Is every dg-coalgebra the colimit of its finite dimensional dg-subcoalgebras?In which categories is every coalgebra a sum of its finite-dimensional subcoalgebras?“Strøm-type” model structure on chain complexes?Is there a model category structure on non-negatively graded commutative chain algebras?Resolutions by Adapted Class of Objects and Model CategoriesMaps between sets and coalgebrasBounded dg algebra vs unbounded dg algebrasTensor product of coaugmented conilpotent coalgebrasBuilding conilpotent coalgebras from co-square-zero-extensionsLimits in subcategories of Powerset-coalgebrasIs $textDGA^-$ a monoidal model category?
Is every dg-coalgebra the colimit of its finite dimensional dg-subcoalgebras?
In which categories is every coalgebra a sum of its finite-dimensional subcoalgebras?“Strøm-type” model structure on chain complexes?Is there a model category structure on non-negatively graded commutative chain algebras?Resolutions by Adapted Class of Objects and Model CategoriesMaps between sets and coalgebrasBounded dg algebra vs unbounded dg algebrasTensor product of coaugmented conilpotent coalgebrasBuilding conilpotent coalgebras from co-square-zero-extensionsLimits in subcategories of Powerset-coalgebrasIs $textDGA^-$ a monoidal model category?
$begingroup$
I saw this result in A Model Category Structure for Differential Graded Coalgebras by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $mathbbZ$-graded?.
Thanks.
at.algebraic-topology kt.k-theory-and-homology model-categories coalgebras dg-categories
edited Apr 29 at 18:01
David White
13.2k465105
asked Apr 29 at 17:24
Victor TCVictor TC
1064
$endgroup$
add a comment |
$begingroup$
I saw this result in A Model Category Structure for Differential Graded Coalgebras by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $mathbbZ$-graded?.
Thanks.
at.algebraic-topology kt.k-theory-and-homology model-categories coalgebras dg-categories
edited Apr 29 at 18:01
David White
13.2k465105
asked Apr 29 at 17:24
Victor TCVictor TC
1064
$endgroup$
$begingroup$
The result the OP is pointing to in Getzler-Goerss is Corollary 1.6.
$endgroup$
– David White
Apr 29 at 18:02
$begingroup$
@david Indeed, and I also expect to extend this proposition to differential $mathbbZ$-graded coalgebras by using the previous lemmas of the article. I hope an artifice might suffice.
$endgroup$
– Victor TC
2 days ago
add a comment |
$begingroup$
I saw this result in A Model Category Structure for Differential Graded Coalgebras by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $mathbbZ$-graded?.
Thanks.
at.algebraic-topology kt.k-theory-and-homology model-categories coalgebras dg-categories
edited Apr 29 at 18:01
David White
13.2k465105
asked Apr 29 at 17:24
Victor TCVictor TC
1064
$endgroup$
I saw this result in A Model Category Structure for Differential Graded Coalgebras by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $mathbbZ$-graded?.
Thanks.
at.algebraic-topology kt.k-theory-and-homology model-categories coalgebras dg-categories
at.algebraic-topology kt.k-theory-and-homology model-categories coalgebras dg-categories
edited Apr 29 at 18:01
David White
13.2k465105
asked Apr 29 at 17:24
Victor TCVictor TC
1064
edited Apr 29 at 18:01
David White
13.2k465105
asked Apr 29 at 17:24
Victor TCVictor TC
1064
edited Apr 29 at 18:01
David White
13.2k465105
edited Apr 29 at 18:01
David White
13.2k465105
edited Apr 29 at 18:01
David White
13.2k465105
13.2k465105
asked Apr 29 at 17:24
Victor TCVictor TC
1064
asked Apr 29 at 17:24
Victor TCVictor TC
1064
asked Apr 29 at 17:24
Victor TCVictor TC
1064
1064
$begingroup$
The result the OP is pointing to in Getzler-Goerss is Corollary 1.6.
$endgroup$
– David White
Apr 29 at 18:02
$begingroup$
@david Indeed, and I also expect to extend this proposition to differential $mathbbZ$-graded coalgebras by using the previous lemmas of the article. I hope an artifice might suffice.
