Free operad over a monoid object The 2019 Stack Overflow Developer Survey Results Are InAn interpretation of this construction giving an operad from a bialgebra?What is the free monoidal category generated by a monoid?Unitalization internal to monoidal categoriesCorrespondence between operads and $infty$-operads with one objectCorrespondence between operads and monads requires tensor distribute over coproduct?understanding the definition of $infty$-operad of module objectsReference for “multi-monoidal categories”When is a quasicategory over $N(Delta)^op$ a planar $infty$-operad?An interpretation of this construction giving an operad from a bialgebra?$H$-space structure on coloured algebrasTwo monoidal structures and copowering

Free operad over a monoid object



The 2019 Stack Overflow Developer Survey Results Are InAn interpretation of this construction giving an operad from a bialgebra?What is the free monoidal category generated by a monoid?Unitalization internal to monoidal categoriesCorrespondence between operads and $infty$-operads with one objectCorrespondence between operads and monads requires tensor distribute over coproduct?understanding the definition of $infty$-operad of module objectsReference for “multi-monoidal categories”When is a quasicategory over $N(Delta)^op$ a planar $infty$-operad?An interpretation of this construction giving an operad from a bialgebra?$H$-space structure on coloured algebrasTwo monoidal structures and copowering










4












$begingroup$


Let $mathcalO$ be an operad in the monoidal category $M$. Then $mathcalO(1)$ together with the morphisms
$$mathcalO(1)otimes mathcalO(1)to mathcalO(1)$$
and the unit $eta:1to mathcalO(1)$ is a monoid object. Moreover, a morphism $varphi:mathcalOto mathcalO'$ of operads induces a morphism $mathcalO(1)to mathcalO(1)$ of monoid objects. Therefore, we get a forgetful functor
$$mathrmOperads(M)to mathrmMonoids(M).$$
If we work with coloured operads, we get a functor from coloured $M$-operads to $M$-enriched categories. Conversely, if $T$ is a monoid object, we can build an operad by
$$mathcalO_T(r) := T^otimes r.$$
and the structure maps
$$T^otimes rotimes bigotimes_i=1^r T^otimes k_ito T^otimes (k_1+dotsb+k_r)$$
as follows: Let $Delta:Tto T^otimes k$ be the diagonal (existence is clear if $otimes$ is the categorical product). Then
$$Totimes T^otimes kstackrelDeltaotimes mathrmidtoT^otimes kotimes T^otimes k cong (T^otimes 2)^otimes k to T^otimes k.$$
In $mathbfSet$, this just means $t(t_1,dotsc,t_k)=(tt_1,dotsc,tt_k)$.
It should be clear that this construction gives us an operad. Now two problems/questions:



  1. Does the morphism $Delta$ always exist in the general setting? It is obviously not the same as
    $$Tcong Totimes 1^otimes (k-1)stackrelmathrmidotimes eta^otimes (k-1)to T^otimes k.$$

  2. Obviously $mathcalO_T(1)cong T$, but it seems to be not true that $mathcalO_T$ is the free operad over the monoid object $T$. Is there another construction for the “free” operad over $T$?









share|cite|improve this question









$endgroup$







  • 5




    $begingroup$
    The Free operads over $T$ is alot simpler than this: it has $mathcalO(1)=T$ and all the $mathcalO(n)$ for $n>1$ are the unit.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 2




    $begingroup$
    @SimonHenry Should $mathcalO(n)$ be initial object of the category, actually?
    $endgroup$
    – Najib Idrissi
    1 hour ago







  • 2




    $begingroup$
    Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 1




    $begingroup$
    @SimonHenry That's true, thanks.
    $endgroup$
    – Najib Idrissi
    1 hour ago















