Results relying on higher derived algebraic geometryApplications of derived categories to “Traditional Algebraic Geometry”Complicating an Example by Toen (motivations for DAG)Derived global functions on (derived) stacks $BG$ and $G/G$Derived algebraic geometry: how to reach research level math?DG categories in algebraic geometry - guide to the literature?What is the relationship between connective and nonconnective derived algebraic geometry?Derived noncommutative geometry includes derived, or spectral algebraic geometry?Linear $infty$-categories $mathrmQC(X)$ and $mathrmPerf(X)$ of a “derived” stack $mathrmX$How would you organize a cycle of seminars aimed at learning together some basics of Derived Algebraic Geometry?Is every algebraic space a 1-geometric stack?relative spectrum in derived algebraic geometry
Results relying on higher derived algebraic geometry
Applications of derived categories to “Traditional Algebraic Geometry”Complicating an Example by Toen (motivations for DAG)Derived global functions on (derived) stacks $BG$ and $G/G$Derived algebraic geometry: how to reach research level math?DG categories in algebraic geometry - guide to the literature?What is the relationship between connective and nonconnective derived algebraic geometry?Derived noncommutative geometry includes derived, or spectral algebraic geometry?Linear $infty$-categories $mathrmQC(X)$ and $mathrmPerf(X)$ of a “derived” stack $mathrmX$How would you organize a cycle of seminars aimed at learning together some basics of Derived Algebraic Geometry?Is every algebraic space a 1-geometric stack?relative spectrum in derived algebraic geometry
$begingroup$
Are there any results in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures but whose most natural (known) proof uses derived $n$-Artin stacks for $n>1$? We are using Toen--Vezzosi terminology.
ag.algebraic-geometry higher-category-theory derived-algebraic-geometry
asked 15 hours ago
schematic_boischematic_boi
332120
$endgroup$
|
show 16 more comments
$begingroup$
Are there any results in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures but whose most natural (known) proof uses derived $n$-Artin stacks for $n>1$? We are using Toen--Vezzosi terminology.
ag.algebraic-geometry higher-category-theory derived-algebraic-geometry
asked 15 hours ago
schematic_boischematic_boi
332120
$endgroup$
4
$begingroup$
Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count?
$endgroup$
– Charles Rezk
8 hours ago
1
$begingroup$
@CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures?
$endgroup$
– schematic_boi
8 hours ago
3
$begingroup$
If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent.
$endgroup$
– Sam Gunningham
7 hours ago
3
$begingroup$
To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $infty$-stacks?
$endgroup$
– Sam Gunningham
7 hours ago
1
$begingroup$
I am very confused by this question. Would the solution to the Weibel conjecture count?
$endgroup$
– Denis Nardin
5 hours ago
|
show 16 more comments
$begingroup$
Are there any results in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures but whose most natural (known) proof uses derived $n$-Artin stacks for $n>1$? We are using Toen--Vezzosi terminology.
ag.algebraic-geometry higher-category-theory derived-algebraic-geometry
asked 15 hours ago
schematic_boischematic_boi
332120
$endgroup$
Are there any results in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures but whose most natural (known) proof uses derived $n$-Artin stacks for $n>1$? We are using Toen--Vezzosi terminology.
ag.algebraic-geometry higher-category-theory derived-algebraic-geometry
ag.algebraic-geometry higher-category-theory derived-algebraic-geometry
asked 15 hours ago
schematic_boischematic_boi
332120
asked 15 hours ago
schematic_boischematic_boi
332120
edited 4 hours ago
schematic_boi
asked 15 hours ago
schematic_boischematic_boi
332120
asked 15 hours ago
schematic_boischematic_boi
332120
asked 15 hours ago
schematic_boischematic_boi
332120
332120
4
$begingroup$
Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count?
$endgroup$
– Charles Rezk
8 hours ago
1
$begingroup$
@CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures?
$endgroup$
– schematic_boi
8 hours ago
3
$begingroup$
If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent.
