# Elliptic complexes on $G_{2}$ and $\mathrm{Spin}(7)-$manifolds

Xue Zhang<EMAIL_ADDRESS>_Department of Mathematical Sciences, Tsinghua
University, Beijing, China_

###### Abstract

There are three types of generalized de Rham complexes arising from
representations of holonomy group on a smooth Riemannian manifold, two of
which are dual to each other. We prove that a complex is elliptic if and only
if its generator satisfies an algebraic condition. As applications, we give
all elliptic complexes on a Riemannian manifold with exceptional holonomy.

Keywords Elliptic complex, holonomy, $G-$structure, representation

###### Contents

  1. 1 Introduction
  2. 2 Prime and complete prime spaces
  3. 3 Generalized de Rham complex
  4. 4 Complexes and adjoint representations
  5. 5 Complexes and representations

##  1 Introduction

It is well known that a de Rham complex on a smooth manifold $M$ is the
cochain complex $(A^{*}(M),d)$ of differential forms on $M$ with the exterior
derivative as its differential. If $M$ carries a Riemannian metric with
special holonomy, each $A^{k}(M)$ possesses a direct sum decomposition. For
instance, $A^{k}\otimes\mathbb{C}$ can split into $\oplus_{p+q=k}A^{p,q}$ with
respect to a $\mathrm{U}(n)-$action and there are Dolbeault complexes on
complex manifold. In Berger’s classification [2] of holonomy groups of a
nonsymmetric, irreducible Riemannian manifold there are two exceptional
holonomy groups, $G_{2}$ in $7$ dimensions and $\mathrm{Spin}(7)$ in $8$
dimensions. A fundamental question is that, for a $G_{2}$ or
$\mathrm{Spin}(7)-$manifold, if there exists a sequence of subbundles
$E_{i}\subset\Lambda^{i}(M),i=0,1,2,\cdots,$ such that the spaces of smooth
sections composes an elliptic complex (ref. [1])

$\displaystyle 0\longrightarrow\Gamma(E_{0})\stackrel{{\scriptstyle
D_{0}}}{{\longrightarrow}}\Gamma(E_{1})\stackrel{{\scriptstyle
D_{1}}}{{\longrightarrow}}\Gamma(E_{2})\stackrel{{\scriptstyle
D_{2}}}{{\longrightarrow}}\cdots,$ (1)

where the boundary operator $D_{i}=p_{i}\circ d$ is a composition of
differential and the projection $p_{i}:A^{i}(M)\rightarrow\Gamma(E_{i})$ with
respect to a Riemannian metric. This kind of complexes (1) can be divided into
three categories due to the condition $D_{i+1}D_{i}=0$

  * •

type I

$\displaystyle 0\rightarrow A^{0}\stackrel{{\scriptstyle
d}}{{\rightarrow}}A^{1}\stackrel{{\scriptstyle
d}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle
d}}{{\rightarrow}}A^{k-1}\stackrel{{\scriptstyle
D_{0}}}{{\rightarrow}}\Gamma(E_{0}^{\perp})\stackrel{{\scriptstyle
D_{1}}}{{\rightarrow}}\Gamma(E_{1}^{\perp})\stackrel{{\scriptstyle
D_{2}}}{{\rightarrow}}\Gamma(E_{2}^{\perp})\stackrel{{\scriptstyle
D_{3}}}{{\rightarrow}}\cdots,$ (2)

  * •

type II

$\displaystyle 0\rightarrow\Gamma(F)\stackrel{{\scriptstyle
D_{0}}}{{\rightarrow}}\Gamma(E_{0}^{\perp})\stackrel{{\scriptstyle
D_{1}}}{{\rightarrow}}\Gamma(E_{1}^{\perp})\stackrel{{\scriptstyle
D_{2}}}{{\rightarrow}}\Gamma(E_{2}^{\perp})\stackrel{{\scriptstyle
D_{3}}}{{\rightarrow}}\cdots,$ (3)

  * •

type III

$\displaystyle 0\rightarrow\Gamma(E_{0})\stackrel{{\scriptstyle
D_{0}}}{{\rightarrow}}\Gamma(E_{1})\stackrel{{\scriptstyle
D_{1}}}{{\rightarrow}}\Gamma(E_{2})\stackrel{{\scriptstyle
D_{2}}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle
d}}{{\rightarrow}}A^{n-1}\stackrel{{\scriptstyle
d}}{{\rightarrow}}A^{n}\rightarrow 0,$ (4)

where $E_{i}=E_{0}\cdot\Lambda^{i}\subset\Lambda^{k+i}$ for $i=0,1,2,\cdots$
and $F\subset\Lambda^{k-1}.$ Type I and type III are dual to each other. We
say that (2) and (3) are generated by $E_{0}$ and $(F,E_{0})$ respectively. We
write the complex by its generator if there is no ambiguity.

For example, given a compact, oriented Riemannian $4n-$manifold $M,$ the Hodge
star $*$ on $A^{2n}(M)$ satisfies $*^{2}=id.$ The middle bundle
$\Lambda^{2n}(M)$ of $2n-$forms decompose orthogonally as
$\Lambda^{2n}=\Lambda^{+}\oplus\Lambda^{-}.$ There are two elliptic complexes
of type I

$\displaystyle 0\longrightarrow A^{0}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}A^{1}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\cdots\stackrel{{\scriptstyle
d}}{{\longrightarrow}}A^{2n-1}\stackrel{{\scriptstyle
D}}{{\longrightarrow}}A^{\pm}\longrightarrow 0$ (5)

generated by $\Lambda^{\pm}.$ In particular, when the dimension is four, (5)
becomes $A^{0}\rightarrow A^{1}\rightarrow A^{\pm},$ which is a deformation
complex of instanton equation after twisting an adjoint bundle. Donaldson [7,
8, 9] developed the instanton equation as an important tool for studying four-
dimensional topology.

Salamon [15] and Reyes-Carrión [4] generalized the notion of instanton to high
dimensional Riemannian manifolds with special holonomy groups. Reyes-Carrión
showed in [4] the that: Suppose $G\subset\mathrm{SO}(n)$ is one of
$\mathrm{SU}(2)_{+},\mathrm{SU}(m),\mathrm{Sp}(m),G_{2},$ $\mathrm{Spin}(7)$
and the holonomy group of a Riemannian $n-$manifold $(M,g)$ reduces to the
normalizer of $G,$ then the complex generated by $\Lambda(\mathfrak{g})$ is
elliptic. Here $\Lambda(\mathfrak{g})$ is a subbundle of $\Lambda^{2}(M)$
associated with the adjoint representation of $G.$

A key point in example (5) and Reyes-Carrión’s theorem is that the fibers of
generators of complex are invariant under the action of holonomy group. Based
on it, we are concerned in this paper with the generators arising from sub-
representations of exterior algebra of holonomy groups. We introduce the
notion of locally quasi-flat and complete prime. The condition $D^{2}=0$ for
$(E,D)$ and $(F,E,D)$ is equivalent to that $E$ is locally quasi-flat. The
elliptic conditions for $(E,D)$ and $(F,E,D)$ are equivalent to that $E$ and
$(F,E)$ are complete prime. We state our main results:

######  Theorem A.

Let $G$ a connected subgroup of $\mathrm{SO}(n)$ and let $(M,g,G,Q)$ be a
closed Riemannian $n-$manifold with a compatible torsion-free $G-$ structure
$Q.$ Suppose $\mathcal{F}\subset\Lambda^{k-1}(\mathbb{R}^{n})$ and
$\mathcal{E}\subset\Lambda^{k}(\mathbb{R}^{n})$ are invariant subspaces under
the action of $G.$ Set $F=Q\times_{G}\mathcal{F}\subset\Lambda^{k-1}(M)$ and
$E=Q\times_{G}\mathcal{E}\subset\Lambda^{k}(M).$ Then $E$ is locally quasi-
flat. Consequently,

  * (i)

If $\mathcal{E}$ is complete prime, then the sequence $(E,D)$ is an elliptic
complex.

  * (ii)

If $(\mathcal{F},\mathcal{E})$ is complete prime, then the sequence $(F,E,D)$
is an elliptic complex.

Apply Theorem A to $G_{2}$ and $\mathrm{Spin}(7)-$manifolds, we have the
following two theorems

######  Theorem B.

Let $(M,g,\varphi)$ be a compact $G_{2}-$manifold. Then the following
complexes on $M$ are elliptic

$\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda_{7}^{4}\oplus\Lambda_{27}^{4}\rightarrow\Lambda_{14}^{5}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda_{7}^{2}\rightarrow\langle\varphi\rangle\oplus\Lambda_{7}^{3}\rightarrow\langle*\varphi\rangle\rightarrow
0,$ $\displaystyle
0\rightarrow\langle\varphi\rangle\oplus\Lambda_{7}^{3}\rightarrow\langle*\varphi\rangle\oplus\Lambda_{7}^{4}\rightarrow
0.$

The first one is of type I generated by $\langle*\varphi\rangle.$ The last two
are of type II. Moreover, if $\mathrm{Hol}(g)=G_{2},$ except for the above
three, $(\Lambda_{14}^{2},D)$ and their dual complexes, there is no other
generalized de Rham complexes on $M.$

######  Theorem C.

Let $(M,g,\Omega)$ be a compact $\mathrm{Spin}(7)-$manifold. Then the
following complexes on $M$ are elliptic

$\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda^{\pm}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda_{48}^{3}\rightarrow\Lambda_{27}^{4}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\oplus\Lambda^{-}\rightarrow\Lambda_{8}^{5}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda^{4}\rightarrow\Lambda^{5}\rightarrow\Lambda_{21}^{6}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda_{7}^{2}\rightarrow\Lambda_{8}^{3}\rightarrow\langle\Omega\rangle\rightarrow
0,\ \
0\rightarrow\Lambda_{21}^{2}\rightarrow\Lambda_{48}^{3}\rightarrow\Lambda_{27}^{4}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda_{8}^{3}\rightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\rightarrow
0,\ \
0\rightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\rightarrow\Lambda_{8}^{5}\rightarrow
0.$

The first five are of type I generated by
$\Lambda^{\pm},\Lambda_{8}^{3},\Lambda_{27}^{4}$ and $\Lambda_{7}^{6}$
respectively. The last four are of type II. Morever, if
$\mathrm{Hol}(g)=\mathrm{Spin}(7),$ then except for the above nine,
$(\Lambda_{21}^{2},D)$ and their dual complexes, there is no other generalized
de Rham complexes on $M.$

##  2 Prime and complete prime spaces

###  2.1 Prime forms in $\Lambda^{*}(\mathbb{R}^{n})$

We begin by reviewing some facts about exterior algebra. Let $V$ be a real
vector space and $\Lambda^{*}(V)$ the exterior algebra of $V$ which has a
graded structure

$\Lambda^{*}(V)=\bigoplus_{k=0}^{\mathrm{dim}\,V}\Lambda^{k}(V).$

If $V$ is equipped with an inner product, there is a canonical isomorphism $i$
between $\Lambda^{2}(V)$ and the Lie algebra of the isometric transformation
group of $V.$ For each $\alpha\in\Lambda^{2}(V),$ define the rank of $\alpha$
by $\mathrm{rank}\,\alpha:=\frac{1}{2}\mathrm{rank}\,i(\alpha).$ Note that the
rank does not depend on the inner product of $V.$

In this section, we let $V=\mathbb{R}^{n},n\geq 4$ be the Euclidean space.

