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 $P$.

  • $\begingroup$ 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 $P$. $\endgroup$
    – Tim Campion
    Jun 26, 2018 at 21:00
  • $\begingroup$ It's also probably worth mentioning that the algebras for the operad $(P,\circ)$ are precisely the convex spaces. $\endgroup$
    – Tim Campion
    Jul 6, 2020 at 21:48

3 Answers 3


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.

  • 2
    $\begingroup$ 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. $\endgroup$ Jun 25, 2018 at 8:30
  • $\begingroup$ 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. $\endgroup$
    – Tim Campion
    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.

  • $\begingroup$ I'm a little confused. Isn't there an operad morphism $C \to P$ sending the unique point of $C_n$ to $(1/n,\dots,1/n)$? $\endgroup$
    – Tim Campion
    Jun 25, 2018 at 10:51
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 25, 2018 at 10:58
  • 2
    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Jun 25, 2018 at 11:06
  • 1
    $\begingroup$ 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? $\endgroup$ Jun 25, 2018 at 11:24
  • 2
    $\begingroup$ 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. $\endgroup$ 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 $x$.


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