Let $\mathcal X$ be a presentable $\infty$-category. Then the stabilization $Stab(\mathcal X)$ of $\mathcal X$ is the universal presentable stable category on $\mathcal X$.

Conversely, if $\mathcal A$ is a presentable stable $\infty$-category, then we can ask which presentable $\infty$-categories $\mathcal X$ have $Stab(\mathcal X) \simeq \mathcal A$. There's always at least one such $\mathcal X$, namely $\mathcal A$ itself. In particular, I would like to know an answer to the following:

Question 1: Let $\mathcal A$ be a presentable stable $\infty$-category. Under what conditions does there exist an $\infty$-topos $\mathcal X$ such that $Stab(\mathcal X) \simeq \mathcal A$?

For a closely related question, let $StPr^L$ denote the $\infty$-category of presentable stable $\infty$-categories and left adjoint functors. Let $Logoi$ denote the $\infty$-category of $\infty$-topoi, with geometric morphisms pointing in the direction of their inverse images.

Question 2: Does the functor $Stab : Logoi \to StPr^L$ have a left or right adjoint?

If the answer to Question 2 is affirmative, then one might approach Question 1 by asking for criteria ensuring that the unit / counit of the adjunction is an equivalence. Alternatively, one might wonder

Question 3: Note that the functor $Stab : Pr^L \to StPr^L$ has both a left adjoint $L$ and a right adjoint $R$. For which presentable stable $\infty$-categories $\mathcal A$ is $L\mathcal A$ or $R \mathcal A$ an $\infty$-topos?

Question 4: For example, let $G$ be a compact (even finite, say) Lie group. Is the category $Spt_G$ of genuine $G$-spectra the stabilization of an $\infty$-topos?

  • 1
    $\begingroup$ I don't think $Stab$ has a left adjoint, does it ? For instance, it does not preserve the pullback $Spaces \times_{CMon} CGrp = 0$ (along the free functor and the inclusion) Also, its right adjoint $R$ is the forgetful functor, so $R\mathcal A$ is never a topos unless $\mathcal A$ is, which I think happens iff $\mathcal A = 0$. $\endgroup$ Commented Jul 3, 2022 at 21:12
  • $\begingroup$ @MaximeRamzi Oh you're right -- it's rather that the right adjoint itself has a right adjoint. Hm... so Question 3 seems to be a dead end $\endgroup$
    – Tim Campion
    Commented Jul 3, 2022 at 21:13
  • $\begingroup$ I think there's always a symmetric monoidal "smash product" on $Stab(\mathcal X)$ arising from the cartesian product on $\mathcal X$. It may make sense to take such a symmetric monoidal structure as part of the input when trying to reconstruct a topos $\mathcal X$. $\endgroup$
    – Tim Campion
    Commented Jul 4, 2022 at 14:55
  • 1
    $\begingroup$ @TimCampion stabilizing the cartesian product doesn’t give a symmetric monoidal struture in general - this “multiplicative structure” from differentiating the cartesian product only exhibits Sp(X) as an operad. This is discussed in HA 6.2.4 and also in the thesis of Heuts. $\endgroup$
    – Bbb
    Commented Jul 4, 2022 at 22:41
  • 2
    $\begingroup$ Is even the $1$-categorial version of that known (for $1$-topos and Grothendieck abelian categories)? $\endgroup$
    – user40276
    Commented Nov 10, 2022 at 22:16

2 Answers 2


For any topos $\mathcal{X}$, we have $Stab(\mathcal{X})=Sh_{Sp}(\mathcal{X})$, so our topos admits a symmetric monoidal adjunction with $\mathcal{A}$ (just as in the case $\mathcal{X}=\infty Gpd$). To be precise, the smash product of spectra will yield a tensor product on sheaves of spectra, so $\mathcal{A}$ must be symmetric monoidal. As you observed, the suspension map $\mathcal{X}\to\mathcal{A}$ must preserve the symmetric monoidal structure because it does so for the case of spaces (and $\mathcal{X}=Sh_{\infty Gpd}(\mathcal{X})$). So that's one obstruction.

