Let $X$ be an $E\_\infty$-space (not necessarily grouplike). Let $x \in \pi\_0 X$ be an element; say that $x$ is **strictly commutative** if there is a map of $E\_\infty$-spaces $\mathbb{Z}\_{\geq 0} \to X$ that takes $1 \mapsto x$. (The terminology is abusive, as for an element to be strictly commutative is extra data than a condition.) 


There is also a natural space of strictly commutative elements in $X$, given by the (derived) mapping space (in the homotopy theory of $E\_\infty$-spaces) $\hom(\mathbb{Z}\_{\geq 0}, X)$. I do not know of a simple presentation of $\mathbb{Z}\_{\geq 0}$ as an $E\_\infty$-space (the free $E\_\infty$-space on one object is $\bigsqcup\_{n \geq 0} B \Sigma\_n$), so I am not sure how to write this space down in terms of $X$. If $X$ is grouplike, so that it can be identified with a connective spectrum, then this is the mapping space in spectra $\hom( H \mathbb{Z}, X)$. 


What are examples of strictly commutative elements? For instance, I am interested in the following example: given an $E\_\infty$-ring $R$, what is the space of strictly commutative elements in the infinite loop space $\Omega^\infty R$ with *multiplicative structure*? (Equivalently, what is the space of maps $S^0[\mathbb{Z}\_{\geq 0}] \to R$ in $E\_\infty$-rings?) One reason is that the $E\_\infty$-ring $S^0[\mathbb{Z}\_{\geq 0}]$ is easier to compute with than the free $E\_\infty$-ring on a generator in degree zero, but seems to be less nice formally, and I'd like to know conditions under which an element in $\pi\_0 R$ can be hit by a map from the monoid algebra. 



Strictly commutative elements of $E_\infty$-spaces
In the "easier" grouplike case, as you say, this is related to spaces of units, and Jacob and Neil have mentioned things about $gl\_1$. This thing exhibits strange behaviour, and was an object of close study (along with some serious calculation) a number of years ago.


Here's an example of something that may seem counterintuitive about the interaction between $gl\_1$ and "genuinely commutative" phenomena.


Let $R$ be the graded ring $\mathbb{Z}/2[x]/x^3$, where $|x|=1$. Taking zero differential, we can view this as a commutative DGA, hence giving rise to an $E\_\infty$ ring spectrum. Take $gl\_1(R)$: it's a connective spectrum with homotopy groups $0, \mathbb{Z}/2, \mathbb{Z}/2$, and then zeros.


There are two such spectra: one is equivalent to a product of Eilenberg-Mac Lane spectra, and in the other the class in degree two is a multiple of $\eta$. *It turns out that it's the latter*, and so there is no possibility of describing it in chain-complex terms.


Why? Here's a sketch of the argument.


* Suppose that $gl\_1(R)$ is a product of Eilenberg-Mac Lane spaces.
* Then $\pi\_1$ splits off by a map $\Sigma H\mathbb{Z}/2 \to gl\_1(R)$.
* This is equivalent to an infinite loop map $K(\mathbb{Z}/2,1) \to GL\_1(R)$ which is an equivalence in degree 1.
* This is equivalent to a map of $E\_\infty$ ring spectra $\Sigma^{\infty}\_+ K(\mathbb{Z}/2,1) \to R$ (which hits the class in degree 1).
* Since $R$ is an $H\mathbb{Z}/2$-algebra, this is equivalent to a map of $H\mathbb{Z}/2$-algebras $H\mathbb{Z}/2 \wedge \Sigma^\infty\_+ K(\mathbb{Z}/2,1) \to R$ which is an isomorphism in degree 1.
* On homotopy groups, this is a ring map $H\_\*(K(\mathbb{Z}/2,1); \mathbb{Z}/2) \to R$, where the former has the Pontrjagin product, which is an isomorphism in degree one.
* The class in degree one in $H\_\*(K(\mathbb{Z}/2,1); \mathbb{Z}/2)$ squares to zero.


