$ \newcommand\L{\mathrm{L}} \newcommand\E{\mathrm{E}} \newcommand\H{\mathrm{H}} \newcommand\dR{\mathrm{dR}} $

While discussing an interesting forthcoming preprint [2], I asked Shubhodip Mondal whether there was an example in characteristic \(0\) where Hodge-complete derived de Rham cohomology is not coconnective, i.e., whether there is a map of commutative \(\mathbf{Q}\)-algebras \(k\rightarrow R\) such that \(\H^n(\widehat{\mathrm{dR}}_{R/k})\neq 0\) for some $n<0$. Such examples are known in characteristic $p$ by [1]. He suggested the following example of Bhatt.

Let $R=\mathbf{Q}[x^{1/2^\infty}]/(x)$. It is easy to see that the cotangent complex $\L_{R/\mathbf{Q}}$ of $R$ has Tor-amplitude in $[-1,0]$ as it is a filtered colimit of lci commutative $\mathbf{Q}$-algebras. It follows from the cofiber sequence $\mathbf{Q}\otimes_R\L_{R/\mathbf{Q}}\rightarrow\L_{\mathbf{Q}/\mathbf{Q}}\rightarrow\L_{\mathbf{Q}/R}$ that if we let $R\rightarrow\mathbf{Q}$ be the map which sends $x^{1/2^n}$ to zero for each $n$, then $\L_{\mathbf{Q}/R}$ has Tor-amplitude in $[-2,-1]$. However, an easy exercise shows that in fact that Tor-amplitude of $\L_{\mathbf{Q}/R}$ is in $[-2,-2]$ and moreover that $\L_{\mathbf{Q}/R}\simeq\mathbf{Q}[2]$.

Now, recall the Hodge–de Rham spectral sequence \(\E_2^{s,t}=\H^t(\Lambda^s\L_{\mathbf{Q}/R})\Rightarrow\H^{s+t}(\widehat{\dR}_{\mathbf{Q}/R}).\) The derived exterior powers $\Lambda^s(\mathbf{Q}[2])$ vanish for $s>1$, so the spectral sequence collapses and we find $\H^*(\widehat{\dR}_{\mathbf{Q}/R})\cong\mathbf{Q}[z]/(z^2)$ where $|z|=-1$.

This example is in contrast to the results of Bhatt in [1], which in particular say that if $k\rightarrow R$ is a finite type map of noetherian $\mathbf{Q}$-algebras, then $\widehat{\mathrm{dR}}_{R/k}$ is coconnective.

If we drop the requirement that the commutative $\mathbf{Q}$-algebras $k$ and $R$ are discrete, it is very easy to come up with examples where the Hodge-completed derived de Rham cohomology is not coconnective. For example, a kind of universal example is given by considering $R=\mathbf{Q}[x]/(x^2)$ where $|x|=-1$ (the free cdga over $\mathbf{Q}$ on a class in cohomological degree $-1$) and the map $R\rightarrow\mathbf{Q}$ which sends $x$ to $0$. As above, we have $\L_{\mathbf{Q}/R}\simeq\mathbf{Q}[2]$ and the rest of the computation is the same. This line of thinking led us to a nice example for periodic cyclic homology, described in the next post.


[1] Bhatt, Completions and derived de Rham cohomology, arXiv:1207.6193.

[2] Mondal, forthcoming.