing a degree $n$ morphism, $X$ is smooth of dimension $2n+1$.
So if we consider the subscheme $X_n$ of $X$ consist of degree $n$ morphisms, it has a natural action by $X_1 times X_1=Aut(mathbb P^1) times Aut(mathbb P^1)= PGL(2) times PGL(2)$ by composing morphism from left and right sides.
The question: is $X_n$ affine and connected? What is the quotient $X_n/PGL(2) times PGL(2)$ ?
algebraic-geometry
asked Jan 27 at 6:51
zzyzzy
2,6431420
$endgroup$
Consider the Hom scheme $X=Hom(mathbb P^1, mathbb P^1)$ over an algebraically closed field $k$. Using deformation theory, I think we can show locally at a point representing a degree $n$ morphism, $X$ is smooth of dimension $2n+1$.
So if we consider the subscheme $X_n$ of $X$ consist of degree $n$ morphisms, it has a natural action by $X_1 times X_1=Aut(mathbb P^1) times Aut(mathbb P^1)= PGL(2) times PGL(2)$ by composing morphism from left and right sides.
The question: is $X_n$ affine and connected? What is the quotient $X_n/PGL(2) times PGL(2)$ ?
algebraic-geometry
algebraic-geometry
asked Jan 27 at 6:51
zzyzzy
2,6431420
asked Jan 27 at 6:51
zzyzzy
2,6431420
asked Jan 27 at 6:51
zzyzzy
2,6431420
asked Jan 27 at 6:51
zzyzzy
2,6431420
asked Jan 27 at 6:51
zzyzzy
2,6431420
2,6431420
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To give a map $mathbb{P}^1 to mathbb{P}^1$ of degree $n$ is equivalent to giving a pair of two homogeneous polynomials of degree $n$ in two variables up to constant without common linear factors. This means that $X_n$ is an open subset of
$$
bar{X}_n = mathbb{P}^{2n+1}.
$$
This space $bar{X}_n$ is called the space of quasimaps or the Drinfeld compactification of the space of maps.
Furthermore, it is clear that $bar{X}_n setminus X_n$ is the discriminant (of two polynomials) hypersurface, or equivalently the image of the natural map
$$
bar{X}_{n-1} times mathbb{P}^1 to bar{X}_n
$$
that takes a pair of polynomials of degree $n-1$ and multiplies them by a common linear factor.
All this shows that $X_n$ is smooth and affine.
If $U$ and $V$ are two vector spaces of dimension 2 so that the source of the maps is $P(U)$ and the target is $P(V)$ then
$$
bar{X}_n = mathbb{P}(V otimes S^nU^*).
$$
This isomorphism is $PGL(U) times PGL(V)$-equivariant.
answered Jan 27 at 7:38
SashaSasha
5,178139
$endgroup$
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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$begingroup$
To give a map $mathbb{P}^1 to mathbb{P}^1$ of degree $n$ is equivalent to giving a pair of two homogeneous polynomials of degree $n$ in two variables up to constant without common linear factors. This means that $X_n$ is an open subset of
$$
bar{X}_n = mathbb{P}^{2n+1}.
$$
This space $bar{X}_n$ is called the space of quasimaps or the Drinfeld compactification of the space of maps.
Furthermore, it is clear that $bar{X}_n setminus X_n$ is the discriminant (of two polynomials) hypersurface, or equivalently the image of the natural map
$$
bar{X}_{n-1} times mathbb{P}^1 to bar{X}_n
$$
that takes a pair of polynomials of degree $n-1$ and multiplies them by a common linear factor.
All this shows that $X_n$ is smooth and affine.
If $U$ and $V$ are two vector spaces of dimension 2 so that the source of the maps is $P(U)$ and the target is $P(V)$ then
$$
bar{X}_n = mathbb{P}(V otimes S^nU^*).
$$
This isomorphism is $PGL(U) times PGL(V)$-equivariant.
answered Jan 27 at 7:38
SashaSasha
5,178139
$endgroup$
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
add a comment |
$begingroup$
To give a map $mathbb{P}^1 to mathbb{P}^1$ of degree $n$ is equivalent to giving a pair of two homogeneous polynomials of degree $n$ in two variables up to constant without common linear factors. This means that $X_n$ is an open subset of
$$
bar{X}_n = mathbb{P}^{2n+1}.
$$
This space $bar{X}_n$ is called the space of quasimaps or the Drinfeld compactification of the space of maps.
