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Picard group of singular surface
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I am curious wether the following holds true and the argument is correct. Assume we are given a complex algebraic surface $S$ (projective, if you like) that admits one and only one singular point $sin S$. Let us denote the inclusion of the regular locus of $S$ by $iota$, i.e. $iota: S_reghookrightarrow S, xmapsto x$. This gives us a map on the level of Picard groups $$iota_*: Pic(S_reg)to Pic(S).$$ I claim this map to be surjective: If $L$ is any line bundle on $S_reg$, it comes from a divisor $D$, i.e. $L=mathcal O_S_reg(D)$, and $D$ is locally Cartier by which I mean that on a neighborhood $U^*=Ucap S_reg=Usetminuss$, where $U=U(s)subset S$ is a neighborhood of $s$ in $S$, we can write $Dmid_U^*=f/g$ with holomorphic functions $f,g$. But then $f,g$ and hence $f/g$ (hence $D$, hence $L$) extend over $s$ by Hartogs principle and the fact that the codimension of $s$ in $S$ is two.
My questions:
1) Is this argument ok?
2) If the answer to 1) is yes, then $Pic(S)$ does not depend on the singularity of $S$. How should I understand this?
3) Can we pretend that $s$ is a smooth point for calculating $Pic(S)$?
I am grateful for any kind of comments on this and apologize if this is a stupid question or (parts of it) makes no sense. I am still a beginner in algebraic geometry.
algebraic-geometry singularity-theory picard-scheme
asked Jul 17 at 12:57
James
1929
 |Â
show 1 more comment
up vote
1
down vote
favorite
I am curious wether the following holds true and the argument is correct. Assume we are given a complex algebraic surface $S$ (projective, if you like) that admits one and only one singular point $sin S$. Let us denote the inclusion of the regular locus of $S$ by $iota$, i.e. $iota: S_reghookrightarrow S, xmapsto x$. This gives us a map on the level of Picard groups $$iota_*: Pic(S_reg)to Pic(S).$$ I claim this map to be surjective: If $L$ is any line bundle on $S_reg$, it comes from a divisor $D$, i.e. $L=mathcal O_S_reg(D)$, and $D$ is locally Cartier by which I mean that on a neighborhood $U^*=Ucap S_reg=Usetminuss$, where $U=U(s)subset S$ is a neighborhood of $s$ in $S$, we can write $Dmid_U^*=f/g$ with holomorphic functions $f,g$. But then $f,g$ and hence $f/g$ (hence $D$, hence $L$) extend over $s$ by Hartogs principle and the fact that the codimension of $s$ in $S$ is two.
My questions:
1) Is this argument ok?
2) If the answer to 1) is yes, then $Pic(S)$ does not depend on the singularity of $S$. How should I understand this?
3) Can we pretend that $s$ is a smooth point for calculating $Pic(S)$?
I am grateful for any kind of comments on this and apologize if this is a stupid question or (parts of it) makes no sense. I am still a beginner in algebraic geometry.
algebraic-geometry singularity-theory picard-scheme
asked Jul 17 at 12:57
James
1929
1
Hartog's theorem can not be applied for non-normal surfaces. But, $i_*$ is surjective in your case, since any Cartier divisor is linearly equivalent to one not passing through a given point.
– Mohan
Jul 17 at 15:04
Ok, so the argument I gave only works for normal surfaces and if not, then I move the divisor away from the singular point. Thanks for pointing out! What about the other questions?
– James
Jul 17 at 16:00
The real issue is not that, there is no natural map $i_*$ as you describe. You have $i^*$ going the other way. In general, $i_*$ of a line bundle will not be a line bundle.
– Mohan
Jul 17 at 16:06
Can you explain to me why? For me it is not clear why it should fail to be a line bundle -- we move a divisor $D$ away from this point and $mathcal O(D)$ then gives a line bundle..
– James
Jul 17 at 16:38
2
Here is a standard example in dimension 3 (they exist in dimension 2 also, little more difficult). Take $X$, the hypersurface in $mathbbC^4$ defined by $xy-zw=0$ and $U$ the complement of the origin. Then, $Pic U=mathbbZ$ and direct image of a non-trivial line bundle is never a line bundle. Remember, to define $i_*$, by moving divisors, you must still show that it is well defined upto linear equivalence. Typically, you have a restriction map $i^*$ from $Pic X $ to $Pic U$.
