Combinatorial type construction of the free operad

Question

$\DeclareMathOperator\RT{RT}$I am reading the book "Algebraic operads" by J. L. Loday and B. Vallete. The authors have given a combinatorial construction of the free operad over an $\mathbb{S}$-module $M$ in Section 5.5, pg. 126. I have a question regarding this construction, which I am describing below.

Let $\RT(n)$ denote the set of all nonplanar rooted trees with $n$ leaves such that each vertex has at least one input. For any tree $t \in \RT(n)$, let $\mathrm{vert}(t)$ be the set of all vertices of $t$ and $\mathrm{in}(v)$ denote the number of inputs of the vertex $v \in \mathrm{vert}(t)$.

For any $\mathbb{S}$-module $M=(M(0),M(1),M(2),\ldots)$ with $M(0)=0$ and $t \in \RT(n)$, the treewise tensor product of $M$, denoted $M(t)$, is defined as $$M(t) := \bigotimes_{v ~\in~ \mathrm{vert}(t)} M\big(in(v)\big)$$

The free operad $\mathbb{T}(M)$ over $M$:

The free operad $\mathbb{T}(M)= (\mathbb{T}(M)(0), \mathbb{T}(M)(1), \mathbb{T}(M)(2) ,\ldots)$ over the $\mathbb{S}$-module $M$ is defined as $$\mathbb{T}(M)(n) := \bigoplus_{t ~\in ~\RT(n)} M(t)$$ for all $n \ge 0$.

The authors in Section 7.6, p. 193 mentions that for a given binary quadratic operad $\overline{E} = (0,0,E,0,\ldots)$ the lower degree modules of the free operad $T(\overline{E})$ is described as $$T(\overline{E})(0) = 0, \quad T(\overline{E})(1) = \mathbb{K}, \quad T(\overline{E})(2) = E, \quad T(\overline{E})(3) = 3 (E \otimes E)$$

My question:

I am convinced by the fact that $T(\overline{E})(2) = E$. But how are the authors arriving at the fact that $T(\overline{E})(3) = 3(E \otimes E)?$ According to the definition mentioned above of the free operad we get that $$T(\overline{E})(3) = \bigoplus_{t \in \RT(3)} \overline{E}(t)$$
We know that $\RT(3)$ contains only three elements as shown below:

$\hspace{5cm}$

Then we get $T(\overline{E})(3) = \overline{E}(2) \otimes \overline{E}(2) \bigoplus \overline{E}(2) \otimes \overline{E}(2) \bigoplus \overline{E}(3) = 2 (E \otimes E)$. But how do the authors get $T(\overline{E})(3) = 3(E \otimes E)$?

Edit 1:

Following Todd Trimble's answer if we try to give an explicit description of $T(\overline{E})(4)$. Then we must find the collection of all subsets of $\{1,2,3,4\}$ satisfying the relations mentioned in Todd Trimble's answer which gives us $19$ possibilities: $$\big\{\{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{1,2\}, \{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{1,3\}, \{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{2,3\}, \{1,2,3\}, \{1,2,3,4\}\big\}$$
$$\big\{\{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{1,2\}, \{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{1,4\}, \{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{2,4\}, \{1,2,4\}, \{1,2,3,4\}\big\}$$
$$\big\{\{1,3,4\}, \{1,2,3,4\}\big\}, \big\{\{1,3\}, \{1,3,4\}, \{1,2,3,4\}\big\}, \big\{\{1,4\}, \{1,3,4\}, \{1,2,3,4\}\big\}, \big\{\{3,4\}, \{1,3,4\}, \{1,2,3,4\}\big\}$$
$$\big\{\{2,3,4\}, \{1,2,3,4\}\big\}, \big\{\{2,3\}, \{2,3,4\}, \{1,2,3,4\}\big\}, \big\{\{2,4\}, \{2,3,4\}, \{1,2,3,4\}\big\}, \big\{\{3,4\}, \{2,3,4\}, \{1,2,3,4\}\big\}$$
$$\big\{\{1,2\},\{3,4\},\{1,2,3,4\}\big\}, \big\{\{1,3\},\{2,4\},\{1,2,3,4\}\big\}, \big\{\{1,4\},\{2,3\},\{1,2,3,4\}\big\}$$

The possibilities $\big\{\{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{1,3,4\}, \{1,2,3,4\}\big\}$, and $ \big\{\{2,3,4\}, \{1,2,3,4\}\big\}$ indexes a trivial summand (because $\overline{E}(3)=0$). Therefore, that leaves us with the remaining fifteen, where the treewise tensor products are each isomorphic to $E \otimes E \otimes E$.

Therefore, $T(\overline{E})(4) = 15 (E \otimes E \otimes E)$. Is this correct?

Edit 2:

Following Todd Trimble's answer, I have a small question regarding the construction of the free operad. The set $\{\{1,3\},\{1,2,3\}\}$ corresponds to the following graph

$\hspace{6cm}$

is this correct? If yes, then what about this graph

$\hspace{6cm}$

will this not be present in the $\mathbb{T}(E)(3)?$ It looks like this graph also corresponds to the set $\{\{1,3\},\{1,2,3\}\}$.

Answer 1

There are in fact infinitely many nonplanar rooted tree structures having a given nonempty set of leaves $S$, because for example such a tree can look like a linear stalk of any finite height topped by a corolla.

But in the situation of $\mathbb{S}$-modules $M$ such that $M(0) = 0$ and $M(1) = 0$, the trees that correspond to nontrivial summands of the free operad $T(M)$ can be identified with collections of subsets of $S$ such that

• Each $T$ in the collection has cardinality greater than $1$,

• The collection includes $S$, and

• For subsets $T, T'$ belonging to the collection, either $T \subseteq T'$ or $T' \subseteq T$ or $T \cap T' = \emptyset$.

The way you should think of this is that each $T$ in the collection is the set of leaves lying above a given vertex in the tree (and in this situation each vertex of the tree is uniquely determined by the set of leaves above it).

For the set $S = \lbrace 1, 2, 3\rbrace$, there are four possibilities:

$$\lbrace \lbrace 1, 2\rbrace, \lbrace 1, 2, 3\rbrace\rbrace, \qquad \lbrace \lbrace 1, 3\rbrace, \lbrace 1, 2, 3\rbrace\rbrace, \qquad \lbrace \lbrace 2, 3 \rbrace, \lbrace 1, 2, 3\rbrace\rbrace, \qquad \lbrace\lbrace 1, 2, 3\rbrace\rbrace$$

You've already observed that for the given binary quadratic $\mathbb{S}$-module $E$, the last possibility indexes a trivial summand (because $E(3) = 0$), so that just leaves the first three, where the treewise tensor products are each isomorphic to $E \otimes E$.

So I would say that the trouble with your picture is that you didn't actually present the labels on the leaves (which label inputs or arguments for the operations in the free operad). A side issue is that for the two non-corolla trees in your post, these are isomorphic graphs or in other words indistinguishable in the nonplanar context. But the main issue is that you didn't record the labeling of leaves.