$endgroup$
– Sam Gunningham
7 hours ago
3
$begingroup$
To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $infty$-stacks?
$endgroup$
– Sam Gunningham
7 hours ago
1
$begingroup$
I am very confused by this question. Would the solution to the Weibel conjecture count?
$endgroup$
– Denis Nardin
5 hours ago
|
show 16 more comments
4
$begingroup$
Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count?
$endgroup$
– Charles Rezk
8 hours ago
1
$begingroup$
@CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures?
$endgroup$
– schematic_boi
8 hours ago
3
$begingroup$
If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent.
$endgroup$
– Sam Gunningham
7 hours ago
3
$begingroup$
To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $infty$-stacks?
$endgroup$
– Sam Gunningham
7 hours ago
1
$begingroup$
I am very confused by this question. Would the solution to the Weibel conjecture count?
$endgroup$
– Denis Nardin
5 hours ago
4
4
$begingroup$
Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count?
$endgroup$
– Charles Rezk
8 hours ago
$begingroup$
Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count?
$endgroup$
– Charles Rezk
8 hours ago
1
1
$begingroup$
@CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures?
$endgroup$
– schematic_boi
8 hours ago
$begingroup$
@CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures?
$endgroup$
– schematic_boi
8 hours ago
3
3
$begingroup$
If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent.
$endgroup$
– Sam Gunningham
7 hours ago
$begingroup$
If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent.
$endgroup$
– Sam Gunningham
7 hours ago
3
3
$begingroup$
To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $infty$-stacks?
$endgroup$
– Sam Gunningham
7 hours ago
$begingroup$
To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $infty$-stacks?
$endgroup$
– Sam Gunningham
7 hours ago
1
1
$begingroup$
I am very confused by this question. Would the solution to the Weibel conjecture count?
$endgroup$
– Denis Nardin
5 hours ago
$begingroup$
I am very confused by this question. Would the solution to the Weibel conjecture count?
$endgroup$
– Denis Nardin
5 hours ago
|
show 16 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.
Theorem. Let $X$ be a coherent (quasi-compact and quasi-separated) scheme, let $A$ be a ring complete with respect to an ideal $Isubseteq A$. Then
$$ X(A) to varprojlim_n X(A/I^n+1) $$
is bijective.
Before going into the proof, let us consider the case $X$ is affine. Then
$$
X(A) = rm Hom(Gamma(X, mathcalO_X), A) = varprojlim_n rm Hom(Gamma(X, mathcalO_X), A/I^n+1) = varprojlim_n X(A/I^n+1) . $$
The idea for the general (coherent) case is to replace $Gamma(X, mathcalO_X)$ with $rm Perf(X)$, the category of perfect complexes on $X$.
Slogan. Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".
The second idea may be due to Thomason.
More precisely, we have:
Proposition. Let $X$ and $Y$ be schemes.
(a) If $X$ is affine, then
$$ rm Hom(Y, X) to rm Hom(Gamma(X, mathcalO_X), Gamma(Y, mathcalO_Y)) $$
is bijective.
(b) If $X$ is coherent, then
$$ rm Hom(Y, X) to rm Hom(rm Perf(X), rm Perf(Y)) $$
is an equivalence.
We must specify what (b) means (here is where DAG enters the picture). We consider $rm Perf(X)$ as the symmetric monoidal $infty$-category of perfect complexes on $X$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The $rm Hom$ on the right means the $infty$-groupoid (space) exact $otimes$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $f$ to $f^*$, the pull-back functor.
"Proof" of Theorem. We repeat the proof of the affine case, replacing rings with categories of perfect complexes:
$$
X(A) = rm Hom(rm Perf(X), rm Perf(A)) = varprojlim_n rm Hom(rm Perf(X), rm Perf(A/I^n+1)) = varprojlim_n X(A/I^n+1) . $$
Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.
End remarks.
(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $X$ is coherent in the Theorem.
(2) He also said that the above proof is the only one he is aware of.
(3) Reference for the above (thanks to the user crystalline):
Bhargav Bhatt Algebraization and Tannaka duality arxiv.org/abs/1404.7483.
