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Stream: theory: category theory

Topic: two algebra structures

Matteo Capucci (he/him) (Jan 17 2022 at 15:10):

Let $T, S$ be monads on $\mathcal C$. I'm sure the idea of '$(T,S)$-bialgebra', generalizing bimodules in commutative algebras, has already occurred to someone. By that I mean an object $a : \mathcal C$ equipped with an algebra structure for $T$ and one for $S$, satisfying some compatbility conditions. I guess one also needs some kind of compatibility between $T$ and $S$ (e.g. a distributive law).
Can someone point me to a reference on such objects?

Chad Nester (Jan 17 2022 at 15:13):

The tensor product of algebraic theories might be close to what you're after. I think the idea originates with Freyd .

Ralph Sarkis (Jan 17 2022 at 15:23):

A somewhat opaque (imo) reference is Section 3 in The formal theory of monads II.

Mike Shulman (Jan 17 2022 at 15:57):

I don't think this would generalize bimodules in commutative algebra, since a bimodule has a left action by one ring and a right action by another one, whereas an object of C that's both a T-algebra and an S-algebra would have a left action by both T and S.

Mike Shulman (Jan 17 2022 at 15:58):

This difference is why you would seem to need extra structure like a distributive law. But if you do have a distributive law, so that TS is a monad, then the TS-algebras can be characterized as objects with both a T-algebra and an S-algebra structure that satisfy an additional compatibility diagram (which it's a nice exercise to write down).

Matteo Capucci (he/him) (Jan 17 2022 at 17:38):

Mike Shulman said:

I don't think this would generalize bimodules in commutative algebra, since a bimodule has a left action by one ring and a right action by another one, whereas an object of C that's both a T-algebra and an S-algebra would have a left action by both T and S.

Well a right action of $R$ is a left action of $R^{\mathrm{op}}$, isn't it?

Morgan Rogers (he/him) (Jan 17 2022 at 20:39):

@Matteo Capucci (he/him) sounds like you want an algebra-coalgebra structure, then, but if you're interested specifically in monads rather than endofunctors then you're probably losing something by introducing an op.

Mike Shulman (Jan 17 2022 at 23:09):

Matteo Capucci (he/him) said:

Well a right action of $R$ is a left action of $R^{\mathrm{op}}$, isn't it?

Yes, but only when the tensor product is symmetric. The action of an endofunctor on an object is not a symmetric binary operation.

Matteo Capucci (he/him) (Jan 18 2022 at 08:34):

Morgan Rogers (he/him) said:

Matteo Capucci (he/him) sounds like you want an algebra-coalgebra structure, then, but if you're interested specifically in monads rather than endofunctors then you're probably losing something by introducing an op.

Why coalgebra? $R^{op}$ is just another ring, yielding a moand $R^{op} \times -$ whose algebras are right action of $R$.If $A$ is both an $S \times -$ and an $R^{op} \times -$ algebra (in a compatible way*), then it's an $(S,R)$-bimodule.

*The compatible way is the commutativity of this, where $\sigma$ is the symmetry of $\times$ (or in general the distributive law between the two monads) image.png

Matteo Capucci (he/him) (Jan 18 2022 at 08:36):

Mike Shulman said:

Matteo Capucci (he/him) said:

Well a right action of $R$ is a left action of $R^{\mathrm{op}}$, isn't it?

Yes, but only when the tensor product is symmetric. The action of an endofunctor on an object is not a symmetric binary operation.

You mean the tensor on the category the monads are defined on? Yeah, that's a fair point!

Matteo Capucci (he/him) (Jan 18 2022 at 09:05):

Chad Nester said:

The tensor product of algebraic theories might be close to what you're after. I think the idea originates with Freyd .

This seems to be what I'm looking for! There's only one thing I can't understand: Freyd says T-algebras in the category of S-algebras are always equivalent to S-algebras in the category of T-algebras:
image.png

Matteo Capucci (he/him) (Jan 18 2022 at 09:06):

image.png

Matteo Capucci (he/him) (Jan 18 2022 at 09:08):

But you'd certainly need a distributive law between $T$ and $S$, wouldn't you? Or I wouldn't even be able to lift one monad on the other's category of algebras

Ralph Sarkis (Jan 18 2022 at 09:30):

I think this is because in this paper, $T_1$ and $T_2$ are (Lawvere) theories which can always be combined like this. In modern notation, I would translate the statement to $\mathrm{Mod}(T_1,\mathrm{Mod}(T_2,\mathbf{Set})) \simeq \mathrm{Mod}(T_2,\mathrm{Mod}(T_1,\mathbf{Set})) \simeq \mathrm{Mod}(T_1\otimes T_2,\mathbf{Set})$.
There is a monad corresponding to tensored theory but it doesn't have to be built from $T_1$ and $T_2$ using a distributive law. One example is this paper where the authors show the tensor of the theories for semilattices and convex algebras correspond to a monad obtained via a weak distributive law between $\mathcal{P}$ (for semilattices) and $\mathcal{D}$ (for convex algebras). I believe there are also cases where even weak distributive laws will not work.

