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Are there any connective $E_1$ rings $R$ over $\mathbb{F}_p$ satisfying the following?

  1. $\pi_*(R)$ is a finite dimensional $\mathbb{F}_p$ vector space

  2. $R$ is compact as a module over $R \otimes R^{op}$

  3. $R$ is not concentrated in degree 0

Motivation: I am looking for a smooth proper category with negative Hochschild cohomology and someone suggested $\mathit{Perf}(R)$ with $R$ satisfying the above if it existed. In fact they suggested to look for a finite CW complex $X$ with $H_*(\Omega X,\mathbb{F}_p)$ finite over $\mathbb{F}_p$ (then we can take $R=H_*(\Omega X,\mathbb{F}_p)$) but I have not found any of those either.

Edit: As Shaul Barkan explained to me, such an example cannot come from a space X as described above, because the conditions will imply that $H_*(\Omega X,\mathbb{F}_p)$ is in degree 0.

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    $\begingroup$ Here's an example of a spirit similar to that suggested by Maxime, but more algebrao-geometric. Look at (derived!) coherent sheaves on $\mathbb{P}^1$. The object $O \oplus O(1)[1]$ is a generator of this category, so you can identify it with modules over the endomorphisms of this sheaf. Its endomorphism is also connective (there's no cohomology to $O(-1)$) but it is not discrete. But the condition of $R$ being compact as a bi-module follows from the smoothness of $\mathbb{P}^1$. $\endgroup$
    – S. carmeli
    Nov 3, 2022 at 20:13
  • $\begingroup$ @S.carmeli Thanks! that's a great example. By any chance (can't hurt to ask!) do you know any smooth proper categories over a field with negative Hochschild cohomology? $\endgroup$
    – davik
    Nov 4, 2022 at 15:49
  • $\begingroup$ what do you mean by "negative"? concentrated in non-positive cohomological degrees? or having a non-zero class in such degree? $\endgroup$
    – S. carmeli
    Nov 4, 2022 at 19:21
  • $\begingroup$ having a non-zero homological class $\endgroup$
    – davik
    Nov 6, 2022 at 19:32

1 Answer 1

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Here's an example: consider the $\infty$-category $Fun(\Delta^1,Perf(\mathbb F_p))$. It has two canonical generators $A= \mathbb F_p\to \mathbb F_p$ and $B=\mathbb F_p\to 0$; and I claim that $R= End(A\oplus \Sigma B)$ is an example of what you're looking for (derived endomorphisms - more generally, everything here is "derived").

Note that $\hom(A,B) = \mathbb F_p$, $\hom(B,A)=0$, $\hom(A,A) = \mathbb F_p, \hom(B,B)= \mathbb F_p$.

In particular, $End(A\oplus\Sigma B) = \hom(A,\Sigma B)\oplus \hom(A,A)\oplus \hom(\Sigma B,A)\oplus \hom(\Sigma B,\Sigma B)$.

Note that $\hom(\Sigma B,\Sigma B)\simeq \hom(B,B)$, $\hom(\Sigma B,A) = 0$, so really $End(A\oplus \Sigma B)= \hom(A,A)\oplus \hom(B,B)\oplus \Sigma\hom(A,B)\simeq \mathbb F_p\oplus\mathbb F_p\oplus\Sigma\mathbb F_p$.

In particular, $End(A\oplus\Sigma B)$ is connective and perfect over $\mathbb F_p$, so it remains to argue that it is smooth. But smoothness is Morita invariant, i.e. it only depends on the category of modules.

Now, $A,B$ are generators of $Fun(\Delta^1,Perf(\mathbb F_p))$, so $Perf(End(A\oplus\Sigma B))\simeq Fun(\Delta^1,Perf(\mathbb F_p))$ by the Schwede-Shipley theorem (I'm using implicitly here that $Ind(C^{\Delta^1}) = Ind(C)^{\Delta^1}$ for any stable $C$), so it suffices to argue that $Fun(\Delta^1,Perf(\mathbb F_p))$ is smooth.

Now, for an indexing diagram $C$, $Fun(C,Mod_R)^\omega$ is smooth over $R$ if and only if $R[map_C(-,-)]$ is compact in $Fun(C^{op}\times C,Mod_R)$, and one can show that $map(R[map_C(-,-)], F)\simeq \int_{c\in C}F$, the end of $F$. It follows that $Fun(C,Mod_R)^\omega$ is smooth if and only if $\int_{c\in C}: Fun(C^{op}\times C,Mod_R)\to Mod_R$ preserves filtered colimits.

Recall that ends are a special case of limits over $Tw(C)$, so it suffices (but is not equivalent, as far as I know) for $\lim_{Tw(C)}$ to preserve filtered colimits. In this case, however, $Tw(\Delta^1)$ is a finite $\infty$-category, so this stronger condition is satisfied, and $Fun(\Delta^1,Perf(\mathbb F_p))$ is smooth, hence so is $R$.

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