## Automorphisms of the rooted tree operad

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This follows Ryan Budney's comment to the question asked here.

What is the automorphism group of the rooted tree operad?

(By the rooted tree operad, I just mean the operad with object rooted trees and morphisms given by grafting a root to a leaf).

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Are you interested in the linear operad, or the setwise operad?   Also what is the arity of a tree, is it the number of vertices or the number of leaves?  Also I wouldn't call it the rooted tree operad, I would guess that most people would take this to be a form of the PreLie operad as described by Chapoton and Livernet. – James Griffin Sep 29 2010 at 13:38
arity = number of leaves. – Dr Shello Sep 29 2010 at 14:27

## 1 Answer

I think the answer to the question as literally stated is "the trivial group", but I think there are related inquiries which get into some deep combinatorics.

One way of thinking about the rooted tree operad is that it is the free operad $O(F)$ generated by the Joyal species $F$ (a functor $\mathbb{P} \to Set$ where $\mathbb{P}$ is the groupoid of finite sets ${1, \ldots, n}$ and permutations) where $F(0)$ is empty and $F(n)$ is a singleton for $n \geq 1$. You can think of the element of $F(n)$ as a "sprout" $s_n$ consisting of a root, $n$ leaves, and no other nodes, and then the elements of $O(F)$ are obtained recursively by starting with sprouts and applying grafting operations.

So we're looking at operad automorphisms $\phi: O(F) \to O(F)$. By freeness, the endomorphisms of $O(F)$ are in bijection with natural transformations $\psi: F \to U O(F)$ where $U O(F)$ is the underlying species or permutation representation of $O(F)$. Concretely, to give such a natural transformation is to give a collection of trees $t_n = \psi_n(s_n)$ for all $n \geq 1$ where each $t_n$ must be invariant under permuting the leaves, since the sprout $s_n$ is invariant under such permutations. That's a pretty strong condition on $t_n$, and there are actually precious few such collections.

But now you want more: you want $\phi$ to be an automorphism as well. So each sprout $s_n$ must be in the image of $\phi_n$. But no nonsprout tree $u$ can ever map to $s_n$ under $\phi_n$, because if $u$ is obtained by grafting together more than one sprout $s_k$, then $\phi_n(u)$ is obtained by similarly grafting together more than one tree $t_k$, and this is never a sprout.

So in order for there to exist $u$ such that $\phi_n(u) = s_n$, we must have $u = s_n$. To have that for all $n$ means $\psi(s_n) = s_n$ for all $n$, hence the only operad automorphism is the identity automorphism.

I think a more interesting inquiry is to understand the groupoid of rooted trees and isomorphisms between them. This is an incredibly rich object.

Edit: Let me make my last suggestion more precise. Let's define a rooted tree to be a finite set $X$ equipped with a function $f: X \to X$ and an element $r \in X$ such that $f^{(n)}(X) = {r}$ for sufficiently large $n$. The idea is that $f(x)$ is one step closer to the root than $x$, unless $x$ is the root. Then an isomorphism is a function $\phi: X \to Y$ which preserves the stepping-closer function and the root. It is determined by its restriction to the leaf set.

But even this groupoid isn't that mysterious; it seems automorphism groups are iterated wreath products of symmetric groups. Here is a different but related inquiry which I think is rather more interesting: regarding trees $T$ as posets, describe the category of order-preserving bijections between trees.

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Very interesting. Can you say more about this groupoid of rooted trees? Also, what are examples of: - a "simple" operad that has a non-trivial automorphism group? - an operad that has a relatively simple automorphism group? – Dr Shello Sep 29 2010 at 14:34

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1. The groupoid is equivalent to a sum over isomorphism classes of trees of the automorphism groups of class representatives, and each such automorphism group is an iterated wreath products of symmetric groups. A useful picture might be to think of a tree as a hereditarily finite multiset. Then an automorphism of a multiset consists of a permutation of multiple copies of an element together with an automorphism of each element (as a multiset). 2. E.g., a monoid or group can be viewed as an operad where each operation has arity one. Pick a monoid with an interesting automorphism group. (Cont.) – Todd Trimble Sep 30 2010 at 7:38
For example, free groups have interesting automorphism groups; they contain braid groups for instance (cf. Artin representation). 3. I am not absolutely sure, but I think the automorphism group of the operad whose algebras are monoids might be Z mod 2. The nontrivial automorphism would send an operation of arity n, namely a total ordering of the elements 1, 2, ..., n, to the reverse ordering. – Todd Trimble Sep 30 2010 at 7:50

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That's right the automorphism group of the (set-wise) associative operad is Z mod 2. A really important point is that automorphisms of operads induce automorphisms of categories of algebras. In the case of the associative operad the non-trivial automorphism takes an algebra A to its oppositive algebra. – James Griffin Sep 30 2010 at 10:54

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Oh and I think that we should point out that the automorphism group of the operad in the original question is only trivial if we take it to be the set-wise operad. Working instead with the linear version I think we get a group resembling the upper-triangular matrices, but I haven't checked the details. – James Griffin Sep 30 2010 at 10:58
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