4












$begingroup$


Let $mathcalO$ be an operad in the monoidal category $M$. Then $mathcalO(1)$ together with the morphisms
$$mathcalO(1)otimes mathcalO(1)to mathcalO(1)$$
and the unit $eta:1to mathcalO(1)$ is a monoid object. Moreover, a morphism $varphi:mathcalOto mathcalO'$ of operads induces a morphism $mathcalO(1)to mathcalO(1)$ of monoid objects. Therefore, we get a forgetful functor
$$mathrmOperads(M)to mathrmMonoids(M).$$
If we work with coloured operads, we get a functor from coloured $M$-operads to $M$-enriched categories. Conversely, if $T$ is a monoid object, we can build an operad by
$$mathcalO_T(r) := T^otimes r.$$
and the structure maps
$$T^otimes rotimes bigotimes_i=1^r T^otimes k_ito T^otimes (k_1+dotsb+k_r)$$
as follows: Let $Delta:Tto T^otimes k$ be the diagonal (existence is clear if $otimes$ is the categorical product). Then
$$Totimes T^otimes kstackrelDeltaotimes mathrmidtoT^otimes kotimes T^otimes k cong (T^otimes 2)^otimes k to T^otimes k.$$
In $mathbfSet$, this just means $t(t_1,dotsc,t_k)=(tt_1,dotsc,tt_k)$.
It should be clear that this construction gives us an operad. Now two problems/questions:



  1. Does the morphism $Delta$ always exist in the general setting? It is obviously not the same as
    $$Tcong Totimes 1^otimes (k-1)stackrelmathrmidotimes eta^otimes (k-1)to T^otimes k.$$

  2. Obviously $mathcalO_T(1)cong T$, but it seems to be not true that $mathcalO_T$ is the free operad over the monoid object $T$. Is there another construction for the “free” operad over $T$?









share|cite|improve this question









$endgroup$







  • 5




    $begingroup$
    The Free operads over $T$ is alot simpler than this: it has $mathcalO(1)=T$ and all the $mathcalO(n)$ for $n>1$ are the unit.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 2




    $begingroup$
    @SimonHenry Should $mathcalO(n)$ be initial object of the category, actually?
    $endgroup$
    – Najib Idrissi
    1 hour ago







  • 2




    $begingroup$
    Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 1




    $begingroup$
    @SimonHenry That's true, thanks.
    $endgroup$
    – Najib Idrissi
    1 hour ago













4












4








4





$begingroup$


Let $mathcalO$ be an operad in the monoidal category $M$. Then $mathcalO(1)$ together with the morphisms
$$mathcalO(1)otimes mathcalO(1)to mathcalO(1)$$
and the unit $eta:1to mathcalO(1)$ is a monoid object. Moreover, a morphism $varphi:mathcalOto mathcalO'$ of operads induces a morphism $mathcalO(1)to mathcalO(1)$ of monoid objects. Therefore, we get a forgetful functor
$$mathrmOperads(M)to mathrmMonoids(M).$$
If we work with coloured operads, we get a functor from coloured $M$-operads to $M$-enriched categories. Conversely, if $T$ is a monoid object, we can build an operad by
$$mathcalO_T(r) := T^otimes r.$$
and the structure maps
$$T^otimes rotimes bigotimes_i=1^r T^otimes k_ito T^otimes (k_1+dotsb+k_r)$$
as follows: Let $Delta:Tto T^otimes k$ be the diagonal (existence is clear if $otimes$ is the categorical product). Then
$$Totimes T^otimes kstackrelDeltaotimes mathrmidtoT^otimes kotimes T^otimes k cong (T^otimes 2)^otimes k to T^otimes k.$$
In $mathbfSet$, this just means $t(t_1,dotsc,t_k)=(tt_1,dotsc,tt_k)$.
It should be clear that this construction gives us an operad. Now two problems/questions:



  1. Does the morphism $Delta$ always exist in the general setting? It is obviously not the same as
    $$Tcong Totimes 1^otimes (k-1)stackrelmathrmidotimes eta^otimes (k-1)to T^otimes k.$$

  2. Obviously $mathcalO_T(1)cong T$, but it seems to be not true that $mathcalO_T$ is the free operad over the monoid object $T$. Is there another construction for the “free” operad over $T$?









