# What is an Operad? Part 1

/If you browse through the research of your local algebraist, homotopy theorist, algebraic topologist or―well, anyone whose research involves an

*operation*of some type, you might come across the word "operad." But what are*operads*? And what are they good for? Loosely speaking, operads―which come in a*wide*variety of types―keep track of various "flavors" of operations. Historically, they arose from a quest to understand $k$-fold loop spaces in homotopy theory. I'll say a little more about this next time, when we'll look at some examples and applications. In today's post, I simply want to present the*definition*of an operad. To start, what do I mean by "flavors" of operations?Let's think about multiplication on an algebra. An

*algebra*is a vector space $V$ equipped with a bilinear map \begin{align*} m\colon V\times V&\longrightarrow V\\ (a,b)&\longmapsto m(a,b) \end{align*} which we can think of as multiplication, so let's write $a\cdot b$ instead of $m(a,b)$. (And if we want a*unital*algebra, we may ask for an element $1\in V$ so that $1\cdot a=a=a\cdot 1$ for all $a\in V$.) Now, depending on what relations we ask $m$ to satisfy, our algebra $V$ is given different names. For example, if we ask that $$(a \cdot b)\cdot c = a \cdot (b\cdot c)$$ then $V$ is called an**associative algebra**. If we also ask that $$a\cdot b = b\cdot a$$ then $V$ is a**commutative algebra**. The real line $\mathbb{R}$ with the usual multiplication is an example of a commutative algebra. On the other hand, if―and let's now think of $m(a,b)$ as a bracket $[a,b]$―we ask that $$[a,b]=-[b,a] \qquad\text{and}\qquad [a,[b,c]]=[[a,b],c]+[b,[a,c]]$$ then $V$ is called a**Lie algebra.**An example of a Lie algebra is all vectors in $\mathbb{R}^3$ together with the cross product. Behind*each*of these algebras is a particular operad that encodes the "flavor" of the operation $m\colon V\times V\to V$. How so? Let's think of the 2-to-1 map $m$ as a binary tree:More abstractly, let’s call anything with $n$ inputs and 1 output an "$n$-to-1 operation" or an "$n$-ary operation."

There's a way to compose operations. For instance, if $f$ is a 3-ary operation and $g$ is a 4-ary operation, we can combine them to get a 6-ary operation: just use the output of $g$ as one of the inputs of $f$:

Notice I've grafted $g$ on the

*second*leaf of the tree depicting $f$, so I'm calling that composition "$\circ_2$." Also notice how the leaves are relabeled after the composition. And there are two other possibilities, $f\circ_1g$ and $f\circ_3 g.$In general, there are $n$ ways to compose an $m$-ary operation with an $n$-ary operation and the result is always an $m+n-1$-ary operation. An

**operad**is the collection of all such operations, together with the compositions. These compositions should satisfy three very sensible axioms that are easy to understand pictorially, but a mess to write down. So let's look at some pictures! To start, suppose we're given operations $f,g$ and $h$.## Axiom 1:

Composition behaves nicely.

Let's say we want to stick $h$ onto $f\circ_2 g$. How many ways can we do this? There are three options: (case 1) we can stick $h$ somewhere to the left of $g$, or (case 2) we can stick $h$ on $g$ itself, or (case 3) we can stick $h$ somewhere to the right of $g$. The picture below is an example of case 1. Grafting $h$ onto the first leaf of $f\circ_2 g$ gives us the first equals sign. But we could have obtained the tree in the middle (the one with 7 leaves) in a

*different*way: first graft $h$ to the first leaf of $f$,*then*graft $g$ onto the*third*leaf of the new composition, $f\circ_1 h$. And that gives us the second equals sign. Conclusion? $(f\circ_2 g)\circ_1h=(f\circ_1 h)\circ_3g$.Now, what about case 2? Grafting $h$

*on*$g$ (while $g$ is already attached to $f$) should be the same as first grafting $h$ on $g$ (before $g$ is attached to $f$), and then inserting the new operation $h\circ g$ into $f$. For example, $(f\circ_2 g)\circ_3h = f\circ_2(g\circ_2h).$Finally, in case 3, Inserting $h$ to the

*right*of $g$ (while $g$ is already attached to $f$) should be the same as first inserting $h$ on $f$, and then attaching $g$. For example, $(f\circ_2)g\circ_6 h= (f\circ_3h)\circ_2 g.$## Axiom 2:

Permutations behave nicely.

If we like, we can permute the inputs of any $n$-ary operation using elements in the symmetric group $S_n$. For example, if $f$ is a 3-ary operation and $\sigma=(123)\in S_3$, then $\sigma f$ is a new 3-ary operation given by

And what if we want to compose this with a 2-ary operation , $g$ whose inputs have also been permuted? This axiom requires that first twisting $f$ and $g$ (individually) and then composing is the

*same*as first composing and*then*twisting the new tree in an appropriate way. For example if $\tau=(12)\in S_2$ then we must have $\sigma f \circ_2 \tau g=(\sigma\circ_2 \tau)(f\circ_{\sigma(1)}g)$.where $\sigma\circ_2\tau$ is the permutation, "do $\sigma$, but replace 2 by $\tau$."

