Sign convention when commuting shifts and tensor product

In Markl, Schnider, and Stasheff's Operads in algebra, topology, and physics, they give the observation $\mathfrak{s}^{-1} \mathcal{E}nd_V \cong \mathcal{E}nd_{V[-1]}$ (lemma 3.16) as motivation for why the operadic suspension involves the signum representation. I'm having some sign issues when following their calculation for this. The relevant step troubling me is $$\operatorname{Hom}(((V[-1])^{\otimes n})^j, (V[-1])^{i+j}) \cong \operatorname{Hom}((V^{\otimes n})^{j-n}, V^{i+j-1}) \otimes \operatorname{sgn}_n$$ where $V$ is a differential graded algebra and $\operatorname{sgn}_n$ is the signum representation of $\Sigma_n$ concentrated in degree 0. Here, the symmetric group $\Sigma_n$ acts on $n$-fold tensor products by permuting the tensor factors.

I think this isomorphism ought to follow from $V[-1]^{\otimes n} \cong V^{\otimes n}[-n] \otimes \operatorname{sgn}_n$, and I will be convinced if it works when $n = 2$, but I'm having difficulty showing $$V[-1] \otimes V[-1] \cong (V \otimes V)[-2] \otimes \operatorname{sgn}_2$$
as $\Sigma_2$-dgas. The isomorphism needs to be $\Sigma_2$-equivariant, hence the need for $\operatorname{sgn}_2$.

Some ideas that I had: I think the issue is that the isomorphism is not the "obvious" one, i.e., not $v \otimes w \mapsto v \otimes w$. In fact, I can identify two places where signs might need to be adjusted. First, we have the Koszul sign rule for interchange. Let $v \in V^{a-1}$ and $w \in V^{b-1}$. We can think of $v \otimes w$ as an element in $V[-1]^a \otimes V[-1]^b$, and we have: $$w \otimes v = (-1)^{ab} v \otimes w.$$

However, the "same" element $v \otimes w \in V^{a-1} \otimes V^{b-1} \subset V \otimes V$ satisfies a different relation: $$w \otimes v = (-1)^{(a-1)(b-1)} v \otimes w.$$

Somehow the isomorphism ought to send the former relation to the latter, and the signum representation may be needed, since we are interchanging the two tensor factors. But I can't quite see how.

A different place where signs appear is the definition of the differential. In $V[-1] \otimes V[-1]$, with $v, w$ as above, we have $$d(v \otimes w) = dv \otimes w + (-1)^a v \otimes dw.$$

But in $V \otimes V$, we have $$d(v \otimes w) = dv \otimes w + (-1)^{a-1} v \otimes dw.$$

Can someone help me out? I think that I'm misunderstanding sign conventions horribly.

Pick a convention for how you want to think about the shift functor in terms of tensor products. For example, you might pick $V[-1] = -1 \otimes V$ where $-1$ is concentrated in whatever degree it needs to be to agree with your conventions. Then you can obtain an isomorphism
$$-1 \otimes V \otimes -1 \otimes V \cong -1 \otimes -1 \otimes V \otimes V$$
by swapping the two middle factors. Since everything in $-1$ has odd degree, this introduces a sign of $-1$ on the odd parts of the middle $V$.
• Thanks for your answer; thinking of $V[-1]$ as $k[-1] \otimes V$ was helpful. However, I am confused. It seems that you are suggesting that the sgn in the formula does not arise from the $\Sigma_2$-equivariant structure on $V \otimes V$, which is surprising because $\operatorname{sgn} \cong k$ if we ignore the $\Sigma_2$-action. Moreover, sgn acting only on odd degree elements in the middle $V$ was not what I was expecting; it appears on all degrees in $V \otimes V$, no? Am I misinterpreting sgn? – JHF Dec 8 '15 at 18:09
• @JHF: I'm suggesting that the sgn in the formula arises from not fully committing to the picture implied by the sign conventions. In that picture I claim that the above isomorphism is already $\Sigma_2$-equivariant, provided that whenever you talk about the action of $\Sigma_n$ on $V^{\otimes n}$ you are already decorating this action with signs according to the usual sign convention (this is what I mean by working with the braiding suggested by the sign conventions). – Qiaochu Yuan Dec 8 '15 at 18:15
• And also that when you talk about the action of $\Sigma_2$ on $-1 \otimes V \otimes -1 \otimes V$ you should recognize that the categorically most natural action involves commuting $V$ past $-1$ on its way to being commuted past the other copy of $V$. – Qiaochu Yuan Dec 8 '15 at 18:17
• Thanks! I think I see it now. The $\operatorname{sgn}$ from the calculation seems to appear from the fact that I needed to commute $k[-1] \otimes k[-1]$ in $V[-1] \otimes V[-1]$, but not $k[-2]$ in $(V \otimes V)[-2]$. – JHF Dec 8 '15 at 18:57