normal sheaf of length two cycle supported at singularity of curve
Assume that $C$ is a curve with a singular point $c_0$. I would like to know about the normal sheaf $mathcal N_{Z/C}$, where $Z$ is a length two cycle entirely supported at $c_0$ and about $H^1(Z,mathcal N_{Z/C})$.
This question arises, since I would like to compute the tangent space to $operatorname{Hilb}^2(C)$ at the points corresponding to such a $Z$. I saw some computations of this but only in the case when everything was smooth and the normal sheaf is a bundle, not just a sheaf. Here, I don't even get any idea of how the normal sheaf should look like or how to compute the $H^1$. Please help/give hints!
abstract-algebra algebraic-geometry
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
asked Nov 27 at 12:57
user620217
103
add a comment |
Assume that $C$ is a curve with a singular point $c_0$. I would like to know about the normal sheaf $mathcal N_{Z/C}$, where $Z$ is a length two cycle entirely supported at $c_0$ and about $H^1(Z,mathcal N_{Z/C})$.
This question arises, since I would like to compute the tangent space to $operatorname{Hilb}^2(C)$ at the points corresponding to such a $Z$. I saw some computations of this but only in the case when everything was smooth and the normal sheaf is a bundle, not just a sheaf. Here, I don't even get any idea of how the normal sheaf should look like or how to compute the $H^1$. Please help/give hints!
abstract-algebra algebraic-geometry
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
asked Nov 27 at 12:57
user620217
103
First, to speak of (co)normal sheaf you need to specify a subscheme, for a cycle this is just not defined. A length 2 subscheme supported at a singular point $P$ of a variety is determined by a vector in the tangent space to $P$, the answer does depend on this. Second, the answer does depend on the singularity type of the ambient variety (curve).
– Sasha
Nov 27 at 13:04
add a comment |
Assume that $C$ is a curve with a singular point $c_0$. I would like to know about the normal sheaf $mathcal N_{Z/C}$, where $Z$ is a length two cycle entirely supported at $c_0$ and about $H^1(Z,mathcal N_{Z/C})$.
This question arises, since I would like to compute the tangent space to $operatorname{Hilb}^2(C)$ at the points corresponding to such a $Z$. I saw some computations of this but only in the case when everything was smooth and the normal sheaf is a bundle, not just a sheaf. Here, I don't even get any idea of how the normal sheaf should look like or how to compute the $H^1$. Please help/give hints!
abstract-algebra algebraic-geometry
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
asked Nov 27 at 12:57
user620217
103
Assume that $C$ is a curve with a singular point $c_0$. I would like to know about the normal sheaf $mathcal N_{Z/C}$, where $Z$ is a length two cycle entirely supported at $c_0$ and about $H^1(Z,mathcal N_{Z/C})$.
This question arises, since I would like to compute the tangent space to $operatorname{Hilb}^2(C)$ at the points corresponding to such a $Z$. I saw some computations of this but only in the case when everything was smooth and the normal sheaf is a bundle, not just a sheaf. Here, I don't even get any idea of how the normal sheaf should look like or how to compute the $H^1$. Please help/give hints!
abstract-algebra algebraic-geometry
abstract-algebra algebraic-geometry
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
asked Nov 27 at 12:57
user620217
103
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
asked Nov 27 at 12:57
user620217
103
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
edited Dec 12 at 12:30
Arnaud D.
15.7k52343
15.7k52343
asked Nov 27 at 12:57
user620217
103
asked Nov 27 at 12:57
user620217
103
asked Nov 27 at 12:57
user620217
103
103
First, to speak of (co)normal sheaf you need to specify a subscheme, for a cycle this is just not defined. A length 2 subscheme supported at a singular point $P$ of a variety is determined by a vector in the tangent space to $P$, the answer does depend on this. Second, the answer does depend on the singularity type of the ambient variety (curve).
– Sasha
Nov 27 at 13:04
add a comment |
First, to speak of (co)normal sheaf you need to specify a subscheme, for a cycle this is just not defined. A length 2 subscheme supported at a singular point $P$ of a variety is determined by a vector in the tangent space to $P$, the answer does depend on this. Second, the answer does depend on the singularity type of the ambient variety (curve).
– Sasha
Nov 27 at 13:04
First, to speak of (co)normal sheaf you need to specify a subscheme, for a cycle this is just not defined. A length 2 subscheme supported at a singular point $P$ of a variety is determined by a vector in the tangent space to $P$, the answer does depend on this. Second, the answer does depend on the singularity type of the ambient variety (curve).
