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 $V$.

You can avoid explicitly keeping track of permutation representations by using a different action of the symmetric group, namely the one suggested by the sign conventions. More formally, it's cleaner to work with the braiding suggested by the sign conventions as opposed to the usual braiding.

  • $\begingroup$ 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? $\endgroup$ – JHF Dec 8 '15 at 18:09
  • $\begingroup$ @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). $\endgroup$ – Qiaochu Yuan Dec 8 '15 at 18:15
  • $\begingroup$ 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 $V$. $\endgroup$ – Qiaochu Yuan Dec 8 '15 at 18:17
  • $\begingroup$ 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 $(V \otimes V)[-2]$. $\endgroup$ – JHF Dec 8 '15 at 18:57

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