Formality of commutative differential graded algebras
I want to understand the definition of a commutative differential graded algebra (CDGA) to be formal.
Actually I encountered two definitions, but I have trouble with both.
From Wikipedia:
A commutative differential graded algebra $A$, again with $A^0 = mathbb Q$, is called formal if $A$ has a model with vanishing differential. This is equivalent to requiring that the cohomology algebra of $A$ (viewed as a differential algebra with trivial differential) is a model for $A$ (though it does not have to be the minimal model).
From Griffiths, Morgan:
A space is said to be formal if the homotopy type of the DGA of forms on the space is the same as the homotopy type of the cohomology ring of the space. Thus, if $X$ is formal and $mathcal m_X$ is a minimal model for the forms on $X$, then there is a DGA map $mathcal m_Xto (H^*(X),d=0))$ which induces the identity on cohomology.
As far as I know, a CDGA is a model for another CDGA, if they are quasi-isomorphic (i.e. isomorphic in cohomology). A CDGA is a minimal model if it is a model and it is minimal (free and decomposable).
So here are my issues with the definitions:
Wikipedia: To me it seems that the complex $(H^*(A),d=0))$ always is a model of $A$.
Griffiths, Morgan: We already have the existence of an isomorphism $mathcal (H^*(m_X,d_mathcal m))to (H^*(X),d=0))$. But this is not strong enough, we need this to be the identity. But in order to speak about the identity, the spaces must be same, which they are only up to isomorphy.
It seems like I missed a crucial point of the definitions. But what is it?
abstract-algebra algebraic-topology homology-cohomology homotopy-theory rational-homotopy-theory
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
add a comment |
I want to understand the definition of a commutative differential graded algebra (CDGA) to be formal.
Actually I encountered two definitions, but I have trouble with both.
From Wikipedia:
A commutative differential graded algebra $A$, again with $A^0 = mathbb Q$, is called formal if $A$ has a model with vanishing differential. This is equivalent to requiring that the cohomology algebra of $A$ (viewed as a differential algebra with trivial differential) is a model for $A$ (though it does not have to be the minimal model).
From Griffiths, Morgan:
A space is said to be formal if the homotopy type of the DGA of forms on the space is the same as the homotopy type of the cohomology ring of the space. Thus, if $X$ is formal and $mathcal m_X$ is a minimal model for the forms on $X$, then there is a DGA map $mathcal m_Xto (H^*(X),d=0))$ which induces the identity on cohomology.
As far as I know, a CDGA is a model for another CDGA, if they are quasi-isomorphic (i.e. isomorphic in cohomology). A CDGA is a minimal model if it is a model and it is minimal (free and decomposable).
So here are my issues with the definitions:
Wikipedia: To me it seems that the complex $(H^*(A),d=0))$ always is a model of $A$.
Griffiths, Morgan: We already have the existence of an isomorphism $mathcal (H^*(m_X,d_mathcal m))to (H^*(X),d=0))$. But this is not strong enough, we need this to be the identity. But in order to speak about the identity, the spaces must be same, which they are only up to isomorphy.
It seems like I missed a crucial point of the definitions. But what is it?
abstract-algebra algebraic-topology homology-cohomology homotopy-theory rational-homotopy-theory
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
Two complexes can have isomorphic cohomology but not be quasi-isomorphic.
– JHF
Dec 5 '18 at 19:54
@JHF: So what is the real definition of quasi-isomorphy?
– klirk
Dec 5 '18 at 21:18
@klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3times S^2$ and $RP^2 times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent.
– Aleksandar Milivojevic
Dec 6 '18 at 13:21
Ah, I see. thx.
– klirk
Dec 6 '18 at 20:23
add a comment |
I want to understand the definition of a commutative differential graded algebra (CDGA) to be formal.
Actually I encountered two definitions, but I have trouble with both.
From Wikipedia:
A commutative differential graded algebra $A$, again with $A^0 = mathbb Q$, is called formal if $A$ has a model with vanishing differential. This is equivalent to requiring that the cohomology algebra of $A$ (viewed as a differential algebra with trivial differential) is a model for $A$ (though it does not have to be the minimal model).
From Griffiths, Morgan:
A space is said to be formal if the homotopy type of the DGA of forms on the space is the same as the homotopy type of the cohomology ring of the space. Thus, if $X$ is formal and $mathcal m_X$ is a minimal model for the forms on $X$, then there is a DGA map $mathcal m_Xto (H^*(X),d=0))$ which induces the identity on cohomology.
As far as I know, a CDGA is a model for another CDGA, if they are quasi-isomorphic (i.e. isomorphic in cohomology). A CDGA is a minimal model if it is a model and it is minimal (free and decomposable).
So here are my issues with the definitions:
Wikipedia: To me it seems that the complex $(H^*(A),d=0))$ always is a model of $A$.
Griffiths, Morgan: We already have the existence of an isomorphism $mathcal (H^*(m_X,d_mathcal m))to (H^*(X),d=0))$. But this is not strong enough, we need this to be the identity. But in order to speak about the identity, the spaces must be same, which they are only up to isomorphy.
