Definition of dualizing complexChain Homotopy classes as n-homology of a double complex$A_infty$ structure on Ext-algebras well defined?Balanced dualizing complexes according to A. YekutieliDuality of morphisms induced by multiplication of regular normal elementDerived Category.interpretation of homology of “non-commutative Koszul complex”Formal DG-algebraDifferent definitions of derived functorsDualizing complex definition ubiquityA question on some lemmas in Orlov's “Triangulated Categories of Singularities and D-Branes in Landau-Ginzburg Models” (Exts vanishing)

Definition of dualizing complex


Chain Homotopy classes as n-homology of a double complex$A_infty$ structure on Ext-algebras well defined?Balanced dualizing complexes according to A. YekutieliDuality of morphisms induced by multiplication of regular normal elementDerived Category.interpretation of homology of “non-commutative Koszul complex”Formal DG-algebraDifferent definitions of derived functorsDualizing complex definition ubiquityA question on some lemmas in Orlov's “Triangulated Categories of Singularities and D-Branes in Landau-Ginzburg Models” (Exts vanishing)













3












$begingroup$


Sorry for a not research level question asking for a definition but unfortunately I nowhere found a source which explains the construction presented below in a satisfactory way.



This question refers to definition of dualizing complex $omega_A ^bullet$ presented in the Stacks project Stacks as an object satisfying following properties:



(1) $ω^∙_A$ has finite injective dimension,



(2) $H^i(ω^∙_A)$ is a finite A-module for all i, and



(3) $A→RHomA(ω^∙_A,ω^∙_A)$ is a quasi-isomorphism.



My rudimentary question is what is the complex $mathrmRHom_A(ω^∙_A,ω^∙_A)$ and how is is it given in each degree? what is the for example the $k$-degree value of $mathrmRHom_A(ω^∙_A,ω^∙_A)$ concretely?



Naively if I try to imitate the calculations of higher derived functors then I would try following: I would take injective resolution $I^∙_k$ of $ω^∙_A[k]$, apply the functor $mathrmHom(ω^∙_A[k],-)$ to the resolution $I^∙_k toω^∙_A[k]$ and take the first homology. does this give me $R^nmathrmHom_A(ω^∙_A,ω^∙_A)[k]$ or is my approach wrong? in case of $R^nmathrmHom_A(ω^∙_A,ω^∙_A)$ I would do same procedure but take the $n$-th homology.










share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch.
    $endgroup$
    – abx
    8 hours ago










  • $begingroup$
    I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    in Hartshorne $RF$ is described as follows: if $F: A to B$ is a functor then the induced functor $RF: D(A) to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X to I^bullet$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M to I^bullet$ but generally this cannot be expected.
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$?
    $endgroup$
    – Tim Grosskreutz
    8 hours ago
















3












$begingroup$


Sorry for a not research level question asking for a definition but unfortunately I nowhere found a source which explains the construction presented below in a satisfactory way.



This question refers to definition of dualizing complex $omega_A ^bullet$ presented in the Stacks project Stacks as an object satisfying following properties:



(1) $ω^∙_A$ has finite injective dimension,



(2) $H^i(ω^∙_A)$ is a finite A-module for all i, and



(3) $A→RHomA(ω^∙_A,ω^∙_A)$ is a quasi-isomorphism.



My rudimentary question is what is the complex $mathrmRHom_A(ω^∙_A,ω^∙_A)$ and how is is it given in each degree? what is the for example the $k$-degree value of $mathrmRHom_A(ω^∙_A,ω^∙_A)$ concretely?



