### Total Freedom

#### Posted by John Baez

Wow! I just learned an objective reason why sets and vector spaces are special!

Of course we all know math relies heavily on set theory and linear algebra. And if you know category theory, you can say various things about why the categories $\mathsf{Set}$ and $\mathsf{Vect}$ are particularly convenient frameworks for calculation. But I’d never known a *theorem* that picks out these categories, and just a few others.

Briefly: these are categories of algebraic gadgets where *all the objects are free!*

- Keith A. Kearnes, Emil W. Kiss and Agnes Szendrei, Varieties whose finitely generated members are free.

We could call these ‘totally free’ algebraic gadgets.

By ‘algebraic gadgets’ I mean sets equipped with some $n$-ary operations obeying equational laws. Examples include monoids, groups, rings, modules over a fixed ring, Lie algebras over a fixed field, etc.

There are three famous formalisms for studying algebraic gadgets. They all describe the same kinds of algebraic gadgets, so which you use is largely a matter of convenience:

varieties in the sense of universal algebra,

Lawvere theories, and

finitary monads on the category of sets.

Given any variety, or Lawvere theory, or finitary monad on the category of sets, we get a category $\mathsf{C}$ of algebraic gadgets together with a functor

$R : \mathsf{C} \to \mathsf{Set}$

sending each gadget of this kind to its underlying set. And this functor will always have a left adjoint

$L : \mathsf{Set} \to \mathsf{C}$

sending any set to the free gadget on that set.

Steven Givant and later Kearnes, Kiss and Szendrei completely classified the varieties for which every object in $\mathsf{C}$ is isomorphic to one of the form $L(S)$ for some set $S$. We could call these the **totally free** varieties:

- sets
- pointed sets
- modules over some fixed division ring $F$
- affine spaces over some fixed division ring $F$.

That’s all!

A commutative division ring is called a ‘field’, and a module over a field is called a ‘vector space’. The quaternions are a nice example of a noncommutative division ring. Anyone who has studied modules over the quaternions knows that these act a lot like vector spaces, in part because they’re all free.

An affine space over a field is, poetically speaking, just a vector space that has forgotten its origin. If you pick any point in the affine space and call it $0$, you get a vector space. You can’t take linear combinations of points in an affine space, just ‘affine combinations’. These are $n$-ary operations that obey a bunch of equational laws that I’m too lazy to list. But when your affine space comes from a vector space in the way I just described, these affine combinations can be written as linear combinations:

$a_1 v_1 + \cdots + a_n v_n$

where the coefficients sum to one: $a_1 + \cdots + a_n = 1$. So, for example, if you have two points $v_1, v_2$ in an affine space you can get all the points on the line through them by taking affine combinations $av_1 + (1-a)v_2$.

We can also do all this stuff with a division ring replacing a field! In fact, ‘field’ used to mean ‘division ring’, and the noncommutative ones were called ‘skew fields’.

Now, you’ll notice from what I said that a module over a division ring $F$ is just the same as a *pointed* affine space over $F$ — that is, an affine space over $F$ equipped with a chosen point. Similarly a pointed set is just a set with a chosen point. So we can list the totally free varieties in a more enlightening way:

- sets
- pointed sets
- affine spaces over a division ring
- pointed modules over a division ring.

And this should remind us of the ‘field with one element’. We don’t know what the field with one element is, exactly, but we know that the modules of this mythical beast should be pointed sets. There are lots of reasons for that. Here we see another: it would unify the above classification!

Suppose we could go back in time, redefine ‘field’ to include division rings and also one extra thing: the ‘field with one element’, $F_{\mathrm{un}}$. We wouldn’t even need to know what $F_{\mathrm{un}}$ is, just that that affine spaces over it are sets. Then this would be the complete classification of totally free varieties:

- affine spaces over a field
- vector spaces over a field.

That would be nice.

But of course, it’s sort of trivial that every set is the free set on some set. In this case our category $\mathsf{C}$ of algebraic gadgets is just $\mathsf{Set}$ itself, and $R : \mathsf{Set} \to \mathsf{Set}$ is the identity functor, so $L: \mathsf{Set} \to \mathsf{Set}$ and the monad $T = R L$ are also the identity.

This raises the question of ‘relativising’ the ideas I’ve been talking about, by replacing $\mathsf{Set}$ with some other category $\mathsf{X}$.

**Puzzle.** Give me as many interesting examples as you can of categories $\mathsf{X}$ and monads $T: \mathsf{X} \to \mathsf{X}$ such that every $T$-algebra is free.

Of course we can always take $T$ to be the identity, but that’s boring.

A very interesting small step would be to stick with $\mathsf{X} = \mathsf{Set}$ but drop the requirement that $T$ be finitary. This lets us talk about algebraic gadgets with *infinitary* operations. Are there any interesting ones where every $T$-algebra is free?

But I’m more interested in other categories $\mathsf{X}$. For example, what if $\mathsf{X}$ itself is $\mathsf{Vect}$, or the category of algebras of some other finitary monad on $\mathsf{Set}$?

## Re: Total Freedom

On Mathstodon Oscar Cunningham wrote: