# Graded commutativity of the $n$th Browder bracket

Let $$\mathcal{O}$$ be a topological operad and $$X$$ an algebra over $$\mathcal{O}$$. Then $$H_*(X)$$ is an algebra (in the category of $$\mathbb{Z}$$-graded $$R$$-modules) over $$H_*(\mathcal{O})$$. Each $$e\in H_n(\mathcal{O}(2))$$ gives us a graded product $$H_p(X)\otimes H_q(X)\to H_{p+q+n}(X),a\otimes b\mapsto e(a\otimes b).$$ Now $$\mathfrak{S}_2$$ acts on $$H_*(\mathcal{O}(2))$$ on the right and on $$H_*(X)^{\otimes 2}$$ by $$\tau_*(a\otimes b) = (-1)^{ab}(b\otimes a)$$. By equivariance, we get $$e(b\otimes a) = (-1)^{ab} (\tau^*e)(a\otimes b).$$ Now consider the special operad $$\mathcal{C}_{n+1}$$ of little $$(n+1)$$-cubes. Here, $$\mathcal{C}_{n+1}(2)\simeq \mathbb{S}^n$$, so its $$n$$th homology is generated by the fundamental class $$s$$ of the submanifold $$\mathbb{S}^n\to \mathrm{Conf}_2(\mathbb{R}^{n+1}), x\mapsto (0,x).$$ Now it is clear that $$\tau^*s=(-1)^{n+1}s$$ as $$x\mapsto (x,0)$$ is homotopic to $$x\mapsto (0,-x)$$. Cohen defines the Browder bracket by $$[a,b]:=s(a\otimes b)$$. By the above argumentation, we would get $$[b,a]=s(b\otimes a)=(-1)^{ab}(\tau^*s)(a\otimes b)= (-1)^{ab+n+1}[a,b].$$ However, in both Cohen’s paper The homology of $$\mathcal{C}_{n+1}$$-spaces (1976) and in May’s Operads, algebras and modules, the sign is different, namely $$[b,a] = (-1)^{ab+1+n(a+b+1)}[a,b].$$ It is clear that they don’t coincide for $$|a|+|b|$$ odd. What am I missing?

Both papers choose a normalization that is different from yours: they define $$[a,b] = (-1)^{na+1} s(a \otimes b).$$ This is definition 5.7 in Cohen's paper that you mentioned.
The reason for this convention is that it is more consistent with writing the Browder bracket as a binary operation, rather than a function applying on the left: we should think of this as the image of $$a \otimes s \otimes b$$ under the composite which first twists the two factors and then acts. If you don't do this, then you start to get into sign mishaps when you verify things like the Jacobi identity because the triple bracket $$[a,[b,c]]$$ involves $$s (a \otimes s(b\otimes c))$$ and introduces an unexpected sign when moving $$s$$ across $$a$$. Similar remarks apply to the derivation property for the bracket acting on products..