$endgroup$
– Victor TC
2 days ago
add a comment |
$begingroup$
The result the OP is pointing to in Getzler-Goerss is Corollary 1.6.
$endgroup$
– David White
Apr 29 at 18:02
$begingroup$
@david Indeed, and I also expect to extend this proposition to differential $mathbbZ$-graded coalgebras by using the previous lemmas of the article. I hope an artifice might suffice.
$endgroup$
– Victor TC
2 days ago
$begingroup$
The result the OP is pointing to in Getzler-Goerss is Corollary 1.6.
$endgroup$
– David White
Apr 29 at 18:02
$begingroup$
The result the OP is pointing to in Getzler-Goerss is Corollary 1.6.
$endgroup$
– David White
Apr 29 at 18:02
$begingroup$
@david Indeed, and I also expect to extend this proposition to differential $mathbbZ$-graded coalgebras by using the previous lemmas of the article. I hope an artifice might suffice.
$endgroup$
– Victor TC
2 days ago
$begingroup$
@david Indeed, and I also expect to extend this proposition to differential $mathbbZ$-graded coalgebras by using the previous lemmas of the article. I hope an artifice might suffice.
$endgroup$
– Victor TC
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. The category of $mathbbZ$-coalgebras is locally presentable, and objects are filtered colimits of finite dimensional subobjects. See the appendix to Coalgebraic models for combinatorial model categories by Ching and Riehl. See also Lemma 5.2 of Model Structures for Coalgebras by Drummond-Cole and Hirsh. This paper of Adamek and Porst might also be helpful.
By the way, the main result of the Getzler-Goerss paper you cite is generalized in Corollary 6.3.5 of A necessary and sufficient condition for induced model structures by Hess, Kedziorek, Riehl, and Shipley. It works for any $mathbbZ$-graded coalgebras over any commutative ring $R$.
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
$endgroup$
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
1
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
add a comment |
$begingroup$
For coassociative dg-coalgebras over any field $k$ the answer is positive, because:
Let $C$ be a $mathbb Z$-graded coalgebra and $Dsubset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^grsubset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $Dsubset D^gr$ and $D^gr$ is a finite-dimensional graded subcoalgebra of $C$.
Let $(C,d)$ be a dg-coalgebra and $Dsubset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^dg=D+d(D)subset C$. Then $Dsubset D^dg$ and $D^dg$ is a finite-dimensional dg-subcoalgebra of $C$.
Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.
Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.
EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $Dotimes_k D$ is not a submodule of the tensor product $Cotimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph.
One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=mathfrak g=k[[z]],d/dz$, the Lie algebra of vector fields on the formal disk.
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
$endgroup$
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
1
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
add a comment |
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
);
);
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%2fmathoverflow.net%2fquestions%2f330299%2fis-every-dg-coalgebra-the-colimit-of-its-finite-dimensional-dg-subcoalgebras%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. The category of $mathbbZ$-coalgebras is locally presentable, and objects are filtered colimits of finite dimensional subobjects. See the appendix to Coalgebraic models for combinatorial model categories by Ching and Riehl. See also Lemma 5.2 of Model Structures for Coalgebras by Drummond-Cole and Hirsh. This paper of Adamek and Porst might also be helpful.
By the way, the main result of the Getzler-Goerss paper you cite is generalized in Corollary 6.3.5 of A necessary and sufficient condition for induced model structures by Hess, Kedziorek, Riehl, and Shipley. It works for any $mathbbZ$-graded coalgebras over any commutative ring $R$.
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
$endgroup$
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
1
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
add a comment |
$begingroup$
Yes. The category of $mathbbZ$-coalgebras is locally presentable, and objects are filtered colimits of finite dimensional subobjects. See the appendix to Coalgebraic models for combinatorial model categories by Ching and Riehl. See also Lemma 5.2 of Model Structures for Coalgebras by Drummond-Cole and Hirsh. This paper of Adamek and Porst might also be helpful.
By the way, the main result of the Getzler-Goerss paper you cite is generalized in Corollary 6.3.5 of A necessary and sufficient condition for induced model structures by Hess, Kedziorek, Riehl, and Shipley. It works for any $mathbbZ$-graded coalgebras over any commutative ring $R$.
