Suspension operad

Question

There are many candidates for suspension operad in literature. Among them, $\Lambda= \operatorname{End}_{s\mathbb{K}}$ and $\Lambda'=\{s^{1-n}\mathbb{K}\otimes \operatorname{sgn}_n\}_{n\geq 0}$ are typical ones. The operad operation in $\Lambda'$ is defined as follows: $1_m\circ_i 1_n = 1_{m+n-1}$, where $1_m$ denotes the canonical generator of $s^{1-n}\mathbb{K}\otimes \operatorname{sgn}_n$. One can readily see that $\Lambda$ and $\Lambda'$ are not isomorphic as operads. So my question is which one is the "correct" definition of the so-called "suspension operad".

Answer 1

$\Lambda$ is correct. $\Lambda'$ is not an operad because the $\circ_i$ maps are not equivariant with respect to the symmetric group actions. If $\sigma$ is the nontrivial element of $S_2$, then for some element $\tau$ of $S_3$ we would have

$$ -1_3 = -1_2\circ_2 1_2 = \sigma(1_2)\circ_2 1_2 = \tau (1_2\circ_1 1_2) = \tau 1_3 $$ but it is easy to see that $\tau$ is an even permutation.

Answer 2

A bit late of an answer, but there is a very streamlined way of remembering what the signs that appear in the composition of the suspension operad are. Namely, the dg operad $\mathscr S = \operatorname{End}_{s\mathbb k}$ has $$\mathscr S(n) = \hom((s\mathbb k)^n,s\mathbb k)$$

which lives purely in degree $1-n$, and it is one dimensional generated by the unique map $\nu_n$ that sends the element $s\otimes \cdots \otimes s$ to the element $s$. It is also clear that $S_n$ acts through the sign representation, as it simply permutes $s$ with the rule that $(12) (s\otimes s)= - s\otimes s$.

When computing the composition $\nu_n \circ_i\nu_m$ at the element $s\otimes\cdots\otimes s$, you will have to move $i-1$ many $s$ past $\nu_m$ before evaluating them, creating a Koszul sign $(i-1)(m-1)$, and hence you see that $$ \nu_n \circ_i\nu_m = (-1)^{(i-1)(m-1)} \nu_{m+n-1}. $$ In particular, the simplest incarnation of this is the fact $\nu_2$ is an element of degree $-1$ that is associative, in the sense that $$\nu_2\circ_1\nu_2 + \nu_2\circ_2\nu_2 = 0,$$ which will remind you the operad $\Lambda'$ is not a good fit, as in there $\nu_2$ has a homological degree, but is "honestly associative", which causes sign issues with the symmetric group action.