Naively if I try to imitate the calculations of higher derived functors then I would try following: I would take injective resolution $I^∙_k$ of $ω^∙_A[k]$, apply the functor $mathrmHom(ω^∙_A[k],-)$ to the resolution $I^∙_k toω^∙_A[k]$ and take the first homology. does this give me $R^nmathrmHom_A(ω^∙_A,ω^∙_A)[k]$ or is my approach wrong? in case of $R^nmathrmHom_A(ω^∙_A,ω^∙_A)$ I would do same procedure but take the $n$-th homology.










share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch.
    $endgroup$
    – abx
    8 hours ago










  • $begingroup$
    I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    in Hartshorne $RF$ is described as follows: if $F: A to B$ is a functor then the induced functor $RF: D(A) to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X to I^bullet$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M to I^bullet$ but generally this cannot be expected.
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$?
    $endgroup$
    – Tim Grosskreutz
    8 hours ago














3












3








3





$begingroup$


Sorry for a not research level question asking for a definition but unfortunately I nowhere found a source which explains the construction presented below in a satisfactory way.



This question refers to definition of dualizing complex $omega_A ^bullet$ presented in the Stacks project Stacks as an object satisfying following properties:



(1) $ω^∙_A$ has finite injective dimension,



(2) $H^i(ω^∙_A)$ is a finite A-module for all i, and



(3) $A→RHomA(ω^∙_A,ω^∙_A)$ is a quasi-isomorphism.



My rudimentary question is what is the complex $mathrmRHom_A(ω^∙_A,ω^∙_A)$ and how is is it given in each degree? what is the for example the $k$-degree value of $mathrmRHom_A(ω^∙_A,ω^∙_A)$ concretely?



Naively if I try to imitate the calculations of higher derived functors then I would try following: I would take injective resolution $I^∙_k$ of $ω^∙_A[k]$, apply the functor $mathrmHom(ω^∙_A[k],-)$ to the resolution $I^∙_k toω^∙_A[k]$ and take the first homology. does this give me $R^nmathrmHom_A(ω^∙_A,ω^∙_A)[k]$ or is my approach wrong? in case of $R^nmathrmHom_A(ω^∙_A,ω^∙_A)$ I would do same procedure but take the $n$-th homology.










share|cite|improve this question











$endgroup$




Sorry for a not research level question asking for a definition but unfortunately I nowhere found a source which explains the construction presented below in a satisfactory way.



This question refers to definition of dualizing complex $omega_A ^bullet$ presented in the Stacks project Stacks as an object satisfying following properties:



(1) $ω^∙_A$ has finite injective dimension,



(2) $H^i(ω^∙_A)$ is a finite A-module for all i, and



(3) $A→RHomA(ω^∙_A,ω^∙_A)$ is a quasi-isomorphism.



My rudimentary question is what is the complex $mathrmRHom_A(ω^∙_A,ω^∙_A)$ and how is is it given in each degree? what is the for example the $k$-degree value of $mathrmRHom_A(ω^∙_A,ω^∙_A)$ concretely?



Naively if I try to imitate the calculations of higher derived functors then I would try following: I would take injective resolution $I^∙_k$ of $ω^∙_A[k]$, apply the functor $mathrmHom(ω^∙_A[k],-)$ to the resolution $I^∙_k toω^∙_A[k]$ and take the first homology. does this give me $R^nmathrmHom_A(ω^∙_A,ω^∙_A)[k]$ or is my approach wrong? in case of $R^nmathrmHom_A(ω^∙_A,ω^∙_A)$ I would do same procedure but take the $n$-th homology.







ac.commutative-algebra homological-algebra derived-functors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









YCor

30k4 gold badges89 silver badges144 bronze badges




30k4 gold badges89 silver badges144 bronze badges










asked 9 hours ago









Tim GrosskreutzTim Grosskreutz

2581 silver badge7 bronze badges




2581 silver badge7 bronze badges







  • 3




    $begingroup$
    If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch.
    $endgroup$
    – abx
    8 hours ago










  • $begingroup$
    I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    in Hartshorne $RF$ is described as follows: if $F: A to B$ is a functor then the induced functor $RF: D(A) to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X to I^bullet$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M to I^bullet$ but generally this cannot be expected.
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$?
    $endgroup$
    – Tim Grosskreutz
    8 hours ago