– Sasha
Nov 27 at 13:04
First, to speak of (co)normal sheaf you need to specify a subscheme, for a cycle this is just not defined. A length 2 subscheme supported at a singular point $P$ of a variety is determined by a vector in the tangent space to $P$, the answer does depend on this. Second, the answer does depend on the singularity type of the ambient variety (curve).
– Sasha
Nov 27 at 13:04
add a comment |
1 Answer
1
active
oldest
votes
Let me give the simplest example. Let $C subset mathbb{A}^2$ be given by the equation $xy = 0$, and let $Z subset C$ be the subscheme of length $2$ defined by equations
$$
y - ax = x^2 = 0.
$$
I will assume $a ne 0,infty$ (so that the line $y - ax = 0$ is not a component of $C$). Then $Z$ is the scheme-theoretic intersection of $C$ with the line $y - ax = 0$ (because there is an equality of ideals $(y-ax,xy) = (y-ax,ax^2) = (y-ax,x^2)$). This means that $Z subset C$ is a Cartier divisor with equation $y - ax = 0$, hence its structure sheaf has a resolution
$$
0 to O_C stackrel{y-ax}to O_C to O_Z to 0.
$$
In the other words, $I_Z cong O_C$. It follows that the conormal sheaf is given by
$$
I_Z/I_Z^2 cong I_Z otimes_{O_C} O_Z cong O_C otimes_{O_X} O_Z cong O_Z.
$$
Now, what you are really interested in is not $H^i(Z,N_{Z/C})$, but rather $Ext^i(I_Z,O_Z)$ (in this particular case these are the same, but in general these spaces may be different). In this particular case $I_Z = O_C$ implies that
$$
mathcal{H}om(I_Z,O_Z) cong O_Z,
$$
and
$$
mathcal{E}xt^p(I_Z,O_Z) = 0
$$
for $p > 0$ since $I_Z$ is locally free (hence locally projective).
From the local-to global spectral sequence
$$
H^q(C,mathcal{E}xt^p(I_Z,O_Z)) Rightarrow Ext^{p+q}(I_Z,O_Z)
$$
we conclude
$$
Ext^q(I_Z,O_Z) =
H^q(C,mathcal{H}om(I_Z,O_Z)) =
H^q(C,O_Z)
$$
which is equal to $k[x,y]/(y-ax,x^2) cong k^2$ for $q = 0$ and vanishes for $q > 0$.
answered Nov 27 at 13:59
Sasha
4,353139
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Is it better now?
– Sasha
Nov 27 at 22:10
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
1
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
|
show 3 more comments
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1 Answer
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Let me give the simplest example. Let $C subset mathbb{A}^2$ be given by the equation $xy = 0$, and let $Z subset C$ be the subscheme of length $2$ defined by equations
$$
y - ax = x^2 = 0.
$$
I will assume $a ne 0,infty$ (so that the line $y - ax = 0$ is not a component of $C$). Then $Z$ is the scheme-theoretic intersection of $C$ with the line $y - ax = 0$ (because there is an equality of ideals $(y-ax,xy) = (y-ax,ax^2) = (y-ax,x^2)$). This means that $Z subset C$ is a Cartier divisor with equation $y - ax = 0$, hence its structure sheaf has a resolution
$$
0 to O_C stackrel{y-ax}to O_C to O_Z to 0.
$$
In the other words, $I_Z cong O_C$. It follows that the conormal sheaf is given by
$$
I_Z/I_Z^2 cong I_Z otimes_{O_C} O_Z cong O_C otimes_{O_X} O_Z cong O_Z.
$$
Now, what you are really interested in is not $H^i(Z,N_{Z/C})$, but rather $Ext^i(I_Z,O_Z)$ (in this particular case these are the same, but in general these spaces may be different). In this particular case $I_Z = O_C$ implies that
$$
mathcal{H}om(I_Z,O_Z) cong O_Z,
$$
and
$$
mathcal{E}xt^p(I_Z,O_Z) = 0
$$
for $p > 0$ since $I_Z$ is locally free (hence locally projective).
From the local-to global spectral sequence
$$
H^q(C,mathcal{E}xt^p(I_Z,O_Z)) Rightarrow Ext^{p+q}(I_Z,O_Z)
$$
we conclude
$$
Ext^q(I_Z,O_Z) =
H^q(C,mathcal{H}om(I_Z,O_Z)) =
H^q(C,O_Z)
$$
which is equal to $k[x,y]/(y-ax,x^2) cong k^2$ for $q = 0$ and vanishes for $q > 0$.
answered Nov 27 at 13:59
Sasha
4,353139
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Is it better now?
