# Filtered complexes and cohomologically equivalent subcomplexes

Erlend Grong and Francesca Tripaldi  University of Bergen, Department of
Mathematics, P.O. Box 7803, 5020 Bergen, Norway<EMAIL_ADDRESS>Centro
di Ricerca Matematica Ennio De Giorgi, Scuola Normale Superiore, Pisa, Italy
<EMAIL_ADDRESS>
###### Abstract.

Inspired by Rumin’s work on a subcomplex in sub-Riemannian manifolds which is
cohomologically equivalent to the de Rham complex, we present a more general
construction that produces subcomplexes from any filtered cochain complex of
finite depth and still computes the cohomology of the original filtered
complex. A priori these subcomplexes depend not only on the filtration itself,
but also on the choice of additional structures. However, we show that the
construction only depends on the given filtration up to isomorphism. Finally,
we show how such subcomplexes relate to spectral sequences, a cohomological
machinery that arises naturally when considering a filtered complex.

###### Key words and phrases:

Filtered differential complex, base differential, Rumin complex, spectral
sequences.

###### 2020 Mathematics Subject Classification:

58G05, 58A10, 53C15, 58A12

The first author is supported by the grant GeoProCo from the Trond Mohn
Foundation - Grant TMS2021STG02 (GeoProCo). The second author was partially
supported by the Swiss National Science Foundation Grant nr 200020-191978,
Analytic and geometric structures in singular spaces and would also like to
thank the Centro di Ricerca Matematica Ennio De Giorgi and the Scuola Normale
Superiore for the hospitality and support.

##  1\. Introduction

Filtered complexes show up all over mathematics, from differential forms [22,
10], Lie algebra cohomology [23], algebraic geometry [1], K-theory [9],
filtered Riemannian manifolds [36, 4], sub-Riemannian manifolds [31, 33, 34],
BGG-sequences [38, 15, 16] and parabolic geometries [37] just to name a few.
In 1994, Rumin [31, 28] pointed out that the de Rham complex
$(\Omega^{\bullet},d)$ of a contact manifold contains a subcomplex that is
isomorphic at the level of cohomology. The key idea behind this construction
is the introduction of a tensorial map
$d_{0}\colon\Omega^{\bullet}\to\Omega^{\bullet}$ such that $d-d_{0}$ becomes a
nilpotent operator. This idea has been further developed in [33, 32, 34] into
a complex that is well defined on any sub-Riemannian manifold, once some
preliminary choices for the grading and metric are made. Such subcomplex,
which is now known as the Rumin complex, has proven particularly useful in the
sub-Riemannian context for the study of holonomy of partial connections [17,
20, 13], analytic torsions [35, 24, 21], invariants of PDEs on differential
forms [34, 3, 15], sub-Laplacians and geometric invariants on CR-manifold [2,
8], and Sobolev inequalities and $L^{p,q}$-cohomologies [6, 29, 30, 7, 5].

In the current paper, we demonstrate how the Rumin complex appears as a
specific instance of a more general method of generating subcomplexes that can
be applied to any filtered cochain complex $(K,d)$. In order to produce such
subcomplexes, one needs to make certain choices related to the filtration
itself, such as the operator $d_{0}\colon K\to K$ which we call a base
differential of $(K,d)$, along with the subspaces
$\mathcal{T}\subset\mathcal{C}$, which will be referred to as the base
transversal and cotransversal of $d_{0}$. However, the subcomplexes produced
by any of these choices will all be isomorphic to each other, and they are all
compute the same cohomology as the original filtered complex $(K,d)$. Our main
result is as follows.

######  Theorem 1.1.

Let $(K,d)$ be a filtered cochain complex of finite depth with filtration
$K=K_{0}\supset K_{1}\supset K_{2}\supset\cdots\supset K_{S}\supset K_{S+1}=0$
and $dK_{j}\subset K_{j}$ for any $j=0,\ldots,S$. Then there exists a direct
sum decomposition

$K=\mathcal{T}\oplus d\mathcal{T}\oplus\mathcal{P},$

such that $(\mathcal{P},d|_{\mathcal{P}})$ is a subcomplex, $dK_{j}\subset
d\mathcal{T}+K_{j+1}$ for any $j=0,\ldots,S$, and the map

$(\mathcal{T}\cap K_{j})/K_{j+1}\to K_{j}/K_{j+1}\ ,\quad\alpha\bmod
K_{j+1}\mapsto d\alpha\bmod K_{j+1}\ $

is injective. Furthermore, $(\mathcal{P},d|_{\mathcal{P}})$ is isomorphic to
$(K,d)$ at the cohomology level. Finally, if $K=\tilde{\mathcal{T}}\oplus
d\tilde{\mathcal{T}}\oplus\tilde{\mathcal{P}}$ is another such decomposition,
then $(\mathcal{P},d|_{\mathcal{P}})$ and
$(\tilde{\mathcal{P}},d|_{\tilde{\mathcal{P}}})$ are isomorphic as cochain
complexes.

Another cohomological machinery that arises when considering filtered
complexes is that of spectral sequences. In this paper, we prove that a
filtered cochain complex $(K,d)$ of finite depth is a multicomplex, and as
such it has further properties that allow us to describe each page of their
spectral sequence explicitly. In this way, we are also able to clearly state
the relationship between the the subcomplexes $(\mathcal{P},d|_{\mathcal{P}})$
that we introduce and the spectral sequence differential operators and
quotients. In this sense, one can think of such subcomplexes as arising
“naturally” from the filtration given.

The structure of the paper is as follows. In Section 2 we give the general
definition of a base differential $d_{0}\colon K\to K$ of a filtered complex
$(K,d)$, and its relative base transversals and cotransversals
$\mathcal{T}\subset\mathcal{C}$, and together with some of their basic
properties. We then present the explicit construction of the subcomplex
$(\mathcal{P},d|_{\mathcal{P}})$, show how it relates to the Rumin complex,
and then prove Theorem 1.1. In Section 3 we produce several different examples
of filtered cochain complexes of finite depth, and explain how to identify
their base differential and the relative base complex
$(\mathcal{P},d|_{\mathcal{P}})$. In Section 4 we focus on the construction of
the spectral sequence of a given filtered complex $(K,d)$ of finite depth,
given that such $(K,d)$ turns out to be a multicomplex. Finally, in Section 5,
we present the explicit computations behind each page of the spectral sequence
associated to a nonnegative grading on a 3-dimensional filtered Lie group.

##  2\. Subcomplexes of general filtered complexes

###  2.1. Base differential of filtered complexes

In this section, we present a general approach to finding subcomplexes of any
filtered complex, which are isomorphic to the original complex at the level of
cohomology. Our ideas are built on the work of graded forms presented by Rumin
in [32] and [34], and later used in [16].

Let $(K,d)$ be a cochain complex with differential operator $d$. Assume that
we have a sequence of sub-complexes of finite depth, that is

(2.1) $K=K_{0}\supset K_{1}\supset K_{2}\supset\cdots\supset K_{S}\supset
K_{S+1}=0\text{ with }dK_{j}\subset K_{j}\,.$

We introduce the following definitions.

######  Definition 2.1.

We say that a linear operator $d_{0}:K\to K$ is _a base differential_ of the
filtered complex $(K,d)$ if

  1. (i)

$d^{2}_{0}=0$,

  2. (ii)

If $\alpha\in K_{j}$, then $(d-d_{0})\alpha\in K_{j+1}$,

######  Definition 2.2.

Let $d_{0}:K\to K$ be a base differential of $(K,d)$.

  1. (i)

We say that a subspace $\mathcal{T}\subseteq K$ is a base transversal for
$d_{0}$ if it satisfies

$K=\ker d_{0}\oplus\mathcal{T}\quad\text{and}\quad\mathcal{T}\cap
K_{j}=\\{\alpha\in K_{j}\,:\,d_{0}\alpha\in K_{j}\text{ with }d_{0}\alpha\neq
0\\}\,.$

  2. (ii)

We say that a subspace $\mathcal{C}\subseteq K$ is a base cotransversal for
$d_{0}$ with transversal $\mathcal{T}$ if it satisfies

$K=\operatorname{image}d_{0}\oplus\mathcal{C}\quad\text{and}\quad\mathcal{T}\subset\mathcal{C}\
.$

Observe that by definition $d_{0}K_{j}\subset K_{j}$. Given a choice for a
base differential $d_{0}$, one can always construct a corresponding base
transversal $\mathcal{T}$ and cotransversal $\mathcal{C}$.

######  Proposition 2.3.

Any filtered complex of finite depth $(K,d)$ admits a base differential
$d_{0}\colon K\to K$ together with a base cotransversal and transversal
$\mathcal{T}\subset\mathcal{C}$.

###### Proof.

Let us consider a filtered complex of finite depth as in (2.1). Since any
vector subspace admits a complement, we can write

(2.2) $\displaystyle K_{j}=K_{(j)}\oplus K_{j+1}=K_{(j)}\oplus K_{(j+1)}\oplus
K_{j+2}=\oplus_{k=0}^{S-j}K_{(j+k)}\,.$

Using this direct sum decomposition, since $dK_{j}\subset K_{j}$, the action
of the exterior differential $d$ can be expressed as
$d=d_{(0)}+d_{(1)}+\cdots+d_{(S)}$. Moreover, since $(K,d)$ is a complex, we
have

$\displaystyle 0=$ $\displaystyle
d^{2}=\big{(}d_{(0)}+d_{(1)}+\cdots+d_{(S)}\big{)}\big{(}d_{(0)}+d_{(1)}+\cdots+d_{(S)}\big{)}$
$\displaystyle=$ $\displaystyle\underbrace{d_{(0)}^{2}}_{\in
K_{(j)}}+\underbrace{d_{(0)}d_{(1)}+d_{(1)}d_{(0)}}_{\in
K_{(j+1)}}+\underbrace{d_{(0)}d_{(2)}+d_{(1)}^{2}+d_{(2)}d_{(0)}}_{\in
K_{(j+2)}}+\cdots+\underbrace{d_{(S)}^{2}}_{\in K_{(j+2S)}}$

which, in particular, implies $d_{(0)}^{2}=0$ and $(d-d_{(0)})K_{j}\subset
K_{j+1}$. We can then define $d_{0}:=d_{(0)}$.

Since $d_{0}K_{(j)}\subset K_{(j)}$, we can find a complement
$\mathcal{T}_{(j)}$ to $\ker d_{0}\cap K_{(j)}$ in $K_{(j)}$, as well as a
complement $\mathcal{A}_{(j)}$ to $d_{0}K_{(j)}$ in $\ker d_{0}\cap K_{(j)}$.
The subspaces of $K$ given by

$\mathcal{T}=\oplus_{j=0}^{S}\mathcal{T}_{(j)}\ ,\
\mathcal{C}=\mathcal{T}\oplus(\oplus_{j=0}^{S}\mathcal{A}_{(j)})\ \text{ with
}\ \mathcal{T}\subset\mathcal{C}$

will be respectively a base transversal and base cotransversal to the base
differential $d_{0}$. ∎

Once we have a base differential $d_{0}$ and a choice for
$\mathcal{T}\subset\mathcal{C}$, we can define a partial inverse of the base
differential as follows.

######  Definition 2.4.

Let $d_{0}\colon K\to K$ be a base differential of $(K,d)$, and let
$\mathcal{T}\subset\mathcal{C}$ be a base transversal and cotransversal for
$d_{0}$. Then we can define the _inverse_ map $d_{0}^{-1}\colon K\to K$ of
$d_{0}$ as follows

  1. (a)

$\ker d^{-1}_{0}=\mathcal{C}$,

  2. (b)

if $\alpha\in\operatorname{image}d_{0}$, then $d_{0}^{-1}\alpha$ is the unique
element in $\mathcal{T}$ mapped to $\alpha$ by $d_{0}$.

By construction $\mathcal{T}\cap
K_{j}=\\{\alpha\in\mathcal{T}\,:\,d_{0}\alpha\in K_{j}\text{ with
}d_{0}\alpha\neq 0\\}$, and so $d_{0}^{-1}K_{j}\subset K_{j}$. Moreover, by
requiring $\mathcal{T}\subset\mathcal{C}$, it follows that
$(d_{0}^{-1})^{2}=0$. With these definitions in place, we are ready to produce
the subcomplex $(\mathcal{P},d|_{\mathcal{P}})$ corresponding to the base
differential $d_{0}$ and the relative base transversal and cotransversal
$\mathcal{T}\subset\mathcal{C}$.

We introduce the operator

$P=\operatorname{id}-d_{0}^{-1}d-dd_{0}^{-1}.$

Given $S$ as in (2.1), we introduce the notation $P^{\infty}:=P^{S}$.

######  Definition 2.5.

Given a base differential $d_{0}\colon K\to K$ of $(K,d)$ with base
transversal and cotransversal $\mathcal{T}\subset\mathcal{C}$, we define
$\mathcal{P}\subset K$ as

$\displaystyle\mathcal{P}:=\operatorname{image}P^{\infty}\subset K\,.$

Moreover, since $dP=Pd$, we have that $(\mathcal{P},d)$ is a subcomplex of
$(K,d)$ and we will call this _the base subcomplex_ of $d_{0}$.

Since the construction of the operator $P$ depends not only on the base
differential $d_{0}$, but also on the choice of base transversal and
cotransversal for $d_{0}$, one should write
$\mathcal{P}=\mathcal{P}_{d_{0},\mathcal{T}\subset\mathcal{C}}$. However, in
order to avoid cumbersome notation, once the map $d_{0}\colon K\to K$ and the
subspaces $\mathcal{T}\subset\mathcal{C}$ have been fixed, we will simply
write $\mathcal{P}=\operatorname{image}P^{\infty}\subset K$.

######  Theorem 2.6.

Let $(\mathcal{P},d)$ be the base subcomplex of $d_{0}\colon K\to K$ with
transversal and cotransversal $\mathcal{T}\subset\mathcal{C}$.

  1. (a)

If $\alpha\in K$ is a closed form, then $P^{\infty}\alpha$ is in the same
cohomology class. In particular,

$P^{\infty}:K\to\mathcal{P}$

is an isomorphism at the cohomology level.

  2. (b)

The surjection between the two complexes

$K\to K/(\mathcal{T}+d\mathcal{T})$

is an isomorphism at the cohomology level. Furthermore, one can write

$K=\mathcal{T}\oplus d\mathcal{T}\oplus\mathcal{P}\ \ \text{with}\ \ker
P^{\infty}=\mathcal{T}\oplus d\mathcal{T},$

so the map

$\alpha+\mathcal{T}+d\mathcal{T}\longmapsto P^{\infty}(\alpha)$

is a well-defined isomorphism between the complexes
$\big{(}K/(\mathcal{T}+d\mathcal{T}),d\big{)}$ and $(\mathcal{P},d)$.

  3. (c)

One can rewrite $\mathcal{P}\subset K$ as

$\displaystyle\mathcal{P}$ $\displaystyle=\\{\alpha\in
K\,:\,P(\alpha)=\alpha\\}=\ker(dd_{0}^{-1}+d_{0}^{-1}d)=\ker
d_{0}^{-1}\cap\ker(d_{0}^{-1}d)=\\{\alpha\in\mathcal{C}\,:\,d\alpha\in\mathcal{C}\\}\
.$

In particular, $P^{k}=P^{\infty}$ for any $k\geq S$.

###### Proof.

  1. (a)

If $\alpha$ is closed, $P\alpha=\alpha-dd_{0}^{-1}\alpha$. In other words,
$\alpha$ and $P\alpha$ differ by an exact form. Applying this result
iteratively, we have that $\alpha$ and $P^{\infty}\alpha$ are in the same
cohomology class.