######  Definition 1.

We say a $k-$form $\alpha\in\Lambda^{k}(\mathbb{R}^{n})$ is _prime_ if the
multiplication
$\Lambda^{1}(\mathbb{R}^{n})\rightarrow\Lambda^{k+1}(\mathbb{R}^{n}),\lambda\mapsto\alpha\cdot\lambda$
by $\alpha$ is injective. Further, a linear subspace
$E\subset\Lambda^{k}(\mathbb{R}^{n})$ is _prime_ if every non-zero form in $E$
is prime.

Obviously, $E\subset\Lambda^{k}(\mathbb{R}^{n})$ is prime if and only if the
complex

$\displaystyle\Lambda^{k-2}(\mathbb{R}^{n})\stackrel{{\scriptstyle\lambda}}{{\longrightarrow}}\Lambda^{k-1}(\mathbb{R}^{n})\stackrel{{\scriptstyle\lambda}}{{\longrightarrow}}\Lambda^{k}(\mathbb{R}^{n})/E$
(6)

is exact at $\Lambda^{k-1}(\mathbb{R}^{n})$ for every
$\lambda\in\Lambda^{1}(\mathbb{R}^{n})-\\{0\\}.$

######  Example 2.

We consider a prime form $\alpha$ in $\Lambda^{k}(\mathbb{R}^{n}),n\geq 4$ for
some special cases.

  * •

If $\alpha\in\Lambda^{2}(\mathbb{R}^{n}),$ then $\alpha$ is prime if and only
if $\mathrm{rank}\,\alpha\geq 2.$

  * •

When $k=n-2,$ every $\alpha\in\Lambda^{n-2}(\mathbb{R}^{n})$ can be expressed
as

$\alpha=\sum_{i,j}\alpha_{ij}\cdot
e_{1}\cdots\hat{e}_{i}\cdots\hat{e}_{j}\cdots e_{n}$

with $\alpha_{ij}+(-1)^{i+j}\alpha_{ji}=0,$ where $\\{e_{1},\cdots,e_{n}\\}$
is a basis for $\Lambda^{1}(\mathbb{R}^{n})$ and the hat denotes the omission
of that element. Then $\alpha$ is prime if and only if the coefficient matrix
$(\alpha_{ij})$ is invertible.

  * •

When $k=n-1,$ none of elements in $\Lambda^{n-1}(\mathbb{R}^{n})$ are prime.
For each $\alpha\in\Lambda^{n-1}(\mathbb{R}^{n}),$ the multiplication by
$\alpha$ is not injective since

$\mathrm{dim}\,\Lambda^{1}(\mathbb{R}^{n})=n>1=\mathrm{dim}\,\Lambda^{n}(\mathbb{R}^{n}).$

Let $*$ denote the Hodge star operator on $\Lambda^{*}(\mathbb{R}^{n}).$ Given
a form $\alpha\in\Lambda^{n-2k}(\mathbb{R}^{n}),$ define a traceless operator
related to $\alpha$ by

$T_{\alpha}:\
\Lambda^{k}(\mathbb{R}^{n})\rightarrow\Lambda^{k}(\mathbb{R}^{n}),\ \
T_{\alpha}(\beta)=*(\alpha\cdot\beta).$

The operator $T_{\alpha}$ is self-adjoint when $k$ is even and is anti-self-
adjoint when $k$ is odd.

######  Lemma 3.

Suppose $\alpha\in\Lambda^{n-2k}(\mathbb{R}^{n}).$ Then every eigenspace of
$T_{\alpha}$ with non-zero eigenvalue is prime.

###### Proof.

Fix a non-zero eigenvalue $a$ of $T_{\alpha},$ assume
$T_{\alpha}(\beta)=a\cdot\beta$ and $\beta=\lambda\cdot\beta^{\prime}$ for
some $\beta^{\prime}\in\Lambda^{k-1}(\mathbb{R}^{n})$ and
$\lambda\in\Lambda^{1}(\mathbb{R}^{n})$ with $|\lambda|=1.$ Extend $\lambda$
to an orthonormal basis $B=\\{\lambda,e_{2},e_{3},\cdots,e_{n}\\}$ for
$\Lambda^{1}(\mathbb{R}^{n}).$ We can assume that the expression of
$\beta^{\prime}$ with respect to $B$ does not contain $\lambda.$ Then the
expression of $*(\alpha\cdot\lambda\cdot\beta^{\prime})$ does not contain
$\lambda.$ This is a contradiction to $T_{\alpha}(\beta)=a\cdot\beta.$ ∎

######  Example 4.

The eigenspaces $\Lambda^{\pm}\subset\Lambda^{2n}(\mathbb{R}^{4n})$ of star
operator
$*:\Lambda^{2n}(\mathbb{R}^{4n})\rightarrow\Lambda^{2n}(\mathbb{R}^{4n})$
associated with $\pm 1$ are both prime.

The basic action of $\mathrm{SO}(n)$ on $\mathbb{R}^{n}$ induces a
representation on $\Lambda^{*}(\mathbb{R}^{n})$ by

$A(\omega^{i_{1}}\wedge\omega^{i_{2}}\wedge\cdots\wedge\omega^{i_{k}})=A\omega^{i_{1}}\wedge
A\omega^{i_{2}}\wedge\cdots\wedge A\omega^{i_{k}}.$

In particular, $\Lambda^{2}(\mathbb{R}^{n})$ is the adjoint representation of
$\mathrm{SO}(n)$ by identifying $\Lambda^{2}(\mathbb{R}^{n})$ with
$\mathfrak{so}(n)$. The Hodge star $*:\Lambda^{k}\rightarrow\Lambda^{n-k}$
commutes with $\mathrm{SO}(n).$ Let $G$ be a subgroup of $\mathrm{SO}(n).$ The
real representation $\mathbb{R}^{n}$ of $G$ induces an action of $G$ on the
unit sphere $S^{n-1}$ in $\mathbb{R}^{n}.$ We give a simple but useful lemma.

######  Lemma 5.

Let $E\subset\Lambda^{k}(\mathbb{R}^{n})$ be a $G-$invariant subspace. If $G$
acts on $S^{n-1}$ transitively, then $E$ is prime if and only if there exists
$\lambda\in\Lambda^{1}(\mathbb{R}^{n})$ such that the multiplication
$E\rightarrow\Lambda^{k+1}(\mathbb{R}^{n}),\alpha\mapsto\lambda\cdot\alpha$ by
$\lambda$ is injective.

###  2.2 Complete prime space and pair in $\Lambda^{*}(\mathbb{R}^{n})$

Sometimes we denote $\Lambda^{i}(\mathbb{R}^{n})$ by $\Lambda^{i}.$ For a
linear subspace $E$ of $\Lambda^{*},$ let $E\cdot\Lambda^{i}$ denote the
linear subspace of $\Lambda^{*+i}$ generated by $\\{xy|\ x\in
E,y\in\Lambda^{i}\\}.$ Analogously to (6), we introduce the notion of complete
prime to describe the principal symbols of (2) and (3).

######  Definition 6.

Let $E\subset\Lambda^{k}(\mathbb{R}^{n})$ and
$F\subset\Lambda^{k-1}(\mathbb{R}^{n})$ be two linear subspaces for $k\geq 2.$
Set $E_{i}=E\cdot\Lambda^{i}$ for $i=1,2,\cdots,r$ and
$E\cdot\Lambda^{r+1}=\Lambda^{k+r+1}.$

  * (1)

We say that $E$ is _complete prime_ if the complex

$\displaystyle
0\rightarrow\Lambda^{0}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{1}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k-1}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k}/E\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k+1}/E_{1}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k+r}/E_{r}\rightarrow
0$ (7)

is exact for any $\lambda\in\Lambda^{1}-\\{0\\}.$ Denote this complex by
$(E^{\bullet},\lambda).$

  * (2)

We say that the pair $(F,E)$ is _complete prime_ if the complex

$\displaystyle 0\rightarrow
F\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k}/E\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k+1}/E_{1}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k+2}/E_{2}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k+r}/E_{r}\rightarrow
0$ (8)

is exact for any $\lambda\in\Lambda^{1}-\\{0\\}.$ Denote this complex by
$(F,E^{\bullet},\lambda).$

######  Remark 7.

The dual complex of (7) is given by

$\displaystyle 0\rightarrow
E\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}E_{1}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}E_{2}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}E_{r}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{k+r+1}\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\cdots\stackrel{{\scriptstyle\lambda}}{{\rightarrow}}\Lambda^{n}\rightarrow
0$ (9)

which is the principal symbols of (4). One can check that with the same
generator (7) is exact if and only if (9) is exact.

