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?



Graded commutativity of the $n$th Browder bracket
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..


See also [this paper of Xianglong Ni](https://math.mit.edu/research/undergraduate/spur/documents/2017Ni.pdf).