  2. (b)

We will prove the first part of the statement by showing that there are no
closed forms in $\mathcal{T}$. This observation follows from the fact that
$\mathcal{T}=\operatorname{image}d_{0}^{-1}$, and so

$d_{0}^{-1}d|_{\mathcal{T}}=\operatorname{id}|_{\mathcal{T}}+d_{0}^{-1}(d-d_{0})|_{\mathcal{T}},$

which has inverse $\sum_{i=0}^{S}[(-d_{0}^{-1}(d-d_{0})|_{\mathcal{T}}]^{i}$.
Hence, $\mathcal{T}$ is transverse to $d\mathcal{T}$ and $K\to
K/(\mathcal{T}\oplus d\mathcal{T})$ is an isomorphism at the level of
cohomology. For the second part of the statement, we first observe that since
$\mathcal{T}=\operatorname{image}d_{0}^{-1}$,

(2.3) $P(\alpha)=\alpha\mod\mathcal{T}+d\mathcal{T}.$

Hence, if $P^{k}(\alpha)=0$, then we must have
$\alpha\in\mathcal{T}+d\mathcal{T}$, giving us that $\ker
P^{k}\subset\mathcal{T}+d\mathcal{T}$ for any $k>0$. For the inclusion in the
other direction, we first observe that for $\beta\in\mathcal{T}\cap K_{j}$, we
have

$P(\beta)=\beta-d_{0}^{-1}d\beta=-d_{0}^{-1}(d-d_{0})\beta\in\mathcal{T}\cap
K_{j+1},$

which also implies that $P(d\beta)=dP(\beta)\in d\mathcal{T}\cap K_{j+1}$. By
the finiteness of the filtration, it follows that
$P^{\infty}(\beta)=P^{\infty}(d\beta)=0$.

  3. (c)

Introduce $\tilde{\mathcal{P}}=\\{\alpha\in K\,:\,P(\alpha)=\alpha\\}$. If
$\alpha\in\tilde{\mathcal{P}}$, then $P^{\infty}(\alpha)=\alpha$ by
definition, so $\tilde{\mathcal{P}}\subset\mathcal{P}$. Conversely, if
$\alpha\in\mathcal{P}$, then $\alpha=P^{\infty}(\beta)$, and furthermore,
since $P(\beta)-\beta$ is in $\mathcal{T}+d\mathcal{T}$ by (2.3), we have

$P(\alpha)-\alpha=P^{\infty}(P(\beta)-\beta)=0.$

It follows that $P(\alpha)-\alpha=(dd_{0}^{-1}+d_{0}^{-1}d)\alpha=0$. We
observe first that $dd_{0}^{-1}\alpha$ is in $d\mathcal{T}$, while
$d_{0}^{-1}d\alpha$ is in $\mathcal{T}$ and since these spaces are transverse,
both $dd_{0}^{-1}\alpha$ and $d_{0}^{-1}d\alpha$ have to vanish. Furthermore,
since there are no closed forms in $\mathcal{T}$, we must have
$d_{0}^{-1}\alpha=0$ and the result follows.

∎

######  Corollary 2.7.

Given a base differential $d_{0}\colon K\to K$ of $(K,d)$ with transversal and
cotransversal $\mathcal{T}\subset\mathcal{C}$, we have the following direct
sum decomposition for $K$:

$K=\mathcal{T}\oplus
d_{0}\mathcal{T}\oplus\mathcal{P}=\operatorname{image}d_{0}^{-1}\oplus\operatorname{image}d_{0}\oplus\mathcal{P}.$

Finally, $\mathcal{C}=\mathcal{T}\oplus\mathcal{P}$.

###### Proof.

By Theorem 2.6 (b), for any $\alpha\in K_{j}$, we have unique
$\beta_{j},\gamma_{j}\in\mathcal{T}\cap K_{j}$, $\omega_{j}\in\mathcal{P}\cap
K_{j}$, such that $\alpha=\beta_{j}+d\gamma_{j}+\omega_{j}$, and in particular
we have that

$\displaystyle\alpha=\beta_{j}+d_{0}\gamma_{j}+(d-d_{0})\gamma_{j}+\omega_{j}\
\text{ with }\ (d-d_{0})\gamma_{j}\in K_{j+1}\,.$

Repeating this process on $(d-d_{0})\gamma_{j}$, we have that there exist
unique $\beta_{j+1},\gamma_{j+1}\in\mathcal{T}\cap K_{j+1}$ and
$\omega_{j+1}\in\mathcal{P}\cap K_{j+1}$ such that

$\displaystyle\alpha=$
$\displaystyle\beta_{j}+d_{0}\gamma_{j}+\beta_{j+1}+d\gamma_{j+1}+\omega_{j}+\omega_{j+1}$
$\displaystyle=$
$\displaystyle\beta_{j}+\beta_{j+1}+d_{0}(\gamma_{j}+\gamma_{j+1})+(d-d_{0})\gamma_{j+1}+\omega_{j}+\omega_{j+1}\
,\text{ with }(d-d_{0})\gamma_{j+1}\in K_{j+2}\,.$

Proceeding iteratively and using the fact that our filtered complex has finite
length gives us the result.

By Theorem 2.6 (c), we have that $\mathcal{P}\subset\mathcal{C}$, hence
$\mathcal{T}\oplus\mathcal{P}\subset\mathcal{C}$, and the equality follows
from the fact that both $\mathcal{T}$ and $\mathcal{P}$ are transversal to
$\operatorname{image}d_{0}=d_{0}\mathcal{T}$. ∎

######  Remark 2.8.

Theorem 2.6 (b) and Corollary 2.7 imply that for any element $\alpha\in K$,
there exists a unique way of expressing $\alpha$ in terms of elements
$\beta,\gamma\in\mathcal{T}$ and $\omega\in\mathcal{P}$. Using the partial
inverse $d_{0}^{-1}\colon K\to K$, one can also obtain an explicit formulation
of these elements in terms of the given $\alpha\in K$.

  1. (a)

For $\alpha\in K$ given as $\alpha=\beta+d\gamma+\omega\in K$, with
$\beta,\gamma\in\mathcal{T}$, $\omega\in\mathcal{P}$, then

$\omega=P^{\infty}(\alpha)\ ,\
\gamma=\sum_{i=0}^{S}[-d_{0}^{-1}(d-d_{0})]^{i}d^{-1}_{0}\alpha\ ,\
\beta=\alpha-d\gamma-\omega\ .$

This follows from the fact that $\mathcal{C}=\mathcal{T}\oplus\mathcal{P}$, so

$d_{0}^{-1}\alpha=d_{0}^{-1}d\gamma=\gamma+d_{0}^{-1}(d-d_{0})\gamma\
\Longrightarrow\
\gamma=\sum_{i=0}^{S}[-d_{0}^{-1}(d-d_{0})]^{i}d^{-1}_{0}\alpha\,.$

  2. (b)

For $\alpha\in K$ given as $\alpha=\beta+d_{0}\gamma+\omega\in K$, with
$\beta,\gamma\in\mathcal{T}$, $\omega\in\mathcal{P}$, then
$\gamma=d_{0}^{-1}d_{0}\alpha$, and so

$\omega=P^{\infty}(\alpha-
d_{0}\gamma),\qquad\beta=(\operatorname{id}-P^{\infty})(\alpha-d_{0}\gamma)\
.$

  3. (c)

Observe that by Theorem 2.6 (c), there is a unique element $\beta$ in each
equivalence class $\alpha+\mathcal{T}+d\mathcal{T}$ satisfying

$d^{-1}_{0}\beta=0\quad\text{and}\quad d^{-1}_{0}d\beta=0\ .$

  4. (d)

Moreover, the subcomplex $(\mathcal{P},d)$ is itself a filtered complex

$\mathcal{P}=\mathcal{P}_{0}\supset\mathcal{P}_{1}\supset\cdots\supset\mathcal{P}_{S}\supset\mathcal{P}_{S+1}=0\
\text{ with }\ \mathcal{P}_{j}=\mathcal{P}\cap K_{j}\ .$

###  2.2. Analogue of Rumin forms and Hodge decomposition

Given a base differential $d_{0}$ of a filtered complex $(K,d)$ with relative
base transversal and cotransversal $\mathcal{T}\subset\mathcal{C}$, let us
introduce the subspace

$\mathcal{E}_{0}=\ker d_{0}\cap\ker d_{0}^{-1}=\ker
d_{0}\cap\mathcal{C}\subset K\ .$

From the definition of $\mathcal{E}_{0}$, we readily get a direct sum
decomposition of $\ker d_{0}$ as $\ker
d_{0}=\operatorname{image}d_{0}\oplus\mathcal{E}_{0}$. Moreover, the map

$\Pi_{0}=\operatorname{id}-d_{0}d_{0}^{-1}-d_{0}^{-1}d_{0},$

is a projection onto $\mathcal{E}_{0}$ and

$K=(\operatorname{image}d_{0}^{-1})\oplus(\operatorname{image}d_{0})\oplus\mathcal{E}_{0}=\mathcal{T}\oplus
d_{0}\mathcal{T}\oplus\mathcal{E}_{0}\ .$

######  Proposition 2.9.

The map $\Pi_{0}|_{\mathcal{P}}:\mathcal{P}\to\mathcal{E}_{0}$ is a bijection
with inverse $P^{\infty}|_{\mathcal{E}_{0}}:\mathcal{E}_{0}\to\mathcal{P}$. It
follows that, if we define $d_{c}:\mathcal{E}_{0}\to\mathcal{E}_{0}$ as

$d_{c}\alpha=\Pi_{0}(P^{\infty}(d\alpha))=\Pi_{0}(dP^{\infty}(\alpha))\ ,$

then $(\mathcal{E}_{0},d_{c})$ and $(\mathcal{P},d|_{\mathcal{P}})$ are
isomorphic as differential complexes. If we introduce a filtration on
$\mathcal{E}_{0}$ by imposing $\mathcal{E}_{0,j}=\mathcal{E}_{0}\cap K_{j}$,
then $(\mathcal{E}_{0},d_{c})$ and $(\mathcal{P},d|_{\mathcal{P}})$ are
isomorphic as filtered complexes.

The notation $(\mathcal{E}_{0},d_{c})$ is used to reflect Rumin’s notation in
[34].

###### Proof.

We observe that if
$\alpha\in\mathcal{P}=\\{\alpha\in\mathcal{C}\,:\,d\alpha\in\mathcal{C}\\}$,
then

$P^{\infty}(\Pi_{0}(\alpha))=P^{\infty}(\alpha-d^{-1}_{0}d_{0}\alpha)=\alpha,$

since $d_{0}^{-1}d_{0}\alpha\in\mathcal{T}$. On the other hand, if
$\alpha\in\mathcal{E}_{0}$ then

$P^{\infty}(\alpha)=(\operatorname{id}-d_{0}^{-1}d)^{S}\alpha=\alpha\bmod\operatorname{image}d_{0}^{-1},$

so $\Pi_{0}(P^{\infty}\alpha)=\alpha$. Observe that both $\Pi_{0}$ and
$P^{\infty}$ preserve the filtration. It follows that if
$d_{c}\alpha=\Pi_{0}(dP^{\infty}(\alpha))$ then $(\mathcal{E}_{0},d_{c})$ and
$(\mathcal{P},d|_{\mathcal{P}})$ are by definition isomorphic complexes. ∎

We emphasize the following special case. Assume that we have a filtered
complex $(K,d)$ as in (2.1) with a Hilbert space structure, such that every
subcomplex $K_{j}$ is closed in $K$. We can then use the construction in the
proof of Proposition 2.3, to define $K_{j}=K_{(j)}\oplus K_{j+1}$ where
$K_{(j)}$ is the orthogonal complement of $K_{(j+1)}$ in $K_{j}$. We again
define $d_{0}$ as the linear operator such that for $\alpha\in K_{(j)}$, we
have $d_{0}\alpha$ as the projection of $d\alpha$ to $K_{(j)}$. We make the
final assumption that $\ker d_{0}$ and $\operatorname{image}d_{0}$ are both
closed subspaces. Then one can take $\mathcal{T}=(\ker d_{0})^{\perp}$ and
$\mathcal{C}=(\operatorname{image}d_{0})^{\perp}$ as base transversal and
cotransversal for $d_{0}$. It is also possible to consider the formal adjoint
$d_{0}^{*}$ of $d_{0}$ with respect to the inner product of such Hilbert
space. We can then write a Hodge-type decomposition

$K=\mathcal{T}\oplus
d_{0}\mathcal{T}\oplus\mathcal{E}_{0}=\operatorname{image}(d_{0}^{*})\oplus\operatorname{image}(d_{0})\oplus\ker(\Box_{0})\
,\text{ with }\Box_{0}=d_{0}d_{0}^{*}+d_{0}^{*}d_{0}\ ,$

and the subcomplex $(\ker\Box_{0},d_{c})$ is isomorphic to $(K,d)$ at the
cohomology level (see also [16] and [18]).

###  2.3. Invariance under choice of transversal and cotransversal

Let $d_{0}$ be a base differential of the filtered cochain complex $(K,d)$,
with base transversal and cotransversal $\mathcal{T}\subset\mathcal{C}$. Let
$\tilde{\mathcal{T}}\subset\tilde{\mathcal{C}}$ be a different choice of
transversal and cotransversal for the same base differential $d_{0}$, and
define the subcomplexes
$\mathcal{P}=\mathcal{P}_{d_{0},\mathcal{T}\subset\mathcal{C}}$ and
$\tilde{\mathcal{P}}=\mathcal{P}_{d_{0},\tilde{\mathcal{T}}\subset\tilde{\mathcal{C}}}$
relative to these two different choices. The following result holds.

######  Proposition 2.10.

If $P^{\infty}=P^{S}$ is defined with respect to the base transversal and
cotransversal $\mathcal{T}\subset\mathcal{C}$, then
$P^{\infty}\circ\tilde{\Pi}_{0}|_{\tilde{\mathcal{P}}}$ is an isomorphism
between $\tilde{\mathcal{P}}$ and $\mathcal{P}$ as filtered differential
complexes.

###### Proof.

Let $d_{0}^{-1}$ and $\tilde{d}_{0}^{-1}$ be the partial inverses of the base
differential $d_{0}$, constructed using to $\mathcal{T}\subset\mathcal{C}$ and
$\tilde{\mathcal{T}}\subset\tilde{\mathcal{C}}$ respectively. We introduce
$\Pi_{0}=\operatorname{id}-d_{0}d_{0}^{-1}-d_{0}^{-1}d_{0}$ and
$\tilde{\Pi}_{0}=\operatorname{id}-d_{0}\tilde{d}_{0}^{-1}-\tilde{d}_{0}^{-1}d_{0}$
as the relative projections onto $\mathcal{E}_{0}=\ker d_{0}\cap\mathcal{C}$
and $\tilde{\mathcal{E}}_{0}=\ker d_{0}\cap\tilde{\mathcal{C}}$. We first see
that, for any $\alpha\in\tilde{\mathcal{E}}_{0}$

$\tilde{\Pi}_{0}(\Pi_{0}(\alpha))=\tilde{\Pi}_{0}(\alpha-
d_{0}d_{0}^{-1}\alpha)=\alpha\ .$

We have similar relations with $\Pi_{0}$ and $\tilde{\Pi}_{0}$ reversed.
Hence, the following sequence of invertible maps

$\tilde{\mathcal{P}}\stackrel{{\scriptstyle\tilde{\Pi}_{0}}}{{\longrightarrow}}\tilde{\mathcal{E}}_{0}\stackrel{{\scriptstyle\Pi_{0}}}{{\longrightarrow}}\mathcal{E}_{0}\stackrel{{\scriptstyle
P^{\infty}}}{{\longrightarrow}}\mathcal{P},$

are all linear isomorphisms that preserve the filtration induced by the
filtration (2.1) on $K$.

