Poincare duality on the level of complexesIs there a map of spectra implementing the Thom isomorphism?What is...
Poincare duality on the level of complexes
Is there a map of spectra implementing the Thom isomorphism?What is the Hochschild cohomology of the dg category of perfect complexes on a variety?Perfect complexes and RGamma(X,F) without mentioning derived categoriesHow can one interpret homology and Stokes' Theorem via derived categories?Sheaf cohomology with compact supports (and Verdier duality?)Balanced dualizing complexes according to A. YekutieliOn a theorem of Hopkins-Neeman-Thomason on generators of thick subcategories of perfect complexesDerived stack 2-perfect complexes and derived equivalencesWhat are the uses of coefficient systems for arithmetic cohomology theories?sheaf cohomology and deRham cohomology of a stackSheaf cohomology relative to a closed subspace
$begingroup$
The classical Poincare duality is formulated in terms of cohomology groups. I am wondering if we can also formulate it in terms of complexes.
In particular, suppose $mathcal{C}^*$ is a complex of $k$-modules for some coefficient ring $k$ such that its cohomology groups $H^*(mathcal{C^*})$ is some Weil cohomology groups which satisfies some kind of Poincare duality. I am wondering if there is some form of Poincare duality formulated purely in terms of the complex $mathcal{C}^*$ instead of its cohomology groups.
I understand that there is a derived version of Poincare duality, namely the Verdier duality. But Verdier duality starts with a sheaf $mathcal{F}^*$. In my problem I already have its cohomology $RGamma(mathcal{F}^*)$, so I hope to work with the cohomological complex, and not to go back to the sheaf setting.
Any thoughts or reference is highly appreciated.
ag.algebraic-geometry derived-categories sheaf-cohomology
asked 3 hours ago
yue heyue he
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The classical Poincare duality is formulated in terms of cohomology groups. I am wondering if we can also formulate it in terms of complexes.
In particular, suppose $mathcal{C}^*$ is a complex of $k$-modules for some coefficient ring $k$ such that its cohomology groups $H^*(mathcal{C^*})$ is some Weil cohomology groups which satisfies some kind of Poincare duality. I am wondering if there is some form of Poincare duality formulated purely in terms of the complex $mathcal{C}^*$ instead of its cohomology groups.
I understand that there is a derived version of Poincare duality, namely the Verdier duality. But Verdier duality starts with a sheaf $mathcal{F}^*$. In my problem I already have its cohomology $RGamma(mathcal{F}^*)$, so I hope to work with the cohomological complex, and not to go back to the sheaf setting.
Any thoughts or reference is highly appreciated.
ag.algebraic-geometry derived-categories sheaf-cohomology
asked 3 hours ago
yue heyue he
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{infty}$-ring spectrum level
$endgroup$
– Aknazar Kazhymurat
2 hours ago
$begingroup$
I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
$endgroup$
– Will Sawin
1 hour ago
add a comment |
$begingroup$
The classical Poincare duality is formulated in terms of cohomology groups. I am wondering if we can also formulate it in terms of complexes.
In particular, suppose $mathcal{C}^*$ is a complex of $k$-modules for some coefficient ring $k$ such that its cohomology groups $H^*(mathcal{C^*})$ is some Weil cohomology groups which satisfies some kind of Poincare duality. I am wondering if there is some form of Poincare duality formulated purely in terms of the complex $mathcal{C}^*$ instead of its cohomology groups.
I understand that there is a derived version of Poincare duality, namely the Verdier duality. But Verdier duality starts with a sheaf $mathcal{F}^*$. In my problem I already have its cohomology $RGamma(mathcal{F}^*)$, so I hope to work with the cohomological complex, and not to go back to the sheaf setting.
Any thoughts or reference is highly appreciated.
ag.algebraic-geometry derived-categories sheaf-cohomology
asked 3 hours ago
yue heyue he
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The classical Poincare duality is formulated in terms of cohomology groups. I am wondering if we can also formulate it in terms of complexes.
In particular, suppose $mathcal{C}^*$ is a complex of $k$-modules for some coefficient ring $k$ such that its cohomology groups $H^*(mathcal{C^*})$ is some Weil cohomology groups which satisfies some kind of Poincare duality. I am wondering if there is some form of Poincare duality formulated purely in terms of the complex $mathcal{C}^*$ instead of its cohomology groups.
I understand that there is a derived version of Poincare duality, namely the Verdier duality. But Verdier duality starts with a sheaf $mathcal{F}^*$. In my problem I already have its cohomology $RGamma(mathcal{F}^*)$, so I hope to work with the cohomological complex, and not to go back to the sheaf setting.
