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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeMay 14th 2013
   • (edited May 14th 2013)
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   Author: Mirco Richter
   Format: MarkdownItexComing across this: [nCatCafe](https://golem.ph.utexas.edu/category/2007/02/higher_morphisms_of_lie_nalgeb.html) I'm rater concerned now if my understanding of oo-morphisms between Lie oo-algebras is correct: Suppose we have two Lie oo-algebras $L$ and $M$. Then MY PREVIOUS understanding was that an oo-morphism is a sequence of maps $f_i:L^{\otimes i} \to M$ antisymmetric and homogeneous of degree $(i-1)$, satisfying some commutation laws with the higher brackets of $L$ and $M$. In the talk I mentioned above, this was Jims POV back then. As he thought, a $k$-morphism is just a single element $f_k$ of that sequence. But if I understand it right, then what John answered is, that this SEQUENCE is in fact just a 1-morphism from the POV of (oo,1)-categories and that a 2-morphism is a homotopy between two such sequences (Whatever that means). Is that how to see it? === This is important to understand te paper of Urs, Chris and Domenico

   Coming across this:

   nCatCafe

   I’m rater concerned now if my understanding of oo-morphisms between Lie oo-algebras is correct:

   Suppose we have two Lie oo-algebras $L$ and $M$. Then MY PREVIOUS understanding was that an oo-morphism is a sequence of maps

   $f_i:L^{\otimes i} \to M$ antisymmetric and homogeneous of degree $(i-1)$, satisfying some commutation laws with the higher brackets of $L$ and $M$.

   In the talk I mentioned above, this was Jims POV back then. As he thought, a $k$-morphism is just a single element $f_k$ of that sequence.

   But if I understand it right, then what John answered is, that this SEQUENCE is in fact just a 1-morphism from the POV of (oo,1)-categories and that a 2-morphism is a homotopy between two such sequences (Whatever that means).

   Is that how to see it?

   ===

   This is important to understand te paper of Urs, Chris and Domenico

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 14th 2013
   • (edited May 14th 2013)
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   Author: Urs
   Format: MarkdownItexYou seem to have a terminology mixup on the meaning of the "$n$-" prefix here for $n = \infty$. People speak of "$L_\infty$-morphisms" (and maybe rarely say just "$\infty$-morphisms" for these for short) to just mean the evident thing, which you also mean: a homomorphism of $L_\infty$-algebras. So its a [[1-morphism]] of $L_\infty$-algebras, yes. Now $L_\infty$-algebras have their own [[homotopy theory]] (presented for instance by any of the [[model structures for L-infinity algebras]]) and with respect to this there is a notion of [[homotopy]] between two such 1-morphisms of $L_\infty$-algebras. These are the [[2-morphisms]] in the [[(infinity,1)-category]] of $L_\infty$-algebras, yes.

   You seem to have a terminology mixup on the meaning of the “$n$-” prefix here for $n = \infty$.

   People speak of “$L_\infty$-morphisms” (and maybe rarely say just “$\infty$-morphisms” for these for short) to just mean the evident thing, which you also mean: a homomorphism of $L_\infty$-algebras.

   So its a 1-morphism of $L_\infty$-algebras, yes. Now $L_\infty$-algebras have their own homotopy theory (presented for instance by any of the model structures for L-infinity algebras) and with respect to this there is a notion of homotopy between two such 1-morphisms of $L_\infty$-algebras. These are the 2-morphisms in the (infinity,1)-category of $L_\infty$-algebras, yes.

3. • CommentRowNumber3.
   • CommentAuthorMirco Richter
   • CommentTimeMay 15th 2013
   • (edited May 15th 2013)
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   Author: Mirco Richter
   Format: MarkdownItexA 1-morphism is a homomorphism of $L_\infty$-algebras. Ok. But for example in Loday&Valettes book on operads they define $\infty$-morphisms as that sequence of maps I mentioned above and that Jim thought of too. So what bugs me is, if the $f_2: L \otimes L \to M$ map from "Lodays-sequence" is the same as the $2$-morphism you get in the model structure for $L_\infty$-algebras. You know what I mean? I'm searching for the conection of Lodays $\infty$-morphisms (Its prop 10.2.13 in their book) and thoses frome the $(\infty,1)$-category of $L_\infty$-algebras.