Matteo Capucci (he/him) (Jan 18 2022 at 12:11):

ooh I see

Matteo Capucci (he/him) (Jan 18 2022 at 12:13):

Indeed, I now notice Freyd doesn't define $T_1$ and $T_2$ as monads in the paper!
Ralph Sarkis said:

I think this is because in this paper, $T_1$ and $T_2$ are (Lawvere) theories which can always be combined like this.

What's the reason Lawvere theories (= finitary monads..?) can always be combined like this?

Morgan Rogers (he/him) (Jan 18 2022 at 12:14):

Matteo Capucci (he/him) said:

Morgan Rogers (he/him) said:

Matteo Capucci (he/him) sounds like you want an algebra-coalgebra structure, then, but if you're interested specifically in monads rather than endofunctors then you're probably losing something by introducing an op.

Why coalgebra? $R^{op}$ is just another ring, yielding a moand $R^{op} \times -$...

I was pointing out that for most monads, you don't have reference to an object that you can simply take the dual of to obtain a new monad, and the only sensible thing I could see lying around to dualize was the map constituting the algebra. For endofunctors (without extra conditions) an $R$-algebra-$S$-coalgebra isn't an unreasonable thing to think about, and they even give adjunctions from the category of $R$-algebras to the category of $S$-algebras (again, stressing that this is ignoring the monad structure)

Matteo Capucci (he/him) (Jan 18 2022 at 12:15):

Oh I see, that's a fair point

Ralph Sarkis (Jan 18 2022 at 13:00):

Matteo Capucci (he/him) said:

What's the reason Lawvere theories (= finitary monads..?) can always be combined like this?

As I understand it, the tensor operation on theories is very syntactical (Definition 3.2 here is more categorical), the operations symbols and axioms in both theories are put together and we add the appropriate distributivity axioms. There is nothing that prevents it from existing.

The reason (imo) why this does not translate to monads is that while the correspondence theory--monad yields $\mathrm{Mod}(T,\mathbf{Set}) \cong \mathbf{Set}^{T}$, we don't have $\mathrm{Mod}(T,\mathrm{Mod}(S,\mathbf{Set})) \cong (\mathbf{Set}^S)^T$. In fact, with no distributive law, the RHS is not well-defined because $T$ is not a monad on $\mathbf{Set}^S$.

$*$ $\mathbf{C}^T$ denotes the category of Eilenberg--Moore algebra for $T$

Reid Barton (Jan 18 2022 at 13:06):

It might also be worth noting that the tensor product construction also works when $S$ and $T$ are multi-sorted theories, and then the sorts of the tensor product theory are the cartesian product of the sorts of $S$ and the sorts of $T$ (which need not be the same). For example, you can tensor abelian groups and simplicial sets to get simplicial abelian groups.

Reid Barton (Jan 18 2022 at 13:07):

On the other hand, the construction involving a distributive law of monads only makes sense when $S$ and $T$ are monads on the same underlying category (~ have the same "sorts").

Mike Shulman (Jan 18 2022 at 17:03):

Fundamentally, the tensor product of theories and the composite monad arising from a distributive law are different things. A $T\otimes S$ algebra is an object with unrelated $T$-algebra and $S$-algebra structures; hence why it requires no additional data relating $T$ and $S$. Whereas if $\lambda$ is a distributive law between $T$ and $S$, then a $TS$-algebra is an object with $T$-algebra and $S$-algebra structures that satisfy a compatibility condition mediated by $\lambda$. In principle one could try to construct the "tensor product of monads" in a similar way to the tensor product of theories, without using a distributive law, and I would expect it to exist in good cases like when the base category is locally presentable and the monads are accessible.

Nathanael Arkor (Jan 18 2022 at 17:15):

In principle one could try to construct the "tensor product of monads" in a similar way to the tensor product of theories, without using a distributive law, and I would expect it to exist in good cases like when the base category is locally presentable and the monads are accessible.

For those interested, one general framework for tensor products of monads is Lucyshyn-Wright's Bifold algebras and commutants for enriched algebraic theories.

Matteo Capucci (he/him) (Jan 19 2022 at 09:57):

Mike Shulman said:

Fundamentally, the tensor product of theories and the composite monad arising from a distributive law are different things. A $T\otimes S$ algebra is an object with unrelated $T$-algebra and $S$-algebra structures; hence why it requires no additional data relating $T$ and $S$. Whereas if $\lambda$ is a distributive law between $T$ and $S$, then a $TS$-algebra is an object with $T$-algebra and $S$-algebra structures that satisfy a compatibility condition mediated by $\lambda$.

This makes sense, but I'm confused by Freyd's paper then, where he says the $T$-structure is a morphism of $S$-algebras, and thus it satisfies a commutativity/distributivity condition whereby the two structures commute.