share|cite|improve this question









$endgroup$




Let $mathcalO$ be an operad in the monoidal category $M$. Then $mathcalO(1)$ together with the morphisms
$$mathcalO(1)otimes mathcalO(1)to mathcalO(1)$$
and the unit $eta:1to mathcalO(1)$ is a monoid object. Moreover, a morphism $varphi:mathcalOto mathcalO'$ of operads induces a morphism $mathcalO(1)to mathcalO(1)$ of monoid objects. Therefore, we get a forgetful functor
$$mathrmOperads(M)to mathrmMonoids(M).$$
If we work with coloured operads, we get a functor from coloured $M$-operads to $M$-enriched categories. Conversely, if $T$ is a monoid object, we can build an operad by
$$mathcalO_T(r) := T^otimes r.$$
and the structure maps
$$T^otimes rotimes bigotimes_i=1^r T^otimes k_ito T^otimes (k_1+dotsb+k_r)$$
as follows: Let $Delta:Tto T^otimes k$ be the diagonal (existence is clear if $otimes$ is the categorical product). Then
$$Totimes T^otimes kstackrelDeltaotimes mathrmidtoT^otimes kotimes T^otimes k cong (T^otimes 2)^otimes k to T^otimes k.$$
In $mathbfSet$, this just means $t(t_1,dotsc,t_k)=(tt_1,dotsc,tt_k)$.
It should be clear that this construction gives us an operad. Now two problems/questions:



  1. Does the morphism $Delta$ always exist in the general setting? It is obviously not the same as
    $$Tcong Totimes 1^otimes (k-1)stackrelmathrmidotimes eta^otimes (k-1)to T^otimes k.$$

  2. Obviously $mathcalO_T(1)cong T$, but it seems to be not true that $mathcalO_T$ is the free operad over the monoid object $T$. Is there another construction for the “free” operad over $T$?






ct.category-theory monoidal-categories operads






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









FKranholdFKranhold

3577




3577







  • 5




    $begingroup$
    The Free operads over $T$ is alot simpler than this: it has $mathcalO(1)=T$ and all the $mathcalO(n)$ for $n>1$ are the unit.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 2




    $begingroup$
    @SimonHenry Should $mathcalO(n)$ be initial object of the category, actually?
    $endgroup$
    – Najib Idrissi
    1 hour ago







  • 2




    $begingroup$
    Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 1




    $begingroup$
    @SimonHenry That's true, thanks.
    $endgroup$
    – Najib Idrissi
    1 hour ago












  • 5




    $begingroup$
    The Free operads over $T$ is alot simpler than this: it has $mathcalO(1)=T$ and all the $mathcalO(n)$ for $n>1$ are the unit.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 2




    $begingroup$
    @SimonHenry Should $mathcalO(n)$ be initial object of the category, actually?
    $endgroup$
    – Najib Idrissi
    1 hour ago







  • 2




    $begingroup$
    Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise.
    $endgroup$
    – Simon Henry
    1 hour ago







  • 1




    $begingroup$
    @SimonHenry That's true, thanks.
    $endgroup$
    – Najib Idrissi
    1 hour ago







5




5




$begingroup$
The Free operads over $T$ is alot simpler than this: it has $mathcalO(1)=T$ and all the $mathcalO(n)$ for $n>1$ are the unit.
$endgroup$
– Simon Henry
1 hour ago





$begingroup$
The Free operads over $T$ is alot simpler than this: it has $mathcalO(1)=T$ and all the $mathcalO(n)$ for $n>1$ are the unit.
$endgroup$
– Simon Henry
1 hour ago





2




2




$begingroup$
@SimonHenry Should $mathcalO(n)$ be initial object of the category, actually?
$endgroup$
– Najib Idrissi
1 hour ago





$begingroup$
@SimonHenry Should $mathcalO(n)$ be initial object of the category, actually?
$endgroup$
– Najib Idrissi
1 hour ago





2




2




$begingroup$
Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise.
$endgroup$
– Simon Henry
1 hour ago





$begingroup$
Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise.
$endgroup$
– Simon Henry
1 hour ago





1




1




$begingroup$
@SimonHenry That's true, thanks.
$endgroup$
– Najib Idrissi
1 hour ago




$begingroup$
@SimonHenry That's true, thanks.
$endgroup$
– Najib Idrissi
1 hour ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $mathttCom rtimes T$ of the commutative operad $mathttCom$ by $T$.



