"Algebras are to operads as group representations are to groups"

Question

I just read the wikipedia page on operads, and it says:

Algebras are to operads as group representations are to groups

Based on the definition of operads, I cannot immediately see why this is true.

(ps. I assume that by algebra, they mean algebra over a field rather than algebra in the sense of universal algebra, since this is what the article links to)

Answer 1

A group representation $G \to \operatorname{GL}(V)$ maps group elements to automorphisms $V \to V$; it can also be viewed as a map $G \times V \to V$.

An algebra over an operad $O \to \operatorname{End}(X)$ maps $n$-ary operations to endomorphisms $X^{\otimes n} \to X$; it can also be viewed as a collection of maps $O(n) \times X^{\otimes n} \to X$.

Here "endomorphism" is used in the sense of the endomorphism operad (accepting multiple inputs).

In both cases the "abstract" structures (satisfying map-like properties) are realized as actual maps.

See e.g. What is... an Operad? by Jim Stasheff in the Notices of the AMS.

Answer 2

(ps. I assume that by algebra, they mean algebra over a field rather than algebra in the sense of universal algebra, since this is what the article links to)

No, they mean "algebra over an operad". The algebras over an operad $P$ are to the operad $P$ what the group representations of a group $G$ are to the group $G$.

Ricardo Buring already explained why. Let me add a nice example. Consider a group $G$ – or even a monoid instead of a group but it doesn't really matter. Then you can define an operad $P$ by $$P(n) = \begin{cases} G & \text{if } n = 1, \\ \varnothing & \text{otherwise.} \end{cases}$$ The only nontrivial operad structure maps are $P(1) \times P(1) \to P(1)$, which are given by the group multiplication, and $1 \in P(1)$ is the unit of the group. Then an algebra over this operad is precisely the same thing as a representation of the group $G$ (I'll leave that to you as an easy exercise).