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.


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.


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.

  • 9
    $\begingroup$ 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 $n>2$. $\endgroup$ – Charles Rezk Jul 3 '13 at 23:19
  • $\begingroup$ Very interesting! $\endgroup$ – Akhil Mathew Jul 4 '13 at 1:42
  • $\begingroup$ 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. $\endgroup$ – Charles Rezk Jul 19 '13 at 16:50

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 $gl_1(R)$.

  • 2
    $\begingroup$ 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). $\endgroup$ – Akhil Mathew Jul 3 '13 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.