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Stream: learning: questions

Topic: ✔ when is restriction of scalars monadic?

John Baez (May 11 2023 at 02:22):

While discussing math with some people I thought of a question. Suppose you have a ring homomorphism $f: A \to B$. Restriction along $f$ gives a functor

$f^\ast: \mathsf{Mod}(B) \to \mathsf{Mod}(A)$

and this has a left adjoint, often called extension along $f$:

$f_!: \mathsf{Mod}(A) \to \mathsf{Mod}(B)$

Concretely we have $f_!(M) = B \otimes_A M$.

My question: when is the functor $f^\ast: \mathsf{Mod}(B) \to \mathsf{Mod}(A)$ monadic?

John Baez (May 11 2023 at 02:24):

Anyone who doesn't know about these functors can learn about them here.

Reid Barton (May 11 2023 at 06:56):

It seems to be always the case. For example when $A = \mathbb{Z}$, an "algebra over $f^* f_!$" amounts to the usual definition of a module over $B$.

Tobias Fritz (May 11 2023 at 07:08):

Could this be a general fact about algebraic theories? That is, could it be that any morphism of algebraic theories induces a monadic adjunction between their categories of models?

Tobias Fritz (May 11 2023 at 07:08):

I've done some literature search, such as in the book Algebraic Theories, but don't see this mentioned anywhere; maybe it's there but phrased differently than what I'd expect. That being said, there is a complete characterization of when the extension of scalars functor $\mathsf{Mod}(A) \to \mathsf{Mod}(B)$ is comonadic!

Morgan Rogers (he/him) (May 11 2023 at 07:38):

It helps that restriction of scalars has an adjoint on both sides here - it makes checking the conditions in a monadicity theorem much easier. The answer is as Reid said: restriction of scalars is always monadic because it's always faithful (and hence conservative): restriction of scalars preserves the underlying functions so nothing gets identified in the process.

@Tobias Fritz regarding all morphisms of algebraic theories inducing monadic functors, it depends what morphisms of algebraic theories you allow. For single-sorted theories and morphisms which preserve the sort it is true, because a morphism is determined by its underlying function at the sort. However, if the sort of the theory is allowed to be sent to something more interesting (say we allow a functor between their Lawvere theories) we don't have that guarantee any more.

Tobias Fritz (May 11 2023 at 07:44):

Thanks! Yes, I did have the single-sorted case and sort-preserving morphisms in mind.

Morgan Rogers (he/him) (May 11 2023 at 07:50):

Beware that my explanation wasn't complete: there is a reflexive coequalizer condition to check that is non-trivial in general (when the extra adjoint on the right of the "restriction" functor does not exist). Checking that condition is more involved, but on the other hand it would exactly explain if/when monadicity fails for such morphisms.

Tobias Fritz (May 11 2023 at 07:56):

So the monadicity holds as soon as the extra adjoint exists, and possibly in other cases as well, but not in general? What would be an example where it does not hold?

Simon Willerton (May 11 2023 at 08:48):

As I'm banging this particular drum at the moment, let me mention for those not familiar these ideas, that the extension of scalars can be viewed as left Kan extension.

We take $\mathsf{Ab}$ to be the category of abelian groups, and take $\bar A$ to be the one-object $\mathsf{Ab}$-category with $A$ as its hom-object. (On another day I might have notated that by $\mathsf{B} A$.). Then the category of left $A$-modules can be identified with a presheaf-like category of functors:

$$\mathsf{Mod}(A) \cong [\bar A, \mathsf{Ab}].$$

The ring homomorphsim $f\colon A\to B$ corresponds to an $\mathsf{Ab}$-functor $f\colon \bar A\to \bar B$, and composition with $f$ gives the pullback functor

$$f^\ast\colon [\bar B, \mathsf{Ab}]\to [\bar A, \mathsf{Ab}].$$

Left Kan extension along $f$ then gives the extension of scalars

$$f_! \colon [\bar A, \mathsf{Ab}]\to [\bar B, \mathsf{Ab}].$$

Nathanael Arkor (May 11 2023 at 09:08):

Tobias Fritz said:

Could this be a general fact about algebraic theories? That is, could it be that any morphism of algebraic theories induces a monadic adjunction between their categories of models?

The functor induced by a morphism of $S$-sorted algebraic theories always has a left adjoint, and the induced adjunction is monadic.

Nathanael Arkor (May 11 2023 at 09:10):

(Categories of algebras for algebraic theories always have enough colimits to form the left adjoint, as they're locally strongly finitely presentable.)

Tobias Fritz (May 11 2023 at 09:12):

That's what I had suspected, but Morgan's most recent commented seems to suggest otherwise. Do you have a reference for the monadicity?

Nathanael Arkor (May 11 2023 at 09:22):

Corollary 1.5.2 of Manes's thesis A triple miscellany: some aspects of the theory of algebras over a triple is one reference. There are surely more convenient references, though.

Tobias Fritz (May 11 2023 at 09:32):

Interesting to know, but that reference seems pretty hard to find a digital copy of.

Nathanael Arkor (May 11 2023 at 09:37):

The result is described on this nLab page, with a sketch of a proof.

Tobias Fritz (May 11 2023 at 10:10):

Thanks! I guess the relevant result is the corollary at the beginning, with $R$ being the identity functor on $\mathsf{Set}$. Now I wonder what Morgan meant by his "my explanation wasn't complete: there is a reflexive coequalizer condition to check that is non-trivial in general".

