In W.S.Massey's singular homology (Graduate Texts in Mathematics 70 Springer (1980)) there is a formula for the boundary of a double slant product on page 176 $$\partial(\phi\backslash\backslash a\otimes b\otimes g) = (\delta\phi)\backslash\backslash a\otimes b\otimes g + (-1)^{|\phi|}\phi\backslash\backslash\partial(a\otimes b\otimes g)$$ Am I right in calculating that this is missing a minus sign and should be $$\partial(\phi\backslash\backslash a\otimes b\otimes g) = -(\delta\phi)\backslash\backslash a\otimes b\otimes g + (-1)^{|\phi|}\phi\backslash\backslash\partial(a\otimes b\otimes g)?$$ If the minus sign is required here, then it is surely also required subsequently, for example in the formula $$\partial(u\backslash v)=(\delta u)v+(-1)^{|u|}u\backslash\partial(v)$$ which should then be $$\partial(u\backslash v)=-(\delta u)v+(-1)^{|u|}u\backslash\partial(v).$$ I have been for some years puzzled by this as the formulae in the book give the appearance of having signs in the expected places. I would be very happy to be put right on this issue.
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2$\begingroup$ perhaps you can say why you think there is a typo in the sign? lemma 1.5 of these lecture notes has the same sign. $\endgroup$– Carlo BeenakkerJan 3, 2023 at 9:48
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2$\begingroup$ Hi Peter. For what it's worth, I agree with your signs over those of Massey. $\endgroup$– TyroneJan 5, 2023 at 12:21
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$\begingroup$ @CarloBeenakker If you simply follow through the calculation in Massey's book the minus sign I claim appears to be there. So at first I thought the first displayed formula above was a typo. But when its ramifications percolated through the subsequent text I started to question if I had done the calculation correctly. Tyrone agrees with my calculation. The notes you cite maybe do not shed great light because they don't include the nuts and bolts of the proofs. $\endgroup$– Peter KrophollerJan 6, 2023 at 19:56
1 Answer
You and Tyrone are correct that this is a sign mismatch.
Just to be clear: Massey writes this sign for the double slant product because it is the desirable one. It would imply that we have a map of chain complexes $$C^*(Y, G_1) \otimes C_*(X) \otimes C_*(Y) \otimes G_2 \to C_*(X) \otimes G_1 \otimes G_2$$ where this is the tensor product of chain complexes, using the Koszul sign convention.
This is not compatible with the other things that Massey writes. In particular, a couple of lines later he writes that we should take the convention $$(\delta \phi)(b) = (-1)^{|\phi|} \phi(\partial b)$$ (which does not alter the cohomology groups) and this is the source of the difficulty. This definition does not make the pairing $$C^*(Y, G_1) \otimes C_*(Y) \to G_1$$ into a chain map.
Many people (myself included) would prefer to use the sign convention $$(\delta \phi)(b) = (-1)^{|\phi|+1} \phi(\partial b)$$ for the coboundary operator $\delta$ precisely to fix this type of problem. This looks like it violates the Koszul sign convention, but duality has a tendency to do that. If you make this change, I believe Massey's sign for the double slant product works out.
(One possible explanation is this. In the definition of $\partial$ you take an alternating sum of faces, starting at the 0'th and ending at the n'th. The sign convention I've written above for $\delta$ on cochains corresponds to taking an alternating sum over faces, starting at the n'th and ending at the 0'th. Duality tends to interact better with this order reversal of the faces.)