– Sasha
Nov 27 at 22:10
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
1
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
|
show 3 more comments
Let me give the simplest example. Let $C subset mathbb{A}^2$ be given by the equation $xy = 0$, and let $Z subset C$ be the subscheme of length $2$ defined by equations
$$
y - ax = x^2 = 0.
$$
I will assume $a ne 0,infty$ (so that the line $y - ax = 0$ is not a component of $C$). Then $Z$ is the scheme-theoretic intersection of $C$ with the line $y - ax = 0$ (because there is an equality of ideals $(y-ax,xy) = (y-ax,ax^2) = (y-ax,x^2)$). This means that $Z subset C$ is a Cartier divisor with equation $y - ax = 0$, hence its structure sheaf has a resolution
$$
0 to O_C stackrel{y-ax}to O_C to O_Z to 0.
$$
In the other words, $I_Z cong O_C$. It follows that the conormal sheaf is given by
$$
I_Z/I_Z^2 cong I_Z otimes_{O_C} O_Z cong O_C otimes_{O_X} O_Z cong O_Z.
$$
Now, what you are really interested in is not $H^i(Z,N_{Z/C})$, but rather $Ext^i(I_Z,O_Z)$ (in this particular case these are the same, but in general these spaces may be different). In this particular case $I_Z = O_C$ implies that
$$
mathcal{H}om(I_Z,O_Z) cong O_Z,
$$
and
$$
mathcal{E}xt^p(I_Z,O_Z) = 0
$$
for $p > 0$ since $I_Z$ is locally free (hence locally projective).
From the local-to global spectral sequence
$$
H^q(C,mathcal{E}xt^p(I_Z,O_Z)) Rightarrow Ext^{p+q}(I_Z,O_Z)
$$
we conclude
$$
Ext^q(I_Z,O_Z) =
H^q(C,mathcal{H}om(I_Z,O_Z)) =
H^q(C,O_Z)
$$
which is equal to $k[x,y]/(y-ax,x^2) cong k^2$ for $q = 0$ and vanishes for $q > 0$.
answered Nov 27 at 13:59
Sasha
4,353139
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Is it better now?
– Sasha
Nov 27 at 22:10
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
1
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
|
show 3 more comments
Let me give the simplest example. Let $C subset mathbb{A}^2$ be given by the equation $xy = 0$, and let $Z subset C$ be the subscheme of length $2$ defined by equations
$$
y - ax = x^2 = 0.
$$
I will assume $a ne 0,infty$ (so that the line $y - ax = 0$ is not a component of $C$). Then $Z$ is the scheme-theoretic intersection of $C$ with the line $y - ax = 0$ (because there is an equality of ideals $(y-ax,xy) = (y-ax,ax^2) = (y-ax,x^2)$). This means that $Z subset C$ is a Cartier divisor with equation $y - ax = 0$, hence its structure sheaf has a resolution
$$
0 to O_C stackrel{y-ax}to O_C to O_Z to 0.
$$
In the other words, $I_Z cong O_C$. It follows that the conormal sheaf is given by
$$
I_Z/I_Z^2 cong I_Z otimes_{O_C} O_Z cong O_C otimes_{O_X} O_Z cong O_Z.
$$
Now, what you are really interested in is not $H^i(Z,N_{Z/C})$, but rather $Ext^i(I_Z,O_Z)$ (in this particular case these are the same, but in general these spaces may be different). In this particular case $I_Z = O_C$ implies that
$$
mathcal{H}om(I_Z,O_Z) cong O_Z,
$$
and
$$
mathcal{E}xt^p(I_Z,O_Z) = 0
$$
for $p > 0$ since $I_Z$ is locally free (hence locally projective).