  • 3




    $begingroup$
    If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch.
    $endgroup$
    – abx
    8 hours ago










  • $begingroup$
    I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    in Hartshorne $RF$ is described as follows: if $F: A to B$ is a functor then the induced functor $RF: D(A) to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X to I^bullet$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M to I^bullet$ but generally this cannot be expected.
    $endgroup$
    – Tim Grosskreutz
    8 hours ago











  • $begingroup$
    is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$?
    $endgroup$
    – Tim Grosskreutz
    8 hours ago








3




3




$begingroup$
If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch.
$endgroup$
– abx
8 hours ago




$begingroup$
If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch.
$endgroup$
– abx
8 hours ago












$begingroup$
I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand
$endgroup$
– Tim Grosskreutz
8 hours ago





$begingroup$
I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand
$endgroup$
– Tim Grosskreutz
8 hours ago













$begingroup$
in Hartshorne $RF$ is described as follows: if $F: A to B$ is a functor then the induced functor $RF: D(A) to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X to I^bullet$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M to I^bullet$ but generally this cannot be expected.
$endgroup$
– Tim Grosskreutz
8 hours ago





$begingroup$
in Hartshorne $RF$ is described as follows: if $F: A to B$ is a functor then the induced functor $RF: D(A) to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X to I^bullet$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M to I^bullet$ but generally this cannot be expected.
$endgroup$
– Tim Grosskreutz
8 hours ago













$begingroup$
is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$?
$endgroup$
– Tim Grosskreutz
8 hours ago





$begingroup$
is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$?
$endgroup$
– Tim Grosskreutz
8 hours ago











1 Answer
1






active

oldest

votes


















3












$begingroup$

In the derived category, one doesn't want to directly compute say the $k$-th slot of a complex as complexes can have many different quasi-isomorphic representations. That is in the derived category, there is no distinction between say a module M thought out as a complex supported in degree zero and a projective resolution.



Instead, you might want to know what the cohomology is of the dualizing complex. Indeed, in general $h^n( mathbfRtextrmHom_R(L,M)) = textrmHom_D(R)(L,M[n])$ where $R$ is a commutative ring and $L$ and $M$ are complexes of $R$-modules, see SP TAG 0A64, however this is not always easy to apply to an example. Sometimes it is easier to wrap your head around $mathbfRtextrmHom$ by noting that one has derived hom-tensor adjunction, that is $$textrmHom_D(R)(K, mathbfRtextrmHom_R(L,M)) cong textrmHom_D(R)(K otimes_R^mathbfL L, M)$$ for complexes $K,L,$ and $M$. Maybe the derived tensor product feels easier to understand.



However, dualizing complexes for say modules over local rings have specific interpretation via local duality. I'll leave you to find the corresponding global scheme-theoretic versions. If $(R,mathfrakm)$ is a local ring and $omega_R^bullet$ is a dualizing complex, which is only unique up to quasi-isomorphism and shift, so it often assumed to be normalized so that $h^-textrmdim R( omega_R^bullet)$ is the first non-zero cohomology module, then $omega_R = h^-textrmdim Romega_R^bullet$ is a canonical module for $R$. Local duality gives in the complete case a quasi-isomorphism $$mathbfRtextrmHom_R( K, omega_R^bullet) cong textHom_R( mathbfRGamma_mathfrakm(K),E)$$ where $E$ is the injective hull of the residue field. That is,$R$ is Cohen-Macualay if and only if $omega_R^bullet$ has only one non-zero module of support at $-textrmdim R$ and when $R$ is Gorenstein, this module is isomorphic to $R$.