######  Example 8.

Consider $\mathbb{R}^{5}$ with an inner product. Fix a unit vector
$u\in\mathbb{R}^{5}$ and let $W=\langle u\rangle^{\perp}.$ Then the
eigenspaces $E=\Lambda^{\pm}(W)$ of star operator on $\Lambda^{2}(W)$ are both
complete prime, i.e., the sequence $(E^{\bullet},\lambda)$

$\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}/E\rightarrow\Lambda^{3}/(E\cdot\Lambda^{1})\rightarrow
0$ (10)

is exact. To see this, consider its alternating sum of dimensions. Due to
$\mathrm{dim}\,E\cdot\Lambda^{1}=7,$ one can see that

$\displaystyle\mathrm{dim}\,\Lambda^{0}-\mathrm{dim}\,\Lambda^{1}+(\mathrm{dim}\,\Lambda^{2}-\mathrm{dim}\,E)-(\mathrm{dim}\,\Lambda^{3}-\mathrm{dim}\,E\cdot\Lambda^{1})=0.$
(11)

Since $E$ is prime, (10) is exact at $\Lambda^{1}.$ In addition, for any
$\lambda\in\Lambda^{1}-\\{0\\}$ and $\alpha\in\Lambda^{2},$ if
$\lambda\cdot\alpha\in E\cdot\Lambda^{1},$ then there exists $x\in E$ and
$y\in\Lambda^{1}$ such that $\alpha=\lambda\cdot y+x.$ This means (10) is
exact at $\Lambda^{2}/E.$ It follows from (11) that (10) is exact at
$\Lambda^{3}/(E\cdot\Lambda^{1}).$

We next consider an equivalence relation between subspaces of exterior
algebras. Suppose $V_{1},V_{2}$ are two linear spaces and
$F_{i}\subset\Lambda^{*}(V_{i}),i=1,2$ are two linear subspaces. We say that
$F_{1}$ is isomorphic to $F_{2}$ and denote by $F_{1}\cong F_{2}$ if there
exists an isomorphism of algebras
$f:\Lambda^{*}(V_{1})\rightarrow\Lambda^{*}(V_{2})$ such that
$F_{2}=f(F_{1}).$

######  Lemma 9.

Let $f$ be an isomorphism from $\Lambda^{*}(V)$ to $\Lambda^{*}(W).$ Then a
subspace $E$ (resp. pair $(F,E)$) of $\Lambda^{*}(V)$ is complete prime if and
only if $f(E)$ (resp. pair $(f(F),f(E))$) in $\Lambda^{*}(W)$ is complete
prime.

###### Proof.

The isomorphism $f$ induces a family of isomorphisms

$f_{i}:\Lambda^{k+i}(V)/(E\cdot\Lambda^{i}(V))\rightarrow\Lambda^{k+i}(W)/(f(E)\cdot\Lambda^{i}(W)).$

Then for every $\lambda\in\Lambda^{1}(V)-\\{0\\},$ we have an isomorphism
between complexes

$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda\cdot}$$\textstyle{\Lambda^{k-1}(V)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\lambda\cdot}$$\textstyle{\Lambda^{k}(V)/E\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{0}}$$\scriptstyle{\lambda\cdot}$$\textstyle{\Lambda^{k+1}(V)/(E\cdot\Lambda^{1}(V))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\scriptstyle{\lambda\cdot}$$\textstyle{\cdots}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f(\lambda)\cdot}$$\textstyle{\Lambda^{k-1}(W)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f(\lambda)\cdot}$$\textstyle{\Lambda^{k}(W)/f(E)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f(\lambda)\cdot}$$\textstyle{\Lambda^{k+1}(W)/(f(E)\cdot\Lambda^{1}(W))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f(\lambda)\cdot}$$\textstyle{\cdots.}$

The two families of homology groups are isomorphism respectively. Hence,
$(E^{\bullet},\lambda)$ is exact if and only if $(f(E)^{\bullet},f(\lambda))$
is exact. Same proof for pair $(E,F).$ ∎

######  Corollary 10.

Let $G$ be a subgroup of $\mathrm{SO}(n)$ and let $E,F$ be two $G-$invariant
subspaces of $\Lambda^{*}(\mathbb{R}^{n}).$ If $G$ acts on $S^{n-1}$
transitively, then $E$ (resp. $(F,E)$) is complete prime if and only if
$(E^{\bullet},\lambda)$ (reps. $(F,E^{\bullet},\lambda)$) is exact for some
$\lambda\in\Lambda^{1}(\mathbb{R}^{n})-\\{0\\}.$

######  Definition 11.

Let $M$ be a smooth manifold. A subbundle $E$ of $\Lambda^{k}(M)$ is _perfect_
if $E_{x}\cong E_{y}$ for all $x,y\in M.$ Further, $E$ is prime (resp.
complete prime) if $E$ is perfect and $E_{x}\subset\Lambda^{k}(T_{x}^{*}M)$ is
prime (resp. complete prime) for some $x\in M.$ A bundle pair $F,E$ is
complete prime if $E,F$ are both perfect and $(F_{x},E_{x})$ is complete prime
for some $x\in M.$

######  Remark 12.

In general, the rank of $E\cdot\Lambda^{r}$ depends on the algebraic structure
of $E.$ If $E$ is perfect, then $E\cdot\Lambda^{r}(M)$ is also perfect. As a
result, $E\cdot\Lambda^{r}(M)$ is a subbundle of $\Lambda^{*}(M).$

###  2.3 Prime subspace of $\Lambda^{2}(\mathbb{R}^{n})$

A prime space $E\subset\Lambda^{k}(\mathbb{R}^{n})$ is said to be _maximal_ if
there is not a prime space $E^{\prime}\subset\Lambda^{k}(\mathbb{R}^{n})$
containing $E$ except for itself. Here is a natural question that

######  Question 13.

Let $E$ be a maximal prime subspace of $\Lambda^{k}(\mathbb{R}^{n}).$ Is $E$
unique up to isomorphism? If so, what is $\mathrm{dim}\,E$?

In this subsection we only study the case $k=2.$ Consider the Grassmannian
space $\mathrm{Gr}_{2}(\mathbb{R}^{n})$ of rank two. Let $\mathbb{P}(E)$
denote the projective space of $E\subset\Lambda^{2}(\mathbb{R}^{n}).$ The
product of exterior algebra provides an embedding

$j:\mathrm{Gr}_{2}(\mathbb{R}^{n})\longrightarrow\mathbb{P}(\Lambda^{2}(\mathbb{R}^{n})),\
\ \langle\alpha,\beta\rangle\mapsto\mathbb{R}\cdot\alpha\wedge\beta.$

It is obvious that the prime condition of $E$ is equivalent to
$\mathrm{im}\,j\cap\mathbb{P}(E)=\emptyset.$

Given a linear space $V$ and a subset $S\subset V,$ let $\langle x|x\in
S\rangle$ denote the linear subspace generated by elements of $S.$ Fix an
embedding $\mathbb{R}^{n}\hookrightarrow\mathbb{R}^{n+1}$ and suppose
$\Lambda^{1}(\mathbb{R}^{n+1})=\Lambda^{1}(\mathbb{R}^{n})\oplus\langle\bm{e}\rangle.$
As an algebra, $\Lambda^{*}(\mathbb{R}^{n+1})$ is generated by
$\Lambda^{*}(\mathbb{R}^{n})$ and $\bm{e}.$ Given a prime subspace $E$ of
$\Lambda^{2}(\mathbb{R}^{n}),$ one can construct a prime subspace of
$\Lambda^{2}(\mathbb{R}^{n+1})$ by adding $\bm{e}$ to $E.$ For
$0\neq\lambda\in\Lambda^{1}(\mathbb{R}^{n}),$ we denote by
$L(\lambda)=E\oplus\lambda\cdot\Lambda^{1}(\mathbb{R}^{n})$ and
$V_{\lambda,\alpha}=\langle\lambda\cdot\bm{e}+\alpha\rangle.$ We next state
several simple lemmas.

######  Lemma 14.

Suppose $E\subset\Lambda^{2}(\mathbb{R}^{n})$ is prime,
$0\neq\lambda\in\Lambda^{1}(\mathbb{R}^{n})$ and
$\alpha\in\Lambda^{2}(\mathbb{R}^{n}).$ Then $E\oplus
V_{\lambda,\alpha}\subset\Lambda^{2}(\mathbb{R}^{n+1})$ is prime if and only
if $\alpha\notin L(\lambda).$

###### Proof.

Assume $E\oplus V_{\lambda,\alpha}$ is not prime, i.e.,
$\lambda\cdot\bm{e}+\alpha+\beta$ is a product of two one-forms for some
$\beta\in E.$ Then $\alpha+\beta\in\lambda\wedge\Lambda^{1}(\mathbb{R}^{n})$
which is equivalent to $\alpha\in L(\lambda).$ ∎

######  Lemma 15.

Let $E$ be a maximal prime subspace of $\Lambda^{2}(\mathbb{R}^{n}).$ Suppose
$0\neq\lambda\in\Lambda^{1}(\mathbb{R}^{n})$ and
$\alpha_{1},\alpha_{2}\in\Lambda^{2}(\mathbb{R}^{n})-L(\lambda).$ Then
$E+V_{\lambda,\alpha_{1}}+V_{\lambda,\alpha_{2}}\subset\Lambda^{2}(\mathbb{R}^{n+1})$
is prime if and only if $\alpha_{1}-\alpha_{2}\in E.$ At this moment, $E\oplus
V_{\lambda,\alpha_{1}}=E\oplus V_{\lambda,\alpha_{2}}.$

###### Proof.

Assume $E+V_{\lambda,\alpha_{1}}+V_{\lambda,\alpha_{2}}$ is prime. Then
$E+\mathbb{R}(\alpha_{1}-\alpha_{2})\subset\Lambda^{2}(\mathbb{R}^{n})$ is
prime. Since $E$ is maximal, we have $\alpha_{1}-\alpha_{2}\in E.$ Sufficiency
is obvious. ∎

Let $MP(n)$ denote the dimension of a maximal prime subspace of
$\Lambda^{2}(\mathbb{R}^{n}).$ If a prime space $E$ satisfies
$L(\lambda)=\Lambda^{2}(\mathbb{R}^{n}),$ or equivalently,
$\mathrm{dim}\,E=\frac{1}{2}(n-1)(n-2),$ then $E$ can not be extended to a
greater one in $\Lambda^{2}(\mathbb{R}^{n+1}).$ This leads the corollary as
follows.

######  Corollary 16.

If $MP(n)=\frac{1}{2}(n-1)(n-2),$ Then $MP(n+1)=MP(n).$

######  Lemma 17.

Let $E$ be a maximal prime subspace of $\Lambda^{2}(\mathbb{R}^{n}).$ Suppose
$\lambda_{1},\cdots,\lambda_{s}\in\Lambda^{1}(\mathbb{R}^{n})$ are linearly
independent and $\alpha\in\Lambda^{2}(\mathbb{R}^{n}).$ Then $E\oplus
V_{\lambda_{1},\alpha}\oplus\cdots\oplus
V_{\lambda_{s},\alpha}\subset\Lambda^{2}(\mathbb{R}^{n+1})$ is prime if and
only if $\alpha\notin L(\lambda)$ for all
$\lambda\in\langle\lambda_{1},\cdots,\lambda_{s}\rangle.$ In particular, if
$\alpha\notin
E+\langle\lambda_{1},\cdots,\lambda_{s}\rangle\wedge\Lambda^{1}(\mathbb{R}^{n}),$
then $E\oplus V_{\lambda_{1},\alpha}\oplus\cdots\oplus V_{\lambda_{s},\alpha}$
is prime.