  1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $mu : T otimes T to T$ and a comultiplication $Delta : T to T otimes T$ such that $mu$ is associative and unital, $Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $Delta circ mu = mu circ (Delta otimes Delta)$.


  2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $varnothing$ be the initial object of your category and suppose that $varnothing otimes X = varnothing = X otimes varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $mathttO(1) = T$ and $mathttO(n) = varnothing$ for $n neq 1$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
    $endgroup$
    – FKranhold
    29 mins ago












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









6












$begingroup$

Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $mathttCom rtimes T$ of the commutative operad $mathttCom$ by $T$.



  1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $mu : T otimes T to T$ and a comultiplication $Delta : T to T otimes T$ such that $mu$ is associative and unital, $Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $Delta circ mu = mu circ (Delta otimes Delta)$.


  2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $varnothing$ be the initial object of your category and suppose that $varnothing otimes X = varnothing = X otimes varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $mathttO(1) = T$ and $mathttO(n) = varnothing$ for $n neq 1$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
    $endgroup$
    – FKranhold
    29 mins ago
















6












$begingroup$

Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $mathttCom rtimes T$ of the commutative operad $mathttCom$ by $T$.



  1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $mu : T otimes T to T$ and a comultiplication $Delta : T to T otimes T$ such that $mu$ is associative and unital, $Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $Delta circ mu = mu circ (Delta otimes Delta)$.


  2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $varnothing$ be the initial object of your category and suppose that $varnothing otimes X = varnothing = X otimes varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $mathttO(1) = T$ and $mathttO(n) = varnothing$ for $n neq 1$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
    $endgroup$
    – FKranhold
    29 mins ago














6












6








6





$begingroup$

Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $mathttCom rtimes T$ of the commutative operad $mathttCom$ by $T$.



  1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $mu : T otimes T to T$ and a comultiplication $Delta : T to T otimes T$ such that $mu$ is associative and unital, $Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $Delta circ mu = mu circ (Delta otimes Delta)$.


  2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $varnothing$ be the initial object of your category and suppose that $varnothing otimes X = varnothing = X otimes varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $mathttO(1) = T$ and $mathttO(n) = varnothing$ for $n neq 1$.






share|cite|improve this answer











$endgroup$



Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $mathttCom rtimes T$ of the commutative operad $mathttCom$ by $T$.



  1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $mu : T otimes T to T$ and a comultiplication $Delta : T to T otimes T$ such that $mu$ is associative and unital, $Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $Delta circ mu = mu circ (Delta otimes Delta)$.


  2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $varnothing$ be the initial object of your category and suppose that $varnothing otimes X = varnothing = X otimes varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $mathttO(1) = T$ and $mathttO(n) = varnothing$ for $n neq 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









Najib IdrissiNajib Idrissi

2,17211127




2,17211127











  • $begingroup$
    Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
    $endgroup$
    – FKranhold
    29 mins ago

















  • $begingroup$
    Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
    $endgroup$
    – FKranhold
    29 mins ago
















$begingroup$
Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
$endgroup$
– FKranhold
29 mins ago





$begingroup$
Thank you! Am I correct that you mean the compatibility $Deltacirc mu = (muotimes mu)circ (mathrmidotimes betaotimesmathrmid)circ (DeltaotimesDelta)$ where $beta:T^otimes 2to T^otimes 2$ is the symmetric braiding?
$endgroup$
– FKranhold
29 mins ago


















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