Homotopy coherent Invertibility.

Similarly to how $\mathbb{E}_k$-commutative spectra are a homotopy-coherent version of homotopy commutative spectra, encoding commutativity with higher homotopies, we have a similar phenomenon for invertibility: given a monoidal $\infty$-groupoid $\mathcal{C}$, we can speak of both:

  • Objects $A$ of $\mathcal{C}$ for which $[A]\in\pi_0(\mathcal{C})$ is invertible (homotopy invertibility);

  • Objects $A$ of $\mathcal{C}$ which are homotopy-coherently invertible, coming with:

    • A choice of another object $A^{-1}$ of $\mathcal{C}$;

    • Morphisms of the form $A^{-1}\otimes_{\mathcal{C}}A\to1_\mathcal{C}$ and $A\otimes_{\mathcal{C}}A^{-1}\to1_{\mathcal{C}}$;

    • Homotopies filling a diagram of the form

      (along with another homotopy filling a similar diagram for $A\otimes_{\mathcal{C}}A^{-1}\otimes_{\mathcal{C}}A$)

    • and so on.

    (More precisely, a handy way to encode all these higher homotopies is to define a homotopy-coherent invertible object $A$ of a symmetric monoidal $\infty$-category $\mathcal{C}$ to be a symmetric strong monoidal functor $F\colon\Omega^{\infty}\mathbb{S}\to\mathcal{C}$ with $F(1)=A$, where $\mathbb{S}$ is the sphere spectrum.)

Localisation of ring spectra.

Lurie's Higher Algebra, Section 7.2.3 defines the localisation $S^{-1}R$ of an $\mathbb{E}_1$-ring spectrum $R$ at a nice subset $S$ of homogeneous elements of $\pi_*(R)$. This localisation comes also with a natural map $\phi\colon R\to S^{-1}R$ and admits the following universal property (HA

Given an $\mathbb{E}_1$-ring $A$, precomposition with $\phi$ defines a fully faithful map $$\phi^{*}\colon\mathrm{Map}_{\mathsf{Alg}^{(1)}}(S^{-1}R,A)\to\mathrm{Map}_{\mathsf{Alg}^{(1)}}(R,A)$$ with essential image given by those maps of $\mathbb{E}_1$-rings $\psi\colon R\to A$ such that $\psi(s)$ is invertible in $\pi_*A$ for each $s\in S$.

The question.

Is the localisation procedure in HA homotopy coherent in the above sense? I.e. if we take $R$ to be a connective spectrum, so that we may also view it as a symmetric monoidal $\infty$-groupoid via $\Omega^{\infty}$, and let $S$ be a subset of $\pi_0(R)$, is the object in $\Omega^{\infty}(S^{-1}R)$ corresponding to $s\in S$ homotopy-coherently invertible, for each $s\in S$?

If not, is there a suitable homotopy-coherent analogue of Lurie's localisation?


1 Answer 1


The answer is yes - more generally, "coherent invertibility" is just invertibility, which is what makes conditions of the form "such and such things are invertible" extremely practical.

A more precise statement could be something like: the forgetful functor $map(\mathbb S, C)\to C$ realizes an equivalence between the source and the full subgroupoid of $C^\simeq$ spanned by the objects whose class in $\pi_0(C^\simeq)$ is invertible.

In a sense, it all boils down to the following lemma:

Lemma: Let $M$ be an $\mathbb E_\infty$-monoid, and assume $\pi_0(M)$ is a group. Then $M$ is an $\mathbb E_\infty$-group.

Now, for some people, the hypothesis of the lemma is the definition of $\mathbb E_\infty$-groups, so this is only interesting if I give you another definition.

Possible ones, for which the proof of that lemma is more or less complicated are:

1- there exists a map $i: M\to M$ such that $M\xrightarrow{(i,id_M)} M\times M\xrightarrow{\mu} M$ is constant equal to the neutral element;

2- The shear map, $M\times M\xrightarrow{(\mu,pr_2)} M\times M$ ($(x,y)\mapsto (xy,y)$) is an equivalence;

3- There exists a functor $Proj_\mathbb S\to Spaces$ whose underlying $\mathbb E_\infty$-monoid is $M$;

4- There is a spectrum $m$ with $\Omega^\infty m \simeq M$, i.e. $M$ is an infinite loop space.

Because of these things, invertible elements are very well-behaved - e.g. the result from HA you mention becomes an immediate consequence of the adjoint functor theorem. The interesting part of that result is that (under Ore-type conditions, which are there in ordinary algebra too) one can actually describe the homotopy groups of $S^{-1}R$ ! (the existence of something with that universal property works for any $S$, not only nice ones)

  • $\begingroup$ Thanks, Maxime, this is great! $\endgroup$
    – Emily
    Apr 9, 2023 at 22:56
  • $\begingroup$ What is "a map" in the first item $i\colon M\to M$? A map of anima, $E_1$-monoids, or $E_\infty$-monoids? $\endgroup$
    – Z. M
    Apr 10, 2023 at 10:17
  • $\begingroup$ If $M$ is an $\mathbb E_{n+1}$-monoid (including $n=\infty$, where $\infty+1 = \infty$), you can make $i$ into a map of $\mathbb E_n$-monoids. In fact, you can always make it into a map of $\mathbb E_{n+1}$-monoids, but then the souce (or target) has to be tempered with, something like $M^{op}$ $\endgroup$ Apr 10, 2023 at 11:58

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