###### Proof.

Now that $\lambda_{1},\cdots,\lambda_{s}$ are linearly independent, one can
see that $V_{\lambda_{1},\alpha}+\cdots+V_{\lambda_{s},\alpha}$ is a direct
sum. Assume $E\oplus V_{\lambda_{1},\alpha}\oplus\cdots\oplus
V_{\lambda_{s},\alpha}$ is prime. Then $E\oplus\langle\sum
t_{i}(\lambda_{i}\cdot\bm{e}+\alpha)\rangle$ is prime for any
$(t_{1},\cdots,t_{s})\in\mathbb{R}^{s}.$ According to Lemma 14, we have
$\alpha\notin L(\sum t_{i}\lambda_{i}).$ This implies $\alpha\notin
L(\lambda)$ for all $\lambda\in\langle\lambda_{1},\cdots,\lambda_{s}\rangle.$
Vice versa. ∎

######  Corollary 18.

If $MP(n)+\sum_{i=1}^{s}(n-i)<\frac{1}{2}n(n-1),$ then $MP(n+1)\geq MP(n)+s.$

In fact, the lower bound for $MP(n)$ in Corollary 18 is too loose. We use a
subalgebra of $\mathfrak{so}(n)$ to give another lower bound. When $n=2m$ is
even, consider the special unitary group $\mathrm{SU}(m)$ embedding into
$\mathrm{SO}(2m)$ by

$A+B\cdot\sqrt{-1}\mapsto\begin{pmatrix}A&-B\\\ B&A\end{pmatrix}.$

This induces an injective homomorphism
$\mathfrak{su}(m)\rightarrow\mathfrak{so}(2m)$ as

$a+b\cdot\sqrt{-1}\mapsto\begin{pmatrix}a&-b\\\ b&a\end{pmatrix},$

where $a+a^{T}=0,b=b^{T}$ and $\mathrm{trace}(b)=0.$ It is easy to see that
$\mathrm{rank}\,\alpha\geq 4$ for any non-zero
$\alpha\in\mathfrak{su}(m)\subset\mathfrak{s0}(2m).$ Thus,

######  Proposition 19.

For $n\geq 4,$ we have $MP(n)\geq[\frac{n}{2}]^{2}-1.$

##  3 Generalized de Rham complex

Let $(M,g)$ be a Riemannian manifold with a metric $g.$ Suppose that $E$ is a
subbundle of $\Lambda^{k}(M),2\leq k\leq n-2.$ There is an orthogonal
decomposition $\Lambda^{k}(M)=E\oplus E^{\perp}$ with respect to $g.$ Let $p$
be the projection from $\Lambda^{k}(M)$ to $E^{\perp}.$ For the ease of
notations, we use $\Lambda^{k}$ for $\Lambda^{k}(M)$ in this section. We study
the two kinds of complexes (2) and (3). Type I was first mentioned in [4]. We
will seek the elliptic condition.

Let $G$ be a subgroup of $\mathrm{SO}(n)$ with a Lie algebra $\mathfrak{g}.$
Recall from [6] that a $G-$structure $Q$ on $M$ is integrable if it is locally
isomorphic to a flat $G-$structure, i.e., for every point $x\in M,$ there is
an open set $U$ of $x$ such that $Q|_{U}$ has a local section
$(V_{1},\cdots,V_{n})$ consisting of commuting vector fields. Obviously, if
$Q$ is integrable then the associated bundle
$\Lambda(\mathfrak{g}):=Q\times_{G}\mathfrak{g}$ is locally flat, i.e., for
every $x\in M,$ there exists an open set $U$ of $x$ and closed forms
$\alpha_{1},\cdots,\alpha_{s}$ as a $C^{\infty}(U)-$basis for
$\Gamma(U,\Lambda(\mathfrak{g})).$ We introduce a concept of locally quasi-
flat which is weaker than locally flat.

######  Definition 20.

We say that a subbundle $F$ of $\Lambda^{k}(M)$ is _locally quasi-flat_ if
$d\alpha\in\Gamma(M,F\cdot\Lambda^{1})$ for any $\alpha\in\Gamma(M,F).$

As a direct consequence of definition, we have

######  Lemma 21.

If $F\subset\Lambda^{k}(M)$ is locally quasi-flat, then $F\cdot\Lambda^{1}$ is
also locally quasi-flat.

Let $E$ be a locally quasi-flat subbundle of $\Lambda^{k}(M)$ and define a
list of subbundles $\\{E_{i}\\}_{i=0}^{r}$ of $\Lambda^{k+i}$ by
$E_{i}=E\cdot\Lambda^{i}.$ Lemma 21 shows that every $E_{i}$ is locally quasi-
flat. We claim that the differential in (2) and (3) satisfy $D_{i+1}D_{i}=0.$
To see this, for $\alpha\in\Gamma(M,E_{i-1}^{\perp}),$ set
$d\alpha=u_{1}+u_{2}$ with $u_{1}\in\Gamma(M,E_{i})$ and
$u_{2}\in\Gamma(M,E_{i}^{\perp}).$ Using $du_{1}+du_{2}=0,$ we have
$du_{2}=-du_{1}\in\Gamma(M,E_{i}\cdot\Lambda^{1})=\Gamma(M,E_{i+1}).$ Then
$D_{i+1}D_{i}\alpha=p_{i+1}dp_{i}d\alpha=p_{i+1}du_{2}=0.$

A simple calculation shows that (7) and (8) are principal symbols of (2) and
(3) respectively. According to the elliptic complex theory [1], a complex is
elliptic if and only if the sequence of its principal symbols is exact outside
the zero section. So far we obtain the following theorem.

######  Theorem 22.

Suppose $M$ is a closed Riemannian manifold, $E$ is a locally quasi-flat
subbundle of $\Lambda^{k}(M)$ and $F$ is a subbundle of $\Lambda^{k-1}(M).$ If
$E$ is complete prime, then $(E,D)$ is an elliptic complex. Similarly, if
$(F,E)$ is complete prime, then $(F,E,D)$ is an elliptic complex.

Specially, we have

######  Corollary 23.

Let $M$ be a closed Riemannian $n-$manifold and $E$ a subbundle of
$\Lambda^{k}(M).$ Then the complex

$\displaystyle 0\longrightarrow A^{0}(M)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}A^{1}(M)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\cdots\stackrel{{\scriptstyle
d}}{{\longrightarrow}}A^{k-1}(M)\stackrel{{\scriptstyle
pd}}{{\longrightarrow}}\Gamma(M,E^{\perp})\longrightarrow 0$ (12)

is elliptic if and only if
$\mathrm{rank}\,E=\sum_{i=0}^{k}(-1)^{k+i}\binom{n}{i}$ and $E$ is prime.

###### Proof.

Fix a point $x\in M.$ For any $\lambda\in\Lambda^{1}(T_{x}^{*}M)-\\{0\\},$
consider the complex

$\displaystyle
0\longrightarrow\Lambda^{0}\stackrel{{\scriptstyle\lambda}}{{\longrightarrow}}\Lambda^{1}\stackrel{{\scriptstyle\lambda}}{{\longrightarrow}}\cdots\stackrel{{\scriptstyle\lambda}}{{\longrightarrow}}\Lambda^{k-1}\stackrel{{\scriptstyle
p(\lambda\cdot\ )}}{{\longrightarrow}}E^{\perp}\rightarrow 0.$ (13)

It is obvious that (13) is exact at $\Lambda^{i}$ for $i\leq k-2.$ By (6) we
know that (13) is exact at $\Lambda^{k-1}$ and $E^{\perp}$ for all $\lambda$
if and only if $E$ is prime and

$(-1)^{k}\mathrm{rank}\,E^{\perp}+\sum_{i=0}^{k-1}(-1)^{i}\mathrm{rank}\,\Lambda^{i}(M)=0.$

The proof is completed. ∎

######  Example 24.

Consider a closed, oriented Riemannian $4n-$manifold $(M,g)$. The Hodge star
$*$ operator induces a splitting of the space of $2n-$forms
$\Lambda^{2n}=\Lambda^{+}\oplus\Lambda^{-}$ into the subspace of self-dual
forms $\Lambda^{+}$ and anti-self-dual forms $\Lambda^{-}.$ According to
Example 4, $\Lambda^{+}$ and $\Lambda^{-}$ are both prime. In addition,

$\mathrm{rank}\,\Lambda^{\pm}=\frac{1}{2}\binom{4n}{2n}=\sum_{i=0}^{2n}(-1)^{i}\binom{4n}{i}.$

Apply Corollary 23, the complex

$A^{0}(M)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}A^{1}(M)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\cdots\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\Lambda^{2n-1}(M)\stackrel{{\scriptstyle
pd}}{{\longrightarrow}}\Lambda^{\pm}\rightarrow 0$

generated by $\Lambda^{\mp}$ is elliptic.

