# The integral cohomology of the de Rham complex

Consider the usual de Rham CDGA $$(\Omega^* Sym^*(V),d)$$ for a free $$\mathbb{Z}_{(p)}$$-module $$V$$. What is known about its cohomology?

It is easy to compute ranks of primary summands in $$H^*(\Omega^* Sym^*(V),d)$$ through the Bockstein spectral sequence or by applying the Kunneth's formula. Meanwhile a truly functorial in $$V$$ description of the cohomology groups is available only in special cases and generally seems to be hopeless (according to the work of Roman Mikhailov).

What is known about the ring structure? Can we realize $$H^*(\Omega^* Sym^*(V))$$ as a quotient algebra of something else? Are there any "cohomology operations" acting here?

• The cohomology modulo $p^n$ is related to the de Rham--Witt complex of ${\rm Sym}^*(V\otimes\mathbf{F}_p)$. Mar 19 at 11:08
• I think I have probably misunderstood the question. Is this just asking about the de-Rham complex of $\mathbb{Z}_p[x_1, x_2, \ldots, x_n]$ over $\mathbb{Z}_p$? Because I think I can compute that by hand. Mar 19 at 12:42
• @David, I'm asking about the integral part $\mathbb{Z}_{(p)}$, or reduction modulo p^n, n>1. Do you able to describe the cohomology classes by forms? It is not clear how to control the additive order of a class after multiplication with another. I can not even imagine in which terms it is possible to describe the ring structure explicitly. Mar 19 at 13:05
• Well, I should go do it before I promise too much. What I want to do is start with a very non-functorial thing: Choosing coordinates on $V$. Then my de Rham complex gets a $\mathbb{Z}^n$ grading, where $x_i$ and $d x_i$ are in degree $e_i$, so I can break it up as a direct sum of finite dimensional complexes. The resulting ring won't be finitely generated -- for example, $H^0$ is just $\mathbb{Z}_p$, but $H^1$ is generated by the classes $\tfrac{1}{p^k} d(x_1^{p^k a_i} x_2^{p^k a_2} \cdots x_n^{p^k a_n})$, with $(a_1, a_2, \ldots, a_n)$ not all divisible by $p$; this class is $p^k$ torsion. Mar 19 at 13:12
• Since $H^1$ is not finitely generated as an $H^0$ module, we have to live with infinitely many generators. But they seem pretty combinatorially tractable. Mar 19 at 13:13

Following David Speyer's suggestion, fix exponents $$e_1,\dots, e_n$$ and consider the subcomplex consisting of sums of those monomials of the form $$\prod_{i=1}^n \left( x_i^{e_i - \epsilon_i} (d x_i)^{\epsilon_i} \right)$$ for $$\epsilon_i \in \{0,1\}$$. The point of doing this is that this subcomplex is stable under differentitation and the de Rham complex is the sum.

I claim that the cohomology of this complex, in degree $$d$$, is $$\left(\mathbb Z/ p^{ \min_{i=1}^n v_p( e_i)}\right)^{ \binom{m-1}{d-1}}$$ where $$m$$ is the number of $$e_i$$s that are nonzero.

To do this, note that this subcomplex is clearly the tensor product from $$i$$ from $$1$$ to $$n$$ of the complex consisting of $$x_i^{e_i}$$ in degree $$0$$ and $$x_i^{e_i-1} dx_i$$ in degree $$1$$ (unless $$e_i=0$$, in which case the $$i$$th complex is only in degree $$0$$). Calculating the derivative, this subcomplex is equivalent $$\mathbb Z_{(p)} \to \mathbb Z_{(p)}$$ with the differential multiplication by $$e_i$$.

Because this is a complex of projective modules, tensoring with it computes the derived tensor product. Now choose some $$i$$ minimizing $$v_p(e_i)$$, and observe that the $$i$$th complex is quasi-isomorphic to $$\mathbb Z/p^{ v_p(e_i)}$$ in degree $$1$$.