   A 1-morphism is a homomorphism of $L_\infty$-algebras. Ok.

   But for example in Loday&Valettes book on operads they define $\infty$-morphisms as that sequence of maps I mentioned above and that Jim thought of too. So what bugs me is, if the

   $f_2: L \otimes L \to M$

   map from “Lodays-sequence” is the same as the $2$-morphism you get in the model structure for $L_\infty$-algebras.

   You know what I mean? I’m searching for the conection of Lodays $\infty$-morphisms (Its prop 10.2.13 in their book) and thoses frome the $(\infty,1)$-category of $L_\infty$-algebras.

4. • CommentRowNumber4.
   • CommentAuthorMirco Richter
   • CommentTimeMay 15th 2013
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   Author: Mirco Richter
   Format: MarkdownItexin at least some model structure on $L_\infty$-algebras, of course.

   in at least some model structure on $L_\infty$-algebras, of course.

5. • CommentRowNumber5.
   • CommentAuthorMirco Richter
   • CommentTimeMay 15th 2013
   • (edited May 15th 2013)
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   Author: Mirco Richter
   Format: MarkdownItexOk, I think the language mismatch is that in Urs, Chris and Domenico's paper what is called $L_\infty$-algebra morphism is the same as what is called $\infty$-morphism in Loday&Valette. Compare (Example 3.15 in the first paper with prop. 10.2.13 in the second) So in fact my first guess was right, a $2$-morphism then can be seen as a homotopy between Lodays $\infty$-morpisms and the latter are called just 'morphisms' now (Because restricting to the first element of the sequence is artificial, I guess) Halleluja! That sounds like some complicated structure, when it comes to details...

   Ok, I think the language mismatch is that in Urs, Chris and Domenico’s paper what is called $L_\infty$-algebra morphism is the same as what is called $\infty$-morphism in Loday&Valette. Compare (Example 3.15 in the first paper with prop. 10.2.13 in the second)

   So in fact my first guess was right, a $2$-morphism then can be seen as a homotopy between Lodays $\infty$-morpisms and the latter are called just ’morphisms’ now (Because restricting to the first element of the sequence is artificial, I guess)

   Halleluja! That sounds like some complicated structure, when it comes to details…

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeMay 15th 2013
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   Author: Urs
   Format: MarkdownItex> a 2-morphism then can be seen as a homotopy between Lodays ∞-morpisms and the latter are called just ’morphisms’ Yes. That's what I said above. The reason to have a term like "$\infty$-morphism" -- what Jim Stasheff originally called "sh-maps" ("strong homotopy maps") of $L_\infty$-algebras -- is that in one of the various equivalent definitions of $L_\infty$-algebras, namley in the operadic definition (and as opposed to the coalgebraic definition), the "obvious" morphisms (the morphisms of algebras over an operad) are more restrictive in that they correspond to just "strict" maps that respect all the Lie structure on the nose and not up to homotopy.

   a 2-morphism then can be seen as a homotopy between Lodays ∞-morpisms and the latter are called just ’morphisms’

   Yes. That’s what I said above.

   The reason to have a term like “$\infty$-morphism” – what Jim Stasheff originally called “sh-maps” (“strong homotopy maps”) of $L_\infty$-algebras – is that in one of the various equivalent definitions of $L_\infty$-algebras, namley in the operadic definition (and as opposed to the coalgebraic definition), the “obvious” morphisms (the morphisms of algebras over an operad) are more restrictive in that they correspond to just “strict” maps that respect all the Lie structure on the nose and not up to homotopy.

