Complexes from Complexes

Abstract

This paper is concerned with the derivation and properties of differential complexes arising from a variety of problems in differential equations, with applications in continuum mechanics, relativity, and other fields. We present a systematic procedure which, starting from well-understood differential complexes such as the de Rham complex, derives new complexes and deduces the properties of the new complexes from the old. We relate the cohomology of the output complex to that of the input complexes and show that the new complex has closed ranges, and, consequently, satisfies a Hodge decomposition, Poincaré-type inequalities, well-posed Hodge–Laplacian boundary value problems, regular decomposition, and compactness properties on general Lipschitz domains.

Appendix 1. Proof of Injectivity/Surjectivity Condition

In this appendix we prove Lemma 2. Let $n>0$ and $0\le k< n$, $1\le m\le n$ be integers. The linear map $s=s^{k,m}$ is given by

$$\begin{aligned} s: {\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^m\mathbb {R}^n \rightarrow {\text {Alt}}^{k+1}\mathbb {R}^n\otimes {\text {Alt}}^{m-1}\mathbb {R}^n \end{aligned}$$

by

$$\begin{aligned}&s(v^{1}\wedge \cdots \wedge v^{k} \otimes v^{k+1}\wedge \cdots \wedge v^{k+m}) \nonumber \\&\quad =\sum _{l=1}^m (-1)^{l-1} v^{k+l}\wedge v^{1}\wedge \cdots \wedge v^{k}\nonumber \\&\quad \otimes v^{k+1}\wedge \cdots \wedge \widehat{v^{k+l}}\wedge \cdots \wedge v^{k+m}. \end{aligned}$$

(72)

Our goal is to show that s is injective if $k\le m-1$ and surjective if $k\ge m-1$.

We begin with some notation. For n and p natural numbers we write [n] for $\{1,\ldots ,n\}$ so $[n]^p$ denotes the set of p-tuples of elements of [n]. We use standard indexing notation, so an element $\sigma \in [n]^p$ can be written $(\sigma _1,\ldots ,\sigma _p)$. The symmetric group $S_n$, the set of permutations of [n], may be viewed as a subset of $[n]^n$. If also $0\le k\le p$, we define

$$\begin{aligned} X(n,p,k) = \{\sigma \in [n]^p:\sigma _1<\cdots<\sigma _k,\ \sigma _{k+1}<\cdots <\sigma _p\}, \end{aligned}$$

the set of p-tuples which are strictly increasing in the first k indices and in the remaining $p-k$ indices. To each $\sigma \in [n]^p$ we may associate

$$\begin{aligned} dx^\sigma := dx^{\sigma _1}\wedge \cdots \wedge dx^{\sigma _{k}}\otimes dx^{\sigma _{k+1}}\wedge \cdots \wedge dx^{\sigma _{n}}\in {\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^{p-k}\mathbb {R}^n, \end{aligned}$$

where the $dx^i$ are the usual basis elements of the dual space of $\mathbb {R}^n$. The $dx^\sigma$ for $\sigma \in X(n,p,k)$ then form the standard basis for ${\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^{p-k}\mathbb {R}^n$.

Turning to the proof of Lemma 2, we first consider the case $m=n-k$. In this case,

$$\begin{aligned} s:{\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^{n-k}\mathbb {R}^n\rightarrow {\text {Alt}}^{k+1}\mathbb {R}^n\otimes {\text {Alt}}^{n-k-1}\mathbb {R}^n \end{aligned}$$

and we wish to show injectivity for $n-2k-1\ge 0$ and surjectivity for $n-2k-1\le 0$. Given a subset $I\subset [n]$ of cardinality k, let $\sigma \in S_n\cap X(n, n, k)$ be the unique element for which $\{\sigma _1,\ldots ,\sigma _k\} = I$ and set $\omega (I) = {\text {sgn}}(\sigma )dx^\sigma \in {\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^{n-k}\mathbb {R}^n$. Let W(n, k) denote the subspace of ${\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^{n-k}\mathbb {R}^n$ spanned by the elements $\omega (I)$ for all subsets I of [n] of cardinality k. It then follows from the definition of s that

$$\begin{aligned} s\omega (I) = (-1)^{k} \sum _{j\in [n]\setminus I}\omega (I\cup \{j\}). \end{aligned}$$