We can also equip a stabilized topos with a natural t-structure (coming from its construction as sheaves of spectra; see Prp in Lurie's SAG) which is right complete and compatible with filtered colimits. The heart of this t-structure is the category of abelian group objects in the underlying $1$-topos $\mathcal{X}^{♡}$. That's another obstruction: $\mathcal{A}$ must admit a right-complete t-structure, and the heart must be a Grothendieck category.


Here is another sort of constraint. I'll write $Sp(\mathcal C)$ instead of $Stab(\mathcal C)$.

Claim: If $\mathcal A \simeq Sp(\mathcal X)$ for a nontrivial [1] $\infty$-topos $\mathcal X$, then for any nontrivial localization $Spectra_L$ of $Spectra$, the localization $\mathcal A_L$ is a nontrivial localization of $\mathcal A$.

For instance, this means that the derived category $D(X)$ of a ring or scheme is never the stabilization of an $\infty$-topos, since $D(X)$ is fixed by the nontrivial localization at $H\mathbb Z$.

Let $\mathcal X$ be a nontrivial $\infty$-topos, and let $x^* : Spaces {}^\to_\leftarrow \mathcal X : x_*$ be the unique geometric morphism to $Spaces$. Let us contemplate the accessible left exact functor $\xi = x_* x^* : Spaces \to Spaces$ (the "shape" of $\mathcal X$). Note that

  1. $\xi$ preserves and reflects the initial object $\emptyset \in Spaces$;

"Preserves" is because if the initial object $x^\ast \emptyset = \emptyset_{\mathcal X}$ had a global section $1 \to \emptyset$, then $\mathcal X$ would be trivial. "Reflects" is because any nonempty space has the terminal space $1$ as a retract, and $\xi$ preserves $1$.

  1. If $X \to Y \leftarrow Z$ are maps in $Spaces$ with empty pullback, then $\xi X \to \xi Y \leftarrow \xi Z$ likewise has empty pullback.

This follows from (1) since $\xi$ is left exact. Therefore

  1. If $X \in Spaces$ is disconnected, then so is $\xi(X)$.

For we can take different connected components of $X$ in (2) above. It now follows that

  1. The functor $Sp(\xi) : Spectra \to Spectra$ induced by $\xi$ is conservative.

This functor is induced by applying $\xi$ levelwise to each $\Omega$-spectrum. To verify the claim, note that if $0 \neq E \in Spectra$, then $E_n$ is disonnected for some $n$ (where we think of $E$ as an $\Omega$-spectrum $(E_n)_{n \in \mathbb Z}$). By (3), $Sp(\xi)(E)_n = \xi(E_n)$ is disconnected, and hence $Sp(\xi)(E) \neq 0$. As an exact, zero-reflecting functor between stable categories, this implies that $Sp(\xi)$ is conservative.

It now follows that

Theorem: Let $\mathcal X$ be a nontrivial $\infty$-topos, and let $x^\ast : Spaces \to \mathcal X$ be the inclusion of constant objects. Then the induced functor $Sp(x^\ast) : Spectra \to Sp(\mathcal X)$ is conservative.

This follows from (4) since postcomposing $Sp(x_\ast)$ results in the conservative functor $Sp(x_\ast x^\ast)$.

Proof of Claim:

Note that there is an adjunction $Sp(x^\ast) \dashv Sp(x_\ast)$. If $Sp(\mathcal X) = Spectra_L \otimes Sp(\mathcal X)$, then the left adjoint $Sp(x^\ast) : Spectra \to Sp(\mathcal X)$ factors through $Spectra_L$, contradicting the Theorem.

[1] An $\infty$-topos $\mathcal X$ is "nontrivial" if it is not equivalent to the the terminal $\infty$-category, i.e. if the map from the initial object to the terminal object of $\mathcal X$ is not an equivalence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.