7. • CommentRowNumber7.
   • CommentAuthorMirco Richter
   • CommentTimeMay 17th 2013
   • (edited May 17th 2013)
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   Author: Mirco Richter
   Format: MarkdownItexBack then Urs mentioned, that no one was able to write down what exactly an $k$-morphism looks like in the $(\infty,1)$-category of $L_\infty$-algebras. Is that still so?

   Back then Urs mentioned, that no one was able to write down what exactly an $k$-morphism looks like in the $(\infty,1)$-category of $L_\infty$-algebras. Is that still so?

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeMay 17th 2013
   • (edited May 17th 2013)
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   Author: Urs
   Format: MarkdownItexI wrote that entry that you are looking at years before I heard about any homotopy theory of these beasts. You shouldn't really be looking at that. I should delete it. But it is true that the homotopical structures for $L_\infty$-algebras are a bit more subtle than one would hope. Notably a good description of $L_\infty$-algebras (with the correct sh-morphisms = "$\infty$-morphisms" between them) as the full subcategory of fibrant objects in a model category was given only fairly recently by Jonathan Pridham, in [this proposition](http://ncatlab.org/nlab/show/model%20structure%20for%20L-infinity%20algebras#LInfinityAlgebraIsFibrantObjectIndgFormalSpace) at _[[model structure for L-infinity algebras]]_. But once we have with this theorem the structure of a [[category of fibrant objects]] on the category of $L_\infty$-algebras, one can use the construction of the [derived hom-space in a category of fibrant objects](http://ncatlab.org/nlab/show/%28infinity%2C1%29-categorical+hom-space#InACategoryOfFibrantObjects) which we describe in [section 3.6.2 of arXiv:1207.0249](http://arxiv.org/abs/1207.0249) to get explicit formulas for the higher morphisms of $L_\infty$-algebras.

   I wrote that entry that you are looking at years before I heard about any homotopy theory of these beasts. You shouldn’t really be looking at that. I should delete it.

   But it is true that the homotopical structures for $L_\infty$-algebras are a bit more subtle than one would hope. Notably a good description of $L_\infty$-algebras (with the correct sh-morphisms = “$\infty$-morphisms” between them) as the full subcategory of fibrant objects in a model category was given only fairly recently by Jonathan Pridham, in this proposition at model structure for L-infinity algebras.

   But once we have with this theorem the structure of a category of fibrant objects on the category of $L_\infty$-algebras, one can use the construction of the derived hom-space in a category of fibrant objects which we describe in section 3.6.2 of arXiv:1207.0249 to get explicit formulas for the higher morphisms of $L_\infty$-algebras.

9. • CommentRowNumber9.
   • CommentAuthorMirco Richter
   • CommentTimeMay 17th 2013
   • (edited May 17th 2013)
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   Author: Mirco Richter
   Format: MarkdownItexNo, don't delete that. Just add a comment, that this is seen different now. If you delete it its just gone, but you can't rule out, that there are people who have better access to these higher things, if they follow the path from its beginning. So maybe its better to just 'finalize' this old talk with what you wrote here just above. And what is maybe more important is the psychological effect of those old talks: Showing followers, that you people once had pretty much the same struggles...

   No, don’t delete that. Just add a comment, that this is seen different now. If you delete it its just gone, but you can’t rule out, that there are people who have better access to these higher things, if they follow the path from its beginning. So maybe its better to just ’finalize’ this old talk with what you wrote here just above.

   And what is maybe more important is the psychological effect of those old talks: Showing followers, that you people once had pretty much the same struggles…

10. • CommentRowNumber10.
    • CommentAuthorcrogers
    • CommentTimeMay 17th 2013
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    Author: crogers
    Format: TextMaybe it's worth mentioning, just in case Mirco didn't know, that it's easy to construct good path objects for Loo-algebras (as objects which are both fibrant and cofibrant in Hinich's model structure for connected dg co-commutative coalgebras). This leads to a nice explicit description for homotopies between 1-morphisms, which has been known for awhile, I guess. See, for example, section 5 of http://arxiv.org/abs/math/0703113.

