How many operad structures are there on the symmetric sequence of simplices / finitely-supported probability measures?

Consider the symmetric sequence $$P_n = \Delta^{n-1}$$ of probability measures on finite sets, with coordinatewise $$\Sigma_n$$-action. There is a natural topological operad structure on $$P$$ given by multiplication. That is,

• Let $$p \in \Delta^{n-1}$$ be given in coordinates by $$p = (p^1,\dots,p^n)$$.

• For each $$i=1,\dots,n$$, let $$p_i \in \Delta^{n_i-1}$$ be given in coordinates by $$p_i = (p_i^1,\dots,p_i^{n_i})$$.

Then the operad composition is given by

$$p \circ (p_1,\dots,p_n) = (p^1 p_1^1, p^1 p_1^2,\dots, p^1 p_1^{n_1}, p_2 p_2^1, \dots, p^n p_n^1, \dots, p^n p_n^{n_n})$$.

For example, $$(1/3,2/3) \circ ((1),(1/4,1/4,1/2)) = (1/3,1/6,1/6,1/3)$$.

(I apologize for the notation -- in particular the tuples on the two sides of the above equation mean different things -- on the left, it's a list of inputs to the $$\circ$$ operation, while on the right it's a list of coordinates. Another way to write this would be $$(p^i)_{i=1}^n \circ ((p_1^j)_{j=1}^{n_1},\dots, (p_n^j)_{j=1}^{n_n}) = (p^i p_i^j)_{1 \leq i \leq n, 1 \leq j \leq n_i}$$.)

We can twist this operad by any convex increasing homeomorphism $$f: [0,1] \to [0,1]$$ as follows. Define $$f_\ast: \Delta^{n-1} \to \Delta^{n-1}$$ by $$f_\ast(x_1,\dots,x_n) = (\frac{f(x_1)}{Z},\dots, \frac{f(x_n)}{Z})$$ where $$Z = f(x_1) + \dots + f(x_n)$$; this is a $$\Sigma_n$$-equivariant homemormophism ( invertibility requires $$f$$ to be convex ), and then set

$$p \circ^f (p_1,\dots, p_n) = f_\ast^{-1}(f_\ast(p) \circ (f_\ast(p_1),\dots, f_\ast(p_n)))$$

Question: Is every topological operad structure on the symmetric sequence $$P$$ of the form $$\circ^f$$ for some convex increasing homeomorphism $$f: [0,1] \to [0,1]$$?

Guessing the answer is "yes" is motivated by the characterization of quasi-arithmetic means.

Background: The operad $$(P,\circ)$$ features explicitly in Leinster's operadic characterization of entropy, and twists $$(P,\circ^f)$$ for certain $$f$$ are used by Baez, Fritz, and Leinster to characterize certain deformations of entropy. Certainly these operads are implicit in countless mathematical pursuits.

Explicitly: The fixed data is the sequence of topological spaces $$(P_0,P_1,\dots)$$ and the $$\Sigma_n$$ action on $$P_n$$ (this is called a symmetric sequence, though you might also call it a topological species). An operad structure on the symmetric sequence $$P$$ is a monoid structure for the substitution monoidal product. That is, it consists of continuous operations $$\circ: P_n \times P_{n_1} \times \dots \times P_{n_n} \to P_{n_1 + \dots + n_n}$$ for each $$n,n_1,\dots,n_n \in \mathbb N$$ satisfying:

• unitality: $$1 \circ p = p = p \circ (1, \dots, 1)$$ for some $$1 \in P_1$$

• associativity: The two ways of associating $$p \circ (p_1,\dots,p_n) \circ (p_{1,1},\dots, p_{n,n_n})$$ are equal.

• $$\Sigma_n$$-equivariance: $$(\sigma \cdot p) \circ (p_1,\dots, p_n) = \sigma' \cdot (p \circ (p_{\sigma^{-1}1},\dots, p_{\sigma^{-1}n}))$$ for each $$\sigma \in \Sigma_n$$, where $$\sigma' \in \Sigma_{n_1 + \dots + n_n}$$ is the image of $$\sigma$$ under the natural inclusion $$\Sigma_n \ltimes (\Sigma_{n_1} \times \dots \times \Sigma_{n_n}) \hookrightarrow \Sigma_{n_1+\dots+n_n}$$.

The question asks to classify operad structures on the symmetric sequence $$P$$.

• In the interest of this question not being a moving target, I've accepted James Griffin's answer below -- I could equally have accepted Neil Strickland's answer, which arguably provides more insight; I opted for the first-to-answer criterion. Of course, it would be interesting to know if some weakening of my original proposed classification does indeed hold -- in particular, as suggested by Peter Lefanu Lumsdaine below, it seems natural to ask if the $\circ^f$ operad structures are at least dense among all operad structures on the symmetric sequence $P$. Jun 26, 2018 at 21:00
• It's also probably worth mentioning that the algebras for the operad $(P,\circ)$ are precisely the convex spaces. Jul 6, 2020 at 21:48

Abusing notation, write m for the uniform distribution on each finite set and define p o p' = m for any p, p' not equal to the identity.

This is possibly a limit of operads of the formula you give where you choose homeomorphisms which tend (pointwise) to a constant function f(x) = c. But I think qualifies as an operad structure not in your suggested classification.