##  4 Complexes and adjoint representations

In this section we mainly review Reyes-Carrion’s results in [4]. Let $M$ be an
oriented Riemannian manifold of dimension $n$ and let $\mathrm{SO}(TM)$ be the
oriented orthonormal frame bundle of $TM.$ The associated adjoint bundle
$\mathfrak{so}(TM)$ of $\mathrm{SO}(TM)$ is isomorphic to $\Lambda^{2}(M)$
canonically. Suppose that $G$ is a subgroup of $\mathrm{SO}(n)$ and $Q$ is a
$G-$structure [6] on $M,$ i.e., $Q$ can be locally expressed as
$(V_{1},\cdots,V_{n})\cdot G$ for some local oriented orthonormal frame
$(V_{1},\cdots,V_{n}).$ There is a subbundle $\Lambda(\mathfrak{g})$ of
$\Lambda^{2}(M)$ related to $Q$ as

$\textstyle{\ \mathrm{ad}Q\
\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\mathfrak{so}(TM)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\
\Lambda(\mathfrak{g})\
\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda^{2}(M),}$

where $\mathfrak{g}$ is the Lie algebra of $G$ and the two isomorphisms are
canonical in the above commutative diagram. If we let
$(V_{1}^{*},\cdots,V_{n}^{*})$ be the dual frame of $(V_{1},\cdots,V_{n}),$
then every section $\xi$ of $\Lambda(\mathfrak{g})$ can be locally written as
$\xi=\sum_{i,j}a_{ij}V_{i}^{*}\wedge V_{j}^{*},$ where $(a_{ij})$ is a local
function taking values in $\mathfrak{g}.$ The embedding
$\imath:G\hookrightarrow\mathrm{SO}(n)$ gives an isomorphism $TM\cong
Q\times_{\imath}\mathbb{R}^{n}.$

######  Proposition 25.

Let $Q$ be a $G-$structure on $M.$ Then $\Lambda(\mathfrak{g})$ is a prime
subbundle of $\Lambda^{2}(M)$ if and only if $\mathrm{rank}\,\eta\geq 4$ for
all $\eta\in\mathfrak{g}-\\{0\\}.$

###### Proof.

Let $\mathcal{G}(Q)$ be the gauge group of $Q.$ For any $x\in M,$ we have

$\textstyle{\ \mathcal{G}(Q)_{x}\
\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\scriptstyle{\cong}$$\textstyle{\mathrm{SO}(T_{x}^{*}M)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\
G\
\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{SO}(n),}$

the two isomorphisms in the above commutative diagram are both unique up to
conjugacy. Observe that $\mathrm{rank}\,A=\mathrm{rank}\,P^{-1}AP$ for all
$A\in\mathfrak{so}(n),P\in\mathrm{SO}(n),$ we obtain that
$\mathrm{rank}\,\eta=\mathrm{rank}\,j_{*}^{-1}(\eta)$ for
$\eta\in\mathfrak{g},$ where
$j_{*}:\Lambda(\mathfrak{g})=\mathrm{ad}Q_{x}\rightarrow\mathfrak{g}$ is the
homomorphism induced by $j.$ According to Example 2, we know that
$\mathrm{rank}\,\eta\geq 4$ if and only if $j_{*}^{-1}(\eta)$ is a prime two-
form. This completes the proof. ∎

Roughly speaking, the smaller the group $G,$ the finer the $G-$structure. Let
$N(G)$ denotes the normalizer of $G$ in $\mathrm{SO}(n).$ One only need to
consider $N(G)-$structures rather than $G-$structures due to the fact that an
$N(G)-$structure $Q$ induces a subbundle
$\hat{\mathfrak{g}}:=Q\times_{N(G)}\mathfrak{g}$ of $\Lambda^{2}.$ Reyes-
Carrion [4] fully discusses the complexes associated with special
$N(G)-$structures. He showed that: If the holonomy of the Levi-Civita
connection reduces $N(G)$ then $\hat{\mathfrak{g}}$ is locally quasi-flat.
Futher, if $G$ is one of
$\mathrm{SU}(2)_{+},\mathrm{SU}(m),\mathrm{Sp}(m),G_{2}$ and
$\mathrm{Spin}(7),$ then $\hat{\mathfrak{g}}$ is complete prime. For a general
reference on special holonomy groups we refer the reader to [15] and [12]. We
next briefly recall two cases for $N(G)=\mathrm{U}(n)$ and
$\mathrm{Sp}(n)\mathrm{Sp}(1)$ that constructed in [4]. For the cases
$N(G)=G_{2}$ and $\mathrm{Spin}(7)$ we will give in §lastsectcr.

######  Example 26 (U(n)-structure).

Let $(M,J)$ be a almost complex $n-$manifold with almost complext structure
$J.$ The complexified vector bundle
$\Lambda_{\mathbb{C}}^{k}(M)=\Lambda^{k}(M)\otimes_{\mathbb{R}}\mathbb{C}$
decomposes as a direct sum of complex vector bundle

$\Lambda_{\mathbb{C}}^{k}(M)=\bigoplus_{p+q=k}\Lambda^{p,q}.$

Define the real vector bundles $[\Lambda^{p,q}]$ for $p\neq q$ and
$[\Lambda^{p,p}]$ by

$[\Lambda^{p,q}]\otimes_{\mathbb{R}}\mathbb{C}=\Lambda^{p,q}\oplus\Lambda^{q,p},\
\ [\Lambda^{p,p}]\otimes_{\mathbb{R}}\mathbb{C}=\Lambda^{p,p}.$

The representation of $U(n)$ on $\Lambda^{2}$ splits into irreducible
representations

$\Lambda^{2}=[\Lambda^{2,0}]\oplus[\Lambda_{0}^{1,1}]\oplus\langle\omega\rangle,$

where $\omega$ is a invariant two-form under the action of $U(n)$ and
$[\Lambda_{0}^{1,1}]$ is the orthogonal complement of $\langle\omega\rangle$
in $[\Lambda^{1,1}],$ which can be identified with $\mathfrak{su}(n).$ Then
the complex

$0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\langle\omega\rangle\oplus[\Lambda^{2,0}]\rightarrow[\Lambda^{3,0}]\rightarrow[\Lambda^{4,0}]\rightarrow\cdots\rightarrow[\Lambda^{n,0}]\rightarrow
0$

associated with the $U(n)-$structure is elliptic.

######  Example 27 ($\mathrm{Sp}(n)\mathrm{Sp}(1)-$structure).

Some references are Salamon [14, 15]. Let $(J_{1},J_{2},J_{3})$ be an almost
hyperkähler structure on $\mathbb{R}^{4n},n\geq 2$ and let
$\omega_{1},\omega_{2},\omega_{3}$ be the Hermitian forms of $J_{1},J_{2}$ and
$J_{3}.$ Acting on $\Lambda^{4}(\mathbb{R}^{4n}),$ the group
$\mathrm{Sp}(n)\mathrm{Sp}(1)=\mathrm{Sp}(n)\times\mathrm{Sp}(1)/\mathbb{Z}_{2}$
is the stabilizer of the $4-$form

$\Omega=\omega^{1}\wedge\omega^{1}+\omega^{2}\wedge\omega^{2}+\omega^{3}\wedge\omega^{3}.$

Let $(M,g)$ be a compact quaternionic Kähler manifold of dimension $4n\geq 8.$
The representation on the cotangent space has the form

$\Lambda^{1}\cong[V\otimes_{\mathbb{C}}H],$

where $V$ arises from the basic representation $\mathbb{C}^{2n}$ of
$\mathrm{Sp}(n)$ and $H$ arises from the basic representation $\mathbb{C}^{2}$
of $\mathrm{Sp}(1).$ Remark that $V$ and $H$ are not global but
$V\otimes_{\mathbb{C}}H$ is global. The two-form bundle of $M$ splits into

$\Lambda^{2}=\mathfrak{sp}(n)\oplus[\Lambda^{2}V\otimes S^{2}H],$

where $S^{2}H$ denotes the symmetric tensor product of $H.$ Then the complex

$0\rightarrow\Lambda^{0}\rightarrow[V\otimes H]\rightarrow[\Lambda^{2}V\otimes
S^{2}H]\rightarrow[\Lambda^{3}V\otimes
S^{3}H]\rightarrow\cdots\rightarrow[\Lambda^{2n}V\otimes S^{2n}H]\rightarrow
0$

associated with the $\mathrm{Sp}(n)\mathrm{Sp}(1)-$structure is elliptic.

##  5 Complexes and representations

Inspired by D. Salamon [15] and Reyes-Carrion [4], we use representations of
holonomy group to construct complete prime bundles.

Let $G$ be a subgroup of $\mathrm{SO}(n)$ and $Q$ a $G-$structure on a closed
$n-$manifold $M.$ Such a $G-$structure induces an orientation of $M$ and a
Riemannian metric $g$ with $\mathrm{Hol}(g)\subset G.$ Notice that the basic
action of $G$ on $\mathbb{R}^{n}$ induces a representation on
$\Lambda^{*}(\mathbb{R}^{n}).$ We write
$\Lambda^{k}(\mathbb{R}^{n})=\oplus_{i\in I_{k}}W_{i}^{k},$ where each
$W_{i}^{k}\subset\Lambda^{k}$ is an irreducible representation of $G$ and
$I_{k}$ is a finite indexing set. The Hodge star $*$ gives an isometry between
$\Lambda^{k}$ and $\Lambda^{n-k}.$ Let $\Lambda_{i}^{k}=Q\times_{G}W_{i}^{k}.$
Then there is a natural decomposition [12, Proposition 3.5.1]

$\displaystyle\Lambda^{*}(M)=\bigoplus_{k=0}^{n}\bigoplus_{i\in
I_{k}}\Lambda_{i}^{k}.$ (14)

In fact, each $\Lambda_{i}^{k}$ can be constructed by parallel translation.
Let $\mathcal{E}$ be a $G-$invariant subspace of
$\Lambda^{k}(\mathbb{R}^{n}).$ The vector bundle $\Lambda^{*}(M)$ carries a
connection $\nabla$ induced by the Levi-Civita connection of $g.$ Fix a base
point $x\in M$ and identify $T_{x}^{*}M$ with $\mathbb{R}^{n}.$ For each $y\in
M$ and every path $\gamma:[0,1]\rightarrow M$ with $\gamma(0)=x,\gamma(1)=y,$
the parallel transport along $\gamma$ defines an isomorphism
$P_{\gamma}:\Lambda^{*}(T_{x}^{*}M)\rightarrow\Lambda^{*}(T_{y}^{*}M).$ We
claim that the subspace
$P_{\gamma}(\mathcal{E})\subset\Lambda^{k}(T_{y}^{*}M)$ is independent of
$\gamma.$ To see this, consider another path $\gamma^{\prime}\in
C^{\infty}([0,1],M)$ connecting $x$ and $y,$ the loop
$\gamma^{\prime-1}\circ\gamma$ through $x$ determines a transformation
$P_{\gamma^{\prime}}^{-1}\circ P_{\gamma}\in\mathrm{Hol}(g)$ and hence
$P_{\gamma^{\prime}}^{-1}\circ P_{\gamma}(\mathcal{E})=\mathcal{E}$ which
means $P_{\gamma}(\mathcal{E})=P_{\gamma^{\prime}}(\mathcal{E}).$ Then we get
a vector bundle $E=\bigcup_{y\in M}E_{y},$ where
$E_{y}=P_{\gamma}(\mathcal{E}).$