    Maybe it's worth mentioning, just in case Mirco didn't know, that it's easy to construct good path objects for Loo-algebras (as objects which are both fibrant and cofibrant in Hinich's model structure for connected dg co-commutative coalgebras). This leads to a nice explicit description for homotopies between 1-morphisms, which has been known for awhile, I guess. See, for example, section 5 of http://arxiv.org/abs/math/0703113.
11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeMay 17th 2013
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    Author: Urs
    Format: MarkdownItexnow [here](http://ncatlab.org/nlab/show/model+structure+for+L-infinity+algebras#HomotopiesAndDerivedHomSpaces)

    now here

12. • CommentRowNumber12.
    • CommentAuthorMirco Richter
    • CommentTimeMay 17th 2013
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    Author: Mirco Richter
    Format: MarkdownItexOk. Thanks. I'll have a look.

    Ok. Thanks. I’ll have a look.

13. • CommentRowNumber13.
    • CommentAuthorjim_stasheff
    • CommentTimeMay 17th 2013
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    Author: jim_stasheff
    Format: Text@9 Mirco ``And what is maybe more important is the psychological effect of those old talks: Showing followers, that you people once had pretty much the same struggles…'' Amen! some times the scaffolding is more revealing than the polished final version

    @9 Mirco ``And what is maybe more important is the psychological effect of those old talks: Showing followers, that you people once had pretty much the same struggles…''
    
    Amen! some times the scaffolding is more revealing than the polished final version
14. • CommentRowNumber14.
    • CommentAuthorMirco Richter
    • CommentTimeMay 22nd 2013
    • (edited May 22nd 2013)
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    Author: Mirco Richter
    Format: MarkdownItexOk. Now I have more or less an idea how to derive the $(\infty,1)$-morphisms by [derived hom-space in a category of fibrant objects](http://ncatlab.org/nlab/show/%28infinity%2C1%29-categorical+hom-space#InACategoryOfFibrantObjects) Urs mentioned. Am I right, when I assume, that this hasn't been done, yet? The way I see it, is that Urs comment above is a way to reach the goal of defining $2$-morphisms between $L_\infty$-morphisms etcetera, but thats "only" the equation, not a solution. So being lazy, it would be great if someone else already walked that way and came up with formulars for $k$-morpisms in the $(\infty,1)$-category of $L_\infty$-algebrs.

    Ok. Now I have more or less an idea how to derive the $(\infty,1)$-morphisms by derived hom-space in a category of fibrant objects Urs mentioned.

    Am I right, when I assume, that this hasn’t been done, yet? The way I see it, is that Urs comment above is a way to reach the goal of defining $2$-morphisms between $L_\infty$-morphisms etcetera, but thats “only” the equation, not a solution. So being lazy, it would be great if someone else already walked that way and came up with formulars for $k$-morpisms in the $(\infty,1)$-category of $L_\infty$-algebrs.

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2013
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    Author: Urs
    Format: MarkdownItexSometimes finding some formula from some definition is trivial to some person and is an exercise to some other person. Here you might need to tell me in more detail what it is that you did before I know if it "hadn't been done yet".

    Sometimes finding some formula from some definition is trivial to some person and is an exercise to some other person.

    Here you might need to tell me in more detail what it is that you did before I know if it “hadn’t been done yet”.

16. • CommentRowNumber16.
    • CommentAuthorMirco Richter
    • CommentTimeMay 22nd 2013
    • (edited May 22nd 2013)
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    Author: Mirco Richter
    Format: MarkdownItex> Here you might need to tell me in more detail what it is that you did before I know if it “hadn’t been done yet”. What I did ? Don't understand. I'm looking for formulas for the $k$-morpisms in the $(\infty,1)$-category of $L_\infty$-algebras. Like in the simplicial enriched presentation: for any two $L_\infty$-algebras $L$ and $M$ a $2$-morpism $f_2:\Delta[1]\in Hom_{(\infty,1)-L_\infty}(L,M)$ between the morphisms $d_0(f_2)$ and $d_1(f_2)$ is ... (something) And similar in any other presentation. Don't know how to reformulate this queation, because to me it is pretty clear: How can we present the $k$-morphisms of the $(\infty,1)$-category of $L_\infty$-algebra.

