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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeMay 19th 2012
   • (edited May 20th 2012)
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   Author: Mirco Richter
   Format: MarkdownItexRight now I don't understand exactly what an $(\infty,1)$-monad is, but I'm working on it. To give myself a motivating goal I have the following question: Suppose $M_\bullet$ is a quasi category and $T$ an endofunctor that (together with additional data, that I don't know of right now) is a $(\infty,1)$-monad. Are the $T$-algebras strong homotopy ($A_{\infty}$) algebras? If this is true, it would be good to have a reference to get into it... Actually I'm reading in Luries "Derived Algebraic..."

   Right now I don’t understand exactly what an $(\infty,1)$-monad is, but I’m working on it. To give myself a motivating goal I have the following question:

   Suppose $M_\bullet$ is a quasi category and $T$ an endofunctor that (together with additional data, that I don’t know of right now) is a $(\infty,1)$-monad. Are the $T$-algebras strong homotopy ($A_{\infty}$) algebras?

   If this is true, it would be good to have a reference to get into it… Actually I’m reading in Luries “Derived Algebraic…”

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 20th 2012
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   Author: Mike Shulman
   Format: MarkdownItexIt depends on what $T$ is! To paraphrase your question, suppose $M$ is a category and $T$ an endofunctor that is an ordinary monad. Are $T$-algebras monoids? Well, only if $T$ is the free-monoid monad.

   It depends on what $T$ is!

   To paraphrase your question, suppose $M$ is a category and $T$ an endofunctor that is an ordinary monad. Are $T$-algebras monoids? Well, only if $T$ is the free-monoid monad.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMay 20th 2012
   • (edited May 20th 2012)
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   Author: Urs
   Format: MarkdownItexProbably Mirco meant to ask if the algebras are "strong homotopy algebras" in _some_ sense. The answer to that would essentially be "yes". One usually says "strong homotopy" $T$-algebra if $T$ is just a 1-monad / 1-operad and we are regarding it as an $\infty$-monad / $\infty$-operad. For instance we then have [[A-infinity algebra]] but also [[L-infinity algebra]] [[E-infinity algebra]], and so on. More generally, an $\infty$-monad can be more general than an ordinary monad, and so in general an [[infinity-algebra over an (infinity,1)-monad]] is to be thought of as "a strong homotopy algebra over a strong homotopy monad/operad". If you wish.

   Probably Mirco meant to ask if the algebras are “strong homotopy algebras” in some sense. The answer to that would essentially be “yes”.

   One usually says “strong homotopy” $T$-algebra if $T$ is just a 1-monad / 1-operad and we are regarding it as an $\infty$-monad / $\infty$-operad. For instance we then have A-infinity algebra but also L-infinity algebra E-infinity algebra, and so on.

   More generally, an $\infty$-monad can be more general than an ordinary monad, and so in general an infinity-algebra over an (infinity,1)-monad is to be thought of as “a strong homotopy algebra over a strong homotopy monad/operad”. If you wish.

4. • CommentRowNumber4.
   • CommentAuthorzskoda
   • CommentTimeMay 20th 2012
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   Author: zskoda
   Format: MarkdownItexOperads correspond to algebraic/[[finitary monad]]s only. If one takes a monad on sets which is not finitary and looks at it as a monad on a quasicategory and looks at algebras in quasicategory sense, I am not sure if one could make the same $A_\infty$-like description in that case. I mean I am not sure what the $(\infty,1)$-weakening gives when the monad is not algebraic, hence does not reduce to consideration of hierarchies of $n$-operations for various $n$.

   Operads correspond to algebraic/finitary monads only. If one takes a monad on sets which is not finitary and looks at it as a monad on a quasicategory and looks at algebras in quasicategory sense, I am not sure if one could make the same $A_\infty$-like description in that case. I mean I am not sure what the $(\infty,1)$-weakening gives when the monad is not algebraic, hence does not reduce to consideration of hierarchies of $n$-operations for various $n$.