From the local-to global spectral sequence
$$
H^q(C,mathcal{E}xt^p(I_Z,O_Z)) Rightarrow Ext^{p+q}(I_Z,O_Z)
$$
we conclude
$$
Ext^q(I_Z,O_Z) =
H^q(C,mathcal{H}om(I_Z,O_Z)) =
H^q(C,O_Z)
$$
which is equal to $k[x,y]/(y-ax,x^2) cong k^2$ for $q = 0$ and vanishes for $q > 0$.
answered Nov 27 at 13:59
Sasha
4,353139
Let me give the simplest example. Let $C subset mathbb{A}^2$ be given by the equation $xy = 0$, and let $Z subset C$ be the subscheme of length $2$ defined by equations
$$
y - ax = x^2 = 0.
$$
I will assume $a ne 0,infty$ (so that the line $y - ax = 0$ is not a component of $C$). Then $Z$ is the scheme-theoretic intersection of $C$ with the line $y - ax = 0$ (because there is an equality of ideals $(y-ax,xy) = (y-ax,ax^2) = (y-ax,x^2)$). This means that $Z subset C$ is a Cartier divisor with equation $y - ax = 0$, hence its structure sheaf has a resolution
$$
0 to O_C stackrel{y-ax}to O_C to O_Z to 0.
$$
In the other words, $I_Z cong O_C$. It follows that the conormal sheaf is given by
$$
I_Z/I_Z^2 cong I_Z otimes_{O_C} O_Z cong O_C otimes_{O_X} O_Z cong O_Z.
$$
Now, what you are really interested in is not $H^i(Z,N_{Z/C})$, but rather $Ext^i(I_Z,O_Z)$ (in this particular case these are the same, but in general these spaces may be different). In this particular case $I_Z = O_C$ implies that
$$
mathcal{H}om(I_Z,O_Z) cong O_Z,
$$
and
$$
mathcal{E}xt^p(I_Z,O_Z) = 0
$$
for $p > 0$ since $I_Z$ is locally free (hence locally projective).
From the local-to global spectral sequence
$$
H^q(C,mathcal{E}xt^p(I_Z,O_Z)) Rightarrow Ext^{p+q}(I_Z,O_Z)
$$
we conclude
$$
Ext^q(I_Z,O_Z) =
H^q(C,mathcal{H}om(I_Z,O_Z)) =
H^q(C,O_Z)
$$
which is equal to $k[x,y]/(y-ax,x^2) cong k^2$ for $q = 0$ and vanishes for $q > 0$.
answered Nov 27 at 13:59
Sasha
4,353139
edited Nov 27 at 22:10
answered Nov 27 at 13:59
Sasha
4,353139
answered Nov 27 at 13:59
Sasha
4,353139
answered Nov 27 at 13:59
Sasha
4,353139
4,353139
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Is it better now?
– Sasha
Nov 27 at 22:10
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
1
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
|
show 3 more comments
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Is it better now?
– Sasha
Nov 27 at 22:10
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
1
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
1) Because this is what deformation theory produces as the obstruction space. 2) Not quite, the scheme defined by ideal $(y - ax, x^2 - a^3x^3)$ has length 3 (besides double point at the origin it also contains the point $(a^{-3},a^{-2})$. But still it shows that $Z subset C$ is a Cartier divisor, hence $I_Z cong O_C$ and the same argument shows that the tangent space is 2-dimensional.
– Sasha
Nov 27 at 14:33
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Not quite. It is more convenient to use local-to-global spectral sequence. Since $I_Z$ is locally free, local Ext sheaves vanish, and for the local Hom sheaf we have $mathcal{H}om(I_Z,O_Z) cong O_Z$. So, the spectral sequence gives $Ext^i(I_Z,O_Z) = H^i(C,O_Z)$.
– Sasha
Nov 27 at 15:21
Is it better now?
– Sasha
Nov 27 at 22:10
Is it better now?
– Sasha
Nov 27 at 22:10
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
Dear @Sasha, in your first comment to this (great!) answer you are saying that for a cusp one can still consider the scheme $Z$ (from the first comment of @user620217) although it has length three and thus does not correspond to a point of $Hilb^2$. Could you briefly explain to me the reason?
– James
Nov 28 at 11:23
1
1
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
@James: The point is that it is not necessary for $Z$ to be a complete intersection, a LOCALLY complete intersection (lci) is enough for the ideal sheaf to be simple enough (in case of a zero dimensional subscheme in a curve, the ideal is then ivertible). So, if the ideal of the subscheme $Z$ is generated by one equation in a NEIGHBORHOOD of the only point of $Z$ (the third point is far away), then $Z$ is an lci.
– Sasha
Nov 28 at 11:40
|
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First, to speak of (co)normal sheaf you need to specify a subscheme, for a cycle this is just not defined. A length 2 subscheme supported at a singular point $P$ of a variety is determined by a vector in the tangent space to $P$, the answer does depend on this. Second, the answer does depend on the singularity type of the ambient variety (curve).
– Sasha
Nov 27 at 13:04