Hilbert polynomial of the graph and boundedness of Hom scheme?
$begingroup$
Let $X,Y$ be two projective variety over an algebraically closed field, then we know $Hom(X,Y)$ has a scheme structure by considering the embedding as a open subscheme of $Hilb(X times Y)$ using the graph.
We know for each fixed polynomial $P$, the Hilbert scheme with Hilbert polynomial $P$ is projective hence finite type. And it seems that the graph of any $f:X -> Y$ is isomorphic to $X$ hence has the same Hilbert polynomial viewed as a closed subscheme of $X times Y$. However, Hom scheme can have infinitely many component (for example consider $End(E)$ where $E$ is an ellptic curve). So where does my intuition goes wrong? If $X,Y$ are both smooth curves, can we decide all of the connected components?
algebraic-geometry
asked Jan 20 at 0:43
zzyzzy
2,6071420
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be two projective variety over an algebraically closed field, then we know $Hom(X,Y)$ has a scheme structure by considering the embedding as a open subscheme of $Hilb(X times Y)$ using the graph.
We know for each fixed polynomial $P$, the Hilbert scheme with Hilbert polynomial $P$ is projective hence finite type. And it seems that the graph of any $f:X -> Y$ is isomorphic to $X$ hence has the same Hilbert polynomial viewed as a closed subscheme of $X times Y$. However, Hom scheme can have infinitely many component (for example consider $End(E)$ where $E$ is an ellptic curve). So where does my intuition goes wrong? If $X,Y$ are both smooth curves, can we decide all of the connected components?
algebraic-geometry
asked Jan 20 at 0:43
zzyzzy
2,6071420
$endgroup$
1
$begingroup$
Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $mathbb{P}^2$. However, you could have also embedded $X$ into $mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $mathbb{P}^2$. So, subvarieties of $mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree.
$endgroup$
– Ariyan Javanpeykar
Jan 20 at 15:05
add a comment |
$begingroup$
Let $X,Y$ be two projective variety over an algebraically closed field, then we know $Hom(X,Y)$ has a scheme structure by considering the embedding as a open subscheme of $Hilb(X times Y)$ using the graph.
We know for each fixed polynomial $P$, the Hilbert scheme with Hilbert polynomial $P$ is projective hence finite type. And it seems that the graph of any $f:X -> Y$ is isomorphic to $X$ hence has the same Hilbert polynomial viewed as a closed subscheme of $X times Y$. However, Hom scheme can have infinitely many component (for example consider $End(E)$ where $E$ is an ellptic curve). So where does my intuition goes wrong? If $X,Y$ are both smooth curves, can we decide all of the connected components?
algebraic-geometry
asked Jan 20 at 0:43
zzyzzy
2,6071420
$endgroup$
Let $X,Y$ be two projective variety over an algebraically closed field, then we know $Hom(X,Y)$ has a scheme structure by considering the embedding as a open subscheme of $Hilb(X times Y)$ using the graph.
We know for each fixed polynomial $P$, the Hilbert scheme with Hilbert polynomial $P$ is projective hence finite type. And it seems that the graph of any $f:X -> Y$ is isomorphic to $X$ hence has the same Hilbert polynomial viewed as a closed subscheme of $X times Y$. However, Hom scheme can have infinitely many component (for example consider $End(E)$ where $E$ is an ellptic curve). So where does my intuition goes wrong? If $X,Y$ are both smooth curves, can we decide all of the connected components?
algebraic-geometry
algebraic-geometry
asked Jan 20 at 0:43
zzyzzy
2,6071420
asked Jan 20 at 0:43
zzyzzy
2,6071420
asked Jan 20 at 0:43
zzyzzy
2,6071420
asked Jan 20 at 0:43
zzyzzy
2,6071420
asked Jan 20 at 0:43
zzyzzy
2,6071420
2,6071420
1
$begingroup$
Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $mathbb{P}^2$. However, you could have also embedded $X$ into $mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $mathbb{P}^2$. So, subvarieties of $mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree.
$endgroup$
– Ariyan Javanpeykar
Jan 20 at 15:05
add a comment |
1
$begingroup$
Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $mathbb{P}^2$. However, you could have also embedded $X$ into $mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $mathbb{P}^2$. So, subvarieties of $mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree.
$endgroup$
– Ariyan Javanpeykar
Jan 20 at 15:05
1
1
$begingroup$
Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $mathbb{P}^2$. However, you could have also embedded $X$ into $mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $mathbb{P}^2$. So, subvarieties of $mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree.