Morgan Rogers (he/him) (May 11 2023 at 11:13):

Okay great! I was just being extra careful because I hadn't checked it, but I am reassured that someone knew where it had been checked.

Mike Shulman (May 11 2023 at 14:19):

I often find it easier to show a functor is monadic by directly comparing its domain to the category of algebras for the induced monad, rather than by mucking around with split or reflexive coequalizers. In this case an $f^*f_!$-algebra is an $A$-module with a map $B\otimes _A M \to M$ satisfying some conditions, or equivalently an $A$-bilinear map $(B,M) \to M$ satisfying some conditions. If you forget about $A$-bilinearity and just regard this map as $\mathbb{Z}$-bilinear, the conditions are precisely those that make it define a $B$-module, and then the extra $A$-bilinearity is equivalent to saying that the given $A$-module structure is the one induced from this $B$-module structure via $f$.

John Baez (May 11 2023 at 18:27):

Thanks, everyone! I felt that I could figure it out, but I decided it would be more fun to ask, and I'm glad I did because I got a mulitplicity of interesting answers. If I'd tried to work it out, I would probably have taken Mike's approach, since I still fear the monadicity theorem.

John Baez (May 11 2023 at 18:28):

But I'm really excited by the generalization to Lawvere theories.

John Baez (May 11 2023 at 20:31):

Morgan Rogers (he/him) said:

It helps that restriction of scalars has an adjoint on both sides here - it makes checking the conditions in a monadicity theorem much easier. The answer is as Reid said: restriction of scalars is always monadic because it's always faithful (and hence conservative)

It almost sounds like you're claiming faithful $\implies$ conservative, so I just want to mention for beginners that this is not true. Let $C$ be the walking arrow, a category with only one three morphisms, namely the nonidentity morphism $f: x \to y$ and the identities on $x$ and $y$. Let $D$ be the walking isomorphism, a category with only four morphisms, namely three already mentioned and also an inverse $f^{-1} : y \to x$. There is a functor $F: C \to D$ that includes $C$ in $D$ in the obvious way. This is faithful, but it's not conservative, since $F(f)$ is an isomorphism while $f$ is not.

Morgan Rogers (he/him) (May 12 2023 at 06:13):

The first draft of the quoted message contained the explanation "because categories of modules are balanced and have equalizers", which are conditions ensuring the implications between faithfulness and conservativity. However, it's not true that an arbitrary category of models of an algebraic theory is balanced (cf the injective epimorphism from Z to Q in the category of rings) so one needs to check conservativity more directly in the more general case suggested by @Tobias Fritz

John Baez (May 12 2023 at 15:10):

I figured you were relying on some extra conditions present in this case.

John Baez (May 12 2023 at 17:52):

Simon Willerton said:

We take $\mathsf{Ab}$ to be the category of abelian groups, and take $\bar A$ to be the one-object $\mathsf{Ab}$-category with $A$ as its hom-object. (On another day I might have notated that by $\mathsf{B} A$.). Then the category of left $A$-modules can be identified with a presheaf-like category of functors:

$$\mathsf{Mod}(A) \cong [\bar A, \mathsf{Ab}].$$

The ring homomorphsim $f\colon A\to B$ corresponds to an $\mathsf{Ab}$-functor $f\colon \bar A\to \bar B$, and composition with $f$ gives the pullback functor

$$f^\ast\colon [\bar B, \mathsf{Ab}]\to [\bar A, \mathsf{Ab}].$$

Left Kan extension along $f$ then gives the extension of scalars

$$f_! \colon [\bar A, \mathsf{Ab}]\to [\bar B, \mathsf{Ab}].$$

Nice! So I bet right Kan extension gives [[coextension of scalars]].

John Baez (May 12 2023 at 17:54):

I had never thought about coextension until recently. I just added something to that nLab article pointing out a case where extension and coextension of scalars agree: when your $f$ is an inclusion of group algebras $k[H] \to k[G]$ and $H$ is a group of finite index in $G$. But I figure there must be some much more general theorems about when they agree. So I hope someone adds those to the article.

John Baez (May 12 2023 at 18:03):

Besides the case of modules of rings, what are some other cases where restriction along a map of Lawvere theories has a right adjoint ("coextension") as well as the usual left adjoint ("extension")?

Mike Shulman (May 12 2023 at 18:07):

Well, this is not a lot more general, but instead of $k$ being a field I would expect it to suffice for it to be a ring in which the index $[G:H]\in \mathbb{N}$ is invertible. We can also make sense of that condition when we replace $\rm Ab$ by any semiadditive monoidal category, by saying that the sum of $[G:H]$ copies of the identity morphism of the unit object is an isomorphism. Kate Ponto and I used a condition like that to deduce a generalization of the orbit-counting theorem from the linearity of traces (around Theorem 4.12). We didn't prove that such a condition makes extension and coextension of scalars coincide, but it seems a natural conjecture.

Nathanael Arkor (May 12 2023 at 22:49):

John Baez said:

Besides the case of modules of rings, what are some other cases where restriction along a map of Lawvere theories has a right adjoint ("coextension") as well as the usual left adjoint ("extension")?

I believe this was unsolved problem (2) in Lawvere's Some Algebraic Problems in the context of Functorial Semantics of Algebraic Theories. Lawvere states, in the TAC reprint of his thesis, that Wraith made considerable progress on this problem in Algebraic theories (presumably in §3 ibid.).

Notification Bot (May 14 2023 at 00:39):

John Baez has marked this topic as resolved.