This was previously asked and bountied at math.stackexchange with no response.

Let $C(\mathbb{R}^2,\mathbb{R})$ be the space of all continuous functions $\mathbb{R}^2\rightarrow \mathbb{R}$ with the compact-open topology. I'm interested in analyzing subspaces of $C(\mathbb{R}^2,\mathbb{R})$ determined by first-order theories, especially finite equational theories. Specifically, for $T$ a set of first-order sentences in the language of a single binary operation, let $T_\mathbb{R}$ be the subspace of continuous $f$ such that $(\mathbb{R};f)\models T$. I'm curious about how $T$ affects the purely topological properties of $T_\mathbb{R}$. For example, when is $T_\mathbb{R}$ connected? etc.

However, even in very concrete cases, I'm having trouble understanding what $T_\mathbb{R}$ looks like. Letting $C$ and $A$ be the usual statements of commutativity and associativity, I think an answer to the following question would clear things up immensely:

Question: is $\{C\}_\mathbb{R}\cong\{A\}_\mathbb{R}$?

$\{C\}_\mathbb{R}$ is pretty tame since as Eric Wofsey observed it's a vector subspace of $C(\mathbb{R}^2,\mathbb{R})$. However, $\{A\}_\mathbb{R}$ seems much weirder. For example, I don't even know whether $\{A\}_\mathbb{R}$ is connected.

  • $\begingroup$ My guess is not. Since associativity is equivalent to more general forms, one can make from certain associative f some simple polynomial functions g which are also associative (for example x+y+c, xyc, and g(x,y)= f(x,f(y,c)) more generally), but is unclear that there is a path from constant functions to addition, for example. Gerhard "Is Not Seeing Much Connection" Paseman, 2020.04.24. $\endgroup$ Apr 24 '20 at 23:57
  • $\begingroup$ @GerhardPaseman I also suspect the answer is negative (I said so in the MSE version), but I don't see any way to prove that. $\endgroup$ Apr 25 '20 at 0:01
  • $\begingroup$ Since the equivalence sign is ambiguous to me, you might add that you mean homeomorphic, otherwise people might suggest that scalar multiplication distinguishes the two. Gerhard "Make The Context More Local" Paseman, 2020.04.24. $\endgroup$ Apr 25 '20 at 0:25
  • 1
    $\begingroup$ In $\{A\}_\mathbb{R}$, are the functions $f(x,y)=x$ and $f(x,y)=y$. isolated points? Or is there a path between the two? $\endgroup$
    – Matt F.
    Apr 27 '20 at 17:10
  • 1
    $\begingroup$ @arsmath It's "$\cong$," not "$=$" - I'm asking if the corresponding topological spaces are homeomorphic, not if every commutative operation is associative. $\endgroup$ Apr 27 '20 at 18:51

The following result from Section 6.2 of János Aczél's classic book Lectures on Functional Equations and their Applications seems to be closely related, and the long list of references on the solutions of the associativity equation (p.253/4) may provide further entry points into the literature. (Although that list is way out of date, with the book being from 1966.)

Theorem: Let $f \in C(\mathbb{R}^2,\mathbb{R})$ satisfy the associativity equation $$f(x,f(y,z)) = f(f(x,y),z).$$ Suppose moreover that $f$ is cancellative, i.e. that $$ f(x,y) = f(x,z) \quad \Rightarrow \quad y = z, $$ and similarly in the other argument. Then $f$ is conjugate to addition $+ : \mathbb{R}^2 \to \mathbb{R}$ via a monotone homeomorphism. In particular, $f$ is commutative.

It seems plausible that this result can be used to prove that if your theory $T$ consists of associativity and the cancellativity laws, then $T_{\mathbb{R}}$ is homemorphic to the space of monotone homeomorphisms $\mathbb{R} \to \mathbb{R}$. But I haven't worked out the details.

  • 1
    $\begingroup$ +1 - although that doesn't answer the question, it's a very neat observation (and the reference is great, I'd never heard of that book). $\endgroup$ Apr 27 '20 at 16:32

Here are some functions in the connected component of the associative function $f(x,y)=x$.

Proposition: If $f(x,y)=g(x)$ and $f$ is associative, then there is an interval $[a,b]$ (possibly infinite on either side) such that $$g(x)=x \text{ for all }x\in [a,b]$$ $$g(x)\in[a,b]\text{ for all }x\in \mathbf{R}$$ Furthermore all such $g$ produce associative $f$.

Proof: By associativity, $g(g(x))=g(x)$. So if $y$ is in the range of $g$ then $g(y)=y$. Then let $a=\inf(range(g))$ and let $b=\sup(range(g))$. The other direction is trivial. $\square$

So in any case, the connected component of $f(x,y)=x$ is substantial. As an easy example, the function $f(x,y)=x$ is the limit of $f_n(x,y)=\max(n,x)$.

It seems plausible to me that these associative functions are disconnected from the ones that depend on both variables. If so, the associative subspace would be disconnected.

  • 1
    $\begingroup$ I think there is a path between the projection functions $(x,y)\mapsto x$ and $(x,y) \mapsto y$. Consider the one-parameter family of continuous associative functions $f_t(x,y)=\max(\min(x,\tan(\pi t/2)),y)$ with $−1<t<1$, which form a path between $f_{−1}(x,y):=y$ and $f_1(x,y):=\max(x,y)$. By symmetry there is another path between $f_1(x,y)$ and $(x,y)↦x$, so we may take the composition of these two paths. $\endgroup$
    – pregunton
    Mar 21 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.