A counterexample for the de Rham filtration on HP
Here is an example where the de Rham filtration on periodic cyclic homology is not exhaustive, motivated by the example of Bhatt and Mondal I describe in the previous post.
The close connection between de Rham cohomology and periodic cyclic homology realized by this filtration goes back at least to work of Feigin–Tsygan [3], Kassel [4], Loday [5], and Toen–Vezzosi [9] in the case of $\mathbf{Q}$algebras and has been studied more recently in general by Bhatt–Morrow–Scholze in [2], myself in [1], Moulinos–Robalo–Toën in [7], and Raksit in [8]. The associated spectral sequence however already appears in work of Majadas [6] from 1996.
There is a complete filtration on negative cyclic homology $\mathrm{F}^\star\mathrm{HC}^(R/k)$ and periodic cyclic homology $\mathrm{F}^\star\mathrm{HP}(R/k)$ with associated graded pieces given by \(\mathrm{gr}^n\simeq\widehat{\mathrm{dR}}_{R/k}^{\geq n}[2n]\) or \(\mathrm{gr}^n\simeq\widehat{\mathrm{dR}}_{R/k}[2n]\), respectively. In each case, an additional assumption on the cotangent complex guarantees that the filtrations are exhaustive as well. Specifically, this happens if $\mathrm{L}_{R/k}$ has Toramplitude in $[0,1]$.
Thomas Nikolaus and I were wondering whether or not, especially in characteristic zero, the filtration is always exhaustive. The example below shows that it is not.
Let $R$ be the free animated commutative $\mathbf{Q}$algebra on a single element $x$ of degree $2$.^{1} Thus, $\pi_*R\cong\mathbf{Q}[x]$ where $x=2$. The cotangent complex $\mathrm{L}_{R/\mathbf{Q}}$ is equivalent to $R[2]$ while the cotangent complex $\mathrm{L}_{\mathbf{Q}/R}$ is equivalent to $\mathbf{Q}[3]$ as one sees using the conormal sequence for the sequence $\mathbf{Q}\rightarrow R\rightarrow\mathbf{Q}$ of animated commutative rings.
It follows that \(\pi_*\widehat{\mathrm{dR}}_{\mathbf{Q}/R}\cong\mathbf{Q}[w]\) where $w=2$ and that $\pi_*\mathrm{HH}(\mathbf{Q}/R)\cong\mathbf{Q}[z]$ where $z=4$.
The Tate spectral sequence collapses because Hochschild homology $\mathrm{HH}(R/k)$ is concentrated in even degrees. We see that if $v$ is a degree $2$ generator of $\mathbf{Z}^{\mathrm{h}S^1}$, then $\pi_0\mathrm{HP}(\mathbf{Q}/R)$ is equivalent to $\mathbf{Q}[[vz]]$. On the other hand, one checks that \(\pi_0\mathrm{F}_{n}\mathrm{HP}(\mathbf{Q}/R)\cong \mathbf{Q}\cdot\{1,\ldots,(vz)^n\}\subseteq\mathbf{Q}[[vz]].\) It follows that the colimit is \(\pi_0\mathrm{F}_{\infty}\mathrm{HP}(\mathbf{Q}/R)\cong\mathbf{Q}[vz]\subseteq\mathbf{Q}[[vz]].\) Therefore, the de Rham filtration on $\mathrm{HP}$ is not exhaustive.
For those who would like to avoid thinking about the exact ring structure on $\pi_0\mathrm{HP}(\mathbf{Q}/R)$, one can see from the HKR filtration that $\pi_0\mathrm{HP}(\mathbf{Q}/R)$ is uncountable, but that \(\pi_0\mathrm{F}_{\infty}\mathrm{HP}(\mathbf{Q}/R)\) is countable.
I do not know at the moment whether or not this kind of behavior is possible for a map of ordinary commutative rings $k\rightarrow R$, but I would be very interested if there is such an example.
Finally, let me mention that in characteristic $0$ there is a canonical map \(\mathrm{F}^\star_\mathrm{H}\widehat{\mathrm{dR}}_{R/k}\rightarrow\mathrm{F}^\star_{\mathrm{HKR}}\mathrm{HP}(R/k).\) By base change up to $k^{\mathrm{t}S^1}$, we get a filtered map \(\mathrm{F}^\star_{\mathrm{H}}\widehat{\mathrm{dR}}_{R/k}\otimes_kk^{\mathrm{t}S^1}\rightarrow\mathrm{F}^\star_{\mathrm{HKR}}\mathrm{HP}(R/k).\)
Added 15 December 2020: I previously claimed that this map is an equivalence on associated graded pieces, but this is not true in general as \(\Lambda^s\mathrm{L}_{R/k}\otimes_kk^{\mathrm{t}S^1}\rightarrow(\Lambda^s\mathrm{L}_{R/k})^{\mathrm{t}S^1}\) is not typically an equivalence. Nor is it true that in general $R\otimes_kk^{\mathrm{t}S^1}\simeq R^{\mathrm{t}S^1}$. In particular, the map above does not become an equivalence upon completing the lefthand side in general. Thanks to Thomas Nikolaus for pointing this out.
Added 15 December 2020: contemplation of the previous remark leads one to conclude that there is an easier example of the phenomenon outlined in this post, but one must leave the world of discrete rings altogether. Indeed, if $R$ is as above, so that $\pi_*R\cong\mathbf{Q}[x]$ where $x=2$, then the de Rham filtration $\mathrm{F}^\star\mathrm{HP}(R/R)$ on $\mathrm{HP}(R/R)\simeq R^{\mathrm{t}S^1}$ is not exhaustive! In fact, this is basically the same example as the one above given that \(R\simeq\widehat{\mathrm{dR}}_{R/R}\rightarrow\widehat{\mathrm{dR}}_{\mathbf{Q}/R}\) is an equivalence.
References
[1] Antieau, Periodic cyclic homology and derived de Rham cohomology, Ann. KTheory 4 (2019), no. 3, 505519.
[2] Bhargav Bhatt, Matthew Morrow, and Peter Scholze, Topological Hochschild homology and integral padic Hodge theory, Publ. Math. Inst. Hautes Études Sci. 129 (2019), 199–310.
[3] Feĭgin, Tsygan, Cyclic homology of algebras with quadratic relations, universal enveloping algebras and group algebras in Ktheory, arithmetic and geometry (Moscow, 1984–1986), 210–239, Lecture Notes in Math., 1289, Springer, Berlin, 1987.
[4] Kassel, Cyclic homology, comodules, and mixed complexes, J. Algebra 107 (1987), no. 1, 195–216.
[5] Loday, Cyclic homology, second ed., vol. 301, SpringerVerlag, Berlin, 1998.
[6] Majadas, Derived de Rham complex and cyclic homology, Mathematica Scandinavica 79, no. 2 (1996), pp. 176188.
[7] Moulinos, Robalo, and Toën, A universal HKR theorem, arXiv:1906.00118.
[8] Raksit, Hochschild homology and the derived de Rham complex revisited, arXiv:2007.025760.
[9] Toën, Vezzosi, Algèbres simpliciales S1équivariantes, théorie de de Rham et théorèmes HKR multiplicatifs, Compos. Math. 147 (2011), no. 6, 1979–2000.
Footnotes

We work with homological gradings in this post. ↩