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Stream: theory: category theory

Topic: tensor product of EM-algebras

Eigil Rischel (Mar 18 2021 at 11:19):

We can define the tensor product of abelian groups as follows: it is the unique closed monoidal structure on $Ab$ so that the free abelian group functor $\mathbb{Z}[-] : Set \to Ab$ is strong monoidal.

The above makes perfect sense if we replace "abelian group" with "$M$-algebra" for some monad $M$. Has anyone studied this "tensor product"?

I'm particularly interested in the case where $M$ is something like the "discrete distribution" monad, whose algebras are convex sets. But it seems that this definition probably works for any algebraic structure?

Peter Arndt (Mar 18 2021 at 11:48):

I think Section 6.4 of Martin Brandenburg: Tensor categorical foundations of algebraic geometry gives a good summary and further references.

Peter Arndt (Mar 18 2021 at 11:51):

Actually all of chapter 6 is good to read, I guess. You need a "strength" for your monad, but if you look at monads on $Set$ you always have one.

Peter Arndt (Mar 18 2021 at 11:52):

See the last example in Section 4 of this nLab page for an example from probability theory.

Eigil Rischel (Mar 18 2021 at 11:57):

Thanks!

Eigil Rischel (Mar 18 2021 at 12:12):

Yes, this looks exactly like what I was looking for

Tom Hirschowitz (Mar 18 2021 at 12:33):

He does in particular mention Seal's Tensors, monads and actions, which I think is a good reference.

Paolo Perrone (Mar 18 2021 at 12:39):

You need the strength of the monad to be commutative, I think. See also this nlab page.

Jules Hedges (Mar 18 2021 at 12:43):

When your strong monad is commutative then its Kleisli category also gets a monoidal product (a fact that I use all the time!)... anyone know whether it's the restriction of the tensor product or the cartesian product from the EM category? I think the cartesian product...... which makes me wonder what you get if you restrict this tensor product to free algebras...

Peter Arndt (Mar 18 2021 at 13:17):

I don't know which monoidal product on the Kleisli category you are thinking of, but the tensor product on the EM category _does_ restrict to the Kleisli category. That's the content of Prop. 6.4.6 of Brandenburg.

Eigil Rischel (Mar 18 2021 at 14:00):

I believe Jules is talking about the tensor product on the Kleisli category which extends the structure on the base category. The statement that the free functor $T: C \to Alg(T)$ is strong monoidal should be exactly equivalent to this, namely that $T(X \otimes Y) = TX \otimes TY$.

Eigil Rischel (Mar 18 2021 at 14:00):

(Indeed extending this structure on the Kleisli category is exactly the thing I was thinking about)

Peter Arndt (Mar 18 2021 at 14:18):

Then yes, it's the restriction from the tensor product of EM algebras (by the propositions I cited above).

John Baez (Mar 18 2021 at 16:23):

Jules was asking whether this tensor product on the Kleisli category of a commutative strong monad is the restriction of the cartesian product on the Eilenberg-Moore category, and nobody has come out and said the answer is no, but I think the answer I'm hearing here is "no".

Joe Moeller (Mar 18 2021 at 16:27):

Right, I believe the tensor product this stuff gives in the case of abelian groups is the "tensor product of abelian groups", which is neither cartesian nor cocartesian. Ab has biproducts, direct sum.

Mike Shulman (Mar 18 2021 at 19:43):

One possible source of confusion is that if you regard the objects of the Kleisli category as being the objects of the category on which the monad acts, rather than as the free algebras thereon, then the monoidal structure of the Kleisli category does coincide on objects with the given monoidal structure on the underlying category, which is often (e.g. for Set) cartesian. Put differently, the left adjoint to the forgetful functor from the Eilenberg-Moore category is strong monoidal (though the forgetful functor itself is only lax monoidal), and the Kleisli category its its full image.