Let $\\{e_{i}\\}$ be an orthogonal frame field on $M$ and $\\{\omega^{i}\\}$
the dual orthogonal frame field. A well known fact is that the differential
can be written as $d=\sum_{i=1}^{n}\omega^{i}\wedge\nabla_{e_{i}}.$ It is
important to note that $\nabla$ can reduce to a connection on $E$ since
$\mathrm{Hol}(g)$ preserves $E.$ That is, for all $\xi\in\Gamma(E),$ we have
$\nabla\xi\in\Gamma(E\otimes\Lambda^{1}).$ This implies that $E$ is locally
quasi-flat. Besides, since parallel transport along a path induces an
isomorphism between algebras, then $E$ (resp. $(F,E)$) is complete prime if
and only if $\mathcal{E}$ (resp. $(\mathcal{F},\mathcal{E})$) is complete
prime. Using Theorem 22, we have the following theorem

######  Theorem A.

Let $G$ a connected subgroup of $\mathrm{SO}(n)$ and let $(M,g,G,Q)$ be a
closed Riemannian $n-$manifold with a compatible torsion-free $G-$ structure
$Q.$ Suppose $\mathcal{F}\subset\Lambda^{k-1}(\mathbb{R}^{n})$ and
$\mathcal{E}\subset\Lambda^{k}(\mathbb{R}^{n})$ are two invariant subspaces
under the action of $G.$ Set $F=Q\times_{G}\mathcal{F}\subset\Lambda^{k-1}(M)$
and $E=Q\times_{G}\mathcal{E}\subset\Lambda^{k}(M).$ Then $E$ is locally
quasi-flat. Consequently,

  * (i)

If $\mathcal{E}$ is complete prime, then the sequence $(E,D)$ is an elliptic
complex.

  * (ii)

If $(\mathcal{F},\mathcal{E})$ is complete prime, then the sequence $(F,E,D)$
is an elliptic complex.

We next discuss two examples with exceptional holonomy groups. For more
details on compact manifolds with exceptional holonomy, see [15] and [11].

###  $G_{2}-$manifold

Let $(x_{1},x_{2},\cdots,x_{7})$ be coordinates on $\mathbb{R}^{7}.$ For
simplicity write $\omega^{ij\cdots l}$ to denote the product $dx_{i}\wedge
dx_{j}\wedge\cdots\wedge dx_{l}.$ Define a $G_{2}-$invariant $3-$form on
$\mathbb{R}^{7}$ by

$\varphi=\omega^{123}+\omega^{145}+\omega^{167}+\omega^{246}-\omega^{257}-\omega^{347}-\omega^{356}.$

It follows from [15, Lemma 11.4] that the representation
$\Lambda^{*}(\mathbb{R}^{7})$ of $G_{2}$ splits orthogonally into components
as

$\displaystyle\Lambda^{1}=\Lambda_{7}^{1},\ \
\Lambda^{2}=\Lambda_{7}^{2}\oplus\Lambda_{14}^{2},\ \
\Lambda^{3}=\Lambda_{1}^{3}\oplus\Lambda_{7}^{3}\oplus\Lambda_{27}^{3},$
$\displaystyle\Lambda^{4}=\Lambda_{1}^{4}\oplus\Lambda_{7}^{4}\oplus\Lambda_{27}^{4},\
\ \Lambda^{5}=\Lambda_{7}^{5}\oplus\Lambda_{14}^{5},\ \
\Lambda^{6}=\Lambda_{7}^{6},$

where $\Lambda_{l}^{k}$ corresponds to an irreducible representation of
$G_{2}$ of dimension $l.$ The prime forms $\varphi$ and $*\varphi$ induce
canonical isomorphisms between $\Lambda_{7}^{k}$ for $k=1,\cdots,6$ by

$\displaystyle\Lambda_{7}^{2}=*(*\varphi\cdot\Lambda^{1}),\
\Lambda_{7}^{3}=*(\varphi\Lambda^{1}),\ \Lambda_{7}^{4}=\varphi\Lambda^{1},\
\Lambda_{7}^{5}=*\varphi\cdot\Lambda^{1},\ \Lambda_{7}^{6}=*\Lambda^{1}.$ (15)

######  Proposition 28.

These spaces $\Lambda_{i}^{k}$ satisfy the following relations

$\displaystyle\Lambda_{7}^{2}\cdot\Lambda^{1}=\Lambda^{3},\ \
\Lambda_{14}^{2}\cdot\Lambda^{1}=\Lambda_{7}^{3}\oplus\Lambda_{27}^{3},$
$\displaystyle\Lambda_{7}^{3}\cdot\Lambda^{1}=\Lambda^{4},\ \
\Lambda_{27}^{3}\cdot\Lambda^{1}=\Lambda_{7}^{4}\oplus\Lambda_{27}^{4},$
$\displaystyle\Lambda_{7}^{4}\cdot\Lambda^{1}=\Lambda_{27}^{4}\cdot\Lambda^{1}=\Lambda^{5},$
$\displaystyle\Lambda_{7}^{5}\cdot\Lambda^{1}=\Lambda_{14}^{5}\cdot\Lambda^{1}=\Lambda^{6}.$

We introduce an important lemma about $G_{2}.$ See [3] or [13].

######  Lemma 29 ([3, Theorem III]).

Let $\mathbb{R}^{7}$ be the basic irreducible of $G_{2}.$ Then $G_{2}$ acts on
$S^{6}$ transitively.

Using Lemma 29 and Lemma 5, we have

######  Proposition 30.

All $G_{2}-$invariant prime subspaces in $\Lambda^{*}(\mathbb{R}^{7})$ are as
follows

$\Lambda_{7}^{2},\ \ \Lambda_{14}^{2},\ \ \Lambda_{1}^{3},\ \
\Lambda_{7}^{3},\ \ \Lambda_{1}^{3}\oplus\Lambda_{7}^{3},\ \ \Lambda_{1}^{4}.$

Reyes-Carrion showed in [4] that the complex $(\Lambda_{14}^{2},D)$ generated
by the adjoint representation of $G_{2}$ is elliptic. We develop Theorem B
that lists all elliptic complexes on a $G_{2}-$manifold generated by
$G_{2}-$invariant subspaces of $\Lambda^{*}(\mathbb{R}^{7}).$

######  Theorem B.

Let $(M,g,\varphi)$ be a compact $G_{2}-$manifold. Then the following
complexes on $M$ are elliptic

$\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda_{7}^{4}\oplus\Lambda_{27}^{4}\rightarrow\Lambda_{14}^{5}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda_{7}^{2}\rightarrow\langle\varphi\rangle\oplus\Lambda_{7}^{3}\rightarrow\langle*\varphi\rangle\rightarrow
0,$ $\displaystyle
0\rightarrow\langle\varphi\rangle\oplus\Lambda_{7}^{3}\rightarrow\langle*\varphi\rangle\oplus\Lambda_{7}^{4}\rightarrow
0.$

The first one is of type I generated by $\langle*\varphi\rangle.$ The last two
are of type II. Moreover, if $\mathrm{Hol}(g)=G_{2},$ except for the above
three, $(\Lambda_{14}^{2},D)$ and their dual complexes, there is no other
generalized de Rham complexes on $M.$

Proof of Theorem B.
We divide the proof into three parts. By (15), let $\\{e_{i}^{k}\\}_{i=1}^{7}$
be a basis for $\Lambda_{7}^{k},k=2,3,4,5,$ where
$e_{i}^{2}=*(*\varphi\cdot\omega^{i}),e_{i}^{3}=*(\varphi\cdot\omega^{i}),e_{i}^{4}=\varphi\cdot\omega^{i},e_{i}^{5}=(*\varphi)\cdot\omega^{i}.$

  * (1)

According to Theorem A, $(\langle*\varphi\rangle,D)$ is elliptic if we show
that $\langle*\varphi\rangle$ is complete prime. Using Lemma 29 and Corollary
10, it suffices to show that the complex

$\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda_{7}^{4}\oplus\Lambda_{27}^{4}\rightarrow\Lambda_{14}^{5}\rightarrow
0$ (16)

is exact for some $\lambda\in\Lambda^{1}-\\{0\\}.$ Set $\psi=*\varphi.$
Observe that the alternating sum of dimension of (16) equals zero. It suffices
to show that
$p(\lambda\cdot):\Lambda_{7}^{4}\oplus\Lambda_{27}^{4}\rightarrow\Lambda_{14}^{5}$
is surjective for some $\lambda\in\Lambda^{1}-\\{0\\}.$ This is equivalent to
$\mathrm{dim}\,\lambda\Lambda^{4}\cap\Lambda_{7}^{5}=1$ since
$\lambda(\Lambda_{7}^{4}\oplus\Lambda_{27}^{4})=\lambda\Lambda^{4}.$

Claim: Take $\lambda=\omega^{1},$ we have
$\omega^{1}\Lambda^{4}\cap\Lambda_{7}^{5}=\langle\omega^{1}\psi\rangle.$
Assume $\beta=\sum_{I}b_{I}\omega^{I}\in\Lambda^{4}$ such that
$\omega^{1}\beta\in\Lambda_{7}^{5},$ where the index $I$ does not contain $1.$
The basis $\\{e_{i}^{5}\\}_{i=1}^{7}$ for $\Lambda_{7}^{5}$ is given by

$\displaystyle e_{1}^{5}=\omega^{14567}+\omega^{12367}+\omega^{12345},\ \
e_{2}^{5}=\omega^{24567}-\omega^{12357}+\omega^{12346},$ $\displaystyle
e_{3}^{5}=\omega^{34567}-\omega^{12356}-\omega^{12347},\ \
e_{4}^{5}=\omega^{23467}+\omega^{13457}-\omega^{12456},$ $\displaystyle
e_{5}^{5}=\omega^{23567}+\omega^{13456}-\omega^{12457},\ \
e_{6}^{5}=\omega^{23456}-\omega^{13567}+\omega^{12467},$ $\displaystyle
e_{7}^{5}=\omega^{23457}-\omega^{13467}-\omega^{12567}.$

Set $\omega^{1}\beta=\sum_{i=1}^{7}a_{i}e_{i}^{5}.$ We can see immediately
that $a_{i}=0$ for $i\geq 2.$ Hence,

$\omega^{1}\beta=a_{1}(\omega^{14567}+\omega^{12367}+\omega^{12345})=a_{1}\omega^{1}\psi.$