To conclude, it is sufficient to observe that for any
$\alpha\in\tilde{\mathcal{P}}$, we have

$\displaystyle P^{\infty}(\Pi_{0}(\tilde{\Pi}_{0}(\alpha)))=$ $\displaystyle
P^{\infty}(\Pi_{0}(\alpha-\tilde{d}_{0}^{-1}d_{0}\alpha))=P^{\infty}(\alpha-\tilde{d}_{0}^{-1}d_{0}\alpha-
d_{0}d_{0}^{-1}(\alpha-\tilde{d}_{0}^{-1}d_{0}\alpha))$ $\displaystyle=$
$\displaystyle
P^{\infty}(\alpha-\tilde{d}_{0}^{-1}d_{0}\alpha)=P^{\infty}\circ\tilde{\Pi}_{0}(\alpha)\
.$

∎

###  2.4. Invariance under choice of base differential

We now look into the base differentials themselves. Let $d_{0}$ be a base
differential for a filtered complex $(K,d)$, with base transversal and
cotransversal $\mathcal{T}\subset\mathcal{C}$, and with
$\mathcal{P}=\mathcal{P}_{d_{0},\mathcal{T}\subset\mathcal{C}}$. Let
$\tilde{d}_{0}$ be a different choice of base differential of the same
filtered complex $(K,d)$, with similar associated objects
$\tilde{\mathcal{T}}\subset\tilde{\mathcal{C}}$ and $\tilde{\mathcal{P}}$. We
then have the following result

######  Proposition 2.11.

  1. (a)

$\mathcal{T}\subset\mathcal{C}$ are also base transversals and cotransversals
for $\tilde{d}_{0}$.

  2. (b)

If $P^{\infty}=P^{S}$ is defined with respect to $d_{0}$ and
$\mathcal{T}\subset\mathcal{C}$, then
$P^{\infty}\circ\tilde{\Pi}_{0}|_{\tilde{\mathcal{P}}}$ is an isomorphism of
differential complexes between $\tilde{\mathcal{P}}$ and $\mathcal{P}$.

###### Proof.

  1. (a)

Let us introduce the notation $\tilde{d}_{0}=d_{0}-N$, where the map
$N=(d-\tilde{d}_{0})-(d-d_{0})$ maps $K_{j}$ into $K_{j+1}$. It follows that
$\hat{N}:=d_{0}^{-1}N$ is nilpotent with $\hat{N}^{S+1}=0$, meaning that
$A:=\operatorname{id}-\hat{N}$ is invertible. If we consider the projection
map $d_{0}d_{0}^{-1}$ applied to the image of $\tilde{d}_{0}$, we have that
for any $\alpha\in K$

$\displaystyle
d_{0}d_{0}^{-1}(\tilde{d}_{0}\alpha)=d_{0}d_{0}^{-1}(d_{0}\alpha-N\alpha)=d_{0}\alpha-
d_{0}\hat{N}\alpha=d_{0}(\alpha-\hat{N}\alpha)$

and so we get

$\displaystyle d_{0}d^{-1}_{0}|_{\operatorname{image}(\tilde{d}_{0})}$
$\displaystyle:\operatorname{image}\tilde{d}_{0}\to\operatorname{image}d_{0}\
,\ \tilde{d}_{0}\alpha\longmapsto d_{0}A\alpha\ .$

The map given by the composition

$\mathcal{T}\stackrel{{\scriptstyle
A^{-1}}}{{\longrightarrow}}\mathcal{T}\stackrel{{\scriptstyle\tilde{d}_{0}}}{{\longrightarrow}}\tilde{d}_{0}\mathcal{T}\stackrel{{\scriptstyle
d_{0}d_{0}^{-1}}}{{\longrightarrow}}\operatorname{image}d_{0}\ ,$

simplifies into $d_{0}\colon\mathcal{T}\to\mathrm{image\,}d_{0}$, since for
any $\alpha\in K$ we have
$d_{0}d_{0}^{-1}\tilde{d}_{0}A^{-1}\alpha=d_{0}AA^{-1}\alpha=d_{0}\alpha$.
Since $d_{0}$ is an invertible map from $\mathcal{T}$ to
$\operatorname{image}d_{0}$, this implies that $\tilde{d}_{0}|_{\mathcal{T}}$
must be invertible as well. Furthermore, by definition of base transversal for
$d_{0}$, we have

$\displaystyle\mathcal{T}\cap K_{j}=$ $\displaystyle\\{\alpha\in
K_{j}:d_{0}\alpha\in K_{j}\text{ with }d_{0}\alpha\neq 0\\}=\\{\alpha\in
K_{j}:d_{0}\alpha=d_{0}d_{0}^{-1}\tilde{d}_{0}A^{-1}\alpha\in K_{j}\text{ with
}d_{0}\alpha\neq 0\\}$ $\displaystyle=$ $\displaystyle\\{\alpha\in
K_{j}:\tilde{d}_{0}A^{-1}\alpha\in K_{j}\text{ with
}\tilde{d}_{0}A^{-1}\alpha\neq 0\\}=\\{\alpha=A\beta\in
K_{j}:\tilde{d}_{0}\beta\in K_{j}\text{ with $\tilde{d}_{0}\beta\neq 0\\}$}$

In other words, $\\{\beta\in K_{j}:\tilde{d}_{0}\beta\in K_{j}\text{ with
}\tilde{d}_{0}\beta\neq 0\\}=A^{-1}(\mathcal{T}\cap K_{j})=\mathcal{T}\cap
K_{j}$, and so $\mathcal{T}$ is a base transversal for $\tilde{d}_{0}$ as
well. Finally, by construction we have $\mathcal{T}\subset\mathcal{C}$.
Moreover the map

$K\to\operatorname{image}\tilde{d}_{0},\qquad\alpha\mapsto\tilde{d}_{0}A^{-1}d_{0}^{-1}\alpha,$

will be a projection onto $\operatorname{image}\tilde{d}_{0}$ with kernel
$\ker d_{0}^{-1}=\mathcal{C}$, implying that $\mathcal{C}$ is transverse to
$\operatorname{image}\tilde{d}_{0}$.

  2. (b)

By (a) and Proposition 2.10, it is sufficient to prove the result in the case
where $\mathcal{T}\subset\mathcal{C}$ equals
$\tilde{\mathcal{T}}\subset\tilde{\mathcal{C}}$. Continuing with the notation
in (a), we observe that $\tilde{d}_{0}^{-1}=A^{-1}d_{0}^{-1}$. Furthermore, if
$\tilde{P}=\operatorname{id}-\tilde{d}_{0}^{-1}d-d\tilde{d}_{0}^{-1}$, then

$\displaystyle\tilde{\mathcal{P}}$
$\displaystyle=\\{\tilde{P}(\alpha)=\alpha\,:\,\alpha\in K\\}=\\{\alpha\in
K\,:\,\tilde{d}_{0}^{-1}\alpha=0,\tilde{d}_{0}^{-1}d\alpha=0\\}$
$\displaystyle=\\{\alpha\in
K\,:\,A^{-1}d_{0}^{-1}\alpha=0,A^{-1}d_{0}^{-1}d\alpha=0\\}=\mathcal{P}$

since $A^{-1}$ is an invertible map of $\mathcal{T}$ to itself. The result
follows.

∎

###  2.5. Proof of Theorem 1.1

The existence of a splitting $K=\mathcal{T}\oplus
d\mathcal{T}\oplus\mathcal{P}$ follows from the existence of a base
differential $d_{0}$ of $(K,d)$ as shown in Proposition 2.3. The property
$dK_{j}\subset d\mathcal{T}+K_{j+1}$ is a direct consequence of the fact that
for any $\alpha\in K_{j}$

$\displaystyle d\alpha$ $\displaystyle=d_{0}\alpha\bmod
K_{j+1}=d_{0}\beta\bmod K_{j+1}=d\beta\bmod K_{j+1}\ ,\text{ where
}\beta=d^{-1}_{0}d_{0}\alpha\in\mathcal{T}\ .$

Conversely, assume that we have a decomposition $K=\mathcal{T}\oplus
d\mathcal{T}\oplus\mathcal{P}$ as described in Theorem 1.1, corresponding to a
base differential. Using the direct sum decomposition
$K_{j}=\oplus_{i=0}^{S-j}K_{(j+i)}$ presented in (2.2) and the fact that
$dK_{j}\subset K_{j}$, the action of the exterior differential $d$ can be
expressed as $d=d_{(0)}+d_{(1)}+\cdots+d_{(S)}$. Then by our assumptions on
$\mathcal{T}$, for any $\alpha\in K_{j}$ there exists a unique
$\beta\in\mathcal{T}$ such that

$d\alpha=d_{(0)}\alpha\bmod K_{j+1}=d\beta\bmod K_{j+1}=d_{(0)}\beta\bmod
K_{j+1}.$

Hence, $\mathcal{T}$ is a base transversal for the base differential
$d_{(0)}$. By repeating the same reasoning of Corollary 2.7, we get that
$K=\mathcal{T}\oplus d_{(0)}\mathcal{T}\oplus\mathcal{P}$, and so
$\mathcal{C}=\mathcal{T}\oplus\mathcal{P}$ gives a base cotransversal for
$d_{(0)}$.

Once we define a partial inverse map
$d_{(0)}^{-1}\colon\mathrm{image}\,d_{(0)}\to\mathcal{T}$, the map
$P^{\infty}=(\operatorname{id}-d_{(0)}^{-1}d-dd_{(0)}^{-1})^{S}\colon
K\to\mathcal{P}$ gives us an isomorphism at the level of cohomology. Finally,
the subcomplex $(\mathcal{P},d|_{\mathcal{P}})$ we have obtained in this way
is isomorphic to any other base complex by Proposition 2.11.

##  3\. Examples

We consider the following examples of filtered complexes $(K,d)$ and their
relative base complexes $(\mathcal{P},d)$.

###  3.1. Manifolds with a negative filtration of the tangent bundle

Let $M$ be a connected manifold. Assume that we have a filtration on the
tangent bundle

$TM=TM_{-s}\supset TM_{-s+1}\supset\cdots\supset TM_{-2}\supset TM_{-1}\supset
TM_{0}=0\,,$

where we are assuming $TM_{-s-k}=TM_{-s}=TM$ for any $k\geq 0$. This gives us
a filtration

$\bigwedge TM=(\bigwedge TM)_{-S}\supset(\bigwedge
TM)_{-S+1}\supset\cdots\supset(\bigwedge TM)_{-1}\supset(\bigwedge
TM)_{0}\supset 0\ ,$ (3.1) $\displaystyle\text{where
}S=\sum_{j=1}^{s}j\left(\operatorname{rank}(TM)_{-j}-\operatorname{rank}(TM)_{-j+1}\right)\,,$

defined by imposing $\chi_{1}\wedge\chi_{2}\in(\bigwedge TM)_{-i-j}$, if
$\chi_{1}\in(\bigwedge TM)_{-i}$ and $\chi_{2}\in(\bigwedge TM)_{-j}$. We
remark that from this convention $\mathbb{R}=\bigwedge^{0}TM$ is in
$(\bigwedge TM)_{0}$, but not in $(\bigwedge TM)_{-1}$.

By duality, we introduce $(\bigwedge T^{\ast}M)_{j}$ defined as the subspace
of the elements in $\bigwedge T^{*}M$ that vanish on $(\bigwedge TM)_{-j+1}$,
for any $j\geq 1$. We will denote by $\Omega_{j}$ the space of all smooth
sections of $(\bigwedge T^{\ast}M)_{j}$. Just like in the case of the
filtration on $\bigwedge TM$, $\Omega_{0}$ is the space of all smooth forms,
so that $0$-forms $\Omega^{0}=C^{\infty}(M)$ will belong to $\Omega_{0}$ but
not $\Omega_{1}$.

It is worth mentioning that, by definition, the filtration on $\bigwedge TM$
gives rise to a dual decreasing filtration on $\bigwedge T^{*}M$

$\displaystyle\bigwedge T^{*}M=(\bigwedge T^{*}M)_{0}\supset(\bigwedge
T^{*}M)_{1}\supset\cdots\supset(\bigwedge T^{*}M)_{S}\supset(\bigwedge
T^{*}M)_{S+1}=0\,,$

where $S$ is the index given in (3.1). This filtration then naturally extends
to the space of smooth forms as

$\displaystyle\Omega=\Omega_{0}\supset\Omega_{1}\supset\cdots\supset\Omega_{S}\supset\Omega_{S+1}=0\,.$

We introduce the following crucial assumption on our filtration:

(A) $\text{if }X\in\mathfrak{X}_{-i}\text{ and }Y\in\mathfrak{X}_{-j}\ ,\text{
then }[X,Y]\in\mathfrak{X}_{-i-j}$

where we denote by $\mathfrak{X}_{-j}$ the space of smooth sections of
$TM_{-j}$.

######  Lemma 3.1.

If (A) holds, then for $j=0,1,\dots,S$, we have
$d\Omega_{j}\subset\Omega_{j}$.

###### Proof.

For $j=0$, the claim is trivial. Let $\alpha\in\Omega_{j}$ be a $k$-form with
$j\geq 1$ and prove that also $d\alpha\in\Omega_{j}$. For any $k+1$ smooth
vector fields $X_{1},\ldots,X_{k+1}$ on $M$, the formula for the exterior
differential on $\alpha$ is given by

(3.2) $\displaystyle d\alpha(X_{1}\wedge\cdots\wedge X_{k+1})=$
$\displaystyle\sum_{1\leq i\leq
k+1}(-1)^{i-1}X_{i}\big{(}\alpha(X_{1}\wedge\cdots\wedge\hat{X}_{i}\wedge\cdots\wedge
X_{k+1})\big{)}+$ $\displaystyle+$ $\displaystyle\sum_{1\leq i<l\leq
k+1}(-1)^{i+l}\alpha\big{(}[X_{i},X_{l}]\wedge
X_{1}\wedge\cdots\wedge\hat{X}_{i}\wedge\cdots\wedge\hat{X}_{l}\wedge\cdots\wedge
X_{k+1}\big{)}\,,$

where the hat-symbol is used to indicate the terms omitted.

If $X_{1}\wedge\cdots\wedge X_{k+1}\in\mathfrak{X}_{-j+1}$, then for any
$i=1,\ldots,k+1$, we have $X_{i}\in\mathfrak{X}_{-n_{i}}$ with $n_{i}\geq 1$,
and so $X_{1}\wedge\cdots\wedge\hat{X}_{i}\wedge\cdots\wedge
X_{k+1}\in\mathfrak{X}_{-m_{i}}$ with $-m_{i}-n_{i}\geq-j+1$. This readily
implies that $-m_{i}\geq-j+2$, that is
$\mathfrak{X}_{-m_{i}}\subset\mathfrak{X}_{-j+1}$, and so the first addend of
(3.2) vanishes, since by assumption $\alpha\in\Omega_{j}$.