John Baez (Mar 18 2021 at 19:51):

Yes, that can easily fool people! For those who are blown away by the generality of Mike's comment: you can think of the category of free abelian groups and homomorphisms between them as having sets as objects and homomorphisms between the free abelian groups on those sets as morphisms. The trick is to use a set as a stand-in for the free abelian group on that set.

In these terms, the tensor product of free abelian groups amounts to the cartesian product of sets. However, this tensor product is not the cartesian product in the category of free abelian groups.

Jules Hedges (Mar 18 2021 at 22:04):

Ah, right. Whereas the cartesian product of groups is the .... what? Is it the free product? (I never use groups as my go-to example because I don't know much about groups...)

John Baez (Mar 18 2021 at 22:07):

The so-called "free product" of groups is their coproduct.

The cartesian product of groups is just what people call the product, $G \times H$: it's the group consisting of ordered pairs $(g,h)$.

It helps to remember that right adjoints preserve products, so the underlying set of the cartesian product of groups has to be the cartesian product of their underlying sets.

Jules Hedges (Mar 18 2021 at 22:07):

On the other hand my go-to example is the powerset monad. Its kleisli category is Rel, and its monoidal product that it inherits from the powerset monad's commutative strength is the monoidal product of Rel given by cartesian product of the underlying sets. On the other hand the EM category of powerset is complete semilattices (aka complete lattices + join-preserving functions, unless I got things back to front). The cartesian product of complete semilattices is probably the obvious pointwise one, but I wonder what the monoidal product of complete semilattices would be

John Baez (Mar 18 2021 at 22:10):

By the way, I hadn't been talking about groups; I was talking about abelian groups, because then there's a well-known tensor product of abelian groups, which is different from the coproduct and the product (which happen to be the same).

I was pointing out that the tensor product of free abelian groups is different from the product.

For free abelian groups we have

$\mathbb{Z}^m + \mathbb{Z}^n \cong \mathbb{Z}^{m+n}$

$\mathbb{Z}^m \times \mathbb{Z}^n \cong \mathbb{Z}^{m+n}$

$\mathbb{Z}^m \otimes \mathbb{Z}^n \cong \mathbb{Z}^{mn}$

and Mike had pointing out that the tensor product is not the product, but it does involve the product of the underlying sets: $m \times n = mn$.

Jules Hedges (Mar 18 2021 at 22:10):

Oh, I see what happened: you said that something is not the product in the category of free abelian groups, and then I switched to asking about the product of mere groups, which I guess changes everything

Jules Hedges (Mar 18 2021 at 22:11):

Right.... in other words we're doing category theory and these things are properties of the category, not of the objects themselves

John Baez (Mar 18 2021 at 22:12):

By the way, if you don't like abelian groups, you might like vector spaces, and then you can just cross out all those $\mathbb{Z}$'s and replace them with $\mathbb{R}$'s. It all works the same way.

Jules Hedges (Mar 18 2021 at 22:13):

Except then you have the red herring that all vector spaces are free, I guess

Jules Hedges (Mar 18 2021 at 22:13):

Uh, all finite dimensional vector spaces?

Joe Moeller (Mar 18 2021 at 22:14):

Then replace vector spaces with modules, and switch $\mathbb R$ for... $R$.

John Baez (Mar 18 2021 at 22:14):

All vector spaces are free, because every vector space has a basis. (Well, if you assume the axiom of choice.)

John Baez (Mar 18 2021 at 22:15):

Yeah, the general thing I'm talking about here is "free modules of a commutative ring $R$": you can cross out those $\mathbb{Z}$'s and write $R$'s.

Jules Hedges (Mar 18 2021 at 22:15):

Is there a free vector space monad on Set, and its kleisli and EM categories are the same?

John Baez (Mar 18 2021 at 22:16):

Yes, and yes: they're the same if you assume the axiom of choice.

The free vector space on a set S is the vector space consisting of finite linear combinations of elements of S. S will be a basis for this vector space.

John Baez (Mar 18 2021 at 22:17):

If you don't assume the axiom of choice, you can't prove every vector space has a basis, so you can't prove they're all free on some set. I've never gone down that dark and scary road.

