Previously asked on math.stackexchange, but perhaps this is more appropriate for MathOverflow:


Let $X$ be a spectrum. I've heard that the action of the Steenrod algebra on $H^\*(X; \mathbb{F}\_p)$ and the action of the Dyer-Lashof algebra on the homology of the associated infinite loop space, $H\_\*(\Omega^{\infty}X; \mathbb{F}\_p)$, are "dual" in some sense. 


To my knowledge, this has something to do with Koszul duality, but I've never seen this spelled out in detail. Anyone willing to enlighten me on this? Or point me toward a helpful reference? 



Dyer-Lashof algebra and Steenrod algebra "duality"
The original paper on Koszul algebras, [Stewart Priddy, Koszul resolutions. Trans. Amer. Math. Soc. 152 (1970) 39–60], was, in essence, written to explain this example. Well almost: he was considering the Steenrod algebra and the Lambda algebra. The Dyer Lashof algebra is pretty much the Lambda algebra with some unstable side conditions. 


Haynes Miller makes the connection very explicitly in [Miller, Haynes, A spectral sequence for the homology of an infinite delooping. Pacific J. Math. 79 (1978), no. 1, 139–155]. This influenced other papers of his, and also my work on the Whitehead conjecture.


An alternative delooping spectral sequence was defined in my paper [The McCord model for the tensor product of a space and a commutative ring spectrum. Categorical decomposition techniques in algebraic topology (Isle of Skye, 2001), 213–236, Progr. Math., 215, Birkhäuser, Basel, 2004.] The geometric objects involve the Lie operad, and the associated cohomology operations lead to Steenrod operations.


Dually, a looping spectral sequence was studied here: [Kuhn, Nicholas; McCarty, Jason, The mod 2 homology of infinite loopspaces. Algebr. Geom. Topol. 13 (2013), no. 2, 687–745.] Now the geometric objects involve the commutative operad, and the associated homology operations lead to Dyer-Lashof operations. 


The geometric underpinnings of this sort of situation have been studied by Arone and Ching.


So yeah, you have heard right.