Likewise, if $X_{i}\in\mathfrak{X}_{-n_{i}}$ and
$X_{l}\in\mathfrak{X}_{-n_{l}}$ with both $n_{i},n_{l}\geq 1$, then the
$(k-1)$-vector field
$X_{1}\wedge\cdots\wedge\hat{X}_{i}\wedge\cdots\wedge\hat{X}_{l}\wedge\cdots\wedge
X_{k+1}$ will belong to $\mathfrak{X}_{-m_{i,l}}$ with
$-m_{i,l}\geq-j+1+n_{i}+n_{l}$. By assumption (A) we have that
$[X_{i},X_{l}]\in\mathfrak{X}_{-n_{i}-n_{l}}$, and so

$[X_{i},X_{l}]\wedge
X_{1}\wedge\cdots\wedge\hat{X}_{i}\wedge\cdots\wedge\hat{X}_{l}\wedge\cdots\wedge
X_{k+1}\in\mathfrak{X}_{-m_{i,l}-n_{i}-n_{l}}\subset\mathfrak{X}_{-j+1}$

since $-m_{i,l}-n_{i}-n_{l}\geq-j+1+n_{i}+n_{l}-n_{i}-n_{l}\geq-j+1$, which
implies that also the second addend of (3.2) vanishes, which concludes the
proof. ∎

Given the decreasing filtration over the space of smooth forms, it is possible
to express $\Omega$ as a direct sum decomposition
$\Omega=\Omega_{(0)}\oplus\cdots\oplus\Omega_{S}$, where the subspaces
$\Omega_{(j)}$ are taken so that $\Omega_{j}=\Omega_{(j)}\oplus\Omega_{j+1}$.
We observe that if $\alpha\in\Omega_{j}$, then

$d(f\alpha)=fd\alpha\mod\Omega_{j+1}.$

Hence, we can define a $C^{\infty}(M)$-linear map
$d_{0}\colon\Omega_{j}\to\Omega_{j}$ such that, for any
$\alpha\in\Omega_{(j)}$,

$d_{0}\alpha:=d\alpha\mod\Omega_{j+1}\ \text{ so that }\
d_{0}\alpha\in\Omega_{(j)}\,,\text{ and }\ (d-d_{0})\alpha\in\Omega_{j+1}\ .$

By tensoriality, the map $d_{0}\colon\Omega\to\Omega$ induces a vector bundle
morphism $\check{d}_{0}\colon\bigwedge T^{*}M\to\bigwedge T^{*}M$ for which
$\check{d}_{0}^{2}=0$. Moreover, we have that for any $\chi\in(\bigwedge
T^{*}M)_{j}$, $(\check{d}-\check{d}_{0})\chi\in(\bigwedge T^{*}M)_{j+1}$,
where $\check{d}\colon\bigwedge T^{*}M\to\bigwedge T^{*}M$ is the vector
bundle morphism induced by the exterior derivative $d\colon\Omega\to\Omega$.

The map $\check{d}_{0}$ is therefore a base differential of the complex
$(\bigwedge T^{*}M,\check{d})$, for which it is possible to construct
$\check{\mathcal{T}}\subset\check{\mathcal{C}}\subset\bigwedge T^{*}M$ a base
transversal and cotransversal so that

$\displaystyle\bigwedge
T^{*}M=\ker\check{d}_{0}\oplus\check{\mathcal{T}}=\mathrm{image\,}\check{d}_{0}\oplus\check{\mathcal{C}}\,,\,\check{\mathcal{T}}\cap(\bigwedge
T^{*}M)_{j}=\\{\chi\in(\bigwedge T^{*}M)_{j}:\check{d}_{0}\chi\in(\bigwedge
T^{*}M)_{j}\text{ with }\check{d}_{0}\chi\neq 0\\}.$

It is therefore sufficient to take the spaces of smooth sections
$\mathcal{T}\subset\mathcal{C}$ of
$\check{\mathcal{T}}\subset\check{\mathcal{C}}$ to obtain a base transversal
and cotransversal for $d_{0}\colon\Omega\to\Omega$. Indeed, the map
$d_{0}\colon\Omega\to\Omega$ is a base differential for the de Rham complex
$(\Omega,d)$, since for any $\alpha\in\Omega_{j}$, we have
$(d-d_{0})\alpha=(d-\check{d}_{0})\alpha\in\Omega_{j+1}$.

These definitions lead to well-defined base complexes
$(\mathcal{P},d|_{\mathcal{P}})$ as in Theorem 1.1, and
$(\mathcal{E}_{0},d_{c})$ as in subsection 2.2. In particular, the latter
subcomplex coincides with the complex defined by Rumin in [32] and [34], where

  1. $\bullet$

grading of the tangent bundle is defined so that $TM_{-j-1}$ is spanned by
$TM_{-j}$ brackets of vector fields with one taking value in $TM_{-j}$ and the
other in $TM_{-1}$;

  2. $\bullet$

the subspaces $\mathcal{T}$ and $\mathcal{C}$ are defined by taking the
orthogonal complements with respect to an auxiliary Riemannian metric on $M$
of $\ker d_{0}$ and $\mathrm{image\,}d_{0}$ respectively.

###  3.2. Compact Riemannian manifolds and Hodge decomposition

Let $(M,g)$ be a compact Riemannianian manifold, and let $(\Omega,d)$ be the
de Rham complex of smooth differential forms. By the Hodge decomposition
theorem, we have

(3.3)
$\Omega=\operatorname{image}d\oplus\operatorname{image}d^{*}\oplus\mathcal{H},$

where $\mathcal{H}=\ker(dd^{*}+d^{*}d)$ is the space of harmonic forms, that
is forms in the kernel of the Hodge-Laplacian, and
$d^{*}\colon\Omega\to\Omega$ denotes the codifferential.

Let us take the trivial filtration over the space of smooth forms,
$\Omega=\Omega_{0}\supset\Omega_{1}=0$. In this case, we have that $d_{0}=d$,
since $d_{0}\alpha=d\alpha\mod\Omega_{0}=d\alpha$ for any
$\alpha\in\Omega_{0}$. If we take $\mathcal{T}=\operatorname{image}d^{*}$ and
$\mathcal{C}=\operatorname{image}d^{*}\oplus\mathcal{H}$, then
$P^{\infty}=P=\mathrm{id}-d^{*}d-dd^{*}$ is the projection onto
$\mathcal{P}=\mathcal{H}$. In the resulting base complex
$(\mathcal{P},d|_{\mathcal{P}})=(\mathcal{H},d|_{\mathcal{H}})$, the
differential $d|_{\mathcal{H}}\colon\mathcal{H}\to\mathcal{H}$ is the trivial
map, and so Theorem 1.1 implies that each cohomology class is uniquely
represented by a harmonic form, as expected.

On non-compact manifolds, one can extend such construction by considering
$L^{p}$-forms and obtain a similar result in the cases where we have an
$L^{p}$-Hodge decomposition. See [11, 12, 26, 14] for details on when such a
decomposition exists.

###  3.3. Compact Riemannian manifolds with non-positive filtration of the
tangent bundle

We can combine the approaches of Section 3.1 and Section 3.2 to consider non-
positive filtrations. Consider a compact Riemannian manifold $(M,g)$ of
dimension $n$ and assume we have a filtration of the tangent bundle
$TM=TM_{-s}\supset TM_{-s+1}\supset\cdots\supset TM_{-1}\supset TM_{0}\supset
0$. We still require that assumption (A) holds for the vector fields on $M$.

Using the Riemannian metric, one can now express $TM$ as a direct sum
decomposition by defining orthogonal complements, that is
$TM_{-j}=TM_{(-j)}\oplus_{\perp}TM_{-j+1}$ for $j=-1,0,\dots,s$, so that

$\displaystyle
TM=TM_{(-s)}\oplus_{\perp}TM_{(-s+1)}\oplus_{\perp}\cdots\oplus_{\perp}TM_{(-1)}\oplus_{\perp}TM_{0}\,.$

This orthogonal decomposition naturally extends to the space of smooth vector
fields on one hand

$\displaystyle\mathfrak{X}=\mathfrak{X}_{(-s)}\oplus_{\perp}\mathfrak{X}_{(-s+1)}\oplus_{\perp}\cdots\oplus_{\perp}\mathfrak{X}_{(-1)}\oplus_{\perp}\mathfrak{X}_{(0)}\,,$

and to the exterior algebra $\bigwedge TM$ on the other

$\displaystyle\bigwedge TM=(\bigwedge TM)_{(-S)}\oplus_{\perp}(\bigwedge
TM)_{(-S+1)}\oplus_{\perp}\cdots\oplus_{\perp}(\bigwedge
TM)_{(-1)}\oplus_{\perp}(\bigwedge TM)_{(0)}\,,$

by imposing that if $X_{i}\in TM_{(-j_{i})}$ then $X_{1}\wedge\cdots\wedge
X_{k}\in(\bigwedge TM)_{(-\sigma)}$ where $\sigma=\sum_{i=1}^{k}j_{i}$.

By repeating the same reasoning as in subsection 3.1, by duality we obtain a
decreasing filtration on $\bigwedge T^{*}M$ and $\Omega$. In this particular
case, one can define $(\bigwedge T^{*}M)_{(j)}$ as those elements in
$\bigwedge T^{*}M$ that vanish on $(\bigwedge TM)_{(-r)}$ for any $r\neq j$,
and then extend it to the space of smooth forms by imposing $\Omega_{(j)}$ to
be the space of smooth sections of $(\bigwedge T^{*}M)_{(j)}$. As a result, we
obtain an $L^{2}$-orthogonal decomposition of the space of smooth forms

$\Omega=\Omega_{(0)}\oplus_{\perp}\Omega_{(1)}\oplus_{\perp}\cdots\oplus_{\perp}\Omega_{(S-1)}\oplus_{\perp}\Omega_{(S)}\,,$

where $S=\sum_{j=1}^{s}j(\operatorname{rank}\Omega_{(j)})$ coincides with the
index (3.1).

Since assumption (A) still holds, we have that $d(\oplus_{k\geq
j}\Omega_{(k)})\subset\oplus_{k\geq j}\Omega_{(k)}$ (see Lemma 3.1), and we
can therefore express the exterior differential in terms of its action with
respect to this orthogonal decomposition, that is $d=d_{(0)}+\cdots+d_{(s)}$,
where $d_{(r)}(\Omega_{(j)})\subset\Omega_{(j+r)}$.

Once we define $d_{0}=d_{(0)}$, we readily obtain that it is a base
differential of the de Rham complex $(\Omega,d)$ (see also the proof of
Proposition 2.3). If we also take its formal adjoint $d_{0}^{*}=d_{(0)}^{*}$
and consider

$\mathcal{T}=(\ker
d_{0})^{\perp}=\operatorname{image}d_{0}^{*},\qquad\mathcal{C}=(\operatorname{image}d_{0})^{\perp}=\ker
d_{0}^{*}\,,$

we obtain a base transversal and cotransversal of $d_{0}$, since
$\mathcal{T}\subset\mathcal{C}$. Observe that now $d_{0}$ and $d_{0}^{-1}$ are
no longer tensorial operators.

Similarly to what was mentioned in Section 3.2, we can extend the same
construction to $L^{p}$-forms on non-compact manifolds, provided that the
exterior differential operator and its adjoint are closed operators on
$L^{p}$-forms. Again, we refer to [19, Section 1.4] for conditions for when
such requirements are satisfied.

###  3.4. Forms vanishing at a point

Let $(M,\nabla)$ be an affine manifold of dimension $n$ and let $o\in M$ be a
fixed point. For $i$ vector fields $X_{1},\dots,X_{i}\in\mathfrak{X}$, we
define the $i$-th covariant derivative iteratively by imposing
$\nabla_{X_{1}}^{1}=\nabla_{X_{1}}$ and

$\nabla_{X_{1},\dots,X_{i}}^{i}=\nabla_{X_{1}}\nabla_{X_{2},\dots,X_{i}}^{i-1}-\nabla_{\nabla_{X_{1}}X_{2},\dots,X_{i}}^{i-1}-\cdots-\nabla_{X_{2},\dots,\nabla_{X_{1}}X_{i}}^{i-1}\,\text{
for }i\geq 2\,.$

We note that the operator $\nabla_{X_{1},\dots,X_{i}}^{i}$ is tensorial in all
arguments.

Let us define the filtration
$\Omega=\Omega_{0}\supset\Omega_{1}\supset\Omega_{2}\supset\cdots$ over the
space of smooth forms, where $\Omega_{j}$ denotes the space of all forms
$\alpha$ satisfying

(3.4) $\nabla_{X_{1},\dots,X_{i}}^{i}\alpha|_{o}=0\ ,\ \text{for any
$X_{1},\dots,X_{i}\in\mathfrak{X}$, and $i\leq j-\deg(\alpha)-1$}$

with the convention that (3.4) reads as $\alpha|_{o}=0$ in the case $i=0$, and
we have no conditions when $j\leq\deg(\alpha)$.

This filtration is independent of the connection $\nabla$. In fact, it is
straightforward to verify that for any local coordinate system
$(x^{1},\ldots,x^{n})$ defined on a neighbourhood $U$ of the given point $o\in
M$, one can express any $k$-form $\alpha\in\Omega_{j}$ via the local chart
$(U,\varphi)$ with $\varphi\colon U\to\mathbb{R}^{n}$ as

$\displaystyle(\varphi^{-1})^{\sharp}\alpha|_{U}=\sum_{i_{1}<i_{2}<\cdots<i_{k}}\alpha_{i_{1},i_{2},\ldots,i_{k}}dx^{i_{1}}\wedge
dx^{i_{2}}\wedge\cdots\wedge dx^{i_{k}}\,,$

where each $\alpha_{i_{1},i_{2},\ldots,i_{k}}\in C^{\infty}(\varphi(U))$ is a
function that has a zero at $\varphi(o)$ of degree at least $j-\deg(\alpha)$.

Let $S$ be fixed and define $\Omega_{S}=\oplus_{j\geq S}\Omega_{j}$ so that
$\Omega_{S+1}=0$, in order to obtain a filtration of finite depth
$\Omega=\Omega_{0}\supset\Omega_{1}\supset\cdots\supset\Omega_{S}\supset\Omega_{S+1}=0$.
The following proposition can readily be obtained by applying the explicit
formula of the exterior derivative (3.2).

######  Proposition 3.2.

The finite filtration
$\Omega=\Omega_{0}\supset\cdots\supset\Omega_{S}\supset\Omega_{S+1}=0$ as
defined in (3.4) is a filtered differential complex, that is,
$d\Omega_{j}\subset\Omega_{j}$ for $j=0,\ldots,S$.

The subspaces $\Omega_{j}$ that make up this filtration can easily be re-
expressed in terms of jet spaces. Indeed, let us denote by $\pi_{j}$ the
projection map that sends an arbitrary $k$-form $\alpha$ to its $(j-k-1)$-jet
at the point $o\in M$, where we are imposing $\pi_{j}(\alpha)=0$ if $j-k<0$.
We then have that $\Omega_{j}=\ker\pi_{j}$ for any $j\in\mathbb{N}$, as well
as the series of inclusions $\ker\pi_{j}\subset\ker\pi_{j-1}$, and so
$\Omega_{j}=\ker\pi_{j}$ for $j=0,\ldots,S-1$, and $\Omega_{S}=\oplus_{j\geq
S}\Omega_{j}=\oplus_{j\geq S}\ker\pi_{j}$.

Similarly as in the previous examples of differential complexes filtered, one
can express $\Omega$ as a direct sum decomposition
$\Omega=\Omega_{(0)}\oplus\Omega_{(1)}\oplus\cdots\oplus\Omega_{(S-1)}\oplus\Omega_{S}$,
where the subspaces $\Omega_{(j)}$ are defined so that
$\Omega_{j}=\Omega_{(j)}\oplus\Omega_{j+1}$ for each $j=0,\ldots,S-1$. For any
$\alpha\in\Omega_{(j-1)}$, we define $d_{0}\alpha$ as the component of
$d\alpha$ that still belongs to $\Omega_{(j)}$, and impose $d_{0}$ to be the
zero map on $\Omega_{S}$ to obtain a base differential for the de Rham complex
$(\Omega,d)$. We can then take $\mathcal{T}$ to be some finite dimensional
complement of $\ker d_{0}\cap(\Omega_{(0)}\oplus\cdots\oplus\Omega_{(S-1)})$
and $\mathcal{C}=\Omega_{S}\oplus\tilde{\mathcal{C}}$, where
$\tilde{\mathcal{C}}$ is a finite dimensional complement of
$\operatorname{image}d_{0}$ in $\Omega_{(0)}\oplus\cdots\oplus\Omega_{(S-1)}$
containing $\mathcal{T}$. This gives us $\mathcal{T}\subset\mathcal{C}$ a base
transversal and cotransversal for this $d_{0}$.

