Why are Stein manifolds/spaces the analog of affine varieties/schemes in algebraic geometry?Stein Manifolds and Affine VarietiesAre all parametrizations via polynomials algebraic varieties?Are all manifolds affine?Why can I divide an affine variety by the action of the general linear group?Chern classes of ideal sheaf of an analytic subsetAlgebraic varieties in “mixed” affine spacesAlgebraic spaces which are automatically schemes“Affine” algebraic spacesKahler manifolds and algebraic varietiesThe cone of curves of complex projective manifolds with an algebraic torus action
Why are Stein manifolds/spaces the analog of affine varieties/schemes in algebraic geometry?
Stein Manifolds and Affine VarietiesAre all parametrizations via polynomials algebraic varieties?Are all manifolds affine?Why can I divide an affine variety by the action of the general linear group?Chern classes of ideal sheaf of an analytic subsetAlgebraic varieties in “mixed” affine spacesAlgebraic spaces which are automatically schemes“Affine” algebraic spacesKahler manifolds and algebraic varietiesThe cone of curves of complex projective manifolds with an algebraic torus action
$begingroup$
I presume this is a GAGA-style result, but I cannot find a reference.
ag.algebraic-geometry reference-request complex-geometry
asked May 22 at 15:48
John RachedJohn Rached
29716
$endgroup$
add a comment |
$begingroup$
I presume this is a GAGA-style result, but I cannot find a reference.
ag.algebraic-geometry reference-request complex-geometry
asked May 22 at 15:48
John RachedJohn Rached
29716
$endgroup$
add a comment |
$begingroup$
I presume this is a GAGA-style result, but I cannot find a reference.
ag.algebraic-geometry reference-request complex-geometry
asked May 22 at 15:48
John RachedJohn Rached
29716
$endgroup$
I presume this is a GAGA-style result, but I cannot find a reference.
ag.algebraic-geometry reference-request complex-geometry
ag.algebraic-geometry reference-request complex-geometry
asked May 22 at 15:48
John RachedJohn Rached
29716
asked May 22 at 15:48
John RachedJohn Rached
29716
asked May 22 at 15:48
John RachedJohn Rached
29716
asked May 22 at 15:48
John RachedJohn Rached
29716
asked May 22 at 15:48
John RachedJohn Rached
29716
29716
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbbC^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.
edited May 25 at 13:22
Andrej Bauer
32.3k481174
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
$endgroup$
3
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
1
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrmX$ sheaf cohomology vanishes, $mathrmH^n(mathrmX,-)=0$ for $ngeqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
$endgroup$
5
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
2
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
5
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbbC^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.
edited May 25 at 13:22
Andrej Bauer
32.3k481174
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
$endgroup$
3
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
1
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
add a comment |
$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbbC^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.
edited May 25 at 13:22
Andrej Bauer
32.3k481174
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
$endgroup$
3
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
1
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
add a comment |
$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbbC^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.
edited May 25 at 13:22
Andrej Bauer
32.3k481174
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
$endgroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbbC^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.
edited May 25 at 13:22
Andrej Bauer
32.3k481174
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
edited May 25 at 13:22
Andrej Bauer
32.3k481174
edited May 25 at 13:22
Andrej Bauer
32.3k481174
edited May 25 at 13:22
Andrej Bauer
32.3k481174
32.3k481174
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
answered May 22 at 17:08
Donu ArapuraDonu Arapura
25.9k268128
25.9k268128
3
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
1
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
add a comment |
3
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
1
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
3
3
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
$begingroup$
In particular, the analytification of a complex algebraic variety is a Stein space.
$endgroup$
– Earthliŋ
May 23 at 10:31
1
1
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@Earthliŋ: You probably mean affine algebraic variety.
$endgroup$
– Jérôme Poineau
May 25 at 20:45
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
$begingroup$
@JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"...
$endgroup$
– Earthliŋ
May 25 at 21:50
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrmX$ sheaf cohomology vanishes, $mathrmH^n(mathrmX,-)=0$ for $ngeqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
$endgroup$
5
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
2
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
5
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrmX$ sheaf cohomology vanishes, $mathrmH^n(mathrmX,-)=0$ for $ngeqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
$endgroup$
5
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
2
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
5
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrmX$ sheaf cohomology vanishes, $mathrmH^n(mathrmX,-)=0$ for $ngeqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
$endgroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrmX$ sheaf cohomology vanishes, $mathrmH^n(mathrmX,-)=0$ for $ngeqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
answered May 22 at 16:12
Mere ScribeMere Scribe
1,0762820
1,0762820
5
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
2
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
5
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
add a comment |
5
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
2
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
5
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
5
5
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
$begingroup$
There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= mathbbC^*$ and cohomology with coefficients in $mathbbZ$ is not zero, but $X$ is Stein.
$endgroup$
– Donu Arapura
May 25 at 13:29
2
2
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
$begingroup$
@DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $mathcalO_X$-modules or which have no finiteness properties.
$endgroup$
– Mere Scribe
May 25 at 13:38
5
5
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
$begingroup$
OK, but I'll leave the comment, since not everyone would understand the hidden assumptions.
$endgroup$
– Donu Arapura
May 25 at 13:45
add a comment |
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