Furthermore, it is clear that $bar{X}_n setminus X_n$ is the discriminant (of two polynomials) hypersurface, or equivalently the image of the natural map
$$
bar{X}_{n-1} times mathbb{P}^1 to bar{X}_n
$$
that takes a pair of polynomials of degree $n-1$ and multiplies them by a common linear factor.
All this shows that $X_n$ is smooth and affine.
If $U$ and $V$ are two vector spaces of dimension 2 so that the source of the maps is $P(U)$ and the target is $P(V)$ then
$$
bar{X}_n = mathbb{P}(V otimes S^nU^*).
$$
This isomorphism is $PGL(U) times PGL(V)$-equivariant.
answered Jan 27 at 7:38
SashaSasha
5,178139
$endgroup$
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
add a comment |
$begingroup$
To give a map $mathbb{P}^1 to mathbb{P}^1$ of degree $n$ is equivalent to giving a pair of two homogeneous polynomials of degree $n$ in two variables up to constant without common linear factors. This means that $X_n$ is an open subset of
$$
bar{X}_n = mathbb{P}^{2n+1}.
$$
This space $bar{X}_n$ is called the space of quasimaps or the Drinfeld compactification of the space of maps.
Furthermore, it is clear that $bar{X}_n setminus X_n$ is the discriminant (of two polynomials) hypersurface, or equivalently the image of the natural map
$$
bar{X}_{n-1} times mathbb{P}^1 to bar{X}_n
$$
that takes a pair of polynomials of degree $n-1$ and multiplies them by a common linear factor.
All this shows that $X_n$ is smooth and affine.
If $U$ and $V$ are two vector spaces of dimension 2 so that the source of the maps is $P(U)$ and the target is $P(V)$ then
$$
bar{X}_n = mathbb{P}(V otimes S^nU^*).
$$
This isomorphism is $PGL(U) times PGL(V)$-equivariant.
answered Jan 27 at 7:38
SashaSasha
5,178139
$endgroup$
To give a map $mathbb{P}^1 to mathbb{P}^1$ of degree $n$ is equivalent to giving a pair of two homogeneous polynomials of degree $n$ in two variables up to constant without common linear factors. This means that $X_n$ is an open subset of
$$
bar{X}_n = mathbb{P}^{2n+1}.
$$
This space $bar{X}_n$ is called the space of quasimaps or the Drinfeld compactification of the space of maps.
Furthermore, it is clear that $bar{X}_n setminus X_n$ is the discriminant (of two polynomials) hypersurface, or equivalently the image of the natural map
$$
bar{X}_{n-1} times mathbb{P}^1 to bar{X}_n
$$
that takes a pair of polynomials of degree $n-1$ and multiplies them by a common linear factor.
All this shows that $X_n$ is smooth and affine.
If $U$ and $V$ are two vector spaces of dimension 2 so that the source of the maps is $P(U)$ and the target is $P(V)$ then
$$
bar{X}_n = mathbb{P}(V otimes S^nU^*).
$$
This isomorphism is $PGL(U) times PGL(V)$-equivariant.
answered Jan 27 at 7:38
SashaSasha
5,178139
edited Jan 27 at 15:50
answered Jan 27 at 7:38
SashaSasha
5,178139
answered Jan 27 at 7:38
SashaSasha
5,178139
answered Jan 27 at 7:38
SashaSasha
5,178139
5,178139
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
add a comment |
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
Thank you very much! Is there similiar result for $Hom(mathbb P^n, mathbb P^m)$?
$endgroup$
– zzy
Jan 27 at 15:18
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
And I am also quite interested in the action of $PGL_2 times PGL_2$.
$endgroup$
– zzy
Jan 27 at 15:23
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
@zzy: I added a description of the group action. The formula $mathbb{P}(V otimes S^nU^*)$ still gives a compactification of the space of maps from $mathbb{P}(U)$ to $mathbb{P}(V)$ for $U$ and $V$ of any dimension, but the boundary has higher codimension, so the space of maps is no loner affine.
$endgroup$
– Sasha
Jan 27 at 15:54
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
$begingroup$
Thank you again!
$endgroup$
– zzy
Jan 27 at 18:36
add a comment |
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