Poincare duality on the level of complexes
$begingroup$
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 Mar 12 at 3:49
yue heyue he
311
$endgroup$
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 Mar 12 at 3:49
yue heyue he
311
$endgroup$
$begingroup$
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
Mar 12 at 4:29
$begingroup$
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
Mar 12 at 5:20
$begingroup$
@WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
$endgroup$
– yue he
Mar 13 at 0:15
$begingroup$
I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
$endgroup$
– Will Sawin
Mar 13 at 0:22
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 Mar 12 at 3:49
yue heyue he
311
$endgroup$
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 Mar 12 at 3:49
yue heyue he
311
asked Mar 12 at 3:49
yue heyue he
311
asked Mar 12 at 3:49
yue heyue he
311
asked Mar 12 at 3:49
yue heyue he
311
asked Mar 12 at 3:49
yue heyue he
311
311
$begingroup$
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
Mar 12 at 4:29
$begingroup$
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
Mar 12 at 5:20
$begingroup$
@WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
$endgroup$
– yue he
Mar 13 at 0:15
$begingroup$
I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
$endgroup$
– Will Sawin
Mar 13 at 0:22
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
Mar 12 at 4:29
$begingroup$
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
Mar 12 at 5:20
$begingroup$
@WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
$endgroup$
– yue he
Mar 13 at 0:15
$begingroup$
I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
$endgroup$
– Will Sawin
Mar 13 at 0:22
$begingroup$
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
Mar 12 at 4:29
$begingroup$
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
Mar 12 at 4:29
$begingroup$
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
Mar 12 at 5:20
$begingroup$
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
Mar 12 at 5:20
$begingroup$
@WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
$endgroup$
– yue he
Mar 13 at 0:15
$begingroup$
@WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
$endgroup$
– yue he
Mar 13 at 0:15
$begingroup$
I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
$endgroup$
– Will Sawin
Mar 13 at 0:22
$begingroup$
I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
$endgroup$
– Will Sawin
Mar 13 at 0:22
add a comment |
1 Answer
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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). Moreover, without the orientability assumption this gives Poincaré duality with local coefficients (since $Ewedge M^{-TM}$ can be interpreted as $E$-homology with coefficients in the local system $xmapsto Ewedge S^{-T_xM}$).
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 Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
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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). Moreover, without the orientability assumption this gives Poincaré duality with local coefficients (since $Ewedge M^{-TM}$ can be interpreted as $E$-homology with coefficients in the local system $xmapsto Ewedge S^{-T_xM}$).
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 Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
$endgroup$
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). Moreover, without the orientability assumption this gives Poincaré duality with local coefficients (since $Ewedge M^{-TM}$ can be interpreted as $E$-homology with coefficients in the local system $xmapsto Ewedge S^{-T_xM}$).
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 Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
$endgroup$
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). Moreover, without the orientability assumption this gives Poincaré duality with local coefficients (since $Ewedge M^{-TM}$ can be interpreted as $E$-homology with coefficients in the local system $xmapsto Ewedge S^{-T_xM}$).
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 Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
$endgroup$
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). Moreover, without the orientability assumption this gives Poincaré duality with local coefficients (since $Ewedge M^{-TM}$ can be interpreted as $E$-homology with coefficients in the local system $xmapsto Ewedge S^{-T_xM}$).
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 Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
edited Mar 12 at 9:04
answered Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
answered Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
answered Mar 12 at 6:22
Denis NardinDenis Nardin
9,06223564
9,06223564
add a comment |
add a comment |
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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
Mar 12 at 4:29
$begingroup$
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
Mar 12 at 5:20
$begingroup$
@WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
$endgroup$
– yue he
Mar 13 at 0:15
$begingroup$
I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
$endgroup$
– Will Sawin
Mar 13 at 0:22