As an example of how this is used, let's show a quick proof that when $R$ is a local Noetherian domain, $textrmAnn H_mathfrakm^i(R)$ is not zero for $i < textrmdim R$. Indeed, you just need to find an element $r$ so that $r h^-i(omega_R^bullet) = 0$ by local duality. Each $ h^-i(omega_R^bullet)$ is finitely generated, so if we localize at the unique minimal prime, we end up with a dualizing complex over a field. The latter lives in exactly one degree, $-textrmdim R$.



Finally, an important computational note is that when $S$ is a regular ring and $R = S/I$ then $omega_R^bullet := mathbfRtextrmHom_S(R,S)$ is a dualizing complex for $R$. Also, often dualizing complexes are used to control the study of singularities. Once one has such, they are most often studied by viewing them in various commutative diagrams and taking homology, so one almost never needs to know what the actual supporting modules are but much more interested in the cohomology of the complex.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago














Your Answer








StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f335211%2fdefinition-of-dualizing-complex%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

In the derived category, one doesn't want to directly compute say the $k$-th slot of a complex as complexes can have many different quasi-isomorphic representations. That is in the derived category, there is no distinction between say a module M thought out as a complex supported in degree zero and a projective resolution.



Instead, you might want to know what the cohomology is of the dualizing complex. Indeed, in general $h^n( mathbfRtextrmHom_R(L,M)) = textrmHom_D(R)(L,M[n])$ where $R$ is a commutative ring and $L$ and $M$ are complexes of $R$-modules, see SP TAG 0A64, however this is not always easy to apply to an example. Sometimes it is easier to wrap your head around $mathbfRtextrmHom$ by noting that one has derived hom-tensor adjunction, that is $$textrmHom_D(R)(K, mathbfRtextrmHom_R(L,M)) cong textrmHom_D(R)(K otimes_R^mathbfL L, M)$$ for complexes $K,L,$ and $M$. Maybe the derived tensor product feels easier to understand.



However, dualizing complexes for say modules over local rings have specific interpretation via local duality. I'll leave you to find the corresponding global scheme-theoretic versions. If $(R,mathfrakm)$ is a local ring and $omega_R^bullet$ is a dualizing complex, which is only unique up to quasi-isomorphism and shift, so it often assumed to be normalized so that $h^-textrmdim R( omega_R^bullet)$ is the first non-zero cohomology module, then $omega_R = h^-textrmdim Romega_R^bullet$ is a canonical module for $R$. Local duality gives in the complete case a quasi-isomorphism $$mathbfRtextrmHom_R( K, omega_R^bullet) cong textHom_R( mathbfRGamma_mathfrakm(K),E)$$ where $E$ is the injective hull of the residue field. That is,$R$ is Cohen-Macualay if and only if $omega_R^bullet$ has only one non-zero module of support at $-textrmdim R$ and when $R$ is Gorenstein, this module is isomorphic to $R$.



As an example of how this is used, let's show a quick proof that when $R$ is a local Noetherian domain, $textrmAnn H_mathfrakm^i(R)$ is not zero for $i < textrmdim R$. Indeed, you just need to find an element $r$ so that $r h^-i(omega_R^bullet) = 0$ by local duality. Each $ h^-i(omega_R^bullet)$ is finitely generated, so if we localize at the unique minimal prime, we end up with a dualizing complex over a field. The latter lives in exactly one degree, $-textrmdim R$.



Finally, an important computational note is that when $S$ is a regular ring and $R = S/I$ then $omega_R^bullet := mathbfRtextrmHom_S(R,S)$ is a dualizing complex for $R$. Also, often dualizing complexes are used to control the study of singularities. Once one has such, they are most often studied by viewing them in various commutative diagrams and taking homology, so one almost never needs to know what the actual supporting modules are but much more interested in the cohomology of the complex.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago
















3












$begingroup$

In the derived category, one doesn't want to directly compute say the $k$-th slot of a complex as complexes can have many different quasi-isomorphic representations. That is in the derived category, there is no distinction between say a module M thought out as a complex supported in degree zero and a projective resolution.



