# Topological universal algebra: what is a variety?

Very roughly, universal algebra is the study of those classes of algebraic structures which can be defined via a set of equations; such a class is called a variety. Of course there is far more to the subject than this, but this is the essential starting point; and (in my opinion) the essential first result is the HSP theorem, which characterizes varieties in an often more tractable way:

HSP theorem: $\mathcal{V}$ is a variety iff $\mathcal{V}$ is closed under homomorphic images, substructures, and arbitrary-arity products.

I'm interested in whether there is a similar characterization of varieties over a space.

An algebra $A$ with underlying set $X$ is compatible with a topology $\tau$ on $X$ if all the operations of $A$ are continuous in the sense of $\tau$. (We can also look at compatibility up to homotopy, but that doesn't obviously make things cleaner here, so I'll use the more restrictive version here, although I'm interested in each.) For instance, the ring of real numbers $(\mathbb{R}; +, \times, 0, 1)$ is compatible with the usual topology on the reals, and by contrast there is no group compatible with the usual topology on $S^2$. Vastly more generally, Adams' Hopf invariant one theorem states that there is no unital magma structure compatible with the usual topology on $S^n$ unless $n\in\{0, 1, 3, 7\}$.

Walter Taylor has studied algebras compatible with given topologies; among other things, he extended Adams' theorem to show that if $n\not\in\{0, 1, 3, 7\}$, then $S^n$ doesn't admit any nontrivial algebraic structure at all, in a precise sense. He also studied the satisfiability problem for algebras over a topological space, from a computability-theoretic perspective.

I'm interested in a different aspect of "topological universal algebra": we can also ask about those classes of algebras with underlying set $X$, compatible with $\tau$, which are defined by equations; and we can define a $(X, \tau)$-variety as such a class of algebras. This is a perfectly reasonable notion; however, each of the operations H, S, P are terribly behaved in this context! My main question here is:

Q1.1. Is there a "nice" alternate characterization of $(X, \tau)$-varieties, for instance in terms of a fixed family of operations which build new $(X, \tau)$-algebras from old ones?

This is hopelessly broad, though, so a reasonable thing to do is look at a more restrictive case:

Q1.2. Is there a "nice" alternate characterization of algebras compatible with the reals (with the usual topology)?

Note that already on $\mathbb{R}$, we see interesting structure - e.g. Taylor showed that the problem of deciding which finite equational theories are compatible with $\mathbb{R}$ is undecidable.

Even this, though, seems intractable to me - there are individual equations whose corresponding variety I don't understand at all! Take our language to consist of a single binary function symbol, "$*$". I don't have a good sense of the class of algebras on $\mathbb{R}$ in which $*$ is commutative. However, at least I know a little about the natural topology on this class - it's path-connected, by taking weighted averages of operations: $$a*_{p}b={a*_1b\over p}+{a*_2b\over 1-p}.$$ An equation that by contrast I know nothing about is associativity:

Q2.1. What is a good description of the class of continuous associative binary operations on $\mathbb{R}$?

It's easy to check that weighted averages no longer preserve associativity in general. And this brings me to my final, most-concrete question:

Q2.2. Is there any interesting way to combine two continuous associative binary operations on $\mathbb{R}$ and get a third?

Of course "interesting" is a vague term here - I certainly want to rule out trivialities like constant maps and the projection maps, as well as completely ad-hoc constructions. I could try to make this precise (e.g. asking for continuity with respect to the natural topology on the space of such operations), but rather than do that it feels more natural to leave this subjective.

Basically: as natural a notion as it appears to me, I have absolutely no idea what a topological variety is, and would like to.

• The "algebraic-topology" tag here is a guess on my part; feel free to remove it if it seems inappropriate. – Noah Schweber May 28 '17 at 20:49
• I'm curious, why the downvote? – Noah Schweber May 28 '17 at 20:50
• I have opinions (not answers) on all of these that I will share later. In brief, you might have X as part of the similarity type, like R-modules over a ring R. To combine two associative operations and expect a third leads to hyperassociativity. Finally, if you expect any homogeneity from the topology (and a reasonable collection of operations), this lands you into normal spaces if you have T_1 (or something like that, maybe T_0 gives T_7/2). Gerhard "More On These Subjects Later" Paseman, 2017.05.28. – Gerhard Paseman May 28 '17 at 22:30
• @GerhardPaseman I'm not familiar with hyperassociativity (and googling is giving me lots of stuff about psychology), what is that? Also, what do you mean by "you might have X as part of the similarity type"? (The analogy with $R$-modules seems incorrect to me, since an $R$-module won't have domain $R$ in general, but I could be missing something.) As for the topology, I am definitely happy to stick to "very nice" spaces like $\mathbb{R}$, but I would also be interested in weirder spaces, or even natural non-$T_2$ spaces like Zariski topologies. I'm eager to hear your thoughts! – Noah Schweber May 28 '17 at 22:43
• I'm not sure the term "variety" is appropriate here --- in universal algebra this refers to the class of all algebras of a given type which satisfy some set of equations. Whereas apparently you are interested in algebraic operations defined on a single space, which seems very different. – Nik Weaver May 29 '17 at 1:01

## 1 Answer

This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. (In my universe, universal algebra is synonymous with algebra.)

Nevertheless, it is clear that this question is about varieties, so I will write about them.

Let $Var$ be the category of varieties. The morphisms between objects are the homomorphisms between the clones of the varieties. One object in $Var$ is the variety ${\mathcal V}_{\mathbb R}$ generated by the algebra whose underlying set is $\mathbb R$ and whose operations are all continuous operations on $\mathbb R$.

Now let $\mathcal U$ be any other variety. Each morphism $\mathcal U\to {\mathcal V}_{\mathbb R}$ corresponds to a way of equipping $\mathbb R$ with compatible continuous operations defining a $\mathcal U$-structure on $\mathbb R$.

If you want to know `what kinds of algebras are compatible with the reals', then you want to know the principal ideal in $Var$ defined by ${\mathcal V}_{\mathbb R}$: i.e. $({\mathcal V}_{\mathbb R}]:=\{{\mathcal U}\;|\;\exists \varphi(\varphi\colon {\mathcal U}\to {\mathcal V}_{\mathbb R}\;\textrm{is a hom})\}$. Moreover, if you want to know all ${\mathcal U}$-structures on $\mathbb R$, you want to know all homomorphisms $\varphi\colon {\mathcal U}\to {\mathcal V}_{\mathbb R}$.

(The last sentence of the previous paragraph shows that a better object than $({\mathcal V}_{\mathbb R}]$ for this question is the slice category $Var/{\mathcal V}_{\mathbb R}$, since this identifies not just the types of algebras definable on $\mathbb R$ by continuous operations, but also their realizations.)

Taylor showed that there is no algorithm to determine if a finitely presented variety belongs to $({\mathcal V}_{\mathbb R}]$. He did this by interpreting the undecidable satisfiability problem for Diophantine equations into this problem.

On the other hand, Taylor shows that there is an easy algorithm to determine whether a variety $\mathcal U$ defined by a finite set $\Sigma$ of simple equations belongs to $({\mathcal V}_{\mathbb R}]$. (An equation is simple if each side has at most one function symbol.) He shows that such ${\mathcal U}$ belongs to $({\mathcal V}_{\mathbb R}]$ iff $\mathcal U$ has a 2-element model.

From Taylor's work it is easy to see that it is possible to equip $\mathbb R$ with a continuous commutative binary operation, since $x*y=y*x$ is a simple equation that has a $2$-element model. But Taylor's work does not tell us all ways of equipping $\mathbb R$ with a continuous commutative binary operation.

Taylor's work says nothing about the associative law, because it is not simple.

Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, consider constant operations $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

Regarding Q2.2: There are obstacles to constructing a third associative binary operation from two given ones, if you intend to construct the third by some form of composition. For example, the projections $\ell(x,y)=x$ and $r(x,y)=y$ are both associative, but the only binary operations you can get from these by composition are $\ell(x,y)$ and $r(x,y)$ themselves. Similarly, $\max(x,y)$ and $\min(x,y)$ are associative, but the only binary operations in the clone they generate are $\max, \min, \ell, r$. So any general construction would have to trivialize in these situations.

Nevertheless, here is something. Suppose you are given continuous, associative, binary operations $x+y$ and $x*y$. Assume in addition that (i) addition is commutative, and (ii) multiplication distributes over addition on both sides. Then $x\circ y:= x+y+x*y$ is a third continuous, associative, binary operation.

• Are there infinitely many nonisomorphic associative continuous operations on R? – Emil Jeřábek May 29 '17 at 19:15
• Hmm, actually, there are: there are even continuum many nonisomorphic continuous t-norms on $[0,1]$, which I believe extend to the whole R. Now, these are classified by the Mostert-Shields theorem; is a similar classification possible for all continuous associative operations on R? – Emil Jeřábek May 29 '17 at 19:19
• Great observation! This seems like a potentially achievable project. – Keith Kearnes May 29 '17 at 19:35