• May I suggest modifying your potential classification by asking that the symmetric sub-sequence consisting of the simplex boundaries is an operad ideal. There is probably a nicer way of stating the above, but a face of a simplex has a zero in it, and you probably want that zero to act trivially in some sense. Jun 25, 2018 at 8:30
• Having an additional boundary condition is an interesting suggestion. I was worried for a minute that $P_0 = \emptyset$ might mess things up, but it is in fact the case that $(P,\circ^f)$ has the property that the boundaries form an operad ideal -- in fact, for any $k$, the codimension-$k$ corners form an operad ideal. Interestingly, Neil Strickland's example also has this property, so it's not a strong enough condition to cut down to just the $\circ^f$ operad structures. Jun 25, 2018 at 11:25

Put $Q_n=\{(x\in [0,1]^n:\text{max}(x_1,\dotsc,x_n)=1\}$. Then there is an isomorphism $Q\to P$ of symmetric sequences given by $x\mapsto x/\sum_ix_i$. Define $$p \circ (p_1,\dots,p_n) = (\min(p^1,p_1^1), \min(p^1,p_1^2),\dots, \min(p^1 ,p_1^{n_1}),\min(p_2,p_2^1), \dots,\min(p^n,p_n^1), \dots,\min(p^n,p_n^{n_n})).$$ This gives an operad structure on $Q$. Let $C$ be the commutative operad, so that $C_n$ is a singleton for all $n$. There is an operad morphism $C\to Q$ sending the unique point of $C_n$ to $(1,\dotsc,1)$, but there is no operad morphism $C\to (P,\circ^f)$, so $Q\not\simeq(P,\circ^f)$ as operads.

• I'm a little confused. Isn't there an operad morphism $C \to P$ sending the unique point of $C_n$ to $(1/n,\dots,1/n)$? Jun 25, 2018 at 10:51
• No, that's not an operad morphism. Try some small examples. This is essentially the fact that the group operation on the fundamental group is not associative before passing to homotopy. Jun 25, 2018 at 10:58
• Oh of course -- you need to look at $f \circ (g_1,\dots, g_n)$ where the arities of the $g_i$ are not all the same. Jun 25, 2018 at 11:06
• Like James Griffin’s example below, this structure is a limit of operad structures that do appear in your classification, if I’m not mistaken: it’s the limit as $n \to \infty$ of $(P, \circ^f)$ where $f$ is the $n$th power map, just as the “max” operation is a limit of the $n$th-power means. So perhaps the structures you suggest might be dense among all operad structures? Jun 25, 2018 at 11:24
• The examples so far arise from different choices of structure on the set of non-negative real numbers. For the $\circ_f$ example it is a rig with the structure maps conjugates by f of the usual one. For Neil's example, min and max are the operations. My example is too degenerate to say much about. Jun 25, 2018 at 13:20

I just want to verify that

Fact: If $$f: [0,1] \to [0,1]$$ is any increasing homeomorphism, then

$$f_\ast: \Delta^{n-1} \to \Delta^{n-1}, (x_i)_i \mapsto (f(x_i)/(f(x_1) + \dots + f(x_n)))_i$$

is a ($$\Sigma_n$$-equivariant) homeomorphism.

It is not necessary for $$f$$ to be convex as I had initially asserted.

Proof: Let $$\vec y = (y_i)_i \in \Delta^{n-1}$$, and assume without loss of generality that $$y_1 = \max_i y_i$$ (and in particular $$y_1 \neq 0$$). Because $$f_\ast$$ is a continuous map between compact Hausdorff spaces, it suffices to show that $$\vec y$$ has a unique inverse under $$f_\ast$$.

Extend $$f_\ast$$ to a map $$[0,1]^n \setminus (0,\dots,0) \to \Delta^{n-1}$$ by using the same formula as above. Note that for $$\vec x \in [0,1]^n$$, we have $$f_\ast(x_1,\dots,x_n) = (y_1,\dots,y_n)$$ if and only if $$x_1 \neq 0$$ and $$f(x_i)/f(x_1) = y_i/y_1$$ for all $$i$$. That is, defining $$\nu: (0,1] \to [0,1]^n$$ by $$\nu_i(x) = f^{-1}(y_i f(x_1)/y_1)$$, we have $$f_\ast(\vec x) = \vec y$$ if and only if $$x_1 \neq 0$$ and $$\vec x = \nu(x_1)$$. (The formula for $$\nu$$ makes sense because $$y_i \leq y_1$$ and $$f(x_1) \leq 1$$).

We now claim that there is a unique $$x \in (0,1]$$ such that $$\nu(x) \in \Delta^{n-1}$$, i.e. such that $$\sum_i \nu_i(x) = 1$$, which is equivalent to our claim that $$\vec y$$ has a unique inverse under $$f_\ast: \Delta^{n-1} \to \Delta^{n-1}$$. This holds because the function $$(0,1] \to [0,\infty), x \mapsto \sum_i \nu_i(x)$$ is strictly increasing and continuous, tending to 0 as $$x \to 0$$, and with a maximum value at $$x=1$$ of $$\nu_1(1) + \nu_2(1)+ \dots + \nu_n(1) = 1 + \nu_2(1) + \dots + \nu_n(1) \geq 1$$. So by the intermediate value theorem, this function attains the value 1, and because it is strictly increasing it attains this value for a unique $$x$$.