Instead, you might want to know what the cohomology is of the dualizing complex. Indeed, in general $h^n( mathbfRtextrmHom_R(L,M)) = textrmHom_D(R)(L,M[n])$ where $R$ is a commutative ring and $L$ and $M$ are complexes of $R$-modules, see SP TAG 0A64, however this is not always easy to apply to an example. Sometimes it is easier to wrap your head around $mathbfRtextrmHom$ by noting that one has derived hom-tensor adjunction, that is $$textrmHom_D(R)(K, mathbfRtextrmHom_R(L,M)) cong textrmHom_D(R)(K otimes_R^mathbfL L, M)$$ for complexes $K,L,$ and $M$. Maybe the derived tensor product feels easier to understand.



However, dualizing complexes for say modules over local rings have specific interpretation via local duality. I'll leave you to find the corresponding global scheme-theoretic versions. If $(R,mathfrakm)$ is a local ring and $omega_R^bullet$ is a dualizing complex, which is only unique up to quasi-isomorphism and shift, so it often assumed to be normalized so that $h^-textrmdim R( omega_R^bullet)$ is the first non-zero cohomology module, then $omega_R = h^-textrmdim Romega_R^bullet$ is a canonical module for $R$. Local duality gives in the complete case a quasi-isomorphism $$mathbfRtextrmHom_R( K, omega_R^bullet) cong textHom_R( mathbfRGamma_mathfrakm(K),E)$$ where $E$ is the injective hull of the residue field. That is,$R$ is Cohen-Macualay if and only if $omega_R^bullet$ has only one non-zero module of support at $-textrmdim R$ and when $R$ is Gorenstein, this module is isomorphic to $R$.



As an example of how this is used, let's show a quick proof that when $R$ is a local Noetherian domain, $textrmAnn H_mathfrakm^i(R)$ is not zero for $i < textrmdim R$. Indeed, you just need to find an element $r$ so that $r h^-i(omega_R^bullet) = 0$ by local duality. Each $ h^-i(omega_R^bullet)$ is finitely generated, so if we localize at the unique minimal prime, we end up with a dualizing complex over a field. The latter lives in exactly one degree, $-textrmdim R$.



Finally, an important computational note is that when $S$ is a regular ring and $R = S/I$ then $omega_R^bullet := mathbfRtextrmHom_S(R,S)$ is a dualizing complex for $R$. Also, often dualizing complexes are used to control the study of singularities. Once one has such, they are most often studied by viewing them in various commutative diagrams and taking homology, so one almost never needs to know what the actual supporting modules are but much more interested in the cohomology of the complex.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago














3












3








3





$begingroup$

In the derived category, one doesn't want to directly compute say the $k$-th slot of a complex as complexes can have many different quasi-isomorphic representations. That is in the derived category, there is no distinction between say a module M thought out as a complex supported in degree zero and a projective resolution.



Instead, you might want to know what the cohomology is of the dualizing complex. Indeed, in general $h^n( mathbfRtextrmHom_R(L,M)) = textrmHom_D(R)(L,M[n])$ where $R$ is a commutative ring and $L$ and $M$ are complexes of $R$-modules, see SP TAG 0A64, however this is not always easy to apply to an example. Sometimes it is easier to wrap your head around $mathbfRtextrmHom$ by noting that one has derived hom-tensor adjunction, that is $$textrmHom_D(R)(K, mathbfRtextrmHom_R(L,M)) cong textrmHom_D(R)(K otimes_R^mathbfL L, M)$$ for complexes $K,L,$ and $M$. Maybe the derived tensor product feels easier to understand.