###  3.5. Graded Lie algebras

Let $\mathfrak{g}$ be a Lie algebra of dimension $n$ and let
$\rho:\mathfrak{g}\to\mathfrak{gl}(V)$ be a representation. On the space of
linear forms

$K=\oplus_{k=0}^{n}K^{k}\ ,\quad K^{k}=\wedge^{k}\mathfrak{g}^{*}\otimes V\ ,$

one can define the operator $\tilde{d}:K^{k}\to K^{k+1}$ by imposing that, for
any $\alpha\in K^{k}$ and for any $X_{1},\ldots,X_{k+1}$

$\displaystyle\tilde{d}\alpha(X_{1},\dots,X_{k+1})$ $\displaystyle=\sum_{1\leq
i\leq k+1}(-1)^{i-1}\rho(X_{i})\alpha(X_{1},\dots,\hat{X}_{i},\dots,X_{k+1})$
$\displaystyle\qquad+\sum_{1\leq
i<l<k+1}(-1)^{i+l}\alpha([X_{i},X_{l}],X_{1},\dots,\hat{X}_{i},\dots,\hat{X}_{l},\dots,X_{k+1})\
.$

where the hat-symbol is used to indicate the terms that are omitted. Just like
the exterior differential defined in (3.2), this map $\tilde{d}$ satisfies
$\tilde{d}^{2}=0$.

Assume that we have a non-positive grading, that is a direct sum decomposition
$\mathfrak{g}=\mathfrak{g}_{0}\oplus\mathfrak{g}_{-1}\oplus\cdots\oplus\mathfrak{g}_{-s}$,
where

(3.5) $\displaystyle\text{ if }X\in\mathfrak{g}_{-k}\text{ and
}Y\in\mathfrak{g}_{-l}\ ,\ \text{ then
}[X,Y]\in\mathfrak{g}_{-k-l}\oplus\mathfrak{g}_{-k-l+1}\oplus\cdots\oplus\mathfrak{g}_{0}\
.$

We can also introduce an inner product on $\mathfrak{g}$ such that this
decomposition is given by orthogonal subspaces. Then a finite filtration of
$\mathfrak{g}$ can be obtained by taking
$\mathfrak{g}=\oplus_{j=0}^{-s}\mathfrak{g}_{j}\supset\oplus_{j=0}^{-s+1}\mathfrak{g}_{j}\supset\cdots\supset\mathfrak{g}_{-1}\oplus\mathfrak{g}_{0}\supset\mathfrak{g}_{0}\supset
0$.

By repeating the same construction presented in subsections 3.1 and 3.3, we
obtain that this orthogonal decomposition naturally extends to the space
$\bigwedge\mathfrak{g}^{\ast}\otimes V$:

$\displaystyle K^{k}=\wedge^{k}\mathfrak{g}^{\ast}\otimes V=K^{k}_{(0)}\oplus
K^{k}_{(1)}\oplus\cdots\oplus K^{k}_{(S-1)}\oplus K^{k}_{S}\ ,\text{ for each
}k=0,\ldots,n\,,$

where $S$ is again defined using the same formula as in (3.1).

Our assumption of working with a non-positive grading (3.5) being equivalent
to assumption (A) implies that we have a graded decomposition of the operator
$\tilde{d}=\tilde{d}_{(0)}\oplus\tilde{d}_{(1)}\oplus\cdots\oplus\tilde{d}_{(s)}$,
for which $\tilde{d}_{(0)}^{2}=0$ (again, it is sufficient to follow the steps
presented in subsection 3.3). Therefore, the map
$\tilde{d}_{0}:=\tilde{d}_{(0)}$ is a base differential for the complex
$(\bigwedge\mathfrak{g}^{\ast}\otimes V,\tilde{d})$, and it is sufficient to
take $\mathcal{T}:=(\ker\tilde{d}_{0})^{\perp}$ and
$\mathcal{C}=(\operatorname{image}\tilde{d}_{0})^{\perp}$ as the relative base
transversal and cotransversal.

##  4\. Spectral methods for the filtered subcomplex

In many of the examples considered in Section 3, not only do we have a finite
depth filtration of our complex $(K,d)$ as in (2.1), but the complex is also
given an additional grading $(K^{k})$, $k=0$,$1$, $\dots$. This is indeed the
case for our examples in Section 3, where an element belongs to $K^{k}$ if it
is a $k$-form. We observe that the grading satisfies

(4.1) $dK^{k}\subseteq K^{k+1},\qquad K^{k}\subset K_{k},\qquad
k\in\mathbb{N}_{0}\ .$

A very natural question arises about the spectral sequences that can be
constructed from such a filtration. In this section we will see that, just
like in the simpler case of Carnot groups [25], any filtered complex of finite
depth is also a multicomplex. As shown in [27], the extra algebraic structure
of a multicomplex allows us to obtain an explicit formulation of the
differentials that arise from the spectral sequence obtained from this finite
filtration. In this way, we are also able to compare these differentials with
the projection map $P^{\infty}$ introduced in Section 2.

###  4.1. Spectral sequence associated to a multicomplex

We will first include the basic definitions that we will use for the remainder
of the section.

######  Definition 4.1.

A multicomplex, also called a twisted (co)chain complex, is a
$(\mathbb{Z},\mathbb{Z})$-graded $k$-module $C$ equipped with maps
$d_{i}\colon C\to C$ for $i\geq 0$ of bidegree $|d_{i}|=(i,1-i)$ such that

(4.2) $\displaystyle\sum_{i+j=l}d_{i}d_{j}=0\ \text{ for all }\ l\geq 0\,.$

For $C$ a multicomplex and $(a,b)\in\mathbb{Z}\times\mathbb{Z}$, we write
$C_{a,b}$ for the $k$-module in bidegree $(a,b)$.

Given a multicomplex $C$, one could consider different possible total
complexes, however we will focus on the following one.

######  Definition 4.2.

For a multicomplex $C$, its associated total complex $\mathrm{Tot}\,C$ is the
chain complex with

$\displaystyle\left(\mathrm{Tot}\,C\right)_{k}=\bigg{(}\prod_{\begin{subarray}{c}a+b=k\\\
a\leq
0\end{subarray}}C_{a,b}\bigg{)}\oplus\bigg{(}\bigoplus_{\begin{subarray}{c}a+b=k\\\
a>0\end{subarray}}C_{a,b}\bigg{)}=\bigg{(}\bigoplus_{\begin{subarray}{c}a+b=k\\\
b\leq
0\end{subarray}}C_{a,b}\bigg{)}\oplus\bigg{(}\prod_{\begin{subarray}{c}a+b=k\\\
b>0\end{subarray}}C_{a,b}\bigg{)}\,.$

The differential on $\mathrm{Tot}\,C$ is given for an arbitrary element
$c\in\big{(}\mathrm{Tot\,C}\big{)}_{k}$ by

(4.3) $\displaystyle(dc)_{a}=\sum_{i\geq 0}d_{i}(c)_{a-i}\,,$

where $(c)_{a}$ denotes the projection of
$c\in\big{(}\mathrm{Tot}\,C\big{)}_{k}$ to
$C_{a,\bullet}=\prod_{b,a+b=k}C_{a,b}$.

In general, when working with $(\mathrm{Tot}\,C)_{k}$ it is not always
possible to consider the direct product total complex $\prod_{a+b=k}C_{a,b}$,
as the formula (4.3) may involve infinite sums. However, since in this paper
we are working on a complex of finite depth, the associated total complex
takes the simpler form

(4.4)
$\displaystyle\big{(}\mathrm{Tot}\,C\big{)}_{k}=\bigoplus_{\begin{subarray}{c}a+b=k\\\
a\geq 0\end{subarray}}C_{a,b}\ ,\ \text{ with differential
}(dc)_{a}=\sum_{i=0}^{S}d_{i}(c)_{a-i}\,.$

######  Definition 4.3.

Given a multicomplex $C$ and its associated total complex $\mathrm{Tot}\,C$,
we can define for each degree $k$ the following subcomplexes

$\displaystyle\big{(}\mathcal{F}_{p}\mathrm{Tot\,}C\big{)}_{k}=\bigoplus_{\begin{subarray}{c}a+b=k\\\
a\geq p\end{subarray}}C_{a,b}\,.$

By definition, as the value of $p$ varies, the subcomplexes
$\mathcal{F}_{p}\mathrm{Tot}\,C$ form a filtration of $\mathrm{Tot}\,C$ of
finite depth:

$\displaystyle\mathrm{Tot}\,C=\mathcal{F}_{0}\mathrm{Tot}\,C\supset\mathcal{F}_{1}\mathrm{Tot}\,C\supset\cdots\supset\mathcal{F}_{S}\mathrm{Tot}\,C\supset\mathcal{F}_{S+1}\mathrm{Tot}\,C=0\,.$

Let us now consider the spectral sequence associated with this filtered
complex $\mathcal{F}_{p}\mathrm{Tot}\,C$. For $r\geq 0$, the $r^{th}$-page of
the spectral sequence is a bigraded module
$E_{r}^{p,\bullet}(\mathrm{Tot}\,C)$ with a map $\delta_{r}$ of bidegree
$(r,1-r)$ for which $\delta_{r}\circ\delta_{r}=0$. Moreover, we have that the
spaces $E_{r}^{p,\bullet}(\mathrm{Tot}\,C)$ can be expressed as the quotients

$\displaystyle
E_{r}^{p,\bullet}(\mathrm{Tot}\,C)\cong\mathcal{Z}_{r}^{p,\bullet}(\mathrm{Tot}\,C)/\mathcal{B}^{p,\bullet}_{r}(\mathrm{Tot}\,C)$

where the $r$-cycles are given by

$\displaystyle\mathcal{Z}_{r}^{p,\bullet}(\mathrm{Tot}\,C):=\left\\{c\in\mathcal{F}_{p}\mathrm{Tot}\,C\,:\,d\alpha\in\mathcal{F}_{p+r}\mathrm{Tot}\,C\right\\}$

and the $r$-boundaries are given by

$\displaystyle\begin{cases}\mathcal{B}_{0}^{p,\bullet}(\mathrm{Tot}\,C):=\mathcal{Z}_{0}^{p+1,\bullet}(\mathrm{Tot}\,C)\
\text{ and }\\\
\mathcal{B}_{r}^{p,\bullet}(\mathrm{Tot}\,C):=\mathcal{Z}_{r-1}^{p+1,\bullet}(\mathrm{Tot}\,C)+d\mathcal{Z}_{r-1}^{p-(r-1),\bullet}(\mathrm{Tot}\,C)\
\text{ for }r\geq 1\,.\end{cases}$

Given an element $z\in\mathcal{Z}_{r}^{p,\bullet}(\mathrm{Tot}\,C)$, we will
denote by $[x]_{r}$ its image in $E^{p,\bullet}_{r}(\mathrm{Tot}\,C)$, so that

$\displaystyle\delta_{r}([x]_{r})=[dx]_{r}\ ,\ \text{ for any }[x]_{r}\in
E^{p,\bullet}_{r}(\mathrm{Tot}\,C)\,.$

Expanding on the expressions for
$\mathcal{Z}_{r}^{p,\bullet}(\mathrm{Tot}\,C)$ and
$\mathcal{B}_{r}^{p,\bullet}(\mathrm{Tot}\,C)$, one can introduce the
following definition.

######  Definition 4.4.

Let $x\in C_{p,\bullet}$ and let $r\geq 1$. We define the graded submodules
$Z_{r}^{p,\bullet}$ and $B^{p,\bullet}_{r}$ of $C_{p,\bullet}$ as follows:

(4.7) $\displaystyle x\in Z_{r}^{p,\bullet}\ \Longleftrightarrow$
$\displaystyle\begin{array}[]{c}\text{ for }1\leq j\leq r-1\ \text{ there
exists }z_{p+j}\in C_{p+j,\bullet}\ \text{ such that }\\\ d_{0}x=0\ \text{ and
}\ d_{m}x=\sum_{i=0}^{m-1}d_{i}z_{p+m-i}\ \text{ for all }\ 1\leq m\leq
r-1\end{array}$ (4.11) $\displaystyle x\in B_{r}^{p,\bullet}\
\Longleftrightarrow$ $\displaystyle\begin{array}[]{c}\text{ for }1\leq k\leq
r-1\ \text{ there exists }c_{p-k}\in C_{p-k,\bullet}\ \text{ such that}\\\
x=\sum_{k=0}^{r-1}d_{k}c_{p-k}\ \text{ and }\\\
0=\sum_{k=l}^{r-1}d_{k-l}c_{p-k}\ \text{ for }\ 1\leq l\leq r-1\,.\end{array}$

As shown already in [27] and [25], the following holds.

######  Proposition 4.5.

The map

$\displaystyle\psi\colon\mathcal{Z}_{r}^{p,\bullet}(\mathrm{Tot\,}C)/\mathcal{B}_{r}^{p,\bullet}(\mathrm{Tot}\,C)\longrightarrow
Z_{r}^{p,\bullet}/B_{r}^{p,\bullet}$

sending $[x]_{r}$ to the class $[(x)_{p}]$ is a well-defined isomorphism.
Moreover, under this isomorphism, the $r^{th}$-differential of the spectral
sequence corresponds to the following map

$\displaystyle\partial_{r}\colon Z_{r}^{p,\bullet}/B_{r}^{p,\bullet}\to
Z_{r}^{p+r,\bullet}/B_{r}^{p+r,\bullet}\ ,\
\partial_{r}\big{(}[x]\big{)}=\bigg{[}d_{r}x-\sum_{i=1}^{r-1}d_{i}z_{p+r-i}\bigg{]}$

where $x\in Z_{r}^{p,\bullet}$ and the elements $z_{p+j}\in C_{p+j,\bullet}$
satisfy (4.7).

###  4.2. Filtered complexes $(K,d)$ of finite depth as multicomplexes

We apply the formalism of multicomplexes to a differential complex $(K,d)$
having both a filtration $K_{0}\supset K_{1}\supset\cdots\supset K_{S}\supset
K_{S+1}=0$ and a grading $K=K^{0}\oplus\cdots\oplus K^{n}$ satisfying (4.1).

######  Proposition 4.6.

Any filtered complex $(K,d)$ as described above is a multicomplex with maps
$d_{(i)}\colon K\to K$ of bidegree $|d_{(i)}|=(i,1-i)$ with $i=0,\ldots,S$.

