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 $\sigma f \circ_2 \tau g=(\sigma\circ_2 \tau)(f\circ_{2}g)$.         

where $\sigma\circ_2\tau$ is the permutation, "do $\sigma$, but replace 2 by $\tau$."

Axiom 3: There's a unit (that 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:

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

In fact, the $\mathcal{O}(n)$ need not be sets. They can be vector spaces, or topological spaces, or chain complexes, or any kind of object where "$\mathcal{O}(n)\times\mathcal{O}(m)$" makes sense. (In general, an operad can be defined in any 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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