    Here you might need to tell me in more detail what it is that you did before I know if it “hadn’t been done yet”.

    What I did ? Don’t understand. I’m looking for formulas for the $k$-morpisms in the $(\infty,1)$-category of $L_\infty$-algebras. Like in the simplicial enriched presentation: for any two $L_\infty$-algebras $L$ and $M$ a $2$-morpism $f_2:\Delta[1]\in Hom_{(\infty,1)-L_\infty}(L,M)$ between the morphisms $d_0(f_2)$ and $d_1(f_2)$ is … (something)

    And similar in any other presentation.

    Don’t know how to reformulate this queation, because to me it is pretty clear: How can we present the $k$-morphisms of the $(\infty,1)$-category of $L_\infty$-algebra.

17. • CommentRowNumber17.
    • CommentAuthorMirco Richter
    • CommentTimeMay 22nd 2013
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    Author: Mirco Richter
    Format: MarkdownItexOr do you want to know what I'm actually working on, to give me a presentation of the $(\infty,1)$-category of $L_\infty$-algebras that you tink is best suited in my situation?

    Or do you want to know what I’m actually working on, to give me a presentation of the $(\infty,1)$-category of $L_\infty$-algebras that you tink is best suited in my situation?

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeMay 22nd 2013
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    Author: Urs
    Format: MarkdownItexI had maybe misunderstood you in #14 as saying that you had worked out some formulas for yourself and were wondering if it had been done before. If not, then I am not sure what you are asking.

    I had maybe misunderstood you in #14 as saying that you had worked out some formulas for yourself and were wondering if it had been done before.

    If not, then I am not sure what you are asking.

19. • CommentRowNumber19.
    • CommentAuthorMirco Richter
    • CommentTimeMay 28th 2013
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    Author: Mirco Richter
    Format: MarkdownItexMaybe we come back to that later. ... Previously, in understanding $L_\infty$-morphisms the question arise which coalgebra to look at. Seems like here are two different choice that appear in the literature: 1.) The reduced, cocommutative, associative coalgebra, which, as a vector space, equals the graded symmetric tensor algebra $S(V)$ but without the "degree zero part". If I don't mix things up here, this is not a cofree coalgebra. 2.) The reduced cofree, cocommutative, associative coalgebra on the underlying graded vector space of the $L_\infty$-algebra. As far as I know this is more complicated and is build on a subvector space of the completetion $\hat{T}V$ of the (graded) tensor algebra $TV$. The vector space of formal power series, i.e a product, not a direct sum. Seems just like a subtle difference, but in attempt to introduce $L_\infty$-algebra morphisms in a clean way, the question arises which of them to take.

    Maybe we come back to that later. …

    Previously, in understanding $L_\infty$-morphisms the question arise which coalgebra to look at. Seems like here are two different choice that appear in the literature:

    1.) The reduced, cocommutative, associative coalgebra, which, as a vector space, equals the graded symmetric tensor algebra $S(V)$ but without the “degree zero part”. If I don’t mix things up here, this is not a cofree coalgebra.

    2.) The reduced cofree, cocommutative, associative coalgebra on the underlying graded vector space of the $L_\infty$-algebra. As far as I know this is more complicated and is build on a subvector space of the completetion $\hat{T}V$ of the (graded) tensor algebra $TV$. The vector space of formal power series, i.e a product, not a direct sum.

    Seems just like a subtle difference, but in attempt to introduce $L_\infty$-algebra morphisms in a clean way, the question arises which of them to take.