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
$endgroup$
3
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
add a comment |
$begingroup$
The derived moduli stack of perfect complexes $RPerf$ is a derived Artin stack which admits a filtration by open sub stacks $RPerf^[a,b]$. The latter is a derived $(b-a+1)$-Artin stack. See:
Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/
This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see:
Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1
answered 4 hours ago
crystallinecrystalline
36636
$endgroup$
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.
Theorem. Let $X$ be a coherent (quasi-compact and quasi-separated) scheme, let $A$ be a ring complete with respect to an ideal $Isubseteq A$. Then
$$ X(A) to varprojlim_n X(A/I^n+1) $$
is bijective.
Before going into the proof, let us consider the case $X$ is affine. Then
$$
X(A) = rm Hom(Gamma(X, mathcalO_X), A) = varprojlim_n rm Hom(Gamma(X, mathcalO_X), A/I^n+1) = varprojlim_n X(A/I^n+1) . $$
The idea for the general (coherent) case is to replace $Gamma(X, mathcalO_X)$ with $rm Perf(X)$, the category of perfect complexes on $X$.
Slogan. Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".
The second idea may be due to Thomason.
More precisely, we have:
Proposition. Let $X$ and $Y$ be schemes.
(a) If $X$ is affine, then
$$ rm Hom(Y, X) to rm Hom(Gamma(X, mathcalO_X), Gamma(Y, mathcalO_Y)) $$
is bijective.
(b) If $X$ is coherent, then
$$ rm Hom(Y, X) to rm Hom(rm Perf(X), rm Perf(Y)) $$
is an equivalence.
We must specify what (b) means (here is where DAG enters the picture). We consider $rm Perf(X)$ as the symmetric monoidal $infty$-category of perfect complexes on $X$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The $rm Hom$ on the right means the $infty$-groupoid (space) exact $otimes$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $f$ to $f^*$, the pull-back functor.
"Proof" of Theorem. We repeat the proof of the affine case, replacing rings with categories of perfect complexes:
$$
X(A) = rm Hom(rm Perf(X), rm Perf(A)) = varprojlim_n rm Hom(rm Perf(X), rm Perf(A/I^n+1)) = varprojlim_n X(A/I^n+1) . $$
Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.
End remarks.
(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $X$ is coherent in the Theorem.
(2) He also said that the above proof is the only one he is aware of.
(3) Reference for the above (thanks to the user crystalline):
Bhargav Bhatt Algebraization and Tannaka duality arxiv.org/abs/1404.7483.
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
$endgroup$
3
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
add a comment |
$begingroup$
Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.
Theorem. Let $X$ be a coherent (quasi-compact and quasi-separated) scheme, let $A$ be a ring complete with respect to an ideal $Isubseteq A$. Then
$$ X(A) to varprojlim_n X(A/I^n+1) $$
is bijective.
Before going into the proof, let us consider the case $X$ is affine. Then
$$
X(A) = rm Hom(Gamma(X, mathcalO_X), A) = varprojlim_n rm Hom(Gamma(X, mathcalO_X), A/I^n+1) = varprojlim_n X(A/I^n+1) . $$
The idea for the general (coherent) case is to replace $Gamma(X, mathcalO_X)$ with $rm Perf(X)$, the category of perfect complexes on $X$.
Slogan. Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".
The second idea may be due to Thomason.
More precisely, we have:
Proposition. Let $X$ and $Y$ be schemes.
(a) If $X$ is affine, then
$$ rm Hom(Y, X) to rm Hom(Gamma(X, mathcalO_X), Gamma(Y, mathcalO_Y)) $$
is bijective.
(b) If $X$ is coherent, then
$$ rm Hom(Y, X) to rm Hom(rm Perf(X), rm Perf(Y)) $$
is an equivalence.
We must specify what (b) means (here is where DAG enters the picture). We consider $rm Perf(X)$ as the symmetric monoidal $infty$-category of perfect complexes on $X$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The $rm Hom$ on the right means the $infty$-groupoid (space) exact $otimes$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $f$ to $f^*$, the pull-back functor.
