Mixing
Compute normal sheaf from equations
$begingroup$
Assume that we are given $Xsubset Ysubset Z$ locally complete intersections, given by $X: f_1,dots, f_r$, $Y: g_1,dots, g_s$ and we assume that $Z$ is some ambient $mathbb A^n$ or $mathbb P^n$. We can then write down the exact sequence of ideal sheaves $$0to I_{Y/Z}to I_{X/Z}to I_{X/Y}to 0$$ and over an open set $U$, $$I_{Y/Z}(U)=(g_1,dots, g_s)subset k[x_1,dots,x_n],\ I_{X/Z}(U)=(f_1,dots,f_r)subset k[x_1,dots,x_n],\ I_{X/Y}(U)=(f_1,dots,f_r)subsetfrac{k[x_1,dots,x_n]}{(g_1,dots,g_r)}.$$ This seems to be very easy. The exact sequence gives rise to an exact sequence of normal sheaves $$0to N_{X/Y}to N_{X/Z}to N_{Y/Z}$$ which doesn't seem to be that well understood. Can't we just deduce what the maps are on the normal sheaves just because we know them on the ideal sheaves? At least I would expect it to be possible. Unfortunately, I don't really know how to translate the explicit description for the ideal sheaves to one for the normal sheaves... Can anyone help me with that?
algebraic-geometry
$endgroup$
add a comment |
$begingroup$
Assume that we are given $Xsubset Ysubset Z$ locally complete intersections, given by $X: f_1,dots, f_r$, $Y: g_1,dots, g_s$ and we assume that $Z$ is some ambient $mathbb A^n$ or $mathbb P^n$. We can then write down the exact sequence of ideal sheaves $$0to I_{Y/Z}to I_{X/Z}to I_{X/Y}to 0$$ and over an open set $U$, $$I_{Y/Z}(U)=(g_1,dots, g_s)subset k[x_1,dots,x_n],\ I_{X/Z}(U)=(f_1,dots,f_r)subset k[x_1,dots,x_n],\ I_{X/Y}(U)=(f_1,dots,f_r)subsetfrac{k[x_1,dots,x_n]}{(g_1,dots,g_r)}.$$ This seems to be very easy. The exact sequence gives rise to an exact sequence of normal sheaves $$0to N_{X/Y}to N_{X/Z}to N_{Y/Z}$$ which doesn't seem to be that well understood. Can't we just deduce what the maps are on the normal sheaves just because we know them on the ideal sheaves? At least I would expect it to be possible. Unfortunately, I don't really know how to translate the explicit description for the ideal sheaves to one for the normal sheaves... Can anyone help me with that?
algebraic-geometry
$endgroup$
add a comment |
$begingroup$
Assume that we are given $Xsubset Ysubset Z$ locally complete intersections, given by $X: f_1,dots, f_r$, $Y: g_1,dots, g_s$ and we assume that $Z$ is some ambient $mathbb A^n$ or $mathbb P^n$. We can then write down the exact sequence of ideal sheaves $$0to I_{Y/Z}to I_{X/Z}to I_{X/Y}to 0$$ and over an open set $U$, $$I_{Y/Z}(U)=(g_1,dots, g_s)subset k[x_1,dots,x_n],\ I_{X/Z}(U)=(f_1,dots,f_r)subset k[x_1,dots,x_n],\ I_{X/Y}(U)=(f_1,dots,f_r)subsetfrac{k[x_1,dots,x_n]}{(g_1,dots,g_r)}.$$ This seems to be very easy. The exact sequence gives rise to an exact sequence of normal sheaves $$0to N_{X/Y}to N_{X/Z}to N_{Y/Z}$$ which doesn't seem to be that well understood. Can't we just deduce what the maps are on the normal sheaves just because we know them on the ideal sheaves? At least I would expect it to be possible. Unfortunately, I don't really know how to translate the explicit description for the ideal sheaves to one for the normal sheaves... Can anyone help me with that?
algebraic-geometry
$endgroup$
Assume that we are given $Xsubset Ysubset Z$ locally complete intersections, given by $X: f_1,dots, f_r$, $Y: g_1,dots, g_s$ and we assume that $Z$ is some ambient $mathbb A^n$ or $mathbb P^n$. We can then write down the exact sequence of ideal sheaves $$0to I_{Y/Z}to I_{X/Z}to I_{X/Y}to 0$$ and over an open set $U$, $$I_{Y/Z}(U)=(g_1,dots, g_s)subset k[x_1,dots,x_n],\ I_{X/Z}(U)=(f_1,dots,f_r)subset k[x_1,dots,x_n],\ I_{X/Y}(U)=(f_1,dots,f_r)subsetfrac{k[x_1,dots,x_n]}{(g_1,dots,g_r)}.$$ This seems to be very easy. The exact sequence gives rise to an exact sequence of normal sheaves $$0to N_{X/Y}to N_{X/Z}to N_{Y/Z}$$ which doesn't seem to be that well understood. Can't we just deduce what the maps are on the normal sheaves just because we know them on the ideal sheaves? At least I would expect it to be possible. Unfortunately, I don't really know how to translate the explicit description for the ideal sheaves to one for the normal sheaves... Can anyone help me with that?