Any thoughts or reference is highly appreciated.
ag.algebraic-geometry derived-categories sheaf-cohomology
ag.algebraic-geometry derived-categories sheaf-cohomology
asked 3 hours ago
yue heyue he
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
yue heyue he
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
yue heyue he
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
yue heyue he
161
asked 3 hours ago
yue heyue he
161
161
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
yue he is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{infty}$-ring spectrum level
$endgroup$
– Aknazar Kazhymurat
2 hours ago
$begingroup$
I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
$endgroup$
– Will Sawin
1 hour ago
add a comment |
$begingroup$
maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{infty}$-ring spectrum level
$endgroup$
– Aknazar Kazhymurat
2 hours ago
$begingroup$
I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
$endgroup$
– Will Sawin
1 hour ago
$begingroup$
maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{infty}$-ring spectrum level
$endgroup$
– Aknazar Kazhymurat
2 hours ago
$begingroup$
maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{infty}$-ring spectrum level
$endgroup$
– Aknazar Kazhymurat
2 hours ago
$begingroup$
I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
$endgroup$
– Will Sawin
1 hour ago
$begingroup$
I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
$endgroup$
– Will Sawin
1 hour ago
add a comment |
1 Answer
1
active
oldest
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$begingroup$
One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology)
$$M^{-TM}congmathbb{D}(Sigma^∞_+M)$$
Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $Sigma^infty_+$ is the suspension spectrum and $mathbb{D}$ is Spanier-Whitehead duality (duality in the category of spectra). Concretely this mean that for every $E_infty$-ring spectrum $E$ such that $TM$ is $E$-orientable (i.e. such that the Thom isomorphism holds) we have an equivalence of $E$-modules
$$Sigma^{-dim M}Ewedge Sigma^infty_+Mcong Ewedge M^{-TM}cong mathbb{D}_E(Ewedge Sigma^∞_+M)$$
Specializing in the case of $E=HR$, the Eilenberg-MacLane spectrum associated to a ring $R$, we have that the category of $E$-modules is exactly the derived category of $R$, and $HRwedgeSigma^∞_+Mcong C_*(M;R)$, so the formula above gives an equivalence in $mathscr{D}(R)$
$$C^*(M;R):=mathbb{D}_R(C_*(M;R))cong C_*(M;R)[-dim M]$$
You can generalize this to your heart's content, to compact manifolds with boundaries and to Verdier duality (working with local systems of spectra, or even constructible sheaves of spectra), but I hope this is enough to show the power of working in the stable homotopy category (i.e. the category of spectra).
One nice thing about this approach is that the proof of Atiyah duality is nice and geometric, as you can see from Charles Rezk's notes I linked above, and the "hard work" is outsourced to the Thom isomorphism (which is still not that hard)
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
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add a comment |
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$begingroup$
One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology)
$$M^{-TM}congmathbb{D}(Sigma^∞_+M)$$
Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $Sigma^infty_+$ is the suspension spectrum and $mathbb{D}$ is Spanier-Whitehead duality (duality in the category of spectra). Concretely this mean that for every $E_infty$-ring spectrum $E$ such that $TM$ is $E$-orientable (i.e. such that the Thom isomorphism holds) we have an equivalence of $E$-modules
$$Sigma^{-dim M}Ewedge Sigma^infty_+Mcong Ewedge M^{-TM}cong mathbb{D}_E(Ewedge Sigma^∞_+M)$$
Specializing in the case of $E=HR$, the Eilenberg-MacLane spectrum associated to a ring $R$, we have that the category of $E$-modules is exactly the derived category of $R$, and $HRwedgeSigma^∞_+Mcong C_*(M;R)$, so the formula above gives an equivalence in $mathscr{D}(R)$
$$C^*(M;R):=mathbb{D}_R(C_*(M;R))cong C_*(M;R)[-dim M]$$
You can generalize this to your heart's content, to compact manifolds with boundaries and to Verdier duality (working with local systems of spectra, or even constructible sheaves of spectra), but I hope this is enough to show the power of working in the stable homotopy category (i.e. the category of spectra).