– Mohan
Jul 17 at 20:10
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am curious wether the following holds true and the argument is correct. Assume we are given a complex algebraic surface $S$ (projective, if you like) that admits one and only one singular point $sin S$. Let us denote the inclusion of the regular locus of $S$ by $iota$, i.e. $iota: S_reghookrightarrow S, xmapsto x$. This gives us a map on the level of Picard groups $$iota_*: Pic(S_reg)to Pic(S).$$ I claim this map to be surjective: If $L$ is any line bundle on $S_reg$, it comes from a divisor $D$, i.e. $L=mathcal O_S_reg(D)$, and $D$ is locally Cartier by which I mean that on a neighborhood $U^*=Ucap S_reg=Usetminuss$, where $U=U(s)subset S$ is a neighborhood of $s$ in $S$, we can write $Dmid_U^*=f/g$ with holomorphic functions $f,g$. But then $f,g$ and hence $f/g$ (hence $D$, hence $L$) extend over $s$ by Hartogs principle and the fact that the codimension of $s$ in $S$ is two.
My questions:
1) Is this argument ok?
2) If the answer to 1) is yes, then $Pic(S)$ does not depend on the singularity of $S$. How should I understand this?
3) Can we pretend that $s$ is a smooth point for calculating $Pic(S)$?
I am grateful for any kind of comments on this and apologize if this is a stupid question or (parts of it) makes no sense. I am still a beginner in algebraic geometry.
algebraic-geometry singularity-theory picard-scheme
asked Jul 17 at 12:57
James
1929
I am curious wether the following holds true and the argument is correct. Assume we are given a complex algebraic surface $S$ (projective, if you like) that admits one and only one singular point $sin S$. Let us denote the inclusion of the regular locus of $S$ by $iota$, i.e. $iota: S_reghookrightarrow S, xmapsto x$. This gives us a map on the level of Picard groups $$iota_*: Pic(S_reg)to Pic(S).$$ I claim this map to be surjective: If $L$ is any line bundle on $S_reg$, it comes from a divisor $D$, i.e. $L=mathcal O_S_reg(D)$, and $D$ is locally Cartier by which I mean that on a neighborhood $U^*=Ucap S_reg=Usetminuss$, where $U=U(s)subset S$ is a neighborhood of $s$ in $S$, we can write $Dmid_U^*=f/g$ with holomorphic functions $f,g$. But then $f,g$ and hence $f/g$ (hence $D$, hence $L$) extend over $s$ by Hartogs principle and the fact that the codimension of $s$ in $S$ is two.
My questions:
1) Is this argument ok?
2) If the answer to 1) is yes, then $Pic(S)$ does not depend on the singularity of $S$. How should I understand this?
3) Can we pretend that $s$ is a smooth point for calculating $Pic(S)$?
I am grateful for any kind of comments on this and apologize if this is a stupid question or (parts of it) makes no sense. I am still a beginner in algebraic geometry.
algebraic-geometry singularity-theory picard-scheme
asked Jul 17 at 12:57
James
1929
asked Jul 17 at 12:57
James
1929
asked Jul 17 at 12:57
James
1929
asked Jul 17 at 12:57
James
1929
1929
1
Hartog's theorem can not be applied for non-normal surfaces. But, $i_*$ is surjective in your case, since any Cartier divisor is linearly equivalent to one not passing through a given point.
– Mohan
Jul 17 at 15:04
Ok, so the argument I gave only works for normal surfaces and if not, then I move the divisor away from the singular point. Thanks for pointing out! What about the other questions?
– James
Jul 17 at 16:00
The real issue is not that, there is no natural map $i_*$ as you describe. You have $i^*$ going the other way. In general, $i_*$ of a line bundle will not be a line bundle.
– Mohan
Jul 17 at 16:06
Can you explain to me why? For me it is not clear why it should fail to be a line bundle -- we move a divisor $D$ away from this point and $mathcal O(D)$ then gives a line bundle..
– James
Jul 17 at 16:38
2
Here is a standard example in dimension 3 (they exist in dimension 2 also, little more difficult). Take $X$, the hypersurface in $mathbbC^4$ defined by $xy-zw=0$ and $U$ the complement of the origin. Then, $Pic U=mathbbZ$ and direct image of a non-trivial line bundle is never a line bundle. Remember, to define $i_*$, by moving divisors, you must still show that it is well defined upto linear equivalence. Typically, you have a restriction map $i^*$ from $Pic X $ to $Pic U$.
– Mohan
Jul 17 at 20:10
 |Â
show 1 more comment
1
Hartog's theorem can not be applied for non-normal surfaces. But, $i_*$ is surjective in your case, since any Cartier divisor is linearly equivalent to one not passing through a given point.
– Mohan
Jul 17 at 15:04
Ok, so the argument I gave only works for normal surfaces and if not, then I move the divisor away from the singular point. Thanks for pointing out! What about the other questions?
– James
Jul 17 at 16:00
The real issue is not that, there is no natural map $i_*$ as you describe. You have $i^*$ going the other way. In general, $i_*$ of a line bundle will not be a line bundle.
