# Ádem relations for the Steenrod and the Dyer–Lashof algebra

In this paper by Nondas Kechagias, the Steenrod algebra and the Dyer–Lashof algebra are compared. The rough difference ist:

1. The Steenrod algebra arises by dividing out the “cohomological” Ádem relations and $$Q^0=1$$.
2. The Dyer–Lashof algebra arises by dividing out the “homological” Ádem relations and $$Q^I=0$$ where $$I$$ has negative excess.

The cohomological Ádem relations are of the form $$Q^iQ^j =\sum_{k=0}^{[i/2]}\binom{j-k-1}{i-2k}Q^{i+j-k}Q^k~~~\text{for }i<2j$$ The homological Ádem relations are of the form $$Q^iQ^j-\sum_{k>0}\binom{k-j-1}{2k-i}Q^{i+j-k}Q^k~~~\text{for }i>2j.$$ In what way are they dual to each other? If I consider the dual Steenrod sqares $$\mathrm{Sq}_i:H_*(X)\to H_{*-i}(X)$$ for $$X$$ of finite type, the should satisfy other Adem relations than the “homological ones”, namely the cohomological ones with the only difference that the composition has to be read contravariantly.

Why do we call both relations Ádem relations?

See Peter May's "A General Algebraic approach to Steenrod Operations". Both arise from considering the inclusion of a Sylow $$p$$-subgroup into $$\Sigma_{p^2}$$. There is a universal $$\mathbb{Z}$$-indexed set of operations with the Steenrod algebra and Dyer-Lashof algebra arising as the quotients acting on the homology of appropriate non-negative (resp., non-positive) complexes.
Here's a nice example: consider the Steenrod algebra acting on the cohomology of graded cocommutative Hopf algebras. (In this, $$Sq^0 \neq 1$$.) One can grade the operations homologically, $$Q^i : Ext^{s,t} \to Ext^{t+s-i,2t}$$ or cohomologically, $$Sq^i : Ext^{s,t} \to Ext^{s+i,2t}$$.
The Dyer-Lashof operation $$Q^i$$ raises the `total degree' $$t-s$$ by $$i$$, and the Steenrod operation does what you expect from the cohomology of groups, for example, if everything is in internal degree $$t=0$$.
It is the same set of operations on $$Ext$$, but in the first grading you get the homological form of the Adem relations and in the second grading you get the cohomological form.
One reason to consider the homological grading is that if $$R$$ is an $$E_\infty$$-ring spectrum, then $$Q^i$$ in the $$E_2$$-term of the Adams spectral sequence $$Ext_{A}(H^*R, F_2) \Rightarrow \pi_* R$$ agrees with the $$Q^i$$ in $$H_* R$$ under the edge homomorphism and the Hurewicz map.