Mike Shulman (Jan 19 2022 at 15:41):

What paper is that?

Nathanael Arkor (Jan 19 2022 at 16:09):

I presume Algebra valued functors in general and tensor products in particular.

Mike Shulman (Jan 19 2022 at 17:03):

Oh, oops!! I messed up; I confused the tensor product of theories with the coproduct of theories. It's the coproduct for which a $T+S$ algebra is an object with unrelated structures. A $T\otimes S$ algebra is one for which all $T$-operations commute with all $S$-operations, which is sort of the opposite extreme, but again doesn't require the data of a distributive law to specify exactly how the structures relate.

Ralph Sarkis (Jan 19 2022 at 17:38):

Matteo Capucci (he/him) said:

This makes sense, but I'm confused by Freyd's paper then, where he says the $T$-structure is a morphism of $S$-algebras, and thus it satisfies a commutativity/distributivity condition whereby the two structures commute.

(Let me briefly recall what a theory is to make sure my subsequent explanation makes sense.) A theory $T$ is a set of operation symbols $\Sigma_T$ each with an arity and a set of equations (we won't need them here). A $T$--algebra in a category $\mathbf{C}$ with finite products is an object $A \in \mathbf{C}_0$ along with, for each $t \in \Sigma_T$ of arity $n$, a morphism $t^A: A^n \to A$ such that the equations are satisfied (again, we can forget about this here). A morphism of $T$--algebras is a morphism $f: A \rightarrow B$ such that $f \circ t^A = t^B \circ f^n$ for any $t \in \Sigma_T$ of arity $n$. Here is this condition as a commutative diagram:
image.png

Ralph Sarkis (Jan 19 2022 at 17:40):

Now, if you have a $T\otimes S$--algebra on $A$, what you really have is a $T$--algebra and an $S$--algebra on $A$ such that for any $t \in \Sigma_T$ of arity $n$ and $s \in \Sigma_S$ of arity $m$, $t^A \circ (s^A)^n = s^A \circ (t^A)^m$ (you get this if you squint a bit at the distributivity axiom). If you draw this as a commutative diagram, you get the following
image.png

or, by rotating it:
image.png

You can recognize these diagrams as stating that $s^A$ is a $T$--algebra homomorphism and $t^A$ is an $S$-algebra homomorphism respectively.

Matteo Capucci (he/him) (Jan 19 2022 at 21:03):

Thank you Ralph!

Matteo Capucci (he/him) (Jan 19 2022 at 21:05):

Then I grant this is weaker than saying $T$ lifts to $$S$-algebras and viceversa, as @Mike Shulman pointed out too?

Matteo Capucci (he/him) (Jan 19 2022 at 21:06):

But then I see the two squares 'work' because $(A^m)^n \cong (A^n)^m$ and so you can commute the algebra structures

Matteo Capucci (he/him) (Jan 19 2022 at 21:11):

Then tell me where I'm wrong here:

1. Every algebraic theory $T:\Sigma_T \to \mathbb N$ gives rise to a moand $M_T(X) = \sum_{\sigma \in \Sigma_T} X^{T(\sigma)}$
2. Every two algebraic theories $T$ and $S$ admit a distributive law $M_TM_S \Rightarrow M_SM_T$ since products distribute over coproducts in $\mathbf{Set}$

Ralph Sarkis (Jan 19 2022 at 21:44):

That's not the right definition of $M_T$ (what would be the monad structure?). The algebras for the functor $M_T$ you gave are indeed the $T$--algebras, but the monad that corresponds to the theory is harder to construct, it is the algebraically free monad on that functor. In $\mathbf{Set}$ it is still quite easy.

Mike Shulman (Jan 19 2022 at 22:09):

No, I don't think that's right either; the monad corresponding to an algebraic theory is not in general algebraically-free on any endofunctor. Note that your $M_T$, as a mere functor, contains no information about the composition of operations in $T$.

Mike Shulman (Jan 19 2022 at 22:10):

With a correct definition of the monad $M_T$ associated to an algebraic theory, then although $T\otimes S$ is an algebraic theory and has an associated monad $M_{T\otimes S}$, the underlying functor of this monad is not the composite $M_S M_T$ (or $M_T M_S$), therefore it does not induce a distributive law.

Ralph Sarkis (Jan 20 2022 at 07:19):

Isn't $M_T$ enough information when there are no equations in the theory? My line of thought was $T\mathrm{-Alg} \cong M_T\mathrm{-Alg} \cong \mathrm{EM}(M_T^*)$ where $M_T^*$ is the algebraically free monad on $M_T$ (hence the second isomorphism).

Mike Shulman (Jan 21 2022 at 19:45):

If there are no equations in the theory, then yes T-algebras should be the algebras for an endofunctor defined in a simliar way, where "$T(\sigma)$" denotes the set of generating operations.