  * (2)

We next prove that the two complexes

$\displaystyle 0\longrightarrow\Lambda_{7}^{2}\stackrel{{\scriptstyle
p_{1}\circ\lambda}}{{\longrightarrow}}\langle\varphi\rangle\oplus\Lambda_{7}^{3}\stackrel{{\scriptstyle
p_{2}\circ\lambda}}{{\longrightarrow}}\langle*\varphi\rangle\longrightarrow
0,$ (17) $\displaystyle
0\longrightarrow\langle\varphi\rangle\oplus\Lambda_{7}^{3}\stackrel{{\scriptstyle
p_{3}\circ\lambda}}{{\longrightarrow}}\langle*\varphi\rangle\oplus\Lambda_{7}^{4}\longrightarrow
0$ (18)

are exact when $\lambda=\omega^{1}.$ By calculation we have
$p_{1}(e_{1}^{2}\cdot\omega^{1})=\frac{3}{7}\varphi$ and
$p_{1}(e_{2i}^{2}\cdot\omega^{1})=\frac{1}{2}e_{2i+1}^{3},p_{1}(e_{2i+1}^{2}\cdot\omega^{1})=-\frac{1}{2}e_{2i}^{3}$
for $i=1,2,3.$ Then $p_{1}\circ\omega^{1}$ is injective. It follows from
$p_{2}(e_{1}^{3}\cdot\omega^{1})=-\frac{4}{7}*\varphi$ that
$p_{2}\circ\omega^{1}$ is surjective. Hence (17) is exact. Similarly,
$p_{3}\circ\omega^{1}$ is isomorphic since
$p_{3}(\varphi\cdot\omega^{1})=e_{1}^{4},p_{3}(e_{1}^{3}\cdot\omega^{1})=-\frac{4}{7}*\varphi,p_{3}(e_{2i}^{3}\cdot\omega^{1})=-\frac{1}{2}e_{2i+1}^{4}$
and $p_{3}(e_{2i+1}^{3}\cdot\omega^{1})=\frac{1}{2}e_{2i}^{4}$ for $i=1,2,3.$
Hence (18) is exact.

  * (3)

When $\mathrm{hol}(g)=G_{2},$ we explain that
$(\Lambda_{14}^{2},D),(\Lambda_{1}^{4},D),(\Lambda_{7}^{2},\Lambda_{27}^{3},D)$
and $(\langle\varphi\rangle\oplus\Lambda_{7}^{3},\Lambda_{27}^{4})$ are the
only complexes of type I and II on $M.$ Notice that a complete prime space
must be prime and a complete prime pair $(F,E)$ implies $F$ is prime. From
Propostion 30, there are only six $G_{2}-$invariant prime spaces. According to
Proposition 28, one can check one by one that
$\Lambda_{7}^{2},\Lambda_{1}^{3},\Lambda_{7}^{3},\Lambda_{1}^{3}\oplus\Lambda_{7}^{3}$
are not complete prime spaces of type I. In addition, except for
$(\Lambda_{7}^{2},\Lambda_{27}^{3},D)$ and
$(\langle\varphi\rangle\oplus\Lambda_{7}^{3},\Lambda_{27}^{4}),$ the rest of
possible complexes of type II are
$p_{4}\lambda:\langle\varphi\rangle\rightarrow\langle*\varphi\rangle$ and
$p_{5}\lambda:\Lambda_{7}^{3}\rightarrow\Lambda_{7}^{4}.$ However, they are
not isomorphic since $p_{4}\lambda=0$ and $p_{5}(e_{i}^{3}\cdot\omega^{i})=0.$
This completes the proof.

###  $\mathrm{Spin}(7)-$manifold

We review from [10, 11] some facts about $\mathrm{Spin}(7)-$structure. Let
$(x_{1},\cdots,x_{8})$ be coordinates of $\mathbb{R}^{8}.$ Define a closed,
self-dual, $\mathrm{Spin}(7)-$invariant $4-$form $\Omega$ on $\mathbb{R}^{8}$
by

$\displaystyle\Omega$ $\displaystyle=$
$\displaystyle\omega^{1234}+\omega^{1256}+\omega^{1278}+\omega^{1357}-\omega^{1368}-\omega^{1458}-\omega^{1467}$
$\displaystyle-\omega^{2358}-\omega^{2367}-\omega^{2457}+\omega^{2468}+\omega^{3456}+\omega^{3478}+\omega^{5678},$

where $\omega^{ij\cdots l}=dx_{i}\wedge dx_{j}\wedge\cdots\wedge dx_{l}.$ As a
representation of $\mathrm{Spin}(7),$ the exterior algebra
$\Lambda^{*}(\mathbb{R}^{8})$ splits orthogonally into irreducible
representations (ref. [15])

$\displaystyle\Lambda^{1}=\Lambda_{8}^{1},\ \
\Lambda^{2}=\Lambda_{7}^{2}\oplus\Lambda_{21}^{2},\ \
\Lambda^{3}=\Lambda_{8}^{3}\oplus\Lambda_{48}^{3},$
$\displaystyle\Lambda^{4}=\Lambda_{1}^{4}\oplus\Lambda_{7}^{4}\oplus\Lambda_{27}^{4}\oplus\Lambda_{35}^{4},\
\ \Lambda^{-}=\Lambda_{35}^{4},$
$\displaystyle\Lambda^{5}=\Lambda_{8}^{5}\oplus\Lambda_{48}^{5},\ \
\Lambda^{6}=\Lambda_{7}^{6}\oplus\Lambda_{21}^{6},\ \
\Lambda^{7}=\Lambda_{8}^{7}.$

where each $\Lambda_{l}^{k}$ is an irreducible representation of dimension
$l.$ We note that $\Lambda_{7}^{2}$ and $\Lambda_{21}^{2}$ are the eigenspaces
of the operator $T_{\Omega}:\Lambda^{2}\rightarrow\Lambda^{2}$ with
eigenvalues $-1$ and $3,$ respectively. Moreover, there are canonical
isomorphisms
$\Lambda_{8}^{1}\cong\Lambda_{8}^{3}\cong\Lambda_{8}^{5}\cong\Lambda_{8}^{7}$
and $\Lambda_{7}^{2}\cong\Lambda_{7}^{4}\cong\Lambda_{7}^{6}$ by a multiple of
$\Omega.$ From [15, Proposition 12.5], the subspace $\Lambda_{7}^{4}$ can be
described in the following way. Write
$\mathbb{R}^{8}=\mathbb{R}\oplus\mathbb{R}^{7}$ and let
$\Lambda_{7}^{3}(\mathbb{R}^{7})$ be the $G_{2}-$invariant subspace. Define a
map $i:\Lambda_{7}^{3}(\mathbb{R}^{7})\rightarrow\Lambda^{+}(\mathbb{R}^{8})$
by $i(\alpha)=\omega^{1}\wedge\alpha+*_{7}\alpha.$ Then $i$ is injective and
we have $\Lambda_{7}^{4}(\mathbb{R}^{8})=\mathrm{image}\,i.$

As a subgroup of $\mathrm{Spin}(7),$ the special unitary group
$\mathrm{SU}(4)$ acts on $\mathbb{R}^{8}=\mathbb{C}^{4}$ in a natural way.
Since $\mathrm{SU}(4)$ acts on $S^{7}$ transitively, we have

######  Lemma 31.

The group $\mathrm{Spin}(7)$ acts on $S^{7}$ transitively.

Using Lemma 5, we have

######  Proposition 32.

All $\mathrm{Spin}(7)-$invariant prime subspaces of
$\Lambda^{*}(\mathbb{R}^{8})$ are as follows

$\Lambda_{7}^{2},\ \Lambda_{21}^{2},\ \Lambda_{8}^{3},\ \Lambda^{-},\
\mathrm{Spin}(7)-\mbox{invariant subspaces of }\Lambda^{+},\ \Lambda_{7}^{6}.$

######  Proposition 33.

These spaces $\Lambda_{i}^{k}$ satisfy the following relations

$\displaystyle\Lambda_{7}^{2}\cdot\Lambda^{1}=\Lambda_{21}^{2}\cdot\Lambda^{1}=\Lambda^{3},$
$\displaystyle\Lambda_{8}^{3}\cdot\Lambda^{1}=(\Lambda_{27}^{4})^{\perp},\ \
\Lambda_{48}^{3}\cdot\Lambda^{1}=(\Lambda_{1}^{4})^{\perp},$
$\displaystyle\Lambda_{7}^{4}\cdot\Lambda^{1}=\Lambda^{\pm}\cdot\Lambda^{1}=\Lambda^{5},\
\ \Lambda_{27}^{4}\cdot\Lambda^{1}=\Lambda_{48}^{5},$
$\displaystyle\Lambda_{8}^{5}\cdot\Lambda^{1}=\Lambda_{48}^{5}\cdot\Lambda^{1}=\Lambda^{6}.$

Reyes-Carrion showed in [4] that the complex $(\Lambda_{21}^{2},D)$ generated
by the adjoint representation of $\mathrm{Spin}(7)$ is elliptic. We develop
Theorem C that lists all elliptic complexes on a $\mathrm{Spin}(7)-$manifold
generated by $\mathrm{Spin}(7)-$invariant subspaces of
$\Lambda^{*}(\mathbb{R}^{8}).$

######  Theorem C.

Let $(M,g,\Omega)$ be a compact $\mathrm{Spin}(7)-$manifold. Then the
following complexes on $M$ are elliptic

$\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda^{\pm}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda_{48}^{3}\rightarrow\Lambda_{27}^{4}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\oplus\Lambda^{-}\rightarrow\Lambda_{8}^{5}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda^{0}\rightarrow\Lambda^{1}\rightarrow\Lambda^{2}\rightarrow\Lambda^{3}\rightarrow\Lambda^{4}\rightarrow\Lambda^{5}\rightarrow\Lambda_{21}^{6}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda_{7}^{2}\rightarrow\Lambda_{8}^{3}\rightarrow\langle\Omega\rangle\rightarrow
0,\ \
0\rightarrow\Lambda_{21}^{2}\rightarrow\Lambda_{48}^{3}\rightarrow\Lambda_{27}^{4}\rightarrow
0,$ $\displaystyle
0\rightarrow\Lambda_{8}^{3}\rightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\rightarrow
0,\ \
0\rightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\rightarrow\Lambda_{8}^{5}\rightarrow
0.$

The first five are of type I generated by
$\Lambda^{\pm},\Lambda_{8}^{3},\Lambda_{27}^{4}$ and $\Lambda_{7}^{6}$
respectively. The last four are of type II. Morever, if
$\mathrm{Hol}(g)=\mathrm{Spin}(7),$ then except for the above nine,
$(\Lambda_{21}^{2},D)$ and their dual complexes, there is no other generalized
de Rham complexes on $M.$

Proof of Theorem C.
This proof is very analogous to that in Thereom B. It is obvious that
$\Lambda^{+},\Lambda^{-}$ and $\Lambda_{7}^{6}$ are complete prime due to
Proposition 32. We divide the proof into three parts.

