# When is a stable $\infty$-category the stabilization of an $\infty$-topos?

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?

• 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 $\mathcal A = 0$. Commented Jul 3, 2022 at 21:12
• @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 Commented Jul 3, 2022 at 21:13
• 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 $\mathcal X$. Commented Jul 4, 2022 at 14:55
• @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.
– Bbb
Commented Jul 4, 2022 at 22:41
• Is even the $1$-categorial version of that known (for $1$-topos and Grothendieck abelian categories)? Commented Nov 10, 2022 at 22:16

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 1.3.2.7 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.