###### Proof.

As already pointed out in (2.2), we can express the complex $K$ as a direct
sum decomposition on each degree. For each degree $k\in\mathbb{N}$, we have
that

$\displaystyle K^{k}=K_{0}^{k}=K_{(0)}^{k}\oplus K_{1}^{k}=K_{(0)}^{k}\oplus
K^{k}_{(1)}\oplus K_{2}^{k}=K_{(0)}^{k}\oplus K^{k}_{(1)}\oplus\cdots\oplus
K_{(S-1)}^{k}\oplus K^{k}_{S}\ .$

This direct sum decomposition of $K$ based on the filtration induces a
decomposition of the original differential $d$, since $dK_{j}^{k}\subset
K_{j}^{k+1}$. Indeed

$\displaystyle d=d_{(0)}+d_{(1)}+\cdots+d_{(S)}\ \text{ where }\
d_{(i)}K_{(j)}^{k}\subset K_{(j+i)}^{k+1}\ ,\ i,j=0,\ldots,S\,.$

Finally, since $(K,d)$ is a complex, once we expand the formula $d^{2}=0$ and
gather all the terms according to their respective subspace $K_{(j)}$ that
they belong to, we get that for any $\alpha\in K_{(j)}^{k}$

$\displaystyle 0=$ $\displaystyle
d^{2}\alpha=\big{(}d_{(0)}+d_{(1)}+\cdots+d_{(S)}\big{)}\big{(}d_{(0)}+d_{(1)}+\cdots+d_{(S)}\big{)}\alpha$
$\displaystyle=$ $\displaystyle\underbrace{d_{(0)}^{2}\alpha}_{\in
K_{(j)}^{k+2}}+\underbrace{d_{(0)}d_{(1)}\alpha+d_{(1)}d_{(0)}\alpha}_{\in
K_{(j+1)}^{k+2}}+\underbrace{d_{(0)}d_{(2)}\alpha+d_{(1)}^{2}\alpha+d_{(2)}d_{(0)}\alpha}_{\in
K_{(j+2)}^{k+2}}+\cdots+\underbrace{d_{(S)}^{2}\alpha}_{\in
K_{(j+2S)}^{k+2}}=\sum_{l=1}^{2S}\sum_{i+j=l}d_{(i)}d_{(j)}\alpha\,.$

We then have that each addend $\sum_{i+j=l}d_{(i)}d_{(j)}\alpha$ belongs to a
different subspace $K_{(j+l)}^{k+2}$ and so, by the direct sum decomposition
$K_{j}^{k+2}=\bigoplus_{l=0}^{S}K_{(j+l)}^{k+2}$, we get that each addend will
necessarily be zero. ∎

######  Corollary 4.7.

Any filtered complex $(K,d)$ of finite depth as in (2.2) admits a spectral
sequence associated to the specific filtration considered, whose differentials
at each step $r$ have the following expression

$\displaystyle\partial_{r}\colon Z_{r}^{p,\bullet}/B_{r}^{p,\bullet}\to
Z_{r}^{p+r,\bullet}/B_{r}^{p+r,\bullet}\ ,\
\partial_{r}\big{(}[x]\big{)}=\bigg{[}d_{(r)}x-\sum_{i=1}^{r-1}d_{(i)}z_{p+r-i}\bigg{]}\
,\ r=1,\ldots,S\,.$

We stress that, given a filtered complex $(K,d)$ of finite depth with base
differential $d_{0}=d_{(0)}\colon K\to K$, once we fix a base transversal and
cotransversal $\mathcal{T}\subset\mathcal{C}$, it is possible to define the
partial inverse $d_{0}^{-1}\colon K\to K$ of $d_{0}$ as in Definition 2.4.
Therefore, using this additional map, it is possible to rephrase the condition
$x\in Z_{r}^{p,\bullet}$ of (4.7) in terms of this operator $d_{0}^{-1}$, and
require the $z_{p+j}\in K_{(p+j)}$ to be elements in $\mathcal{T}$.

In other words, given $x\in Z_{r}^{p,\bullet}$, we have that there exist
$z_{p+j}\in K_{(p+j)}\cap\mathcal{T}$ with $1\leq j\leq r-1$ such that

$\displaystyle z_{p+1}=$ $\displaystyle d_{0}^{-1}d_{(1)}x$ $\displaystyle
z_{p+2}=$ $\displaystyle
d_{0}^{-1}(d_{(2)}x-d_{(1)}z_{p+1})=d_{0}^{-1}d_{(2)}x-d_{0}^{-1}d_{(1)}d_{0}^{-1}d_{(1)}x$
$\displaystyle z_{p+3}=$ $\displaystyle
d_{0}^{(-1)}(d_{(3)}x-d_{(2)}z_{p+1}-d_{(1)}z_{p+2})$ $\displaystyle=$
$\displaystyle
d_{0}^{-1}d_{(3)}x-d_{0}^{-1}d_{(2)}d_{0}^{-1}d_{(1)}x-d_{0}^{-1}d_{(1)}d_{0}^{-1}d_{(2)}x+d_{0}^{-1}d_{(1)}d_{0}^{-1}d_{(1)}d_{0}^{-1}d_{(1)}x\,.$

If we introduce the multi-index notation

$\displaystyle(d_{0}^{-1}d)_{I_{m}^{j}}:=(d_{0}^{-1}d_{(i_{1})})(d_{0}^{-1}d_{(i_{2})})\cdots(d_{0}^{-1}d_{(i_{m})})$

where $I_{m}^{j}=(i_{1},\ldots,i_{m})\in\mathbb{N}_{+}^{m}$ for which
$|I_{m}^{j}|=i_{1}+i_{2}+\cdots+i_{m}=j$, then for any $j=1,\ldots,r-1$, we
have the expression

(4.12) $\displaystyle
z_{p+j}=\sum_{m=1}^{j}(-1)^{m-1}\sum_{I_{m}^{j}}(d_{0}^{-1}d)_{I_{m}^{j}}x\,.$

######  Remark 4.8.

Applying the definition of the subspaces $\mathcal{E}_{0}$ and Proposition
2.9, there is a clear correspondence between the subspaces
$\mathcal{P}=\mathcal{P}_{d_{0},\mathcal{T}\subset\mathcal{C}}$ and the first
page quotients of the spectral sequence, as

$\displaystyle\mathcal{P}=\mathcal{P}_{d_{0},\mathcal{T}\subset\mathcal{C}}\cong\oplus_{p=0}^{S}E^{p,\bullet}_{1}\,.$

Moreover, if we use the explicit expression of the $z_{p+j}\in
K_{(p+j)}\cap\mathcal{T}$ in terms of $x$ given in (4.12), it is possible to
obtain the following expression for the $r^{th}$-differential of the spectral
sequence

$\displaystyle\partial_{r}([x])=\bigg{[}d_{(r)}x-\sum_{i=1}^{r-1}d_{(i)}\bigg{(}\sum_{m=1}^{r-i}(-1)^{m-1}\sum_{I_{m}^{r-i}}(d_{0}^{-1}d)_{I_{m}^{r-i}}\bigg{)}x\bigg{]}$

It is then a straightforward computation to check that the differential
operator $dP^{\infty}$ introduced in section 2 applied to any element $x\in
Z_{1}^{p,\bullet}$ coincides with the sum of the differentials $\partial_{r}$
that appear in the multicomplex spectral sequence generated by the filtration
(2.2). This result is analogous to the main result contained in [25], where
considerations were only limited to the de Rham complex on Carnot groups.

##  5\. Three dimensional Lie groups with negative grading

We will include the following examples of the theory of Section 4 which also
cover non-nilpotent Lie groups. Let us consider the 3-dimensional simply
connected Lie group $G_{\kappa}$, whose Lie algebra
$\mathfrak{g}_{\kappa}=\operatorname{span}_{\mathbb{R}}\\{X,Y,Z\\}$,which can
be identified with the $\mathbb{R}$-module of left-invariant vector fields,
has the following nontrivial bracket relations

$\displaystyle[X,Y]=Z\ ,\ [Y,Z]=\kappa X\ ,\ [X,Z]=-\kappa Y\,.$

Let us stress that $G_{1}$ is isomorphic to $\operatorname{SU}(2)$, $G_{0}$
coincides with the first Heisenberg group, while $G_{-1}$ is the universal
cover group of $\operatorname{SL}(2)$.

In this example, we will be showing the construction of the spectral sequence
associated with the following “sub-Riemannian” grading:

$\displaystyle\mathfrak{g}_{\kappa}=\mathfrak{g}_{-2}\oplus\mathfrak{g}_{-1}\
,\text{ where }\mathfrak{g}_{-1}=\langle X,Y\rangle\text{ and
}\mathfrak{g}_{-2}=\langle Z\rangle.$

Let us denote by $\\{X^{\ast},Y^{\ast},Z^{\ast}\\}$ the dual basis of
$\\{X,Y,Z\\}$ on $\mathfrak{g}_{\kappa}^{\ast}$.

Using this grading and following the same construction described in subsection
3.5, we obtain a decreasing and finite filtration over the space
$\Omega^{\bullet}(G_{\kappa})$ of smooth forms:

$\displaystyle\mathcal{F}_{0}\Omega^{\bullet}(G_{\kappa})=\Omega^{\bullet}(G_{\kappa})\supset\mathcal{F}_{1}\Omega^{\bullet}(G_{\kappa})\supset\mathcal{F}_{2}\Omega^{\bullet}(G_{\kappa})\supset\mathcal{F}_{3}\Omega^{\bullet}(G_{\kappa})\supset\mathcal{F}_{4}\Omega^{\bullet}(G_{\kappa})\supset\mathcal{F}_{5}\Omega^{\bullet}(G_{\kappa})=\\{0\\}$

where

  * •

$\mathcal{F}_{0}\Omega^{\bullet}(G_{\kappa})=\Omega^{\bullet}(G_{\kappa})$;

  * •

$\mathcal{F}_{1}\Omega^{\bullet}(G_{\kappa})=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{X^{\ast},Y^{\ast},Z^{\ast},X^{\ast}\wedge
Y^{\ast},X^{\ast}\wedge Z^{\ast},Y^{\ast}\wedge Z^{\ast},X^{\ast}\wedge
Y^{\ast}\wedge Z^{\ast}\\}$;

  * •

$\mathcal{F}_{2}\Omega^{\bullet}(G_{\kappa})=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{Z^{\ast},X^{\ast}\wedge
Y^{\ast},X^{\ast}\wedge Z^{\ast},Y^{\ast}\wedge Z^{\ast},X^{\ast}\wedge
Y^{\ast}\wedge Z^{\ast}\\}$;

  * •

$\mathcal{F}_{3}\Omega^{\bullet}(G_{\kappa})=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{X^{\ast}\wedge
Z^{\ast},Y^{\ast}\wedge Z^{\ast},X^{\ast}\wedge Y^{\ast}\wedge Z^{\ast}\\}$;

  * •

$\mathcal{F}_{4}\Omega^{\bullet}(G_{\kappa})=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{X^{\ast}\wedge
Y^{\ast}\wedge Z^{\ast}\\}$;

  * •

$\mathcal{F}_{5}\Omega^{\bullet}(G_{\kappa})=\\{0\\}$.

Following the notation adopted in Section 3.5, we are considering the de Rham
complex $(\Omega^{\bullet}(G_{\kappa}),d)$ as our filtered complex
$(K,\tilde{d})$, where $\tilde{d}$ is the exterior derivative and it admits
the splitting $d=d_{0}+d_{(1)}+d_{(2)}$. This can be seen explicitly by
writing down the action of $d$ on an arbitrary 1-form
$\alpha=fX^{\ast}+gY^{\ast}+hZ^{\ast}\in\Omega^{1}(G_{\kappa})=K^{1}$ with
$f,g,h\in C^{\infty}(G_{\kappa})=K^{0}$:

$\displaystyle d\big{(}fX^{\ast}+gY^{\ast}+hZ^{\ast}\big{)}=$ $\displaystyle
df\wedge X^{\ast}+f\,dX^{\ast}+dg\wedge Y^{\ast}+g\,dY^{\ast}+dh\wedge
Z^{\ast}+h\,dZ^{\ast}$ $\displaystyle=$ $\displaystyle-YfX^{\ast}\wedge
Y^{\ast}-ZfX^{\ast}\wedge Z^{\ast}-\kappa fY^{\ast}\wedge
Z^{\ast}+XgX^{\ast}\wedge Y^{\ast}-ZgY^{\ast}\wedge Z^{\ast}+$
$\displaystyle+\kappa gX^{\ast}\wedge Z^{\ast}-hX^{\ast}\wedge
Y^{\ast}+XhX^{\ast}\wedge Z^{\ast}+YhY^{\ast}\wedge Z^{\ast}$ $\displaystyle=$
$\displaystyle\underbrace{-hX^{\ast}\wedge Y^{\ast}}_{d_{0}}+\underbrace{(Xg-
Yf)X^{\ast}\wedge Y^{\ast}+XhX^{\ast}\wedge Z^{\ast}+YhY^{\ast}\wedge
Z^{\ast}}_{d_{(1)}}+$ $\displaystyle+\underbrace{(\kappa g-Zf)X^{\ast}\wedge
Z^{\ast}-(\kappa f+Zg)Y^{\ast}\wedge Z^{\ast}}_{d_{(2)}}\,.$

Finally, the direct sum decomposition associated to such finite filtration is
given by

$\displaystyle\Omega^{\bullet}(G_{\kappa})=K=K_{(0)}\oplus K_{(1)}\oplus
K_{(2)}\oplus K_{(3)}\oplus K_{4}\,,$

where

  * •

$K_{(0)}=K^{0}=C^{\infty}(G_{\kappa})$;

  * •

$K_{(1)}=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{X^{\ast},Y^{\ast}\\}$;

  * •

$K_{(2)}=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{Z^{\ast},X^{\ast}\wedge
Y^{\ast}\\}$;

  * •

$K_{(3)}=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{X^{\ast}\wedge
Z^{\ast},Y^{\ast}\wedge Z^{\ast}\\}$;

  * •

$K_{(4)}=\operatorname{span}_{C^{\infty}(G_{\kappa})}\\{X^{\ast}\wedge
Y^{\ast}\wedge Z^{\ast}\\}$.

We will now proceed to show the computations of the spectral sequence
associated to the multicomplex that arises from this particular filtration,
just as described in the previous section. Notice that in our computations,
for all the pages $r>0$, we will compute the subspaces $Z_{r}^{p,\bullet}$ and
$B_{r}^{p,\bullet}$, and use the isomorphism

$\displaystyle
E_{r}^{p,\bullet}=E_{r}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))=\frac{\mathcal{Z}_{r}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))}{\mathcal{B}_{r}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))}=\frac{Z_{r}^{p,\bullet}}{B_{r}^{p,\bullet}}\,.$

Since the filtration over the space of smooth forms is nontrivial up to $S=4$,
we are going to present and discuss the quotients that appear in the spectral
sequence construction for each $p=0,\ldots,4$. In order to make the
computations more understandable, we are also going to summarise the overall
information at each page in a more pictorial way with the use of diagrams. We
should stress that we chose not to follow the standard convention of such
diagrams, where at each page $r$ the quotient $E^{p,s}_{r}$ is placed at
position $p$ along the horizontal axis and position $s$ along the vertical
one. In this way, each quotient $E_{r}^{p,s}$ would be placed at the point
$(p,s)$ on the Cartesian plane. Instead, in our diagrams the level of the
filtration is encoded along the vertical axis, and the degree of the forms
along the horizontal axis, so that each quotient $E^{p,s}_{r}$ will be placed
at the point $(p+s,-p)$ on the Cartesian plane. Finally, in order to simplify
the notation, we will denote by $\langle\cdot\rangle$ the span over the space
of smooth functions $C^{\infty}(G_{\kappa})$.