"Proof" of Theorem. We repeat the proof of the affine case, replacing rings with categories of perfect complexes:
$$
X(A) = rm Hom(rm Perf(X), rm Perf(A)) = varprojlim_n rm Hom(rm Perf(X), rm Perf(A/I^n+1)) = varprojlim_n X(A/I^n+1) . $$
Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.
End remarks.
(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $X$ is coherent in the Theorem.
(2) He also said that the above proof is the only one he is aware of.
(3) Reference for the above (thanks to the user crystalline):
Bhargav Bhatt Algebraization and Tannaka duality arxiv.org/abs/1404.7483.
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
$endgroup$
3
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
add a comment |
$begingroup$
Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.
Theorem. Let $X$ be a coherent (quasi-compact and quasi-separated) scheme, let $A$ be a ring complete with respect to an ideal $Isubseteq A$. Then
$$ X(A) to varprojlim_n X(A/I^n+1) $$
is bijective.
Before going into the proof, let us consider the case $X$ is affine. Then
$$
X(A) = rm Hom(Gamma(X, mathcalO_X), A) = varprojlim_n rm Hom(Gamma(X, mathcalO_X), A/I^n+1) = varprojlim_n X(A/I^n+1) . $$
The idea for the general (coherent) case is to replace $Gamma(X, mathcalO_X)$ with $rm Perf(X)$, the category of perfect complexes on $X$.
Slogan. Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".
The second idea may be due to Thomason.
More precisely, we have:
Proposition. Let $X$ and $Y$ be schemes.
(a) If $X$ is affine, then
$$ rm Hom(Y, X) to rm Hom(Gamma(X, mathcalO_X), Gamma(Y, mathcalO_Y)) $$
is bijective.
(b) If $X$ is coherent, then
$$ rm Hom(Y, X) to rm Hom(rm Perf(X), rm Perf(Y)) $$
is an equivalence.
We must specify what (b) means (here is where DAG enters the picture). We consider $rm Perf(X)$ as the symmetric monoidal $infty$-category of perfect complexes on $X$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The $rm Hom$ on the right means the $infty$-groupoid (space) exact $otimes$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $f$ to $f^*$, the pull-back functor.
"Proof" of Theorem. We repeat the proof of the affine case, replacing rings with categories of perfect complexes:
$$
X(A) = rm Hom(rm Perf(X), rm Perf(A)) = varprojlim_n rm Hom(rm Perf(X), rm Perf(A/I^n+1)) = varprojlim_n X(A/I^n+1) . $$
Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.
End remarks.
(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $X$ is coherent in the Theorem.
(2) He also said that the above proof is the only one he is aware of.
(3) Reference for the above (thanks to the user crystalline):
Bhargav Bhatt Algebraization and Tannaka duality arxiv.org/abs/1404.7483.
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
$endgroup$
Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.
Theorem. Let $X$ be a coherent (quasi-compact and quasi-separated) scheme, let $A$ be a ring complete with respect to an ideal $Isubseteq A$. Then
$$ X(A) to varprojlim_n X(A/I^n+1) $$
is bijective.
Before going into the proof, let us consider the case $X$ is affine. Then
$$
X(A) = rm Hom(Gamma(X, mathcalO_X), A) = varprojlim_n rm Hom(Gamma(X, mathcalO_X), A/I^n+1) = varprojlim_n X(A/I^n+1) . $$
The idea for the general (coherent) case is to replace $Gamma(X, mathcalO_X)$ with $rm Perf(X)$, the category of perfect complexes on $X$.
Slogan. Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".
The second idea may be due to Thomason.
More precisely, we have:
Proposition. Let $X$ and $Y$ be schemes.
(a) If $X$ is affine, then
$$ rm Hom(Y, X) to rm Hom(Gamma(X, mathcalO_X), Gamma(Y, mathcalO_Y)) $$
is bijective.
(b) If $X$ is coherent, then
$$ rm Hom(Y, X) to rm Hom(rm Perf(X), rm Perf(Y)) $$
is an equivalence.