20. • CommentRowNumber20.
    • CommentAuthorjim_stasheff
    • CommentTimeMay 29th 2013
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    Author: jim_stasheff
    Format: Text@mirco: this is not a cofree coalgebra. not if you want it to be cocommutative but perhaps that was left out inadvertently for issues of cofree coalgebra versus tensor coalgebras I recommend papers of Walter Michaelis he's very careful and precise

    @mirco: this is not a cofree coalgebra.
    
    not if you want it to be cocommutative but perhaps that was left out inadvertently
    
    for issues of cofree coalgebra versus tensor coalgebras
    I recommend papers of Walter Michaelis
    he's very careful and precise
21. • CommentRowNumber21.
    • CommentAuthorMirco Richter
    • CommentTimeJun 1st 2013
    • (edited Jun 1st 2013)
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    Author: Mirco Richter
    Format: MarkdownItex@jim: I'm not sure if I get your answer right. You say that $\bar{S(V)}$ , i.e the reduced graded co(-commutative, -associative, -algebra) IS cofree? I don't think so. At least not in the category of coalgebras. I would say that $\bar{S(V)}$ is cofree in the (sub)category of locally nilpotent coalgebras. And in the context of $L_\infty$-algebras that all we need. Not?

    @jim: I’m not sure if I get your answer right. You say that $\bar{S(V)}$ , i.e the reduced graded co(-commutative, -associative, -algebra) IS cofree?

    I don’t think so. At least not in the category of coalgebras. I would say that $\bar{S(V)}$ is cofree in the (sub)category of locally nilpotent coalgebras. And in the context of $L_\infty$-algebras that all we need. Not?

22. • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 2nd 2013
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    Author: Todd_Trimble
    Format: MarkdownItexIt might help to point to a specific statement of Michaelis regarding this issue (Jim has often mentioned his Handbook of Algebra article). A precise statement and discussion could then be incorporated within the nLab.

    It might help to point to a specific statement of Michaelis regarding this issue (Jim has often mentioned his Handbook of Algebra article). A precise statement and discussion could then be incorporated within the nLab.

23. • CommentRowNumber23.
    • CommentAuthorMirco Richter
    • CommentTimeJun 3rd 2013
    • (edited Jun 3rd 2013)
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    Author: Mirco Richter
    Format: MarkdownItexIn Jonathan Pridhams paper,that Urs referenced indirectly to in #8, Jon. pointed to that, too (Remark 3.12 in my version). At least he mentioned, that the coalgebra used usually in the definition of $L_\I$-algebras is not the cofree coalgebra. I think we should change that. I would like to, (infact I would like to rewrite allot in the nLab entry on $L_\I$-algebras) but all previous attempts have been undone, so I don't try anymore.

    In Jonathan Pridhams paper,that Urs referenced indirectly to in #8, Jon. pointed to that, too (Remark 3.12 in my version). At least he mentioned, that the coalgebra used usually in the definition of $L_\I$-algebras is not the cofree coalgebra.

    I think we should change that. I would like to, (infact I would like to rewrite allot in the nLab entry on $L_\I$-algebras) but all previous attempts have been undone, so I don’t try anymore.

24. • CommentRowNumber24.
    • CommentAuthorMirco Richter
    • CommentTimeJun 3rd 2013
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    Author: Mirco Richter
    Format: MarkdownItex@Urs: Is the definition of a homotopy between $L_\infty$-morphisms in the sense of Pridham written down somewhere IN DETAIL? I mean in all its components like (hypotetically): A homotpoy between two $L_\infty$-morphisms is a sequence of maps of skew symmetric degree $n$ (symmetric degree +1) such that some complicated equations involving both morphisms are satisfied.

    @Urs: Is the definition of a homotopy between $L_\infty$-morphisms in the sense of Pridham written down somewhere IN DETAIL? I mean in all its components like (hypotetically):

    A homotpoy between two $L_\infty$-morphisms is a sequence of maps of skew symmetric degree $n$ (symmetric degree +1) such that some complicated equations involving both morphisms are satisfied.

25. • CommentRowNumber25.
    • CommentAuthorUrs
    • CommentTimeAug 12th 2013
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    Author: Urs
    Format: MarkdownItexthe question also came up on MO [here](http://mathoverflow.net/questions/139175/what-is-a-homotopy-between-l-infty-algebra-morphisms)

    the question also came up on MO here