$endgroup$
– Ariyan Javanpeykar
Jan 20 at 15:05
$begingroup$
Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $mathbb{P}^2$. However, you could have also embedded $X$ into $mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $mathbb{P}^2$. So, subvarieties of $mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree.
$endgroup$
– Ariyan Javanpeykar
Jan 20 at 15:05
add a comment |
1 Answer
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$begingroup$
To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $X times Y$ is $L_X boxtimes L_Y$, where $L_X$ and $L_Y$ are line bundles on $X$ and $Y$. Then if $f colon X to Y$ is a morphism, the restriction of $L_X boxtimes L_Y$ to the graph $Gamma(f)$ of $f$ corresponds to the line bundle $L_X otimes f^*L_Y$ on $X$, hence the Hilbert polynomial of $Gamma(f)$ equals the Hilbert polynomial of $X$ with respect to $L_X otimes f^*L_Y$. In particular, it does depend on $f$.
answered Jan 20 at 8:32
SashaSasha
5,013139
$endgroup$
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
add a comment |
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1 Answer
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$begingroup$
To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $X times Y$ is $L_X boxtimes L_Y$, where $L_X$ and $L_Y$ are line bundles on $X$ and $Y$. Then if $f colon X to Y$ is a morphism, the restriction of $L_X boxtimes L_Y$ to the graph $Gamma(f)$ of $f$ corresponds to the line bundle $L_X otimes f^*L_Y$ on $X$, hence the Hilbert polynomial of $Gamma(f)$ equals the Hilbert polynomial of $X$ with respect to $L_X otimes f^*L_Y$. In particular, it does depend on $f$.
answered Jan 20 at 8:32
SashaSasha
5,013139
$endgroup$
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
add a comment |
$begingroup$
To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $X times Y$ is $L_X boxtimes L_Y$, where $L_X$ and $L_Y$ are line bundles on $X$ and $Y$. Then if $f colon X to Y$ is a morphism, the restriction of $L_X boxtimes L_Y$ to the graph $Gamma(f)$ of $f$ corresponds to the line bundle $L_X otimes f^*L_Y$ on $X$, hence the Hilbert polynomial of $Gamma(f)$ equals the Hilbert polynomial of $X$ with respect to $L_X otimes f^*L_Y$. In particular, it does depend on $f$.
answered Jan 20 at 8:32
SashaSasha
5,013139
$endgroup$
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
add a comment |
$begingroup$
To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $X times Y$ is $L_X boxtimes L_Y$, where $L_X$ and $L_Y$ are line bundles on $X$ and $Y$. Then if $f colon X to Y$ is a morphism, the restriction of $L_X boxtimes L_Y$ to the graph $Gamma(f)$ of $f$ corresponds to the line bundle $L_X otimes f^*L_Y$ on $X$, hence the Hilbert polynomial of $Gamma(f)$ equals the Hilbert polynomial of $X$ with respect to $L_X otimes f^*L_Y$. In particular, it does depend on $f$.
answered Jan 20 at 8:32
SashaSasha
5,013139
$endgroup$
To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $X times Y$ is $L_X boxtimes L_Y$, where $L_X$ and $L_Y$ are line bundles on $X$ and $Y$. Then if $f colon X to Y$ is a morphism, the restriction of $L_X boxtimes L_Y$ to the graph $Gamma(f)$ of $f$ corresponds to the line bundle $L_X otimes f^*L_Y$ on $X$, hence the Hilbert polynomial of $Gamma(f)$ equals the Hilbert polynomial of $X$ with respect to $L_X otimes f^*L_Y$. In particular, it does depend on $f$.
answered Jan 20 at 8:32
SashaSasha
5,013139
answered Jan 20 at 8:32
SashaSasha
5,013139
answered Jan 20 at 8:32
SashaSasha
5,013139
answered Jan 20 at 8:32
SashaSasha
5,013139
5,013139
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
add a comment |
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
$begingroup$
I forget the same line bundle can be different after restriction, thank you very much!
$endgroup$
– zzy
Jan 20 at 15:35
add a comment |
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$begingroup$
Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $mathbb{P}^2$. However, you could have also embedded $X$ into $mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $mathbb{P}^2$. So, subvarieties of $mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree.
$endgroup$
– Ariyan Javanpeykar
Jan 20 at 15:05