However, dualizing complexes for say modules over local rings have specific interpretation via local duality. I'll leave you to find the corresponding global scheme-theoretic versions. If $(R,mathfrakm)$ is a local ring and $omega_R^bullet$ is a dualizing complex, which is only unique up to quasi-isomorphism and shift, so it often assumed to be normalized so that $h^-textrmdim R( omega_R^bullet)$ is the first non-zero cohomology module, then $omega_R = h^-textrmdim Romega_R^bullet$ is a canonical module for $R$. Local duality gives in the complete case a quasi-isomorphism $$mathbfRtextrmHom_R( K, omega_R^bullet) cong textHom_R( mathbfRGamma_mathfrakm(K),E)$$ where $E$ is the injective hull of the residue field. That is,$R$ is Cohen-Macualay if and only if $omega_R^bullet$ has only one non-zero module of support at $-textrmdim R$ and when $R$ is Gorenstein, this module is isomorphic to $R$.



As an example of how this is used, let's show a quick proof that when $R$ is a local Noetherian domain, $textrmAnn H_mathfrakm^i(R)$ is not zero for $i < textrmdim R$. Indeed, you just need to find an element $r$ so that $r h^-i(omega_R^bullet) = 0$ by local duality. Each $ h^-i(omega_R^bullet)$ is finitely generated, so if we localize at the unique minimal prime, we end up with a dualizing complex over a field. The latter lives in exactly one degree, $-textrmdim R$.



Finally, an important computational note is that when $S$ is a regular ring and $R = S/I$ then $omega_R^bullet := mathbfRtextrmHom_S(R,S)$ is a dualizing complex for $R$. Also, often dualizing complexes are used to control the study of singularities. Once one has such, they are most often studied by viewing them in various commutative diagrams and taking homology, so one almost never needs to know what the actual supporting modules are but much more interested in the cohomology of the complex.






share|cite|improve this answer









$endgroup$



In the derived category, one doesn't want to directly compute say the $k$-th slot of a complex as complexes can have many different quasi-isomorphic representations. That is in the derived category, there is no distinction between say a module M thought out as a complex supported in degree zero and a projective resolution.



Instead, you might want to know what the cohomology is of the dualizing complex. Indeed, in general $h^n( mathbfRtextrmHom_R(L,M)) = textrmHom_D(R)(L,M[n])$ where $R$ is a commutative ring and $L$ and $M$ are complexes of $R$-modules, see SP TAG 0A64, however this is not always easy to apply to an example. Sometimes it is easier to wrap your head around $mathbfRtextrmHom$ by noting that one has derived hom-tensor adjunction, that is $$textrmHom_D(R)(K, mathbfRtextrmHom_R(L,M)) cong textrmHom_D(R)(K otimes_R^mathbfL L, M)$$ for complexes $K,L,$ and $M$. Maybe the derived tensor product feels easier to understand.



However, dualizing complexes for say modules over local rings have specific interpretation via local duality. I'll leave you to find the corresponding global scheme-theoretic versions. If $(R,mathfrakm)$ is a local ring and $omega_R^bullet$ is a dualizing complex, which is only unique up to quasi-isomorphism and shift, so it often assumed to be normalized so that $h^-textrmdim R( omega_R^bullet)$ is the first non-zero cohomology module, then $omega_R = h^-textrmdim Romega_R^bullet$ is a canonical module for $R$. Local duality gives in the complete case a quasi-isomorphism $$mathbfRtextrmHom_R( K, omega_R^bullet) cong textHom_R( mathbfRGamma_mathfrakm(K),E)$$ where $E$ is the injective hull of the residue field. That is,$R$ is Cohen-Macualay if and only if $omega_R^bullet$ has only one non-zero module of support at $-textrmdim R$ and when $R$ is Gorenstein, this module is isomorphic to $R$.



As an example of how this is used, let's show a quick proof that when $R$ is a local Noetherian domain, $textrmAnn H_mathfrakm^i(R)$ is not zero for $i < textrmdim R$. Indeed, you just need to find an element $r$ so that $r h^-i(omega_R^bullet) = 0$ by local duality. Each $ h^-i(omega_R^bullet)$ is finitely generated, so if we localize at the unique minimal prime, we end up with a dualizing complex over a field. The latter lives in exactly one degree, $-textrmdim R$.



