## 60.19 Cosimplicial preparations

In this section we compare crystalline cohomology with de Rham cohomology. We follow [Bhatt].

Example 60.19.1. Suppose that $A_*$ is any cosimplicial ring. Consider the cosimplicial module $M_*$ defined by the rule

$M_ n = \bigoplus \nolimits _{i = 0, ..., n} A_ n e_ i$

For a map $f : [n] \to [m]$ define $M_*(f) : M_ n \to M_ m$ to be the unique $A_*(f)$-linear map which maps $e_ i$ to $e_{f(i)}$. We claim the identity on $M_*$ is homotopic to $0$. Namely, a homotopy is given by a map of cosimplicial modules

$h : M_* \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*)$

see Section 60.16. For $j \in \{ 0, \ldots , n + 1\}$ we let $\alpha ^ n_ j : [n] \to [1]$ be the map defined by $\alpha ^ n_ j(i) = 0 \Leftrightarrow i < j$. Then $\Delta [1]_ n = \{ \alpha ^ n_0, \ldots , \alpha ^ n_{n + 1}\}$ and correspondingly $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*)_ n = \prod _{j = 0, \ldots , n + 1} M_ n$, see Simplicial, Sections 14.26 and 14.28. Instead of using this product representation, we think of an element in $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*)_ n$ as a function $\Delta [1]_ n \to M_ n$. Using this notation, we define $h$ in degree $n$ by the rule

$h_ n(e_ i)(\alpha ^ n_ j) = \left\{ \begin{matrix} e_{i} & \text{if} & i < j \\ 0 & \text{else} \end{matrix} \right.$

We first check $h$ is a morphism of cosimplicial modules. Namely, for $f : [n] \to [m]$ we will show that

60.19.1.1
$$\label{crystalline-equation-cosimplicial-morphism} h_ m \circ M_*(f) = \mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*)(f) \circ h_ n$$

The left hand side of (60.19.1.1) evaluated at $e_ i$ and then in turn evaluated at $\alpha ^ m_ j$ is

$h_ m(e_{f(i)})(\alpha ^ m_ j) = \left\{ \begin{matrix} e_{f(i)} & \text{if} & f(i) < j \\ 0 & \text{else} \end{matrix} \right.$

Note that $\alpha ^ m_ j \circ f = \alpha ^ n_{j'}$ where $0 \leq j' \leq n + 1$ is the unique index such that $f(i) < j$ if and only if $i < j'$. Thus the right hand side of (60.19.1.1) evaluated at $e_ i$ and then in turn evaluated at $\alpha ^ m_ j$ is

$M_*(f)(h_ n(e_ i)(\alpha ^ m_ j \circ f) = M_*(f)(h_ n(e_ i)(\alpha ^ n_{j'})) = \left\{ \begin{matrix} e_{f(i)} & \text{if} & i < j' \\ 0 & \text{else} \end{matrix} \right.$

It follows from our description of $j'$ that the two answers are equal. Hence $h$ is a map of cosimplicial modules. Let $0 : \Delta [0] \to \Delta [1]$ and $1 : \Delta [0] \to \Delta [1]$ be the obvious maps, and denote $ev_0, ev_1 : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*) \to M_*$ the corresponding evaluation maps. The reader verifies readily that the compositions

$ev_0 \circ h, ev_1 \circ h : M_* \longrightarrow M_*$

are $0$ and $1$ respectively, whence $h$ is the desired homotopy between $0$ and $1$.

Lemma 60.19.2. With notation as in (60.17.0.5) the complex

$\Omega _{D(0)} \to \Omega _{D(1)} \to \Omega _{D(2)} \to \ldots$

is homotopic to zero as a $D(*)$-cosimplicial module.

Proof. We are going to use the principle of Simplicial, Lemma 14.28.4 and more specifically Lemma 60.16.1 which tells us that homotopic maps between (co)simplicial objects are transformed by any functor into homotopic maps. The complex of the lemma is equal to the $p$-adic completion of the base change of the cosimplicial module

$M_* = \left( \Omega _{P/A} \to \Omega _{P \otimes _ A P/A} \to \Omega _{P \otimes _ A P \otimes _ A P/A} \to \ldots \right)$

via the cosimplicial ring map $P\otimes _ A \ldots \otimes _ A P \to D(n)$. This follows from Lemma 60.6.6, see comments following (60.17.0.2). Hence it suffices to show that the cosimplicial module $M_*$ is homotopic to zero (uses base change and $p$-adic completion). We can even assume $A = \mathbf{Z}$ and $P = \mathbf{Z}[\{ x_ i\} _{i \in I}]$ as we can use base change with $\mathbf{Z} \to A$. In this case $P^{\otimes n + 1}$ is the polynomial algebra on the elements

$x_ i(e) = 1 \otimes \ldots \otimes x_ i \otimes \ldots \otimes 1$

with $x_ i$ in the $e$th slot. The modules of the complex are free on the generators $\text{d}x_ i(e)$. Note that if $f : [n] \to [m]$ is a map then we see that

$M_*(f)(\text{d}x_ i(e)) = \text{d}x_ i(f(e))$

Hence we see that $M_*$ is a direct sum over $I$ of copies of the module studied in Example 60.19.1 and we win. $\square$

Lemma 60.19.3. With notation as in (60.17.0.4) and (60.17.0.5), given any cosimplicial module $M_*$ over $D(*)$ and $i > 0$ the cosimplicial module

$M_0 \otimes ^\wedge _{D(0)} \Omega ^ i_{D(0)} \to M_1 \otimes ^\wedge _{D(1)} \Omega ^ i_{D(1)} \to M_2 \otimes ^\wedge _{D(2)} \Omega ^ i_{D(2)} \to \ldots$

is homotopic to zero, where $\Omega ^ i_{D(n)}$ is the $p$-adic completion of the $i$th exterior power of $\Omega _{D(n)}$.

Proof. By Lemma 60.19.2 the endomorphisms $0$ and $1$ of $\Omega _{D(*)}$ are homotopic. If we apply the functor $\wedge ^ i$ we see that the same is true for the cosimplicial module $\wedge ^ i\Omega _{D(*)}$, see Lemma 60.16.1. Another application of the same lemma shows the $p$-adic completion $\Omega ^ i_{D(*)}$ is homotopy equivalent to zero. Tensoring with $M_*$ we see that $M_* \otimes _{D(*)} \Omega ^ i_{D(*)}$ is homotopic to zero, see Lemma 60.16.1 again. A final application of the $p$-adic completion functor finishes the proof. $\square$

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