It seems like I missed a crucial point of the definitions. But what is it?
abstract-algebra algebraic-topology homology-cohomology homotopy-theory rational-homotopy-theory
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
I want to understand the definition of a commutative differential graded algebra (CDGA) to be formal.
Actually I encountered two definitions, but I have trouble with both.
From Wikipedia:
A commutative differential graded algebra $A$, again with $A^0 = mathbb Q$, is called formal if $A$ has a model with vanishing differential. This is equivalent to requiring that the cohomology algebra of $A$ (viewed as a differential algebra with trivial differential) is a model for $A$ (though it does not have to be the minimal model).
From Griffiths, Morgan:
A space is said to be formal if the homotopy type of the DGA of forms on the space is the same as the homotopy type of the cohomology ring of the space. Thus, if $X$ is formal and $mathcal m_X$ is a minimal model for the forms on $X$, then there is a DGA map $mathcal m_Xto (H^*(X),d=0))$ which induces the identity on cohomology.
As far as I know, a CDGA is a model for another CDGA, if they are quasi-isomorphic (i.e. isomorphic in cohomology). A CDGA is a minimal model if it is a model and it is minimal (free and decomposable).
So here are my issues with the definitions:
Wikipedia: To me it seems that the complex $(H^*(A),d=0))$ always is a model of $A$.
Griffiths, Morgan: We already have the existence of an isomorphism $mathcal (H^*(m_X,d_mathcal m))to (H^*(X),d=0))$. But this is not strong enough, we need this to be the identity. But in order to speak about the identity, the spaces must be same, which they are only up to isomorphy.
It seems like I missed a crucial point of the definitions. But what is it?
abstract-algebra algebraic-topology homology-cohomology homotopy-theory rational-homotopy-theory
abstract-algebra algebraic-topology homology-cohomology homotopy-theory rational-homotopy-theory
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
asked Dec 5 '18 at 17:45
klirkklirk
2,619530
2,619530
Two complexes can have isomorphic cohomology but not be quasi-isomorphic.
– JHF
Dec 5 '18 at 19:54
@JHF: So what is the real definition of quasi-isomorphy?
– klirk
Dec 5 '18 at 21:18
@klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3times S^2$ and $RP^2 times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent.
– Aleksandar Milivojevic
Dec 6 '18 at 13:21
Ah, I see. thx.
– klirk
Dec 6 '18 at 20:23
add a comment |
Two complexes can have isomorphic cohomology but not be quasi-isomorphic.
– JHF
Dec 5 '18 at 19:54
@JHF: So what is the real definition of quasi-isomorphy?
– klirk
Dec 5 '18 at 21:18
@klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3times S^2$ and $RP^2 times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent.
– Aleksandar Milivojevic
Dec 6 '18 at 13:21
Ah, I see. thx.
– klirk
Dec 6 '18 at 20:23
Two complexes can have isomorphic cohomology but not be quasi-isomorphic.
– JHF
Dec 5 '18 at 19:54
Two complexes can have isomorphic cohomology but not be quasi-isomorphic.
– JHF
Dec 5 '18 at 19:54
@JHF: So what is the real definition of quasi-isomorphy?
– klirk
Dec 5 '18 at 21:18
@JHF: So what is the real definition of quasi-isomorphy?
– klirk
Dec 5 '18 at 21:18
@klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3times S^2$ and $RP^2 times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent.
– Aleksandar Milivojevic
Dec 6 '18 at 13:21
@klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3times S^2$ and $RP^2 times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent.
– Aleksandar Milivojevic
Dec 6 '18 at 13:21
Ah, I see. thx.
– klirk
Dec 6 '18 at 20:23
Ah, I see. thx.
– klirk
Dec 6 '18 at 20:23
add a comment |
1 Answer
1
active
oldest
votes
You need to distinguish between maps of complexes and maps of cdgas. I'll work over a field $k$. If $(C,d)$ is a complex of $k$-vector spaces, then it's true that $C$ is quasi-isomorphic to $(H(C),0)$, via taking sections (which we can do because every vector space is a projective module; over more general rings it is not true that every complex is quasi-isomorphic to its cohomology). If $(C,d)$ is in addition a cdga, then $(H(C),0)$ is a cdga as well. These two cdgas are quasi-isomorphic as complexes of vector spaces, but there may be no quasi-isomorphism between them that preserves the multiplicative structure. You can think of this as a derived version of the fact that a given vector space can admit many different algebra structures.
My go-to concrete example of a non-formal dga $A$ appears on page 13 of Hess's Rational Homotopy Theory: A Brief Introduction [https://arxiv.org/pdf/math/0604626.pdf ] as the exterior algebra on three elements $u,v,w$ of degrees $3,3,5$ respectively, and with $d(w)=uv$. You can check that there is no cdga quasi-isomorphism $A to H(A)$.
