Why are Stein manifolds/spaces the analog of affine varieties/schemes in algebraic geometry?Stein Manifolds...
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
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I presume this is a GAGA-style result, but I cannot find a reference.
ag.algebraic-geometry reference-request complex-geometry
asked 10 hours ago
John RachedJohn Rached
1334
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I presume this is a GAGA-style result, but I cannot find a reference.
ag.algebraic-geometry reference-request complex-geometry
asked 10 hours ago
John RachedJohn Rached
1334
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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 10 hours ago
John RachedJohn Rached
1334
$endgroup$
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 10 hours ago
John RachedJohn Rached
1334
asked 10 hours ago
John RachedJohn Rached
1334
asked 10 hours ago
John RachedJohn Rached
1334
asked 10 hours ago
John RachedJohn Rached
1334
asked 10 hours ago
John RachedJohn Rached
1334
1334
add a comment |
add a comment |
2 Answers
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Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbb{C}^N$ 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, use Cartan B mentioned above. (Added: I don't mean to be too pedantic, but Cartan says that $X$ is Stein iff sheaf cohomology vanishes for coherent analytic coefficients; it is certainly not true otherwise.)
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
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I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrm{X}$ sheaf cohomology vanishes, $mathrm{H}^n(mathrm{X},-)=0$ 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 10 hours ago
Mere ScribeMere Scribe
1,0362820
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2 Answers
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2 Answers
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$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbb{C}^N$ 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, use Cartan B mentioned above. (Added: I don't mean to be too pedantic, but Cartan says that $X$ is Stein iff sheaf cohomology vanishes for coherent analytic coefficients; it is certainly not true otherwise.)
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
$endgroup$
add a comment |
$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbb{C}^N$ 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, use Cartan B mentioned above. (Added: I don't mean to be too pedantic, but Cartan says that $X$ is Stein iff sheaf cohomology vanishes for coherent analytic coefficients; it is certainly not true otherwise.)
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
$endgroup$
add a comment |
$begingroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbb{C}^N$ 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, use Cartan B mentioned above. (Added: I don't mean to be too pedantic, but Cartan says that $X$ is Stein iff sheaf cohomology vanishes for coherent analytic coefficients; it is certainly not true otherwise.)
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
$endgroup$
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $mathbb{C}^N$ 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, use Cartan B mentioned above. (Added: I don't mean to be too pedantic, but Cartan says that $X$ is Stein iff sheaf cohomology vanishes for coherent analytic coefficients; it is certainly not true otherwise.)
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
edited 7 hours ago
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
answered 9 hours ago
Donu ArapuraDonu Arapura
25.8k268128
25.8k268128
add a comment |
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrm{X}$ sheaf cohomology vanishes, $mathrm{H}^n(mathrm{X},-)=0$ 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 10 hours ago
Mere ScribeMere Scribe
1,0362820
$endgroup$
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrm{X}$ sheaf cohomology vanishes, $mathrm{H}^n(mathrm{X},-)=0$ 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 10 hours ago
Mere ScribeMere Scribe
1,0362820
$endgroup$
add a comment |
$begingroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrm{X}$ sheaf cohomology vanishes, $mathrm{H}^n(mathrm{X},-)=0$ 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 10 hours ago
Mere ScribeMere Scribe
1,0362820
$endgroup$
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $mathrm{X}$ sheaf cohomology vanishes, $mathrm{H}^n(mathrm{X},-)=0$ 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 10 hours ago
Mere ScribeMere Scribe
1,0362820
answered 10 hours ago
Mere ScribeMere Scribe
1,0362820
answered 10 hours ago
Mere ScribeMere Scribe
1,0362820
answered 10 hours ago
Mere ScribeMere Scribe
1,0362820
1,0362820
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