algebraic-geometry
algebraic-geometry
asked Jan 29 at 14:03
user634617
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let me deal with the affine case. If $X subset Y$ then $I_Y subset I_X$. This means there are functions $h_{i,j}$ such that for each $i$ one has
$$
g_i = sum_j h_{i,j} f_j.
$$
Next, for complete intersections the normal bundles are trivial:
$$
N_{X/Z} = mathcal{O}_X^{oplus r},
qquad
N_{Y/Z} = mathcal{O}_Y^{oplus s}.
$$
The map $N_{X/Z} to N_{Y/Z}vert_X$ is then the map $mathcal{O}_X^{oplus r} to mathcal{O}_X^{oplus s}$ is then just given by the restriction of the matrix $(h_{i,j})$ to $X$.
answered Jan 29 at 15:08
SashaSasha
5,178139
$endgroup$
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
1
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
add a comment |
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1 Answer
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$begingroup$
Let me deal with the affine case. If $X subset Y$ then $I_Y subset I_X$. This means there are functions $h_{i,j}$ such that for each $i$ one has
$$
g_i = sum_j h_{i,j} f_j.
$$
Next, for complete intersections the normal bundles are trivial:
$$
N_{X/Z} = mathcal{O}_X^{oplus r},
qquad
N_{Y/Z} = mathcal{O}_Y^{oplus s}.
$$
The map $N_{X/Z} to N_{Y/Z}vert_X$ is then the map $mathcal{O}_X^{oplus r} to mathcal{O}_X^{oplus s}$ is then just given by the restriction of the matrix $(h_{i,j})$ to $X$.
answered Jan 29 at 15:08
SashaSasha
5,178139
$endgroup$
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
1
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
add a comment |
$begingroup$
Let me deal with the affine case. If $X subset Y$ then $I_Y subset I_X$. This means there are functions $h_{i,j}$ such that for each $i$ one has
$$
g_i = sum_j h_{i,j} f_j.
$$
Next, for complete intersections the normal bundles are trivial:
$$
N_{X/Z} = mathcal{O}_X^{oplus r},
qquad
N_{Y/Z} = mathcal{O}_Y^{oplus s}.
$$
The map $N_{X/Z} to N_{Y/Z}vert_X$ is then the map $mathcal{O}_X^{oplus r} to mathcal{O}_X^{oplus s}$ is then just given by the restriction of the matrix $(h_{i,j})$ to $X$.
answered Jan 29 at 15:08
SashaSasha
5,178139
$endgroup$
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
1
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
add a comment |
$begingroup$
Let me deal with the affine case. If $X subset Y$ then $I_Y subset I_X$. This means there are functions $h_{i,j}$ such that for each $i$ one has
$$
g_i = sum_j h_{i,j} f_j.
$$
Next, for complete intersections the normal bundles are trivial:
$$
N_{X/Z} = mathcal{O}_X^{oplus r},
qquad
N_{Y/Z} = mathcal{O}_Y^{oplus s}.
$$
The map $N_{X/Z} to N_{Y/Z}vert_X$ is then the map $mathcal{O}_X^{oplus r} to mathcal{O}_X^{oplus s}$ is then just given by the restriction of the matrix $(h_{i,j})$ to $X$.
answered Jan 29 at 15:08
SashaSasha
5,178139
$endgroup$
Let me deal with the affine case. If $X subset Y$ then $I_Y subset I_X$. This means there are functions $h_{i,j}$ such that for each $i$ one has
$$
g_i = sum_j h_{i,j} f_j.
$$
Next, for complete intersections the normal bundles are trivial:
$$
N_{X/Z} = mathcal{O}_X^{oplus r},
qquad
N_{Y/Z} = mathcal{O}_Y^{oplus s}.
$$
The map $N_{X/Z} to N_{Y/Z}vert_X$ is then the map $mathcal{O}_X^{oplus r} to mathcal{O}_X^{oplus s}$ is then just given by the restriction of the matrix $(h_{i,j})$ to $X$.
answered Jan 29 at 15:08
SashaSasha
5,178139
answered Jan 29 at 15:08
SashaSasha
5,178139
answered Jan 29 at 15:08
SashaSasha
5,178139
answered Jan 29 at 15:08
SashaSasha
5,178139
5,178139
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
1
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
add a comment |
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
1
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
$begingroup$
That was what I was looking for, thanks! Just two little questions:1) Why are normal bundles of complete intersections trivial? 2) Is now exactness on the right equivalent to $(h_{i,j})$ having full rank?
$endgroup$
– user634617
Jan 29 at 16:05
1
1
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
$begingroup$
1) This is a general thing --- if $X$ is the zero locus of a global section of a vector bundle $E$ on a smooth scheme $Z$ (it is enough to assume it is Cohen-Macaulay) such that the codimension of $X$ equals the rank of $E$, then $N_{X/Z} cong Evert_X$. In your case $E$ is trivial. 2) Yes, exactness on the right is equivalent to $(h_{i,j})$ haveng full rank on $X$.
$endgroup$
– Sasha
Jan 29 at 16:13
add a comment |
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