– Mohan
Jul 17 at 16:06
Can you explain to me why? For me it is not clear why it should fail to be a line bundle -- we move a divisor $D$ away from this point and $mathcal O(D)$ then gives a line bundle..
– James
Jul 17 at 16:38
2
Here is a standard example in dimension 3 (they exist in dimension 2 also, little more difficult). Take $X$, the hypersurface in $mathbbC^4$ defined by $xy-zw=0$ and $U$ the complement of the origin. Then, $Pic U=mathbbZ$ and direct image of a non-trivial line bundle is never a line bundle. Remember, to define $i_*$, by moving divisors, you must still show that it is well defined upto linear equivalence. Typically, you have a restriction map $i^*$ from $Pic X $ to $Pic U$.
– Mohan
Jul 17 at 20:10
1
1
Hartog's theorem can not be applied for non-normal surfaces. But, $i_*$ is surjective in your case, since any Cartier divisor is linearly equivalent to one not passing through a given point.
– Mohan
Jul 17 at 15:04
Hartog's theorem can not be applied for non-normal surfaces. But, $i_*$ is surjective in your case, since any Cartier divisor is linearly equivalent to one not passing through a given point.
– Mohan
Jul 17 at 15:04
Ok, so the argument I gave only works for normal surfaces and if not, then I move the divisor away from the singular point. Thanks for pointing out! What about the other questions?
– James
Jul 17 at 16:00
Ok, so the argument I gave only works for normal surfaces and if not, then I move the divisor away from the singular point. Thanks for pointing out! What about the other questions?
– James
Jul 17 at 16:00
The real issue is not that, there is no natural map $i_*$ as you describe. You have $i^*$ going the other way. In general, $i_*$ of a line bundle will not be a line bundle.
– Mohan
Jul 17 at 16:06
The real issue is not that, there is no natural map $i_*$ as you describe. You have $i^*$ going the other way. In general, $i_*$ of a line bundle will not be a line bundle.
– Mohan
Jul 17 at 16:06
Can you explain to me why? For me it is not clear why it should fail to be a line bundle -- we move a divisor $D$ away from this point and $mathcal O(D)$ then gives a line bundle..
– James
Jul 17 at 16:38
Can you explain to me why? For me it is not clear why it should fail to be a line bundle -- we move a divisor $D$ away from this point and $mathcal O(D)$ then gives a line bundle..
– James
Jul 17 at 16:38
2
2
Here is a standard example in dimension 3 (they exist in dimension 2 also, little more difficult). Take $X$, the hypersurface in $mathbbC^4$ defined by $xy-zw=0$ and $U$ the complement of the origin. Then, $Pic U=mathbbZ$ and direct image of a non-trivial line bundle is never a line bundle. Remember, to define $i_*$, by moving divisors, you must still show that it is well defined upto linear equivalence. Typically, you have a restriction map $i^*$ from $Pic X $ to $Pic U$.
– Mohan
Jul 17 at 20:10
Here is a standard example in dimension 3 (they exist in dimension 2 also, little more difficult). Take $X$, the hypersurface in $mathbbC^4$ defined by $xy-zw=0$ and $U$ the complement of the origin. Then, $Pic U=mathbbZ$ and direct image of a non-trivial line bundle is never a line bundle. Remember, to define $i_*$, by moving divisors, you must still show that it is well defined upto linear equivalence. Typically, you have a restriction map $i^*$ from $Pic X $ to $Pic U$.
– Mohan
Jul 17 at 20:10
 |Â
show 1 more comment
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1
Hartog's theorem can not be applied for non-normal surfaces. But, $i_*$ is surjective in your case, since any Cartier divisor is linearly equivalent to one not passing through a given point.
– Mohan
Jul 17 at 15:04
Ok, so the argument I gave only works for normal surfaces and if not, then I move the divisor away from the singular point. Thanks for pointing out! What about the other questions?
– James
Jul 17 at 16:00
The real issue is not that, there is no natural map $i_*$ as you describe. You have $i^*$ going the other way. In general, $i_*$ of a line bundle will not be a line bundle.
– Mohan
Jul 17 at 16:06
Can you explain to me why? For me it is not clear why it should fail to be a line bundle -- we move a divisor $D$ away from this point and $mathcal O(D)$ then gives a line bundle..
– James
Jul 17 at 16:38
2
Here is a standard example in dimension 3 (they exist in dimension 2 also, little more difficult). Take $X$, the hypersurface in $mathbbC^4$ defined by $xy-zw=0$ and $U$ the complement of the origin. Then, $Pic U=mathbbZ$ and direct image of a non-trivial line bundle is never a line bundle. Remember, to define $i_*$, by moving divisors, you must still show that it is well defined upto linear equivalence. Typically, you have a restriction map $i^*$ from $Pic X $ to $Pic U$.
– Mohan
Jul 17 at 20:10