###  5.0. Page-zero quotients $E_{0}^{p,\bullet}$

At the zeroth page of the spectral sequence for $r=0$, one has the quotients:

$\displaystyle
E_{0}^{p,\bullet}=\frac{\mathcal{Z}_{0}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))}{\mathcal{B}_{0}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))}=\frac{\mathcal{F}_{p}\Omega^{\bullet}(G_{\kappa})}{\mathcal{F}_{p+1}\Omega^{\bullet}(G_{\kappa})}\,.$

$\mathbf{p=0}$

  * •

$E^{0,0}_{0}=\mathcal{F}_{0}\Omega^{0}(G_{\kappa})/\mathcal{F}_{1}\Omega^{0}(G_{\kappa})=K^{0}=\Omega^{0}(G_{\kappa})=\langle
1\rangle$;

  * •

$E_{0}^{0,1}=\mathcal{F}_{0}\Omega^{1}(G_{\kappa})/\mathcal{F}_{1}\Omega^{1}(G_{\kappa})=[0]_{0}$;

  * •

$E_{0}^{0,2}=\mathcal{F}_{0}\Omega^{2}(G_{\kappa})/\mathcal{F}_{1}\Omega^{2}(G_{\kappa})=[0]_{0}$;

  * •

$E_{0}^{0,3}=\mathcal{F}_{0}\Omega^{3}(G_{\kappa})/\mathcal{F}_{1}\Omega^{3}(G_{\kappa})=[0]_{0}$;

$\mathbf{p=1}$

  * •

$E_{0}^{1,-1}=\mathcal{F}_{1}\Omega^{0}(G_{\kappa})/\mathcal{F}_{2}\Omega^{0}(G_{\kappa})=0$;

  * •

$E_{0}^{1,0}=\mathcal{F}_{1}\Omega^{1}(G_{\kappa})/\mathcal{F}_{2}\Omega^{1}(G_{\kappa})=K^{1}/K^{1}_{(2)}=\Omega^{1}(G_{\kappa})/\langle
Z^{\ast}\rangle=\langle[X^{\ast}]_{0},[Y^{\ast}]_{0}\rangle$;

  * •

$E_{0}^{1,1}=\mathcal{F}_{1}\Omega^{2}(G_{\kappa})/\mathcal{F}_{2}\Omega^{2}(G_{\kappa})=[0]_{0}$;

  * •

$E_{0}^{1,2}=\mathcal{F}_{1}\Omega^{3}(G_{\kappa})/\mathcal{F}_{2}\Omega^{3}(G_{\kappa})=[0]_{0}$;

$\mathbf{p=2}$

  * •

$E^{2,-2}_{0}=\mathcal{F}_{2}\Omega^{0}(G_{\kappa})/\mathcal{F}_{3}\Omega^{0}(G_{\kappa})=0$;

  * •

$E_{0}^{2,-1}=\mathcal{F}_{2}\Omega^{1}(G_{\kappa})/\mathcal{F}_{3}\Omega^{1}(G_{\kappa})=\langle
Z^{\ast}\rangle$;

  * •

$E_{0}^{2,0}=\mathcal{F}_{2}\Omega^{2}(G_{\kappa})/\mathcal{F}_{3}\Omega^{2}(G_{\kappa})=K^{2}/K^{2}_{(3)}=\langle\Omega^{2}(G_{\kappa})/\langle
X^{\ast}\wedge Z^{\ast},Y^{\ast}\wedge Z^{\ast}\rangle=\langle[X^{\ast}\wedge
Y^{\ast}]_{0}\rangle$;

  * •

$E_{0}^{2,1}=\mathcal{F}_{2}\Omega^{3}(G_{\kappa})/\mathcal{F}_{3}\Omega^{3}(G_{\kappa})=[0]_{0}$;

$\mathbf{p=3}$

  * •

$E_{0}^{3,-3}=\mathcal{F}_{3}\Omega^{0}(G_{\kappa})/\mathcal{F}_{4}\Omega^{0}(G_{\kappa})=0$;

  * •

$E_{0}^{3,-2}=\mathcal{F}_{3}\Omega^{1}(G_{\kappa})/\mathcal{F}_{4}\Omega^{1}(G_{\kappa})=0$;

  * •

$E_{0}^{3,-1}=\mathcal{F}_{3}\Omega^{2}(G_{\kappa})/\mathcal{F}_{4}\Omega^{2}(G_{\kappa})=K^{2}_{(3)}=\langle
X^{\ast}\wedge Z^{\ast},Y^{\ast}\wedge Z^{\ast}\rangle$;

  * •

$E_{0}^{3,0}=\mathcal{F}_{3}\Omega^{3}(G_{\kappa})/\mathcal{F}_{4}\Omega^{3}(G_{\kappa})=[0]_{0}$;

$\mathbf{p=4}$

  * •

$E_{0}^{4,-4}=\mathcal{F}_{4}\Omega^{0}(G_{\kappa})=0$;

  * •

$E_{0}^{4,-3}=\mathcal{F}_{4}\Omega^{1}(G_{\kappa})=0$;

  * •

$E_{0}^{4,-2}=\mathcal{F}_{4}\Omega^{2}(G_{\kappa})=0$;

  * •

$E_{0}^{4,-1}=\mathcal{F}_{4}\Omega^{3}(G_{\kappa})=K^{4}=\Omega^{3}(G_{\kappa})=\langle
X^{\ast}\wedge Y^{\ast}\wedge Z^{\ast}\rangle$.

These computations can be easily summarised in a more visual way using the
following diagram

where we are also drawing the action of the operators
$\partial_{0}=d_{0}\colon E_{0}^{p,\bullet}\to E^{p,\bullet}_{0}$. Moreover,
let us stress that such operator $\partial_{0}$ has nontrivial action only on
$E_{0}^{2,-1}$, as

$\displaystyle\partial_{0}\colon E^{2,-1}_{0}\to E^{2,0}_{0}\ ,\
\partial_{0}(Z^{\ast})=d_{0}(Z^{\ast})=-X^{\ast}\wedge
Y^{\ast}=-[X^{\ast}\wedge Y^{\ast}]_{0}\,.$

###  5.1. Page-one quotients $E_{1}^{p,\bullet}$

Starting from page $r=1$, we are going to use the isomorphism

$\displaystyle
E_{1}^{p,\bullet}=\frac{\mathcal{Z}_{1}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))}{\mathcal{B}_{1}^{p,\bullet}(\mathrm{Tot}\,\Omega^{\bullet}(G_{\kappa}))}=\frac{Z_{1}^{p,\bullet}}{B_{1}^{p,\bullet}}\,,$

where

$\displaystyle Z_{1}^{p,\bullet}=$ $\displaystyle\\{\alpha\in
K_{(p)}:d_{0}\alpha=0\\}\ \text{ and }B_{1}^{p,\bullet}=\\{\beta\in
K_{(p)}:\exists\ \gamma\in K_{(p)}\text{ such that }d_{0}\gamma=\beta\\}\,.$

$\mathbf{p\neq 2}$
As pointed out before, we have that $d_{0}$ acts as the zero map on any
$K_{(p)}$ with $p\neq 2$, and so $E_{1}^{p,\bullet}=E^{p,\bullet}_{0}$.
$\mathbf{p=2}$
The map $d_{0}\colon K^{1}_{(2)}\to K^{2}_{(2)}$ is an isomorphism, and so

$\displaystyle Z_{1}^{2,-1}$ $\displaystyle=\\{\alpha\in
K^{1}_{(2)}:d_{0}\alpha=0\\}=0\ \text{ and }\ B_{1}^{2,-1}=0$ $\displaystyle
Z_{1}^{2,0}=$ $\displaystyle\\{\alpha\in
K^{2}_{(2)}:d_{0}\alpha=0\\}=K^{2}_{(2)}\ \text{ and }\
B_{1}^{2,0}=K^{2}_{(2)}\,.$

Therefore

  * •

$E^{2,-2}_{1}=E^{2,-2}_{0}=0$;

  * •

$E_{1}^{2,-1}=Z_{1}^{2,-1}/B_{1}^{2,-1}=0$;

  * •

$E_{1}^{2,0}=Z_{1}^{2,0}/B_{1}^{2,0}=[0]_{1}$;

  * •

$E_{1}^{2,1}=E_{0}^{2,1}=[0]_{0}$.

By the properties of the spectral sequence, for any $r>1$,
$E_{r}^{2,\bullet}\subset E_{1}^{2,\bullet}$, and so $E^{2,\bullet}_{r}\cong
0$ for any $r\geq 1$.

In this diagram of page one, we are also drawing the action of the operators
$\partial_{1}\colon E_{1}^{p,\bullet}\to E_{1}^{p+1,\bullet}$, which have
nontrivial action only on $E_{0}^{0,0}$ and $E^{3,-1}_{1}$. Indeed

$\displaystyle\partial_{1}$ $\displaystyle\colon E^{0,0}_{1}\to E_{1}^{1,0}\
,\ \partial_{1}[f]=[d_{(1)}f]\ ,\ \forall\ f\in C^{\infty}(G_{\kappa})\,,$
$\displaystyle\partial_{1}\colon$ $\displaystyle E_{1}^{3,-1}\to E_{1}^{4,-1}\
,\ \partial_{1}[\alpha]=[d_{(1)}\alpha]\ ,\ \forall\ \alpha\in
Z^{3,-1}_{1}\,.$

In particular, for any $f\in C^{\infty}(G_{\kappa})$, we have that

$\displaystyle\partial_{1}[f]=[d_{(1)}f]=[XfX^{\ast}+YfY^{\ast}]=XfX^{\ast}+YfY^{\ast}\
\mathrm{mod}\,\langle Z^{\ast}\rangle\,,$

while for any $\alpha=fX^{\ast}\wedge Z^{\ast}+gY^{\ast}\wedge Z^{\ast}\in
Z_{1}^{3,-1}=Z_{0}^{3,-1}=K_{(3)}^{2}$, we have that

$\displaystyle\partial_{1}\big{[}fX^{\ast}\wedge Z^{\ast}+gY^{\ast}\wedge
Z^{\ast}\big{]}=\big{[}d_{(1)}\big{(}fX^{\ast}\wedge Z^{\ast}+gY^{\ast}\wedge
Z^{\ast}\big{)}\big{]}=(Xg-Yf)X^{\ast}\wedge Y^{\ast}\wedge Z^{\ast}\,.$

###  5.2. Page-two quotients $E_{2}^{p,\bullet}$

By definition, the subspaces $Z_{2}^{p,\bullet},B_{2}^{p,\bullet}\subset
K_{(p)}$ are given by

$\displaystyle Z_{2}^{p,\bullet}=$ $\displaystyle\\{\alpha\in
K_{(p)}:d_{0}\alpha=0\text{ and }\exists\ \beta_{1}\in K_{(p+1)}\text{ s.t.
}d_{1}\alpha=d_{0}\beta_{1}\\}$ $\displaystyle B_{2}^{p,\bullet}=$
$\displaystyle\\{\beta\in K_{(p)}:\exists\ \gamma_{0}\in K_{(p)},\gamma_{1}\in
K_{(p-1)}\text{ s.t. }\beta=d_{0}\gamma_{0}+d_{(1)}\gamma_{1}\text{ and
}d_{0}\gamma_{1}=0\\}$

$\mathbf{p=0}$
The only nontrivial space of forms in $K_{(0)}$ is the space of 0-forms, and
so

$\displaystyle Z_{2}^{0,0}=\\{f\in C^{\infty}(G_{\kappa})\mid d_{0}f=0\text{
and }\exists\,g\in K^{0}_{(1)}\text{ s.t. }d_{(1)}f=d_{0}g\\}\,.$

Since $K^{0}_{(1)}=0$, the condition $f\in Z_{2}^{0,0}$ reads as $d_{(1)}f=0$,
or in other words

$\displaystyle f\in Z_{2}^{0,0}\longleftrightarrow
d_{(1)}f=XfX^{\ast}+YfY^{\ast}=0\longleftrightarrow Xf\equiv Yf\equiv 0\,.$

Moreover, since $Z=[X,Y]$, we have that $Z_{2}^{0,0}=\mathbb{R}$ the constant
functions, and so $E_{2}^{0,0}=Z_{2}^{0,0}=\mathbb{R}$. Finally, we know that
the de Rham cohomology class in degree 0,
$H^{0}(\Omega^{\bullet}(G_{\kappa}),d)$ is isomorphic to $\mathbb{R}$, and so
by the properties of spectral sequences, we readily get that $E_{r}^{0,0}\cong
H^{0}(G_{\kappa},d)\cong\mathbb{R}$ for any $r\geq 2$.
$\mathbf{p=1}$
The only nontrivial space of forms in $K_{(1)}$ is the space of 1-forms, and
so

$\displaystyle Z_{2}^{1,0}=\\{\alpha\in K_{(1)}^{1}:d_{0}\alpha=0\text{ and
}\exists\,\beta_{1}\in K_{(2)}^{1}\text{ s.t.
}d_{(1)}\alpha=d_{0}\beta_{1}\\}\,.$

An arbitrary element $\alpha\in K_{(1)}^{1}$ with $\alpha=fX^{\ast}+gY^{\ast}$
satisfies $d_{0}\alpha=0$, moreover

$\displaystyle d_{(1)}\alpha=(Xg-Yf)X^{\ast}\wedge Y^{\ast}=d_{0}[(Yf-
Xg)Z^{\ast}]\,,\ \text{ where }(Yf-Xg)Z^{\ast}\in K^{1}_{(2)}$

and so $Z_{2}^{1,0}=K^{1}_{(1)}$. On the other hand,

$\displaystyle B_{2}^{1,0}=\\{\beta\in K_{(1)}^{1}:\exists\,\gamma_{0}\in
K_{(1)}^{0},\,\gamma_{1}\in K_{(0)}^{0}\text{ s.t.
}\beta=d_{0}\gamma_{0}+d_{(1)}\gamma_{1}\text{ and }d_{0}\gamma_{1}=0\\}$

and since $K_{(1)}^{0}=0$, we have that an element
$\beta=fX^{\ast}+gY^{\ast}\in B_{2}^{1,0}$ if there exists $h\in K_{(0)}^{0}$
such that

$\displaystyle\beta=d_{(1)}h\text{ and
}d_{0}h=0\longleftrightarrow\beta=d_{(1)}h=XhX^{\ast}+YhY^{\ast}\,.$

Therefore, we have that
$E_{2}^{1,0}=Z_{2}^{1,0}/B^{1,0}_{2}=K_{(1)}^{1}/d_{(1)}K_{(0)}^{0}=\langle[X^{\ast},Y^{\ast}]_{2}\rangle$.
$\mathbf{p=2}$
As already pointed out in the computations of the previous page, we have that
$E_{2}^{2,\bullet}\cong 0$.
$\mathbf{p=3}$
The only nontrivial space of forms in $K_{(3)}$ is the space of 2-forms, and
so

$\displaystyle Z_{2}^{3,-1}=\\{\alpha\in K_{(3)}^{2}:d_{0}\alpha=0\text{ and
}\exists\,\beta_{1}\in K_{(4)}^{2}\text{ s.t.
}d_{(1)}\alpha=d_{0}\beta_{1}\\}\,.$

An arbitrary element $\alpha\in K_{(3)}^{2}$ with $\alpha=fX^{\ast}\wedge
Z^{\ast}+gY^{\ast}\wedge Z^{\ast}$ satisfies $d_{0}\alpha=0$, moreover, since
$K_{(4)}^{2}=0$, we have that $\alpha\in Z_{2}^{3,-1}$ if $d_{(1)}\alpha=0$.
Since also $d_{(2)}\alpha=0$ for any $\alpha\in K_{(3)}^{2}$, we have that
$d\alpha=d_{0}\alpha+d_{(1)}\alpha+d_{(2)}\alpha=d_{(1)}\alpha$, and so the
space $Z_{2}^{3,-1}$ coincides with the space of closed forms in
$K_{(3)}^{2}$. On the other hand,

$\displaystyle\beta\in B_{2}^{3,-1}\longleftrightarrow\exists\ \gamma_{0}\in
K_{(3)}^{1},\gamma_{1}\in K_{(2)}^{1}\text{ s.t.
}\beta=d_{0}\gamma_{0}+d_{(1)}\gamma_{1}\text{ and }d_{0}\gamma_{1}=0\,.$

We know that $K_{(3)}^{1}=0$, however, for any form $\gamma_{1}=fZ^{\ast}\in
K_{(2)}^{1}$, we have $d_{0}\gamma_{1}=-fX^{\ast}\wedge Y^{\ast}\neq 0$, and
so $B_{2}^{3,-1}=0$. Therefore,
$E_{2}^{3,-1}=Z_{2}^{3,-1}/B_{2}^{3,-1}=Z_{2}^{3,-1}=\\{\alpha\in
K_{(3)}^{2}:d\alpha=0\\}=\langle[X^{\ast}\wedge Z^{\ast},Y^{\ast}\wedge
Z^{\ast}]_{2}\rangle$.
$\mathbf{p=4}$
Finally, the only nontrivial space of forms in $K_{(4)}$ is the space of
3-forms, and so