We may thus replace the $$i$$th factor in the tensor product with $$\mathbb Z/p^{ v_p(e_i)}$$ in degree $$1$$. But having done that, our tensor product complex is now isomorphic to $$\left(\mathbb Z/ p^{ \min_{i=1}^n v_p( e_i)}\right)^{ \binom{m-1}{d-1}}$$ in each degree $$d$$, and the differentials vanish because they are divisible by $$p^{ \min_{i=1}^n v_p( e_i)}$$, so the cohomology in degree $$d$$ is also $$\left(\mathbb Z/ p^{ \min_{i=1}^n v_p( e_i)}\right)^{ \binom{m-1}{d-1}}$$, as desired.

More concretely, we can represent the forms as follows. Say $$e_n$$ has the least $$p$$-adic valuation, for simplicity.

Then every $$k$$-form in the variables $$x_1,\dots,x_n$$ of exponent $$(e_1,\dots, e_n)$$ has the form $$g x_n^{e_n} + f x_n^{e_n-1} dx_n$$ where $$f$$ is a $$k-1$$-form and $$g$$ is a $$k$$-form in the variables $$x_{1},\dots, x_{n-1}$$, each of exponent $$(e_1,\dots, e_{n-1})$$. Differentiating, we get

$$d ( g x_n^{e_n} + f x_n^{e_n-1} dx_n)= dg x_n^{e_n} + (-1)^k e_n g x_n^{e_{n}-1} d x_n + df x_n^{e_n-1} dx_n .$$

So the form is closed if and only if $$dg=0$$ and $$e_n g = (-1)^{k-1} df$$ but the second equation implies the first because $$e_n$$ is a nonzero divisor in $$\mathbb Z$$ and $$d^2 =0$$.

So the closed forms are exactly those of the form $$\frac{df}{e_n} x_n^{e_n} + (-1)^{k-1} f x_n^{e_n-1} dx_n$$ for $$f$$ a $$k-1$$-form in $$x_1,\dots, x_{n-1}$$ of exponents $$e_1,\dots, e_{n-1}$$, and this always exists since $$e_n$$ divides $$e_1,\dots, e_{n-1}$$ over the $$p$$-adic numbers.

If $$f$$ is divisibly by $$e_n$$, then such a form is $$d ( f x_n^{e_n})$$ and hence is exact. Conversely, because all the exponents are divisible by $$e_n$$, if such a form is exact then $$f$$ is divisible by $$e_n$$.

So this gives an explicit basis for this module over $$\mathbb Z/p^{ v_p(e_n)}$$, that being the expressions $$\frac{df}{e_n} x_n^{e_n} + (-1)^{k-1} f x_n^{e_n-1} dx_n$$ for monomial $$f$$.

Multiplying two such expressions, one can, in principle, extract the structure constants for the cup / wedge product, but I haven't done this.

• @DavidESpeyer Good point, thanks. Mar 19 at 17:55
• @Will, thank you for writting this down. This is more or less what I meant by the Kunneth's formula. May be I was too vague, but how does this computation helps to understand the multiplicative structure? Mar 19 at 19:22
• Even additively... The tensor product of two complexes $e_1,e_2:Z\to Z$ is abstractly isomorphic to $Z\to Z$ plus a shifted copy $(Z\to Z)[-1]$ with differential $gcd(e_1,e_2)$. Sure, but this splitting is perfectly not canonical (we have to run the Euclid algorithm to do this!). Finally it is not clear which forms corresponds to which classes, how to multiply them etc. Mar 19 at 19:30
• @BadEnglish To represent the forms, for simplicity assuming the $n$th exponent has the least $p$-adic valuation, any cocycle has the form $f x_n^{e_n-1} dx_n - (df/ e_n ) x_n^{e_n}$ for $f$ a polynomial differential form in the first $n-1$ variables. It's a coboundary if and only if $f$ is divisible by $e_n$. Multiplying two expressions of this form should tell you the multiplicative structure. Mar 19 at 19:37
• @BadEnglish Details are now in my answer. Mar 19 at 19:48