Paolo Perrone (Mar 18 2021 at 22:46):

@Jules Hedges This may clarify things a little: the power set monad is an excellent example, but not for tensor products, unfortunately.
The unlucky part is that the semilattice join operation is idempotent, and so for the power set monad there is no difference between morphisms and "bimorphisms" (definition here, think of linear vs bilinear maps in linear algebra). Therefore, for the algebras of the power set monad, the tensor product and the cartesian product coincide. But this is only a special case for this monad! The example of abelian groups is more similar to the general case.

Another "generic" example is given by probability monads: suppose the set $X$ has 3 elements and the set $Y$ has 2 elements. Then the set $X\times Y$ has 6 elements.
Now, $PX$ has dimension 2 (3 elements minus one for normalization, it's a triangle). $PY$ has dimension 1 (it's a line segment). So $PX\times PY$ has dimension 3. Instead, $PX\otimes PY=P(X\times Y)$ has dimension 5, being probabilities over 6 elements.

Jules Hedges (Mar 18 2021 at 22:58):

Got it

Jules Hedges (Mar 18 2021 at 23:04):

So this operation on simplices that takes the cartesian product of the corners, must extend to a tensor product on arbitrary convex spaces... I wonder what it does? I could modify your example by making either of them non-free: what's the tensor product of the whole real line with a triangle? What's the tensor product of a line segment with a disc? Etc. Maybe I should work these out myself...

Mike Shulman (Mar 19 2021 at 00:48):

Paolo Perrone said:

The unlucky part is that the semilattice join operation is idempotent, and so for the power set monad there is no difference between morphisms and "bimorphisms" ... Therefore, for the algebras of the power set monad, the tensor product and the cartesian product coincide.

Idempotence of join does make semilattices special, but not in that way. It's true that idempotence means that every "two-variable morphism" is a bimorphism, but not conversely. An easy way to see that the tensor product of suplattices is not their cartesian product is that the unit of the tensor product must be the free algebra on 1, and the free suplattice on 1 is the 2-element lattice; but the terminal suplattice is the 1-element one (which is also the initial object).

Mike Shulman (Mar 19 2021 at 00:49):

There is a special way to construct the tensor product of suplattices that doesn't apply to abelian groups, though: since suplattices are a $\ast$-autonomous category, we have $A \otimes B = \hom(A,B^{\rm op})^{\rm op}$, where $\hom$ denotes the set of suplattice maps with the pointwise ordering.

Mike Shulman (Mar 19 2021 at 00:53):

Jules Hedges said:

the EM category of powerset is complete semilattices (aka complete lattices + join-preserving functions, unless I got things back to front).

In fact, the category of complete lattices and join-preserving functions is equivalent to the category of complete lattices and meet-preserving functions, via the functor that sends $A$ to $A^{\rm op}$. So it's not possible to get things back to front!

Mike Shulman (Mar 19 2021 at 00:54):

Jules Hedges said:

you said that something is not the product in the category of free abelian groups, and then I switched to asking about the product of mere groups, which I guess changes everything

I'm not sure if you were saying this, but the tensor product isn't the product in the category of mere groups either. Nor in the category of mere abelian groups.

Paolo Perrone (Mar 19 2021 at 08:17):

Ah, right. Thanks for the correction!

Ben Sprott (Mar 21 2021 at 23:55):

Is this relevant here?

Philip Saville (Apr 20 2021 at 21:25):

Paolo Perrone said:

You need the strength of the monad to be commutative, I think. See also this nlab page.

if you only have a _strong_ monad you can define a tensor but it turns out it's only _skew_ monoidal (ie the associator and unitors may not be invertible). Rather nicely, the monoids wrt to this skew monoidal structure turn out to be exactly monoids in the underlying category with compatible algebraic structure, in the form of a T-algebra on the carrier that is compatible with the multiplication. See these slides from CT 2018