Finally, an important computational note is that when $S$ is a regular ring and $R = S/I$ then $omega_R^bullet := mathbfRtextrmHom_S(R,S)$ is a dualizing complex for $R$. Also, often dualizing complexes are used to control the study of singularities. Once one has such, they are most often studied by viewing them in various commutative diagrams and taking homology, so one almost never needs to know what the actual supporting modules are but much more interested in the cohomology of the complex.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









lemillerlemiller

1608 bronze badges




1608 bronze badges











  • $begingroup$
    thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago

















  • $begingroup$
    thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
    $endgroup$
    – Tim Grosskreutz
    1 hour ago










  • $begingroup$
    so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
    $endgroup$
    – Tim Grosskreutz
    1 hour ago
















$begingroup$
thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
$endgroup$
– Tim Grosskreutz
1 hour ago




$begingroup$
thank you for the explanation. one question regarding your remark in first paragraph: yes in the derived category we of course can only talk about equivalence classes up to homotopy. the point is if we take two certain complexes $C^bullet, D^bullet$ representing $[C^bullet]$ and $[D^bullet]$ in $D(A)$ is there a canonical expression for the $k$-th degree of the induced representant of $RHom_A(C^∙,D^∙)$?
$endgroup$
– Tim Grosskreutz
1 hour ago












$begingroup$
namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
$endgroup$
– Tim Grosskreutz
1 hour ago




$begingroup$
namely in analogy to the situation when again $C^bullet, D^bullet$ representing classes $[C^bullet]$ and $[D^bullet]$ then $[C^bullet otimes^L D^bullet]$ is represented by complex $C^bullet otimes^L D^bullet$ with $k$-degree $(C^bullet otimes^L D^bullet)_k = sum_i+j=n C^bullet_i otimes D^bullet_j$. so althought $[C^bullet otimes^L D^bullet]$ is just a class there exist a concrete formula for degree of the "canonical representant".
$endgroup$
– Tim Grosskreutz
1 hour ago












$begingroup$
so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
$endgroup$
– Tim Grosskreutz
1 hour ago





$begingroup$
so the point of my interest is if these exist such concrete formula for each degree in case of $RHom_A(C^∙,D^∙)$ if we fix two concrete representants $C^bullet, D^bullet$? or has $RHom_A(C^∙,D^∙) $ has only be understood as an "still existing object" and not more?
$endgroup$
– Tim Grosskreutz
1 hour ago


















draft saved

draft discarded
















































Thanks for contributing an answer to MathOverflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f335211%2fdefinition-of-dualizing-complex%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

The fall designs the understood secretary. Looking glass Science Shock Discovery Hot Everybody Loves Raymond Smile 곳 서비스 성실하다 Defas Kaloolon Definition: To combine or impregnate with sulphur or any of its compounds as to sulphurize caoutchouc in vulcanizing Flame colored Reason Useful Thin Help 갖다 유명하다 낙엽 장례식 Country Iron Definition: A fencer a gladiator one who exhibits his skill in the use of the sword Definition: The American black throated bunting Spiza Americana Nostalgic Needy Method to my madness 시키다 평가되다 전부 소설가 우아하다 Argument Tin Feeling Representative Gym Music Gaur Chicken 일쑤 코치 편 학생증 The harbor values the sugar. Vasagle Yammoe Enstatite Definition: Capable of being limited Road Neighborly Five Refer Built Kangaroo 비비다 Degree Release Bargain Horse 하루 형님 유교 석 동부 괴롭히다 경제력

Sahara Skak | Bilen | Luke uk diar | NawigatsjuunCommonskategorii: SaharaWikivoyage raisfeerer: Sahara26° N, 13° O

19. јануар Садржај Догађаји Рођења Смрти Празници и дани сећања Види још Референце Мени за навигацијуу