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I asked this question on MSE but I want to ask it again here with some more context sine it received no answers. In Chapter 3 (Algebra) of the book Operads in Algebra, Topology and Physics by Markl, Shnider and Staffesh there is the Lemma 3.16 where the authors state the isomorphism of operads $\mathfrak{s}^{-1}\mathrm{End}_V\cong \mathrm{End}_{\Sigma V}$. For the definition of the operadic suspension $\mathfrak{s}$ and other notation please see my previous question.

But the proof they give is not very explicit, since they give (non-explicit) isomorphisms of graded modules and don't really justify that they are morphisms of operads. I tried to show that and I came across a problem trying to show that the isomorphism commutes with the action of the symmetric group. This is where my MSE questions begins, so I am going to copy it here.

Let $V$ be a graded vector space and $\text{End}_V(n)=\hom(V^{\otimes n},V)$. There is a natural action of the symmetric group $S_n$ on $\text{End}_V(n)$ by permuting the arguments, i.e. if $f\in \text{End}_V(n)$ and $\sigma\in S_n$, $(f\sigma)(v_1\otimes\cdots\otimes v_n)=\varepsilon(\sigma)f(v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)})$, where $\varepsilon(\sigma)$ is the Koszul sign produced by permuting $v_1,\dots,v_n$ via $\sigma$. We can twist this action by the sign of $\sigma$, i.e. we consider the action

$$(f\sigma)(v_1\otimes\cdots\otimes v_n)=(-1)^{\sigma}\varepsilon(\sigma)f(v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)})$$

If $\Sigma V$ is the suspension of $V$, we consider the natural action of the symmetric group on $\text{End}_{\Sigma V}(n)$ (the first I defined, without the twist).

There is a map $\phi:\text{End}_{\Sigma V}(n)\to \text{End}_V(n)$ given by $f\mapsto \Sigma^{-1}\circ f\circ\Sigma^{\otimes n}$ which is indeed an isomorphism of graded modules.

I need to show that $\phi$ commutes wit the action of the symmetric group, where we have the natural action on the domain and the twisted action on the codomain.

I can show this for transpositions of the form $\sigma=(i\ i+1)$. On the one hand,

$$\phi(f\sigma)(v_1\otimes\cdots\otimes v_n)=(-1)^{\sum_{j=1}^n (n-j)v_j}\Sigma^{-1}\circ (f\sigma)(\Sigma v_1\otimes\cdots\otimes \Sigma v_n)=$$

$$(-1)^{\sum_{j=1}^n (n-j)v_j+(v_i-1)(v_{i+1}-1)}\Sigma^{-1}\circ f(\Sigma v_1\otimes\cdots\otimes\Sigma v_{i+1}\otimes\Sigma v_i\otimes\cdots\otimes \Sigma v_n).$$

On the other hand

$$(\phi(f)\sigma) (v_1\otimes\cdots\otimes v_n)=(-1)^{v_iv_{i+1}-1}\Sigma^{-1}\circ f\circ \Sigma^{\otimes n}(v_1\otimes\cdots\otimes v_{i+1}\otimes v_i\otimes\cdots\otimes v_n)=$$

$$(-1)^{v_iv_{i+1}-1+\sum_{j\neq i,i+1}(n-j)v_j +(n-i-1)v_i+(n-i)v_{i+1}}\Sigma^{-1}\circ f(\Sigma v_1\otimes\cdots\otimes \Sigma v_{i+1}\otimes \Sigma v_i\otimes\cdots\otimes \Sigma v_n).$$

Now I just have to check that the signs are the same. Modulo $2$, the sign of the first map is

$$v_iv_{i+1}+v_i+v_{i+1}-1+\sum_{j=1}^n(n-j)v_j=$$ $$v_iv_{i+1}-1+\sum_{j\neq i,i+1}^n(n-j)v_j+(n-i-1)v_i+(n-i)v_{i+1},$$

which indeed coincide with the sign on the second map.

Question: Since these transpositions generate the symmetric group I feel that I should be able to conclude that the action commutes with $\phi$, but I don't know how to do it.

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    $\begingroup$ I would recommend establishing this at the level of universal properties. Writing (SSeq, #) for the monoidal category of symmetric sequences, and embedding S and V as pure unary sequences, we can observe that naturally in X, SSeq(X, End(S # V)) = SSeq( X # S # V, S # V) = SSeq(S^{-1} # X # S # V, V) = SSeq(S^{-1} # X # S, End(V)) = SSeq(X, S # End(V) # S^{-1}) which is indeed the desired desuspension. one only needs the associativity of composition and the universal property of endomorphisms. $\endgroup$ – pupshaw Jul 28 '20 at 18:46
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Your actual Question has nothing to do with operads. Perhaps it is clarifying to consider the following more general setting: let $G$ be a group, $X$ and $Y$ be right $G$-sets, and $f : X \to Y$ be a function. If $g, h \in G$ and $f$ commutes with the actions of $g$ and of $h$ then it commutes with the action of $gh$: $$f(x) \cdot (gh) = (f(x) \cdot g) \cdot h = f(x \cdot g) \cdot h = f((x \cdot g) \cdot h) = f(x \cdot (gh)).$$ So if $f$ commutes with a set of elements which generate $G$ then it commutes with all elements of $G$.

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  • $\begingroup$ You were right, seeing the problem in a more general setting was clarifying. Thank you. $\endgroup$ – Javi Jul 28 '20 at 19:10

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