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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.

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up vote 13 down vote accepted

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.

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Thank you for describing this surprising example! –  Akhil Mathew Jul 4 '13 at 15:39
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Here's something which may be of interest.

Let $E$ be the Lubin-Tate spectrum associated to a formal group of height $n$ over an algebraically closed field of characteristic $p > 0$, and let $X$ be $E_{\infty}$ space of units of $E$, and let $Y$ be the space of maps from $\mathbf{Z} / p \mathbf{Z}$ into $X$ (as $E_{\infty}$-spaces). This is the ``$p$-torsion'' on the space you're asking about (provided that you're willing to restrict our attention to invertibles).

Then $\pi_{n}(Y) \simeq \mathbf{Z} / p \mathbf{Z}$, and the higher homotopy groups of $Y$ vanish. I'd like to conjecture that $Y$ is a $K( \mathbf{Z}/p \mathbf{Z}, n)$ (this has some nice consequences).

If this is true, it tells you a lot about the space $Z$ of maps from $\mathbf{Z}$ into $X$. I think it says that $\pi_{i}(Z) = 0$ for $i \neq 0, n+1$ and that $\pi_{n+1}(Z) \simeq \mathbf{Z}_{p}$. Unfortunately I don't see that it gives you much control over $\pi_0(Z)$, which is what you're asking about.

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I'll note that this conjecture is true for heights $n=1$ and $n=2$. There is a spectral sequence to compute $E_\infty$-ring maps $\mathbb{Z}\to E$, whose $E_2$-term you can describe if you know enough about power operations for $E$. At heights $1$ and $2$ you do, and the spectral sequence collapses nicely. These techniques also give (when $n=1,2$) that $\pi_0 Z$ is precisely the group of roots of unity in $\pi_0E$. Unfortunately, I don't know what to do for $n>2$. –  Charles Rezk Jul 3 '13 at 23:19
    
Very interesting! –  Akhil Mathew Jul 4 '13 at 1:42
    
Minor correction: instead of "precisely the roots of unity", I should say something like "Teichmuller lifts" of units mod p. -1 is funny at p=2. –  Charles Rezk Jul 19 '13 at 16:50
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For the grouplike case, the space $\text{hom}(H\mathbb{Z},X)$ is very often contractible. If we let $E$ denote the wedge of all Morava $K$-theories (including $K(p,0)=H\mathbb{Q}$ for all $p$, but not $K(p,\infty)=H\mathbb{Z}/p$), then a spectrum $W$ is said to be dissonant if $E\wedge W=0$, and harmonic if it is $E$-local. Many popular spectra are harmonic, including all suspension spectra, and $MU$, and all spectra $X$ such that $MU_*(X)$ has finite projective dimension as an $MU_*$-module. If $W$ is dissonant and $X$ is local then $\text{hom}(W,X)=0$. This is not immediately applicable because $H\mathbb{Z}$ is not dissonant, but $H\mathbb{Z}/n$ is dissonant for all $n>0$. It follows that when $X$ is harmonic, the spectrum $\text{hom}(H\mathbb{Z},X)$ has homotopy groups that are uniquely divisible by $n$ for all $n>0$, so it is a rational spectrum. If $X$ is a torsion spectrum it then follows that $\text{hom}(H\mathbb{Z},X)=0$.

For the application you mentioned, we need to take $X=gl_1(R)$ for some $E_\infty$ spectrum $R$. These spectra are quite mysterious and I do not know how close they are to being harmonic. The Bousfield-Kuhn functor/Rezk logarithm tells us that $L_{K(n)}(gl_1(R))=L_{K(n)}R$ when $R$ is connective, but harmonicity is about the difference between $gl_1(R)$ and $L_{K(n)}gl_1(R)$, and I do not know how to approach that. I think that more people should study all kinds of questions about $gl_1(R)$.

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Thanks for pointing this out. If $R$ is $E_n$-local, then the map $gl_1(R) \to L_{n} gl_1(R)$ is an equivalence above dimension $n +1$, so at least the space of strictly commutative elements seems to be truncated (up to rational pieces). –  Akhil Mathew Jul 3 '13 at 14:18
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