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We can view the construction of the real or $p$-adic numbers as the end result of a series of constructions each of which adds more and more structure to the object at hand:

  1. First, we start with the punctual set $\mathrm{pt}=\{1\}$.
  2. Then, we consider the free monoid $\mathbb{N}$ on $\mathrm{pt}$, giving us an addition operation $+$ and an additive unit $0$.
  3. After that, we now add inverses, taking the free group on $\mathbb{N}$ and obtaining the abelian group of integers $\mathbb{Z}$.
  4. Noticing that $\mathbb{Z}$ also carries a multiplicative monoid structure, we then add inverses again except for $0$, the additive unit, and obtain $\mathbb{Q}$.
  5. Finally we consider absolute values on $\mathbb{Q}$, and find that besides the trivial absolute value we have the Euclidean absolute value as well as the $p$-adic ones. Metrically completing with respect to these leads one to the real numbers $\mathbb{R}$ or the $p$-adics $\mathbb{Q}_p$.

One thing I've been curious about for the longest time is how far can we carry the steps above when replacing sets by spaces/homotopy types, and thus replacing the punctual set by the punctual space. I think I understand how steps 1–3 should proceed, although I don't know anything about 4–5. Here's how the homotopy version of steps 1–3 would go I think:

  1. This time we start with the punctual space $*$.

  2. Next, we already find "branching answers" like $\mathbb{Q}_p$ and $\mathbb{R}$ in step 5, as we now have to consider the free $\mathbb{E}_k$-monoid on $*$ for $1\leq k\leq\infty$.

    In the set theory case the answer is the same for all $k$ (e.g. the free commutative monoid on $\mathrm{pt}$ is also $\mathbb{N}$), although now we have different spaces for each $k$. Indeed, the free $\mathbb{E}_k$-monoid on $*$ is given by $\coprod^{\infty}_{n=0}\mathrm{UConf}_n(\mathbb{R}^k)$, where $\mathrm{UConf}_n(\mathbb{R}^k)$ denotes the unordered configuration spaces of $n$ points in $\mathbb{R}^k$.

  3. The counterpart of step 3 in this setting is also clear: we ought to consider free $\mathbb{E}_k$-groups, which turn out to be $\Omega^kS^k$.

Now $\Omega^kS^k$ also has a multiplicative structure, given by composition of loops instead of concatenation, and I think $\coprod^{\infty}_{n=0}\mathrm{UConf}_n(\mathbb{R}^k)$ also has one, although I'm not sure about the details (the $k=2$ case seems to be related to braid multiplication for instance). That said, I'm not sure how one could proceed as in steps 4–5 with similar universal constructions like rings of fractions and metric completions.

Is there a homotopy-theoretic analogue of these constructions (i.e. rings of fractions, absolute values, and completions for $\mathbb{E}_k$-rings), and if so, what would be the homotopy theoretic counterparts of $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{Q}_p$ in this case?

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  • $\begingroup$ Incidentally, a more satisfactory viewpoint would be to consider $\mathbb{N}$ as the free semiring on $\mathrm{pt}$ instead of thinking of it as the free monoid on $\mathrm{pt}$ and then noticing that it also “happens” to carry a multiplicative structure. The consequence of this is that in the homotopy theory setting we should then look for something like “the homotopy theoretic analogue of a ring that is free on the point, whose additive structure is $\mathbb{E}_n$ and whose multiplicative structure is $\mathbb{E}_m$”, so namely $\endgroup$
    – Emily
    Commented Mar 19, 2023 at 0:22
  • $\begingroup$ the free “$(\mathbb{E}_n,\mathbb{E}_m)$-nearring“ on the punctual space. However, the theory of these kind of objects hasn't been developed yet, and it would also require theories of other objects like "lax monoidal $\infty$-categories" which also haven't been developed yet, so I've opted to phrase things in this less satisfactory way in the question for simplicity. $\endgroup$
    – Emily
    Commented Mar 19, 2023 at 0:22
  • $\begingroup$ "4. Noticing that $\mathbb{Z}$ also carries a multiplicative monoid structure" sounds quite informal amidst these categorical constructions. (By the way you don't need the multiplication to pass from $\mathbb{Z}$ to the additive group $\mathbb{Q}$.) $\endgroup$
    – YCor
    Commented Mar 19, 2023 at 11:21
  • $\begingroup$ @YCor Yes definitely. I opted for going with this because $\Omega^k S^k$ has a clear multiplication (and I think the multiplicative structure of $\coprod^{\infty}_{n=0}\mathrm{UConf}_n(\mathbb{R}^k)$ is also known), and the alternative, more satisfactory approach, would use some constructions which haven't been worked out yet in the $\infty$-categorical setting: really we should be viewing $\mathbb{N}$ and $\mathbb{Z}$ as the free (resp. left, right) near(semi)ring on the punctual set, $\endgroup$
    – Emily
    Commented Mar 19, 2023 at 18:11
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    $\begingroup$ Concerning 5, a satisfactory answer is provided by Morava K-theories and the thick subcategory theorem. $\endgroup$ Commented Mar 20, 2023 at 14:00

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