I cannot really state my question in an incredibly precise way as I'm very new to this area, but I hope what I'm asking will be clear. Let $\mathcal{C}$ be the infinity category of sheaves of quasi-coherent spectra on $\mathbb{P}^1$ as a scheme. Then, I think this should be a stable infinity category. Higher Algebra 7.1.2.1 says that if this category is presentable and has a compact generator, it is the module category for some ring ($\mathbb{E}_1$ ring). My question is, are the conditions satisfied in this case? I suspect it shouldn't be because I don't think it should be a module category. However, I'm not sure why the structure sheaf is not a compact generator and I feel that the category of quasi-coherent sheaves ought to be presentable. What is the issue here? Is there actually a ring which represents projective space?
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4$\begingroup$ If I understand the question correctly,$M = \mathcal O \oplus \mathcal O(1)$ should be a compact generator for $\mathcal C$, so $\mathcal C$ is equivalent to modules for $RHom(M,M)$. Note that $RHom(\mathcal O, \mathcal O(-1)) = 0$, so the structure sheaf is not a generator for $\mathcal C$. $\endgroup$– Sam GunninghamCommented Jan 23, 2020 at 5:55
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1$\begingroup$ Does taking homotopy categories recover the equivalence of derived categories of coherent sheaves on $\mathbb{P}^1$ and modules over the Kronecker algebra? Asking both as a sanity check for me, not being an infinity categories person, and because if yes it would be a reason to have suspected the higher categorical statement in advance. $\endgroup$– Matthew PresslandCommented Jan 23, 2020 at 15:58
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$\begingroup$ @SamGunningham Ah I see, you're definitely right about the structure sheaf. but you're saying that there is some ring whose module category is the modules on P1! Can you describe this ring ? (I mean E1 ring everywhere) $\endgroup$– Andy JiangCommented Jan 23, 2020 at 17:36
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$\begingroup$ @davik Its homotopy is rather easy to describe (this is basically the computation of the sheaf cohomology of line bundles on $\mathbb{P}$), is there anything else you want to know about it in particular? $\endgroup$– Denis NardinCommented Jan 23, 2020 at 18:51
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$\begingroup$ @davik The standard answer, due to Beilinson, and as mentioned by Mathhew Pressland above, is that the derived endomorphism ring of $\mathcal O \oplus \mathcal O(1)$ on $\mathbb P^1$ is equivalent to the path algebra of the Kronecker quiver (so in particular is concentrated in degree $0$). I guess you are asking whether this remains true of we work over the sphere spectrum rather than over a field (which is how I have usually seen this stated). I would guess that the answer is yes, but I am by no means certain. $\endgroup$– Sam GunninghamCommented Jan 23, 2020 at 19:01
1 Answer
$\newcommand{\PP}{\mathbf{P}} \newcommand{\QCoh}{\mathrm{QCoh}} \newcommand{\cf}{\mathcal{F}} \newcommand{\cg}{\mathcal{G}} \newcommand{\Map}{\mathrm{Map}} \newcommand{\co}{\mathcal{O}}$ The ordinary abelian category of quasicoherent sheaves on $\PP^1$ is not the category of modules over any ring. We shall prove the following more general result (which in the discrete case is precisely the result of Beilinson's mentioned in the comments): let $R$ be a connective $\mathbf{E}_\infty$-ring, and let $f:\PP^n_R\to \mathrm{Spec}(R)$ denote the flat projective $n$-space over $R$ (see Section 5.4.2 of SAG --- in particular, if $R$ is discrete or a $\mathbf{Q}$-algebra, then this is the usual projective space). Then $\QCoh(\PP^n_R)$ is equivalent to the ($\infty$-)category of right modules over an $\mathbf{E}_1$-ring. For non-higher-categorical people: if $R$ is discrete, then taking homotopy categories recovers Beilinson's result.
(In the course of the proof below, I'll appeal to some results from SAG, which are easy to prove in the classical setting; the argument below is essentially the usual one used to deduce Beilinson's result.)
Since $\QCoh(\PP^n_R)$ is stable and presentable, the Schwede-Shipley theorem you mentioned implies that it suffices to prove that there is a compact generator for $\QCoh(\PP^n_R)$. As in the classical case, the sheaf $\cf = \co \oplus \co(1) \oplus \cdots \oplus \co(n)$ is such a compact generator. Indeed, $\cf$ is obviously compact. Suppose $\cg\in \QCoh(\PP^n_R)$ is such that $\Map(\cf, \cg) = 0$ (in the classical setting, this means derived Hom). Since $\Map(\cf, \cg) = \bigoplus_{k=0}^n f_\ast(\cg\otimes \co(-k))$, the hypothesis on $\cg$ implies that $f_\ast(\cg\otimes \co(k))$ vanishes for all $0\leq k\leq n$. Lemma 7.2.2.2 of SAG implies by induction that $f_\ast(\cg\otimes \co(k))$ vanishes for all $k$. In the classical setting, this lemma from SAG amounts to the statement that there is an exact sequence $$0\to \co\to \co(1)^{\oplus n+1}\to \co(2)^{\binom{n+1}{2}} \to \cdots \to \co(n+1)\to 0.$$ In particular, we conclude that $\Map(\co(k), \cg) = 0$ for all $k$.
By Lemma 5.6.2.2 of SAG, there is a map $\bigoplus_\alpha \co(d_\alpha)\to \cg$ which induces a surjection on $\pi_0$. In the classical setting, this amounts to the statement that $\cg$ can be resolved by line bundles. The above discussion implies that this morphism is null, and therefore that $\pi_0 \cg = 0$. By replacing $\cg$ with a shift, we conclude that $\pi_d \cg = 0$ for all integers $d$, and hence that $\cg = 0$, as desired.
This argument shows that $\QCoh(\PP^n_R) \simeq \mathrm{Mod}(A)$, where $A$ is the $\mathbf{E}_1$-ring $\mathrm{End}\left(\bigoplus_{k=0}^n \co(k)\right)$, just as in the classical setting. When $n=1$, this decomposition amounts to the semiorthogonal decomposition of $\QCoh(\PP^1_R)$: namely, the full subcategory of $\QCoh(\PP^1_R)$ determined by the functor $\mathrm{Mod}(R)\to \QCoh(\PP^1_R)$ sending $\cf$ to $f^\ast(\cf)\otimes \co(1)$ determines a semiorthogonal decomposition $$\mathrm{Mod}(R) \underset{f_\ast(-\otimes \co(-1))}{\stackrel{f^\ast}{\rightleftarrows}} \QCoh(\PP^1_R) \underset{f^\ast(-)\otimes \co(1)}{\stackrel{f_\ast}{\rightleftarrows}} \mathrm{Mod}(R),$$ which is a categorification of the projective bundle formula. In the discrete case, the $\mathbf{E}_1$-ring $A$ is the path algebra of the Beilinson quiver (for general $n$), which when $n=1$ is the Kronecker quiver $\bullet \rightrightarrows \bullet$. The semiorthogonal decomposition above is just restrictions to each of the vertices. (One comment: the equivalence proved above is not monoidal: the tensor product of two representations of the Kronecker quiver is not just the usual tensor product of quasicoherent sheaves on $\PP^1$.)