## Axiom 3: There's a unit

(which behaves nicely).

Lastly, we require the existence of a 1-ary operation, which I'll call 1, that serves as an identity. For instance, $f\circ_3 1=f=1\circ_1 f$.

## The Definition

And there you have it! An operad is a collection of 1,2,3,$\ldots$-ary operations that can be composed and permuted à la the sensible axioms above. Writing this down is a horrible mess, but let's do it for fun. To start, let's think of all $n$-ary operations as forming a set denoted by $\mathcal{O}(n)$. The definition is as follows

**Definition:** An *operad* is a sequence of sets $\mathcal{O}=\{\mathcal{O}(1),\mathcal{O}(2),\mathcal{O}(3)\ldots\}$ together with composition functions
$$\circ_i\colon \mathcal{O}(n)\times \mathcal{O}(m)\to\mathcal{O}(m+n-1)$$
satisfying the following:

- For all $f\in\mathcal{O}(n)$, $g\in\mathcal{O}(m)$, and $h\in\mathcal{O}(p)$, $$ (f\circ _j g)\circ_i h= \begin{cases} (f\circ_i h)\circ_{j+p-1}g &\text{if $1-\leq i \leq j-1$} \\[5pt] f\circ_j(g\circ_{i-j+1}h) &\text{if $j\leq i\leq n+j-1$}\\[5pt] (f\circ_{i-m+1}h)\circ_jg &\text{if $i\geq n+j$} \end{cases} $$
- Each $\mathcal{O}(n)$ has an action of the symmetric group $S_n$, \begin{align*} S_n\times \mathcal{O}(n)&\longrightarrow \mathcal{O}(n)\\ (\sigma,f) &\longmapsto \sigma f \end{align*} such that for all $f\in\mathcal{O}(n)$ and $g\in \mathcal{O}(m)$, and for all $\sigma\in S_n$ and $\tau\in S_m,$ $$\sigma f\circ_i\tau g=(\sigma\circ_i\tau )(f\circ_{\sigma(i)} g) \qquad 1\leq i \leq n$$ where $\sigma\circ_i\tau$ is the block permutation, "replace the $i$th entry of $\sigma$ with $\tau$."
- There exists $1\in\mathcal{O}(1)$ (think of it as the "identity operation") so that for every $n$ and for every $f\in\mathcal{O}(n)$, $$1\circ_1 f=f \circ_i 1=f \qquad 1\leq i \leq n.$$

*symmetric monoidal category*.) In those cases, we ask that the $\circ_i$ and $S_n$ action be the appropriate type of map: linear transformations, or continuous functions, or chain maps, for instance.## Example?

Endomorphisms!

Here's the flagship example of an operad. Fix a vector space $V$ and for each $n=1,2,\ldots$, define
$$\text{End}_V(n):=\text{hom}(V^n,V)$$
to be the vector space of all multilinear maps $V^{n}\to V$. Then $\text{End}=\{\text{End}_V(1), \text{End}_V(2),\ldots \}$ forms an operad called the

**endomorphism operad**. The composition \begin{align*} \circ_i\colon \text{End}_V(n)\times \text{End}_V(m)&\longrightarrow \text{End}_V(n+m-1)\\ (f,g)&\longmapsto f\circ_i g \end{align*} is the map $f\circ_i g\colon V^{n+m-1}\to V$ given by "use the output of $g$ as the $i$th input of $f$." For example, if $n=3$ and $m=2$, then $$(f\circ_2g)(a,b,c,d)=f(a,g(b,c),d).$$ And the action of the symmetric group simply permutes the inputs. Example? If $\sigma=(123)\in S_3$, then $$\sigma f (a,b,c)=f(c,a,b).$$## But wait, there's more!

The reason that End$_V$ is so important is because it helps us regard abstract operations as

*actual*operations. You see,*operads aren't very interesting on their own---*just like groups aren't very interesting until you look at a*representation*of the group (or let them act on something). Operads become useful when each abstract operation is realized as a*concrete*operation, and we do this by assigning each $n$-ary operation $f\in \mathcal{O}(n)$ a map $V^n\to V$. Such an assignment assembles into a collection of maps $$\{\varphi_n\colon \mathcal{O}(n)\to\text{End}_V(n)\},$$ and we ask that each $\varphi_n$ be compatible with the $\circ_i$ and $S_n$ action. This assembly is called an**algebra over $\mathcal{O}$**or an**$\mathcal{O}$-algebra**. Next time, for example, we'll chat about an operad called*Assoc*. An algebra over*that*operad is precisely an associative algebra! We'll look at other examples, too: topological simplices, the associahedra, little $k$-cubes, and some Riemann surfaces all form operads. And the resulting*algebras over*them are quite interesting.Stay tuned!

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