One nice thing about this approach is that the proof of Atiyah duality is nice and geometric, as you can see from Charles Rezk's notes I linked above, and the "hard work" is outsourced to the Thom isomorphism (which is still not that hard)
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
$endgroup$
add a comment |
$begingroup$
One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology)
$$M^{-TM}congmathbb{D}(Sigma^∞_+M)$$
Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $Sigma^infty_+$ is the suspension spectrum and $mathbb{D}$ is Spanier-Whitehead duality (duality in the category of spectra). Concretely this mean that for every $E_infty$-ring spectrum $E$ such that $TM$ is $E$-orientable (i.e. such that the Thom isomorphism holds) we have an equivalence of $E$-modules
$$Sigma^{-dim M}Ewedge Sigma^infty_+Mcong Ewedge M^{-TM}cong mathbb{D}_E(Ewedge Sigma^∞_+M)$$
Specializing in the case of $E=HR$, the Eilenberg-MacLane spectrum associated to a ring $R$, we have that the category of $E$-modules is exactly the derived category of $R$, and $HRwedgeSigma^∞_+Mcong C_*(M;R)$, so the formula above gives an equivalence in $mathscr{D}(R)$
$$C^*(M;R):=mathbb{D}_R(C_*(M;R))cong C_*(M;R)[-dim M]$$
You can generalize this to your heart's content, to compact manifolds with boundaries and to Verdier duality (working with local systems of spectra, or even constructible sheaves of spectra), but I hope this is enough to show the power of working in the stable homotopy category (i.e. the category of spectra).
One nice thing about this approach is that the proof of Atiyah duality is nice and geometric, as you can see from Charles Rezk's notes I linked above, and the "hard work" is outsourced to the Thom isomorphism (which is still not that hard)
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
$endgroup$
add a comment |
$begingroup$
One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology)
$$M^{-TM}congmathbb{D}(Sigma^∞_+M)$$
Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $Sigma^infty_+$ is the suspension spectrum and $mathbb{D}$ is Spanier-Whitehead duality (duality in the category of spectra). Concretely this mean that for every $E_infty$-ring spectrum $E$ such that $TM$ is $E$-orientable (i.e. such that the Thom isomorphism holds) we have an equivalence of $E$-modules
$$Sigma^{-dim M}Ewedge Sigma^infty_+Mcong Ewedge M^{-TM}cong mathbb{D}_E(Ewedge Sigma^∞_+M)$$
Specializing in the case of $E=HR$, the Eilenberg-MacLane spectrum associated to a ring $R$, we have that the category of $E$-modules is exactly the derived category of $R$, and $HRwedgeSigma^∞_+Mcong C_*(M;R)$, so the formula above gives an equivalence in $mathscr{D}(R)$
$$C^*(M;R):=mathbb{D}_R(C_*(M;R))cong C_*(M;R)[-dim M]$$
You can generalize this to your heart's content, to compact manifolds with boundaries and to Verdier duality (working with local systems of spectra, or even constructible sheaves of spectra), but I hope this is enough to show the power of working in the stable homotopy category (i.e. the category of spectra).
One nice thing about this approach is that the proof of Atiyah duality is nice and geometric, as you can see from Charles Rezk's notes I linked above, and the "hard work" is outsourced to the Thom isomorphism (which is still not that hard)
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
$endgroup$
One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology)
$$M^{-TM}congmathbb{D}(Sigma^∞_+M)$$
Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $Sigma^infty_+$ is the suspension spectrum and $mathbb{D}$ is Spanier-Whitehead duality (duality in the category of spectra). Concretely this mean that for every $E_infty$-ring spectrum $E$ such that $TM$ is $E$-orientable (i.e. such that the Thom isomorphism holds) we have an equivalence of $E$-modules
$$Sigma^{-dim M}Ewedge Sigma^infty_+Mcong Ewedge M^{-TM}cong mathbb{D}_E(Ewedge Sigma^∞_+M)$$
Specializing in the case of $E=HR$, the Eilenberg-MacLane spectrum associated to a ring $R$, we have that the category of $E$-modules is exactly the derived category of $R$, and $HRwedgeSigma^∞_+Mcong C_*(M;R)$, so the formula above gives an equivalence in $mathscr{D}(R)$
$$C^*(M;R):=mathbb{D}_R(C_*(M;R))cong C_*(M;R)[-dim M]$$
You can generalize this to your heart's content, to compact manifolds with boundaries and to Verdier duality (working with local systems of spectra, or even constructible sheaves of spectra), but I hope this is enough to show the power of working in the stable homotopy category (i.e. the category of spectra).
One nice thing about this approach is that the proof of Atiyah duality is nice and geometric, as you can see from Charles Rezk's notes I linked above, and the "hard work" is outsourced to the Thom isomorphism (which is still not that hard)
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
edited 11 mins ago
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
answered 36 mins ago
Denis NardinDenis Nardin
8,67723562
8,67723562
add a comment |
add a comment |
yue he is a new contributor. Be nice, and check out our Code of Conduct.
yue he is a new contributor. Be nice, and check out our Code of Conduct.
yue he is a new contributor. Be nice, and check out our Code of Conduct.
yue he is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{infty}$-ring spectrum level
$endgroup$
– Aknazar Kazhymurat
2 hours ago
$begingroup$
I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
$endgroup$
– Will Sawin
1 hour ago