We must specify what (b) means (here is where DAG enters the picture). We consider $rm Perf(X)$ as the symmetric monoidal $infty$-category of perfect complexes on $X$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The $rm Hom$ on the right means the $infty$-groupoid (space) exact $otimes$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $f$ to $f^*$, the pull-back functor.
"Proof" of Theorem. We repeat the proof of the affine case, replacing rings with categories of perfect complexes:
$$
X(A) = rm Hom(rm Perf(X), rm Perf(A)) = varprojlim_n rm Hom(rm Perf(X), rm Perf(A/I^n+1)) = varprojlim_n X(A/I^n+1) . $$
Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.
End remarks.
(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $X$ is coherent in the Theorem.
(2) He also said that the above proof is the only one he is aware of.
(3) Reference for the above (thanks to the user crystalline):
Bhargav Bhatt Algebraization and Tannaka duality arxiv.org/abs/1404.7483.
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
edited 2 hours ago
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
answered 5 hours ago
Piotr AchingerPiotr Achinger
8,84522955
8,84522955
3
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
add a comment |
3
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
3
3
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483.
$endgroup$
– crystalline
5 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
$begingroup$
Thank you! I thought looking at $X$ through the symmetric monoidal $infty$-category $rm Perf(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance.
$endgroup$
– Piotr Achinger
2 hours ago
add a comment |
$begingroup$
The derived moduli stack of perfect complexes $RPerf$ is a derived Artin stack which admits a filtration by open sub stacks $RPerf^[a,b]$. The latter is a derived $(b-a+1)$-Artin stack. See:
Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/
This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see:
Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1
answered 4 hours ago
crystallinecrystalline
36636
$endgroup$
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
add a comment |
$begingroup$
The derived moduli stack of perfect complexes $RPerf$ is a derived Artin stack which admits a filtration by open sub stacks $RPerf^[a,b]$. The latter is a derived $(b-a+1)$-Artin stack. See:
Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/
This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see:
Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1
answered 4 hours ago
crystallinecrystalline
36636
$endgroup$
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
add a comment |
$begingroup$
The derived moduli stack of perfect complexes $RPerf$ is a derived Artin stack which admits a filtration by open sub stacks $RPerf^[a,b]$. The latter is a derived $(b-a+1)$-Artin stack. See:
Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/
This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see:
Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1
answered 4 hours ago
crystallinecrystalline
36636
$endgroup$
The derived moduli stack of perfect complexes $RPerf$ is a derived Artin stack which admits a filtration by open sub stacks $RPerf^[a,b]$. The latter is a derived $(b-a+1)$-Artin stack. See:
Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/
This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see:
Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1
answered 4 hours ago
crystallinecrystalline
36636
answered 4 hours ago
crystallinecrystalline
36636
answered 4 hours ago
crystallinecrystalline
36636
answered 4 hours ago
crystallinecrystalline
36636
36636
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
add a comment |
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff)
$endgroup$
– EBz
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
@EBz you mean that is needed to define the virtual fundamental class to define counting invariants?
$endgroup$
– schematic_boi
4 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
Is it a theorem that $RPerf^[a, b]$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack?
$endgroup$
– schematic_boi
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
$begingroup$
@op, I mean that it is used to do so in a uniform and v natural way. Whether it is "needed" I couldn't say, it's certainly not equivalent to it so I guess not.
$endgroup$
– EBz
3 hours ago
add a comment |
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4
$begingroup$
Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count?
$endgroup$
– Charles Rezk
8 hours ago
1
$begingroup$
@CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures?
$endgroup$
– schematic_boi
8 hours ago
3
$begingroup$
If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent.
$endgroup$
– Sam Gunningham
7 hours ago
3
$begingroup$
To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $infty$-stacks?
$endgroup$
– Sam Gunningham
7 hours ago
1
$begingroup$
I am very confused by this question. Would the solution to the Weibel conjecture count?
$endgroup$
– Denis Nardin
5 hours ago