In the first section of J. P. May’s General algebraic approach to Steenrod operations, May defines for $\pi\subseteq\Sigma_r$ an integer $q\in\mathbb{Z}$ and a commutative ring $\Lambda$, the $\Lambda\pi$-module $\Lambda(q)=\Lambda$ with sign action $\sigma\lambda = (-1)^{qs(\sigma)}\cdot \lambda$ where $(-1)^{s(\sigma)}$ is the sign of $\sigma$.

For a $\Lambda$-chain complex $K$ we consider $K^{\otimes r}(q)= K^{\otimes r}\otimes \Lambda(q)$ with the diagonal action. I assume that this translates to the explicit sign rule for the transposition $\sigma_{i,i+1}$ in $K^{\otimes r}(q)$ $$\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r) = (-1)^{q+|a_i|\cdot |a_{i+1}|}\cdot (a_1\otimes\dotsb \otimes a_{i+1}\otimes a_i\otimes\dotsb\otimes a_r).$$ Am I correct with this? He later considers cycles $a,b\in K_q$ and $c\in K_{q+1}$ with $dc=a-b$. Now let $I$ be the cellular chain complex of the intervall, i.e. $I_1=\Lambda\langle e\rangle$ and $I_0=\Lambda \langle e_0,e_1\rangle$, and $de=e_1-e_0$. We consider the chain map of degree $q$ $$f:I\to K, e\mapsto (-1)^qc, e_1\mapsto a, e_2\mapsto b.$$ This satisfies $dfe=(-1)^qfde$ and hence is a chain map. May now claims that $f^{\otimes r}:I^{\otimes r}\to K^{\otimes r}(q)$ is $R\pi$-equivariant. I don’t see why: We have $$f^{\otimes r}(\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r)) = (-1)^{|a_i|\cdot |a_{i+1}|} f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r),$$ whereas on the other side, we have $$\sigma_{i,i+1}\cdot f^{\otimes r}(a_1\otimes\dotsb\otimes a_r)=(-1)^{q+(|a_i|+q)\cdot(|a_{i+1}|+q)}\cdot f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r).$$ There are equal iff $q\cdot (|a_i|+|a_{i+1}|)$ is even, but this is not necessarily the case. Concretely, if $q=1$ and $\sigma=\sigma_{1,2}$, then $$f^{\otimes 2}(\sigma\cdot e\otimes e_1) = f^{\otimes 2}(e_1\otimes e)=a\otimes (-c)\ne a\otimes c=\sigma\cdot (-c)\otimes a=\sigma\cdot f^{\otimes 2}(e\otimes e_1).$$ What am I missing?

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    $\begingroup$ A totally reflexive comment on any question about signs in algebraic topology: have you read Tyler Lawson's "In which I try to get the signs right for once"? $\endgroup$
    – LSpice
    Jan 6, 2020 at 18:03
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    $\begingroup$ Ah, thatʼs cool! Thank you! It also solves my confusion: I forgot to use the sign rule $(f\otimes g)(a\otimes b) = (-1)^{|g|\cdot |a|} \cdot f(a)\otimes g(b)$. $\endgroup$
    – FKranhold
    Jan 6, 2020 at 21:13
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    $\begingroup$ When in doubt, invoke the Koszul rule of signs: if one symbol passes through another, multiply by $-1$ raised to the product of their degrees. $\endgroup$ Jan 10, 2020 at 15:12

1 Answer 1


@FKranhold You mean I got it right? You had me fooled. I should apologize for leaving that detail to the reader, but let me give two excuses. First, one does not actually need that detail to prove Lemma 1.1(iv). It just answers an obvious question the reader might have about the proof. Second, that paper was from the good old days when things went fast. There was a conference, March 30 to April 4, 1970, for Steenrod's 60th birthday. Its Proceedings were submitted June 15, 1970. No time to polish the writing. I remember Steenrod telling me that mine was the only paper in the Proceedings that he understood. He died the next year.

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    $\begingroup$ Hopefully there is no causal relationship between the last two sentences. $\endgroup$
    – LSpice
    Jan 19, 2020 at 15:56
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    $\begingroup$ @LSpice Thanks, Loren. I did need a laugh. $\endgroup$
    – Peter May
    Jan 19, 2020 at 16:07
  • $\begingroup$ Yes, I think that everything should be fine. I was just confused because I missed the sign rule for $(f\otimes g)(a\otimes b)$. $\endgroup$
    – FKranhold
    Jan 20, 2020 at 22:58

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