In particular, $sW(n,k)\subset W(n,k+1)$. We define an inner product on each W(n, k) by declaring the basis elements $\omega (I)$ to be orthonormal. Then, the adjoint $s^*:W(n,k+1)\rightarrow W(n,k)$ is given by

$$\begin{aligned} s^*\omega (J) = (-1)^{k} \sum _{j\in J} \omega (J\setminus \{j\}), \quad J\subset [n], \ \#J=k+1. \end{aligned}$$

The next result shows the desired injectivity and surjectivity in the case $m=n-k$, but only for the restriction of s to a map from $W^k$ to $W^{k+1}$.

Lemma 12

If $n-2k-1\ge 0$, then $s:W(n,k)\rightarrow W(n,k+1)$ is injective. If $n-2k-1\le 0$, then it is surjective.

Proof

Let J, K be subsets of [n] of cardinality k. Then,

$$\begin{aligned} \langle s\omega (J),s\omega (K)\rangle = {\left\{ \begin{array}{ll} n-k, &{} J=K,\\ 1, &{} \#J\cap K=k-1,\\ 0,&{} \text {else}, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \langle s^*\omega (J),s^*\omega (K)\rangle = {\left\{ \begin{array}{ll} k, &{} J=K,\\ 1, &{} \#J\cap K=k-1,\\ 0,&{} \text {else}. \end{array}\right. } \end{aligned}$$

It follows that

$$\begin{aligned} \langle s\omega (J),s\omega (K)\rangle =\langle s^*\omega (J),s^*\omega (K)\rangle + (n-2k)\langle \omega (J),\omega (K)\rangle , \end{aligned}$$

and, by bilinearity, that

$$\begin{aligned} \langle s\rho ,s\tau \rangle = \langle s^*\rho ,s^*\tau \rangle + (n-2k)\langle \rho ,\tau \rangle ,\quad \rho ,\tau \in W(n,k). \end{aligned}$$

(73)

Taking $\tau =\rho$ and assuming that $n-2k-1\ge 0$, we see that $s\rho =0$ implies $\rho =0$, so s is injective as claimed.

If we replace k by $k+1$ in (73) and assume that $n-2k-1\le 0$, the same argument implies that $s^*:W(n,k+1)\rightarrow W(n,k)$ is injective, and consequently that s is surjective. $\square$

Now we return to general $n\ge 1$, $0\le k<n$, $1\le m\le n$, and the map s acting on all of ${\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^{m}\mathbb {R}^n$. To prove surjectivity, assuming $k\ge m-1$, we must show that s maps onto all of ${\text {Alt}}^{k+1}\mathbb {R}^n\otimes {\text {Alt}}^{m-1}\mathbb {R}^n$. For this it is enough to take an element of the form

$$\begin{aligned} \rho = v^1\wedge \cdots \wedge v^{k+1}\otimes v^{k+2}\wedge \cdots \wedge v^{k+m} \end{aligned}$$

with the $v^i$ belonging to the dual of $\mathbb {R}^n$, and show that $\rho$ is in the range of s.

Let $p=m+k$ and define a linear map from the dual space of $\mathbb {R}^p$ to that of $\mathbb {R}^n$ by $T dx^i = v^i$, $i=1,\ldots ,p$. Then, T induces a linear map

$$\begin{aligned} T_*:{\text {Alt}}^k\mathbb {R}^p\otimes {\text {Alt}}^m\mathbb {R}^p \rightarrow {\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^m\mathbb {R}^n \end{aligned}$$

given by

$$\begin{aligned}&T_*(u^1\wedge \cdots \wedge u^k\otimes u^{k+1}\wedge \cdots \wedge u^{k+m})\\&\quad = (Tu^1\wedge \cdots \wedge Tu^k\otimes Tu^{k+1}\wedge \cdots \wedge Tu^{k+m}). \end{aligned}$$