$\displaystyle Z^{4,-1}_{2}=\\{\alpha\in K^{3}_{(4)}:d_{0}\alpha=0\text{ and
}\exists\,\beta_{1}\in K_{(5)}^{3}\text{ s.t.
}d_{(1)}\alpha=d_{0}\beta_{1}\\}\,,$

and since $K_{(5)}^{3}=K_{(5)}^{4}=0$, we have that
$Z_{2}^{4,-1}=K_{(4)}^{3}$. On the other hand, the condition $\beta\in
B_{2}^{4,-1}$ reads

$\displaystyle\beta\in B_{2}^{4,-1}\longleftrightarrow\exists\,\gamma_{0}\in
K_{(4)}^{2},\gamma_{1}\in K_{(3)}^{2}\text{ s.t.
}\beta=d_{0}\gamma_{0}+d_{(1)}\gamma_{1}\text{ and }d_{0}\gamma_{1}=0\,.$

Since $K_{(4)}^{2}=0$, and $d_{0}\gamma_{1}=0$ for any $\gamma_{1}\in
K_{(3)}^{2}$, we get $B^{4,-1}_{2}=\\{\beta\in
K_{(4)}^{3}:\exists\,\gamma_{1}\in K_{(3)}^{2}\text{ s.t.
}d_{(1)}\gamma_{1}=\beta\\}=\\{(Xg-Yf)X^{\ast}\wedge Y^{\ast}\wedge
Z^{\ast}\\}$.
Therefore,
$E_{2}^{4,-1}=Z_{2}^{4,-1}/B_{2}^{4,-1}=K_{(4)}^{3}/d_{(1)}K_{(3)}^{2}=[X^{\ast}\wedge
Y^{\ast}\wedge Z^{\ast}]_{2}$.
It should be noted that any $\alpha\in Z_{2}^{4,-1}=K_{(4)}^{3}$ is a closed
form, since $d\alpha=0$. If $\alpha$ is also exact, then there exists
$\beta=\beta_{(2)}+\beta_{(3)}\in K^{2}$ such that $d\beta=\alpha$. Here,
$\beta_{(2)}\in K_{(2)}^{2}$ and $\beta_{(3)}\in K_{(3)}^{2}$, where
$\beta_{(2)}=d_{0}\gamma_{(2)}$ for some $\gamma_{(2)}\in K_{(2)}^{1}$ since
the map $d_{0}\colon K_{(2)}^{1}\to K_{(2)}^{2}$ is an isomorphism. Therefore

$\displaystyle\alpha=d\beta=d_{(2)}\beta_{(2)}+d_{(1)}\beta_{(3)}=d_{(2)}d_{0}\gamma_{(2)}+d_{(1)}\beta_{(3)}=-d_{(1)}^{2}\gamma_{(2)}+d_{(1)}\beta_{(3)}\,.$

Here, in the final equality, we use the fact that $(K,d)$ is a multicomplex
and $d_{(0)}d_{(2)}+d_{(1)}^{2}+d_{(2)}d_{(0)}=0$, and that $d_{(2)}\colon
K_{(2)}^{1}\to K_{(4)}^{2}$ is the zero map. Hence, for any exact form
$\alpha\in Z_{2}^{4,-1}$ there exists
$\tilde{\beta}=\beta_{(3)}-d_{(1)}\gamma_{(2)}\in K_{(3)}^{2}$ such that
$\alpha=d_{(1)}\tilde{\beta}$. In other words, $E_{2}^{4,-1}$ is isomorphic to
$H^{3}(\Omega^{\bullet}(G_{\kappa}),d)$, the de Rham cohomology class of
$G_{\kappa}$ in degree 3. In the case where
$H^{3}(\Omega^{\bullet}(G_{\kappa}),d)=0$, for example in the first Heisenberg
group when $\kappa=0$, we have $E_{2}^{4,-1}\cong 0$, whereas if
$H^{3}(\Omega^{\bullet}(G_{\kappa}),d)\cong\mathbb{R}$, for example when
$\kappa>0$, then $E_{2}^{4,-1}\cong\mathbb{R}$. Following the same reasoning
as in the case of $\mathbf{p=0}$, we also have that $E^{4,-1}_{r}\cong
H^{3}(\Omega^{\bullet}(G_{\kappa}),d)$ for any $r\geq 2$.

In this diagram of page two, we are showing the action of the operators
$\partial_{2}\colon E_{2}^{p,\bullet}\to E_{2}^{p+2,\bullet}$. The only
nontrivial map of such operators is $\partial_{2}\colon E_{2}^{1,0}\to
E_{2}^{3,-1}$, so for any $\alpha\in Z_{2}^{1,0}=K_{(1)}^{1}$

$\displaystyle\partial_{2}\big{[}\alpha\big{]}=\big{[}d_{(2)}\alpha-
d_{(1)}\beta_{1}\big{]}\ \text{ for some }\beta_{1}\in K_{(2)}^{1}\text{ s.t.
}d_{(1)}\alpha=d_{0}\beta_{1}\,.$

As already pointed out before, if we fix $\alpha=fX^{\ast}+gY^{\ast}$, it is
sufficient to take $\beta_{1}=(Yf-Xg)Z^{\ast}\in K_{(2)}^{1}$ in order to
obtain $d_{(1)}\alpha=d_{0}\beta_{1}$, hence

$\displaystyle\partial_{2}\big{[}fX^{\ast}+gY^{\ast}\big{]}=$
$\displaystyle\big{[}d_{(2)}(fX^{\ast}+gY^{\ast})-d_{(1)}(Yf-
Xg)Z^{\ast}\big{]}$ $\displaystyle=$ $\displaystyle\big{[}(\kappa
g-Zf)X^{\ast}\wedge Z^{\ast}-(\kappa f+Zg)Y^{\ast}\wedge Z^{\ast}+$
$\displaystyle-X(Yf-Xg)X^{\ast}\wedge Z^{\ast}-Y(Yf-Xg)Y^{\ast}\wedge
Z^{\ast}\big{]}$ $\displaystyle=$ $\displaystyle(\kappa g-Zf-
XYf+X^{2}g)X^{\ast}\wedge Z^{\ast}+(YXg-Zg-Y^{2}f-\kappa f)Y^{\ast}\wedge
Z^{\ast}$

One can also readily check that $\partial_{2}[\alpha]$ is an element of
$E_{2}^{3,-1}$, since

$\displaystyle X(YXg-Zg-Y^{2}f-\kappa f)=Y(\kappa g-Zf-XYf+X^{2}g)\ ,\
\forall\,f,g\in C^{\infty}(G_{\kappa})\,.$

Moreover, the differential $\partial_{2}$ is well-defined on the quotient
$E_{2}^{1,0}$ since for any $\beta\in B_{2}^{1,0}$ we have
$\beta=XhX^{\ast}+YhY^{\ast}$ for some $h\in C^{\infty}(G_{\kappa})$, and so

$\displaystyle\partial_{2}[\beta]=$
$\displaystyle\big{[}d_{(2)}(XhX^{\ast}+YhY^{\ast})-d_{(1)}(YXh-
XYh)Z^{\ast}\big{]}$ $\displaystyle=$ $\displaystyle\big{[}(\kappa Yh-
ZXh)X^{\ast}\wedge Z^{\ast}-(\kappa Xh+ZYh)Y^{\ast}\wedge
Z^{\ast}+XZhX^{\ast}\wedge Z^{\ast}+YZhY^{\ast}\wedge Z^{\ast}\big{]}=[0]\,.$

###  5.3. Page-three quotients $E_{3}^{p,\bullet}$

By definition, we have that the subspaces
$Z_{3}^{p,\bullet},B_{3}^{p,\bullet}\subset K_{(p)}$ are given by

$\displaystyle Z_{3}^{p,\bullet}=$ $\displaystyle\\{\alpha\in K_{(p)}\mid
d_{0}\alpha=0\,,\,\exists\ \beta_{1}\in K_{(p+1)},\beta_{2}\in K_{(p+2)}\text{
s.t.
}d_{(1)}\alpha=d_{0}\beta_{1}\,,\,d_{(2)}\alpha=d_{0}\beta_{2}+d_{(1)}\beta_{1}\\}$
$\displaystyle B_{3}^{p,\bullet}=$ $\displaystyle\\{\beta\in
K_{(p)}\mid\exists\ \gamma_{i}\in K_{(p-i)},\,i=0,1,2\text{ s.t.
}\beta=\sum_{i=0}^{2}d_{(i)}\gamma_{i}\text{ and
}d_{0}\gamma_{1}+d_{(1)}\gamma_{2}=d_{0}\gamma_{2}=0\\}\,.$

$\mathbf{p=0}$
As already pointed out in the computations of the previous page, we have that
$E^{0,0}_{3}=E^{0,0}_{2}=\mathbb{R}$.
$\mathbf{p=1}$
We have already shown that an arbitrary element $\alpha=fX^{\ast}+gY^{\ast}\in
K_{(1)}^{1}$ belongs to $Z^{1,0}_{2}$ since $d_{0}\alpha=0$ and one can take
$\beta_{1}=(Yf-Xg)Z^{\ast}\in K_{(2)}^{1}$ such that $d_{(1)}\alpha=(Xg-
Yf)X^{\ast}\wedge Y^{\ast}=d_{0}\beta_{1}$. Hence $\alpha\in Z^{1,0}_{3}$ if
there exists $\beta_{2}\in K_{(3)}^{1}$ such that
$d_{(2)}\alpha=d_{0}\beta_{2}+d_{(1)}\beta_{1}$, but since $K_{(3)}^{1}=0$,
this means that

$\alpha\in Z^{1,0}_{3}\longleftrightarrow\alpha\in K_{(1)}^{1}\text{ and
}d_{(2)}\alpha=d_{(1)}\beta_{1}\text{ with }d_{(1)}\alpha=d_{0}\beta_{1}\,.$

On the other hand, $\beta\in B_{3}^{1,0}$ if there exist $\gamma_{i}\in
K_{(1-i)}^{0}$, with $i=0,1,2$ such that
$\beta=d_{0}\gamma_{0}+d_{(1)}\gamma_{1}+d_{(2)}\gamma_{2}$ and
$d_{0}\gamma_{1}+d_{(1)}\gamma_{2}=d_{0}\gamma_{2}=0$. Since
$K_{(-1)}^{0}=K_{(1)}^{0}=0$, we have that $\beta\in B_{3}^{1,0}$ if
$\exists\,h\in C^{\infty}(G_{\kappa})$ such that $\beta=d_{(1)}h$ and
$d_{0}h=0$, or in other words $B_{3}^{1,0}=B_{2}^{1,0}$.
To show that $E^{1,0}_{3}=Z_{3}^{1,0}/B_{3}^{1,0}\cong 0$, let us first
consider the exterior differential of an arbitrary form $\alpha\in
Z_{3}^{1,0}$:

$\displaystyle
d\alpha=d_{(0)}\alpha+d_{(1)}\alpha+d_{(2)}\alpha=0+d_{0}\beta_{1}+d_{(2)}\alpha=d\beta_{1}-d_{(1)}\beta_{1}-d_{(2)}\beta_{1}+d_{(2)}\alpha=d\beta_{1}$

since $d_{(2)}\colon K_{(2)}^{1}\to K_{(4)}^{2}$ is the zero map. This means
that the form $d(\alpha-\beta_{1})=0$, that is $\alpha-\beta_{1}$ is a closed
form. We know that $H^{1}(\Omega^{\bullet}(G_{\kappa}),d)=0$ for any $\kappa$,
and so $\alpha-\beta_{1}$ is also exact, that is $\exists\,h\in
C^{\infty}(G_{\kappa})$ such that

$\displaystyle\alpha-\beta_{1}=dh=d_{(1)}h+d_{(2)}h\ \longleftrightarrow\
\alpha=d_{(1)}h\in K_{(1)}^{1}\text{ and }\beta_{1}=d_{(2)}h\in
K_{(2)}^{1}\,.$

We have therefore shown that for any $\alpha\in Z_{3}^{1,0}$, there exists
$h\in K_{(0)}^{0}$ such that $d_{(1)}h=\alpha$, that is
$Z_{3}^{1,0}=B_{3}^{1,0}$.
$\mathbf{p=3}$
By definition, we have that $\alpha\in Z_{3}^{3,-1}$ if $d_{0}\alpha=0$ and
there exist $\beta_{1}\in K_{(4)}^{2}$ and $\beta_{2}\in K_{(5)}^{2}$ such
that $d_{(1)}\alpha=d_{0}\beta_{1}$ and
$d_{(2)}\alpha=d_{0}\beta_{2}+d_{(1)}\beta_{1}$. Again, since
$K_{(4)}^{2}=K_{(5)}^{2}=0$ and $d_{0}\alpha=0$ for any $\alpha\in
K_{(3)}^{2}$, we have that $\alpha\in Z_{3}^{3,-1}$ if $d_{(1)}\alpha=0$. In
other words, $Z^{3,-1}_{3}=Z^{3,-1}_{2}$ is the space of closed forms in
$K_{(3)}^{2}$. On the other hand, we have $\beta\in B^{3,-1}_{3}$ if there
exist $\gamma_{i}\in K_{(3-i)}^{1}$ with $i=0,1,2$ such that
$\beta=d_{0}\gamma_{0}+d_{(1)}\gamma_{1}+d_{(2)}\gamma_{2}$ and
$d_{0}\gamma_{1}+d_{(1)}\gamma_{2}=d_{0}\gamma_{2}=0$. Since $K_{(3)}^{1}=0$,
we have

$\displaystyle\beta\in B^{3,-1}_{3}\longleftrightarrow\exists\,\gamma_{1}\in
K_{(2)}^{1},\,\gamma_{2}\in K_{(1)}^{1}\text{ s.t.
}\beta=d_{(1)}\gamma_{1}+d_{(2)}\gamma_{2}\text{ and
}d_{0}\gamma_{1}+d_{(1)}\gamma_{2}=d_{0}\gamma_{2}=0\,.$

We now want to show that $B_{3}^{3,-1}=Z_{3}^{3,-1}$, and hence
$E^{3,-1}_{3}\cong 0$. Since $H^{2}(\Omega^{\bullet}(G_{\kappa}),d)=0$ for any
$\kappa$, we have that any $\alpha\in Z_{3}^{3,-1}$ is a closed and hence
exact form, that is $\exists\,\beta=\beta_{1}+\beta_{2}\in K^{1}$ such that
$d\beta=\alpha$. Here, we are taking $\beta_{1}\in K_{(1)}^{1}$ and
$\beta_{2}\in K_{(2)}^{1}$, and so

$\displaystyle
d\beta=d\beta_{(1)}+d\beta_{(2)}=d_{(1)}\beta_{(1)}+d_{(2)}\beta_{(1)}+d_{(0)}\beta_{2}+d_{(1)}\beta_{2}=\alpha\in
K_{(3)}^{2}$

which implies that $d_{(1)}\beta_{1}+d_{0}\beta_{2}=0$ since it is an element
of $K_{(2)}$, and $d_{0}\beta_{1}=0$ since $d_{0}\colon K_{(1)}^{1}\to
K_{(1)}^{2}$ is the zero map. For any $\alpha\in Z_{3}^{3,-1}$, it is
therefore sufficient to take $\gamma_{2}=\beta_{1}\in K_{(1)}^{1}$ and
$\gamma_{1}=\beta_{2}\in K_{(2)}^{1}$ to show that $\alpha\in B^{3,-1}_{3}$.

$\mathbf{p=4}$
As already pointed out in the computations of the previous page, we have that
$E_{3}^{4,-1}=E_{3}^{4,-1}\cong H^{3}(\Omega^{\bullet}(G_{\kappa}),d)$.

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