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
$endgroup$
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
1
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
add a comment |
$begingroup$
Yes. The category of $mathbbZ$-coalgebras is locally presentable, and objects are filtered colimits of finite dimensional subobjects. See the appendix to Coalgebraic models for combinatorial model categories by Ching and Riehl. See also Lemma 5.2 of Model Structures for Coalgebras by Drummond-Cole and Hirsh. This paper of Adamek and Porst might also be helpful.
By the way, the main result of the Getzler-Goerss paper you cite is generalized in Corollary 6.3.5 of A necessary and sufficient condition for induced model structures by Hess, Kedziorek, Riehl, and Shipley. It works for any $mathbbZ$-graded coalgebras over any commutative ring $R$.
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
$endgroup$
Yes. The category of $mathbbZ$-coalgebras is locally presentable, and objects are filtered colimits of finite dimensional subobjects. See the appendix to Coalgebraic models for combinatorial model categories by Ching and Riehl. See also Lemma 5.2 of Model Structures for Coalgebras by Drummond-Cole and Hirsh. This paper of Adamek and Porst might also be helpful.
By the way, the main result of the Getzler-Goerss paper you cite is generalized in Corollary 6.3.5 of A necessary and sufficient condition for induced model structures by Hess, Kedziorek, Riehl, and Shipley. It works for any $mathbbZ$-graded coalgebras over any commutative ring $R$.
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
edited Apr 29 at 18:32
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
answered Apr 29 at 18:00
David WhiteDavid White
13.2k465105
13.2k465105
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
1
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
add a comment |
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
1
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
Thank you very much, I will check the references.
$endgroup$
– Victor TC
Apr 29 at 19:08
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
@VictorTC As an additional remark, this is true more in general for coalgebras over cooperads, see Lemmas 4, 5, and Proposition 12 of the article Homotopy theory of unital algebras by B. Le Grignou.
$endgroup$
– Daniel Robert-Nicoud
Apr 29 at 21:36
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
N.B., we work explicitly with conilpotent coalgebras, and I do not know how to extend our method for proving presentability to the full category of coalgebras. So our paper may not be particularly useful for you.
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:14
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
$begingroup$
@DanielRobert-Nicoud do you know how to resolve the seeming discrepancy between Lemma 5 of Le Grignou and Leonid's counterexample below?
$endgroup$
– Gabriel C. Drummond-Cole
Apr 30 at 0:15
1
1
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
$begingroup$
@Gabriel sorry, I forgot to mention that in Brice's article the coalgebras are supposed to be conilpotent (as can be gathered from the start of the proof of Lemma 5). I suppose Leonid's example is non-conilpotent.
$endgroup$
– Daniel Robert-Nicoud
2 days ago
add a comment |
$begingroup$
For coassociative dg-coalgebras over any field $k$ the answer is positive, because:
Let $C$ be a $mathbb Z$-graded coalgebra and $Dsubset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^grsubset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $Dsubset D^gr$ and $D^gr$ is a finite-dimensional graded subcoalgebra of $C$.
Let $(C,d)$ be a dg-coalgebra and $Dsubset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^dg=D+d(D)subset C$. Then $Dsubset D^dg$ and $D^dg$ is a finite-dimensional dg-subcoalgebra of $C$.
Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.
Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.
EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $Dotimes_k D$ is not a submodule of the tensor product $Cotimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph.
One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=mathfrak g=k[[z]],d/dz$, the Lie algebra of vector fields on the formal disk.
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
$endgroup$
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
1
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
add a comment |
$begingroup$
For coassociative dg-coalgebras over any field $k$ the answer is positive, because:
Let $C$ be a $mathbb Z$-graded coalgebra and $Dsubset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^grsubset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $Dsubset D^gr$ and $D^gr$ is a finite-dimensional graded subcoalgebra of $C$.
Let $(C,d)$ be a dg-coalgebra and $Dsubset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^dg=D+d(D)subset C$. Then $Dsubset D^dg$ and $D^dg$ is a finite-dimensional dg-subcoalgebra of $C$.
Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.
Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.
EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $Dotimes_k D$ is not a submodule of the tensor product $Cotimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph.
One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=mathfrak g=k[[z]],d/dz$, the Lie algebra of vector fields on the formal disk.
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
$endgroup$
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
1
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
add a comment |
$begingroup$
For coassociative dg-coalgebras over any field $k$ the answer is positive, because:
Let $C$ be a $mathbb Z$-graded coalgebra and $Dsubset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^grsubset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $Dsubset D^gr$ and $D^gr$ is a finite-dimensional graded subcoalgebra of $C$.
Let $(C,d)$ be a dg-coalgebra and $Dsubset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^dg=D+d(D)subset C$. Then $Dsubset D^dg$ and $D^dg$ is a finite-dimensional dg-subcoalgebra of $C$.
Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.
Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.
EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $Dotimes_k D$ is not a submodule of the tensor product $Cotimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph.
One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=mathfrak g=k[[z]],d/dz$, the Lie algebra of vector fields on the formal disk.
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
$endgroup$
For coassociative dg-coalgebras over any field $k$ the answer is positive, because:
Let $C$ be a $mathbb Z$-graded coalgebra and $Dsubset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^grsubset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $Dsubset D^gr$ and $D^gr$ is a finite-dimensional graded subcoalgebra of $C$.
Let $(C,d)$ be a dg-coalgebra and $Dsubset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^dg=D+d(D)subset C$. Then $Dsubset D^dg$ and $D^dg$ is a finite-dimensional dg-subcoalgebra of $C$.
Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.
Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.
EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $Dotimes_k D$ is not a submodule of the tensor product $Cotimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph.
One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=mathfrak g=k[[z]],d/dz$, the Lie algebra of vector fields on the formal disk.
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
edited 2 days ago
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
answered Apr 29 at 22:04
Leonid PositselskiLeonid Positselski
11.2k13977
11.2k13977
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
1
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
add a comment |
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
1
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
$begingroup$
Thank you!. Apparently you did not need $C$ to be counital, did you?.
$endgroup$
– Victor TC
2 days ago
1
1
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
No, counitality is not necessary. However, there is a problem with the next-to-last paragraph of my answer. I am now adding an edit.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Concerning counitality, it is of course not needed in 1. and 2., but the key question is extending to noncounital coassociative ungraded coalgebras $C$ the standard result that coassociative coalgebras are unions of their finite-dimensional subcoalgebras. The standard arguments that I know are using counitality, but it can be avoided. Alternatively, you can adjoint a counit to $C$ formally, passing from $C$ to the counital coalgebra $C'=koplus C$, represent $C'$ as the union of its finite-dimensional subcoalgebras $D'$, and conclude that $C=C'/k$ is the union of $D=D'bmod k$.
$endgroup$
– Leonid Positselski
2 days ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
Thank you for the clarification. I apologize for my delay in aswering, I tried to better understand your edit. With this result in hand, is it reasonable to conclude that the category of dg coalgebras has all (small) limits?. I mean, by applying dualization on the finite dimensional dg subcoalgebras and assuming the existence of all small colimits in the category of dg algebras.
$endgroup$
– Victor TC
12 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
$begingroup$
The assertion is correct: all small limits exist in the category of dg-coalgebras. I do not immediately see how this follows from all dg-coalgebras being unions of their finite-dimensional subcoalgebras. Maybe one can deduce the former from the latter by proving that the category of dg-coalgebras is locally presentable (locally finitely presentable, in fact).
$endgroup$
– Leonid Positselski
8 hours ago
add a comment |
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.
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%2fmathoverflow.net%2fquestions%2f330299%2fis-every-dg-coalgebra-the-colimit-of-its-finite-dimensional-dg-subcoalgebras%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
$begingroup$
The result the OP is pointing to in Getzler-Goerss is Corollary 1.6.
$endgroup$
– David White
Apr 29 at 18:02
$begingroup$
@david Indeed, and I also expect to extend this proposition to differential $mathbbZ$-graded coalgebras by using the previous lemmas of the article. I hope an artifice might suffice.
$endgroup$
– Victor TC
2 days ago