Clearly, $T_*s = sT_*$ and, letting

$$\begin{aligned} \omega = dx^1\wedge \cdots \wedge dx^{k+1}\otimes dx^{k+2}\wedge \cdots \wedge dx^{k+m}, \end{aligned}$$

we have $T_*\omega = \rho$. Since $\omega \in W(n,k+1)$, the preceding lemma insures that $\omega =s \mu$ for some $\mu \in W(n,k)\subset {\text {Alt}}^k\mathbb {R}^p\otimes {\text {Alt}}^m\mathbb {R}^p$. Therefore,

$$\begin{aligned} \rho = T_*\omega =T_*s\mu = s(T_*\mu ). \end{aligned}$$

This completes the proof of surjectivity.

We now prove the injectivity for general n, k, and m, continuing to write $p=m+k$. For $\sigma \in X(n,p,k)$ let ${{\tilde{\sigma }}}\in [n]^p$ denote the tuple obtained from $\sigma$ by taking its entries in non-decreasing order. For example, if $\sigma =(2,3,1,2)\in X(3,4,2)$ (so increasing on its first 2 and last 2 indices), then ${{\tilde{\sigma }}} = (1,2,2,3)$. Then, we have the direct sum decomposition

$$\begin{aligned} {\text {Alt}}^k\mathbb {R}^n\otimes {\text {Alt}}^m\mathbb {R}^n = \bigoplus _{J\in [n]^p} Y(n,p,k,J), \end{aligned}$$

where

$$\begin{aligned} Y(n,p,k,J)={\text {span}}\{dx^\sigma :\sigma \in X(n,p,k),\ {{\tilde{\sigma }}}=J\}. \end{aligned}$$

Of course, there is a similar decomposition for ${\text {Alt}}^{k-1}\mathbb {R}^n\otimes {\text {Alt}}^{m+1}\mathbb {R}^n$. The two decompositions are compatible with s, in the sense that $sY(n,p,k,J)\subset Y(n,p,k+1,J)$ for the same J. It follows that it is enough to prove that s is injective when restricted to each of the spaces Y(n, p, k, J), $J\in [n]^p$. The p-tuple J consists of entries which appear only once and entries which appear twice. Let l be the number of repeated entries, so that there are $q:=p-2l$ non-repeated entries. Without loss of generality, we may assume that the non-repeated entries are $1,\ldots ,q$ and the repeated entries $q+1,\ldots ,q+l$, i.e.,

$$\begin{aligned} J = (1,2,\ldots ,q,q+1,q+1, q+2, q+2,\ldots ,q+l,q+l). \end{aligned}$$

The space $S_{p-2l}\cap X(p-2l,p-2l,k-l)$ consists of permutations of $[p-2l]$ which are increasing in their first $k-l$ and last $m-l$ indices. If $\rho$ belongs to this space, we define $Q\rho$ as the p-tuple

$$\begin{aligned} Q\rho= & {} (\rho _1,\ldots ,\rho _{k-l},q+1, q+2,\ldots ,q+l,\\&\rho _{k-l+1}, \ldots , \rho _{p-2l},q+1, q+2,\ldots ,q+l). \end{aligned}$$

This defines a bijection of $S_{p-2l}\cap X(p-2l,p-2l,k-l)$ onto $\{\sigma \in X(n,p,k),\ {{\tilde{\sigma }}}=J\}$. Now we consider the spaces spanned by the basis functions $dx^\sigma$ where $\sigma$ varies in one of these two bijective sets. These spaces are precisely $W(p-2l,k-l)$ and Y(n, p, k, J), respectively, and the bijection just established induces an isomorphism $F:Y(n,p,k,J)\rightarrow W(p-2l,k-l)$, given by

$$\begin{aligned} dx^{Q\rho } \mapsto dx^\rho . \end{aligned}$$

It is easy to see that $Fs=sF$. If $\omega \in Y(n,p,k,J)$ and $s\omega =0$, then $F\omega \in W(p-2l,k-l)$ and $sF\omega =0$, so, by Lemma 12, $F\omega =0$, so $\omega =0$. This completes the proof.