  * (1)

We first proof that $\Lambda_{27}^{4}$ and $\Lambda_{8}^{3}$ are complete
prime, this is equivalent to that the two projections
$p_{1}:\omega^{1}\cdot(\langle\Omega\rangle\oplus\Lambda_{7}^{4}\oplus\Lambda^{-})\rightarrow\Lambda_{8}^{5}$
and $p_{2}:\omega^{1}\cdot\Lambda_{48}^{3}\rightarrow\Lambda_{27}^{4}$ are
surjective. Choose a basis $\\{\omega^{k}\cdot\Omega\\}_{k=1}^{8}$ for
$\Lambda_{8}^{5}.$ Observe that if $(\omega^{k}\cdot\Omega,\omega^{1ijuv})=1$
or $(\omega^{k}\cdot\Omega,\omega^{1}\cdot*\omega^{ijuv})=-1,$ then

$p_{1}(\omega^{1}(\omega^{ijuv}-*\omega^{ijuv}))=\frac{1}{7}\omega^{k}\cdot\Omega.$

Hence we have $p_{1}(\omega^{1}\cdot\Lambda^{-})=\Lambda_{8}^{5}.$ We claim
that
Claim: The projection
$p_{2}:\omega^{1}\cdot\Lambda_{48}^{3}\rightarrow\Lambda_{27}^{4}$ is
surjective.
After choosing two bases of $\Lambda_{48}^{3}$ and $\Lambda_{27}^{4},$ the map
$p_{2}$ can be written explicitly in a matrix which is full rank. It is not
difficult but is too long to give here.

  * (2)

Take $\lambda=\omega^{1},$ we next proof that the complexes

$\displaystyle 0\longrightarrow\Lambda_{7}^{2}\stackrel{{\scriptstyle
p_{3}\omega^{1}}}{{\longrightarrow}}\Lambda_{8}^{3}\stackrel{{\scriptstyle
p_{4}\omega^{1}}}{{\longrightarrow}}\langle\Omega\rangle\rightarrow 0,$ (19)
$\displaystyle 0\longrightarrow\Lambda_{8}^{3}\stackrel{{\scriptstyle
p_{5}\omega^{1}}}{{\longrightarrow}}\langle\Omega\rangle\oplus\Lambda_{7}^{4}\longrightarrow
0,$ (20) $\displaystyle
0\longrightarrow\langle\Omega\rangle\oplus\Lambda_{7}^{4}\stackrel{{\scriptstyle
p_{6}\omega^{1}}}{{\longrightarrow}}\Lambda_{8}^{5}\longrightarrow 0,$ (21)
$\displaystyle 0\longrightarrow\Lambda_{21}^{2}\stackrel{{\scriptstyle
p_{7}\omega^{1}}}{{\longrightarrow}}\Lambda_{48}^{3}\stackrel{{\scriptstyle
p_{2}\omega^{1}}}{{\longrightarrow}}\Lambda_{27}^{4}\longrightarrow 0$ (22)

are exact. Choose a basis $\\{\alpha_{i}\\}_{i=1}^{7}$ for $\Lambda_{7}^{2}$
and a basis $\\{\beta_{i}\\}_{i=1}^{7}$ for $\Lambda_{7}^{4}$ as

$\displaystyle\alpha_{1}=\omega^{12}+\omega^{34}+\omega^{56}+\omega^{78},\ \ \
\alpha_{2}=\omega^{13}-\omega^{24}+\omega^{57}-\omega^{68},$
$\displaystyle\alpha_{3}=\omega^{14}+\omega^{23}-\omega^{58}-\omega^{67},\ \ \
\alpha_{4}=\omega^{15}-\omega^{26}-\omega^{37}+\omega^{48},$
$\displaystyle\alpha_{5}=\omega^{16}+\omega^{25}+\omega^{38}+\omega^{47},\ \ \
\alpha_{6}=\omega^{17}-\omega^{28}+\omega^{35}-\omega^{46},$
$\displaystyle\alpha_{7}=\omega^{18}+\omega^{27}-\omega^{36}-\omega^{45},$
$\displaystyle\beta_{1}=\omega^{2467}+\omega^{2458}+\omega^{2368}-\omega^{2357}-\omega^{1358}-\omega^{1367}-\omega^{1457}+\omega^{1468},$
$\displaystyle\beta_{2}=\omega^{3467}+\omega^{3458}+\omega^{2378}+\omega^{2356}+\omega^{1258}+\omega^{1267}+\omega^{1456}+\omega^{1478},$
$\displaystyle\beta_{3}=\omega^{3457}-\omega^{3468}+\omega^{2478}+\omega^{2456}+\omega^{1257}-\omega^{1268}-\omega^{1356}-\omega^{1378},$
$\displaystyle\beta_{4}=\omega^{2345}-\omega^{4567}-\omega^{3568}+\omega^{2578}-\omega^{1238}-\omega^{1247}+\omega^{1346}+\omega^{1678},$
$\displaystyle\beta_{5}=\omega^{4568}-\omega^{3567}+\omega^{2678}+\omega^{2346}-\omega^{1237}+\omega^{1248}-\omega^{1345}-\omega^{1578},$
$\displaystyle\beta_{6}=\omega^{4578}+\omega^{3678}+\omega^{2567}+\omega^{2347}+\omega^{1236}+\omega^{1245}+\omega^{1348}+\omega^{1568},$
$\displaystyle\beta_{7}=\omega^{3578}-\omega^{4678}+\omega^{2568}+\omega^{2348}+\omega^{1235}-\omega^{1246}-\omega^{1347}-\omega^{1567}.$

Calculations show that the projections $p_{3},p_{4},p_{5},p_{6}$ satisfy

$\displaystyle p_{3}(\omega^{1}\alpha_{i})=-\frac{3}{7}*(\Omega\omega^{i+1}),\
\ \ p_{4}(\omega^{1}*(\Omega\omega^{1}))=\frac{1}{2}\Omega,$ $\displaystyle
p_{5}(\omega^{1}*(\Omega\omega^{i+1}))=\frac{1}{2}\beta_{i},\ \ \
p_{6}(\omega^{1}\beta_{i+1})=\frac{4}{7}\Omega\omega^{i+1},$

where $i=1,2,\cdots,7.$ Therefore (19), (20) and (21) are exact. Because
$\Lambda_{8}^{3}$ is prime, we know that
$(\omega^{1}\Lambda_{21}^{2})\cap\Lambda_{8}^{3}=\\{0\\},$ this implies
$p_{7}\circ\omega^{1}$ is injective. Hence (22) is exact.

  * (3)

When $\mathrm{hol}(g)=\mathrm{Spin}(7),$ each $\Lambda_{i}^{k}$ is a
irreducible representation of holonomy group. We check the prime spaces in the
Proposition 32 one by one. Using the relations of $\Lambda_{i}^{k}$ in
Proposition 33, one can see that $\Lambda_{7}^{2}$ is not complete prime.
Besides, $\Lambda_{27}^{4}$ is the only complete prime proper subspace of
$\Lambda^{+}.$ Hence there is no other complete prime spaces of type I.
Finally, except for
$(\Lambda_{7}^{2},\Lambda_{48}^{3}),(\Lambda_{21}^{2},\Lambda_{8}^{3}),(\Lambda_{8}^{3},\Lambda_{27}^{4}\oplus\Lambda^{-})$
and $(\Lambda_{1}^{4}\oplus\Lambda_{7}^{4},\Lambda_{48}^{5}),$ there is no
more complete prime pairs of type II. This completes the proof.

### Comments

Let $(M,g)$ be a Riemannian $n-$manifold with a compatible $G-$structure $Q.$
Let $\mathcal{W}$ be a $G-$invariant (not necessary irreducible) subspace of
$\Lambda^{k}(\mathbb{R}^{n})$ and $W=Q\times_{G}\mathcal{W}$ as in (14). S. S.
Chern showed in [5] that the projection $p:\Lambda^{k}(M)\rightarrow W$
commutes with Laplacian, i.e., $p\Delta=\Delta p,$ and one can define a
refined cohomology $\mathscr{H}(\mathcal{W})=\mathrm{Ker}(\Delta|_{W})$
corresponding to $\mathcal{W}.$ See [12, Chap. 3] for details.

We ignore generators and write an elliptic complex $\mathcal{C}$ as
$\mathcal{W}_{1},\cdots,\mathcal{W}_{n},$ a sequence of subspaces of
$\Lambda^{*}(\mathbb{R}^{n}).$ Note that $\mathcal{C}$ is of type I, I or III.
Denote by $H^{i}(\mathcal{C})$ the $i-$th cohomology group of $\mathcal{C}.$
We propose a conjecture that

######  Conjecture 34.

The Euler number of $\mathcal{C}$ is given by

$\displaystyle\chi(\mathcal{C}):=\sum(-1)^{i}\mathrm{dim}\,H^{i}(\mathcal{C})=\sum(-1)^{i}\mathrm{dim}\,\mathscr{H}(\mathcal{W}_{i}).$

Further, $\chi(\mathcal{C})$ is a topological invariant of $M.$

When $\mathcal{E}\subset\Lambda^{k}$ is a complete prime space, there are two
elliptic complexes generated by $\mathcal{E}.$ Let
$\chi_{1}(\mathcal{E}),\chi_{3}(\mathcal{E})$ denote the Euler numbers of
elliptic complexes of type I and III respectively. If the above conjecture is
true and $k$ is even, then $\chi_{1}$ and $\chi_{3}$ has the following
relation

$\chi_{1}(\mathcal{E})+\chi_{3}(\mathcal{E})=\chi(M).$

Acknowledgements: The author is grateful to Prof. J. Zhou of Tsinghua
University for many useful suggestions. The author is also indebted to H. J.
Fan for the guidance over the past years.

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