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
1
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027400%2fformality-of-commutative-differential-graded-algebras%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
You need to distinguish between maps of complexes and maps of cdgas. I'll work over a field $k$. If $(C,d)$ is a complex of $k$-vector spaces, then it's true that $C$ is quasi-isomorphic to $(H(C),0)$, via taking sections (which we can do because every vector space is a projective module; over more general rings it is not true that every complex is quasi-isomorphic to its cohomology). If $(C,d)$ is in addition a cdga, then $(H(C),0)$ is a cdga as well. These two cdgas are quasi-isomorphic as complexes of vector spaces, but there may be no quasi-isomorphism between them that preserves the multiplicative structure. You can think of this as a derived version of the fact that a given vector space can admit many different algebra structures.
My go-to concrete example of a non-formal dga $A$ appears on page 13 of Hess's Rational Homotopy Theory: A Brief Introduction [https://arxiv.org/pdf/math/0604626.pdf ] as the exterior algebra on three elements $u,v,w$ of degrees $3,3,5$ respectively, and with $d(w)=uv$. You can check that there is no cdga quasi-isomorphism $A to H(A)$.
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
1
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
add a comment |
You need to distinguish between maps of complexes and maps of cdgas. I'll work over a field $k$. If $(C,d)$ is a complex of $k$-vector spaces, then it's true that $C$ is quasi-isomorphic to $(H(C),0)$, via taking sections (which we can do because every vector space is a projective module; over more general rings it is not true that every complex is quasi-isomorphic to its cohomology). If $(C,d)$ is in addition a cdga, then $(H(C),0)$ is a cdga as well. These two cdgas are quasi-isomorphic as complexes of vector spaces, but there may be no quasi-isomorphism between them that preserves the multiplicative structure. You can think of this as a derived version of the fact that a given vector space can admit many different algebra structures.
My go-to concrete example of a non-formal dga $A$ appears on page 13 of Hess's Rational Homotopy Theory: A Brief Introduction [https://arxiv.org/pdf/math/0604626.pdf ] as the exterior algebra on three elements $u,v,w$ of degrees $3,3,5$ respectively, and with $d(w)=uv$. You can check that there is no cdga quasi-isomorphism $A to H(A)$.
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
1
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
add a comment |
You need to distinguish between maps of complexes and maps of cdgas. I'll work over a field $k$. If $(C,d)$ is a complex of $k$-vector spaces, then it's true that $C$ is quasi-isomorphic to $(H(C),0)$, via taking sections (which we can do because every vector space is a projective module; over more general rings it is not true that every complex is quasi-isomorphic to its cohomology). If $(C,d)$ is in addition a cdga, then $(H(C),0)$ is a cdga as well. These two cdgas are quasi-isomorphic as complexes of vector spaces, but there may be no quasi-isomorphism between them that preserves the multiplicative structure. You can think of this as a derived version of the fact that a given vector space can admit many different algebra structures.
My go-to concrete example of a non-formal dga $A$ appears on page 13 of Hess's Rational Homotopy Theory: A Brief Introduction [https://arxiv.org/pdf/math/0604626.pdf ] as the exterior algebra on three elements $u,v,w$ of degrees $3,3,5$ respectively, and with $d(w)=uv$. You can check that there is no cdga quasi-isomorphism $A to H(A)$.
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
You need to distinguish between maps of complexes and maps of cdgas. I'll work over a field $k$. If $(C,d)$ is a complex of $k$-vector spaces, then it's true that $C$ is quasi-isomorphic to $(H(C),0)$, via taking sections (which we can do because every vector space is a projective module; over more general rings it is not true that every complex is quasi-isomorphic to its cohomology). If $(C,d)$ is in addition a cdga, then $(H(C),0)$ is a cdga as well. These two cdgas are quasi-isomorphic as complexes of vector spaces, but there may be no quasi-isomorphism between them that preserves the multiplicative structure. You can think of this as a derived version of the fact that a given vector space can admit many different algebra structures.
My go-to concrete example of a non-formal dga $A$ appears on page 13 of Hess's Rational Homotopy Theory: A Brief Introduction [https://arxiv.org/pdf/math/0604626.pdf ] as the exterior algebra on three elements $u,v,w$ of degrees $3,3,5$ respectively, and with $d(w)=uv$. You can check that there is no cdga quasi-isomorphism $A to H(A)$.
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
answered Dec 5 '18 at 20:41
Matt BoothMatt Booth
563
563
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
1
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
add a comment |
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
1
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
Thank you for linking the example. Its a lot clearer now. So the quasi-isomorphy must also be an isomorphy of CDGA? Do you know what Griffiths and Morgan mean with "identity"?
– klirk
Dec 5 '18 at 21:29
1
1
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A leftarrow tilde{A} to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism').
– Matt Booth
Dec 5 '18 at 22:09
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027400%2fformality-of-commutative-differential-graded-algebras%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
Two complexes can have isomorphic cohomology but not be quasi-isomorphic.
– JHF
Dec 5 '18 at 19:54
@JHF: So what is the real definition of quasi-isomorphy?
– klirk
Dec 5 '18 at 21:18
@klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3times S^2$ and $RP^2 times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent.
– Aleksandar Milivojevic
Dec 6 '18 at 13:21
Ah, I see. thx.
– klirk
Dec 6 '18 at 20:23