# Differential Inclusions involving the curl operator

Nurun Nesha†,1

###### Abstract.

In this article, we study the existence of $\eta\in
W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ satisfying

$\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega,$

where $n\in\mathbb{N},\Omega\subseteq\mathbb{R}^{n}$ is open, bounded and
$E\subseteq\Lambda^{2}.$

†Indian Institute of Science Education and Research Kolkata, Campus Road,
Mohanpur, West Bengal 741246, India<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction and Main Results
  2. 2 Notations
  3. 3 Non Existence of Solution: Dimensionality of E
  4. 4 Restrictions on the Curl Set
  5. 5 Existence of Solution: Dimension 2n-3
  6. 6 The Case of a k-form

##  1\. Introduction and Main Results

In this paper, we study the following differential inclusion problem

$\displaystyle\operatorname{curl}\eta\in E\ \ \textrm{ a.e. in }\Omega$ (1.1)
$\displaystyle\textrm{ and }\int\limits_{\Omega}\eta\neq 0$

where $\Omega\subseteq\mathbb{R}^{n}$ is open, bounded,
$E\subseteq\Lambda^{2}(\mathbb{R}^{n})$, and $n\geq 4$. This problem has been
studied in Bandyopadhyay-Barroso-Dacorogna-Matias [2] and Bandyopadhyay-
Dacorogna-Kneuss [3] in the lower dimensional cases, namely when
$\dim\operatorname{span}E=n-1$ when $n\geq 3,$ $\dim\operatorname{span}E=3,$
when $n=3.$ In this article, we investigate the case when
$\dim\operatorname{span}E\geq n.$ The most fundamental case is, of course, the
gradient case which has received notable attention, in particular, by Bressan-
Flores [4], Cellina [6, 5], Dacorogna-Marcellini [12] and Friesecke [14]. An
extensive study has been done in [10] on this topic. We prove a few existence
as well as non-existence results in this regard. Our main result is the
following which we will prove in section $4.$

######  Theorem 1.1.

Let $n\in\mathbb{N},n\geq 5$ and $E\subseteq\Lambda^{2}(\mathbb{R}^{n})$ be
such that

$\omega\wedge\omega^{\prime}=0\textrm{ for all }\ \omega,\omega^{\prime}\in
E.$

Let $\Omega\subseteq\mathbb{R}^{n}$ be an open, bounded set. Then there exists
$\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ such that

$\displaystyle\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega$
$\displaystyle\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0\textrm{
for all }e\in E$ (1.2) $\displaystyle\textrm{ and
}\int\limits_{\Omega}\eta\neq 0$

if and only if $0\in\operatorname{rico}E$ and $\dim\operatorname{span}E=n-1.$

The next result is about non-existence of solution when
$\dim\operatorname{span}E=n$ which we will discuss in section $3.$

######  Theorem 1.2.

Let $n\in\mathbb{N},n\geq 4$ and $\Omega\subseteq\mathbb{R}^{n}$ be open,
bounded. Then the following differential inclusion problem

$\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega\textrm{ and
}\displaystyle\int\limits_{\Omega}\eta\neq 0$

has no solution $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ if
$\dim\operatorname{span}E=n$ and
$\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0$ for all
$e\in E$.

In [1], Ball-James considered two gradient problem and found that rank of the
difference of two gradients is less than or equal to $1$ and in a similar way
we can show that, in curl case the rank of the difference will be less than or
equal to $2$. So, in section $4,$ we will see the following theorem under
taking the constraint on the set $E$ that rank of difference of any two
elements is less than or equal to $2.$

######  Theorem 1.3.

Let $n\in\mathbb{N}$ and $\Omega\subseteq\mathbb{R}^{n}$ be open, bounded set.
Let $E\subseteq\Lambda^{2}(\mathbb{R}^{n})$ be such that
$\operatorname{rank}[e-f]\leq 2$ for any $e,f\in E,$ in other words, there
exist $x,y\in\mathbb{R}^{n}$ such that $e-f=x\wedge y.$ Then there does not
exist any $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ of the following
problem

$\operatorname{curl}\eta\in E\ \ \textrm{ a.e. in }\Omega$

if $\dim\operatorname{span}E\geq n+1$ and
$\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0$ for all
$e\in E.$

In section $5$, we will give one existence result of solution $\eta\in
W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ at dimension $(2n-3)$ for the
following differential inclusion problem

$\displaystyle\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega,$
$\displaystyle\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0\textrm{
for all }e\in E.$

Finally, in section $6$, we show that

######  Theorem 1.4.

Let $n\in\mathbb{N},$ $1\leq k\leq n-3.$ Suppose
$f:\mathbb{R}^{n}\to\Lambda^{k}(\mathbb{R}^{n})$ be continuous such that
$f(0)\neq 0$ and $f(0)$ is $k$-divisible, i.e., $f(0)=c^{1}\wedge\ldots\wedge
c^{k}$ for some $c^{i}\in\mathbb{R}^{n}\setminus\\{0\\}$ for $i=1,\ldots,k.$
Then

$\dim\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}\neq n-k+1.$

The results of differential inclusion problems can be applied, embracing the
notions due to Cellina [6, 5] and Friesecke [14], to obtain solutions for a
non-convex variational problem. In particular, according to [2], one can show
that:

######  Theorem 1.5.

Let $\Omega\subseteq\mathbb{R}^{n}$ be a bounded, open set, $0\leq k\leq n-1$
and

$f:\Lambda^{k+1}(\mathbb{R}^{n})\to\mathbb{R}_{+}$

be lower semi-continuous. Let

$\inf\left\\{\displaystyle\int\limits_{\Omega}f(\textrm{d}\eta(x))\textrm{d}x:\eta\in
W_{0}^{1,\infty}\left(\Omega;\Lambda^{k}(\mathbb{R}^{n})\right)\right\\}$
(1.3)

and
$K=\left\\{\xi\in\Lambda^{k+1}(\mathbb{R}^{n}):f^{**}(\xi)<f(\xi)\right\\}$,
where $f^{**}$ is the convex envelope of $f.$ Assume that $K$ is connected and
$0\in K.$ If $K$ is bounded and $f^{**}$ is affine on $K$ then (1.3) has a
solution.

Many results as well as applications of differential inclusions can be
observed in Dacorogna-Pisante [8], Dacorogna-Fonseca [11], Dacorogna-
Marcellini [12], Blasi-Pianigiani [13], Sil [16] and Sychev [17].

##  2\. Notations

We gather here some notations which will be used throughout this article. For
more details on exterior algebra and differential forms see [7] and for convex
analysis see [9] or [15].

  * $(1)$

Let $k,n$ be two integers.

    * $\bullet$

We write $\Lambda^{k}(\mathbb{R}^{n})$ (or simply $\Lambda^{k}$) to denote the
vector space of all alternating $k$-linear maps
$f:\underbrace{\mathbb{R}^{n}\times\cdots\times\mathbb{R}^{n}}_{\textrm{k-times}}\to\mathbb{R}.$
For $k=0,$ we set $\Lambda^{0}(\mathbb{R}^{n})=\mathbb{R}.$ Note that
$\Lambda^{k}(\mathbb{R}^{n})=\\{0\\}$ for $k>n$ and, for $k\leq n,$
$\dim\left(\Lambda^{k}(\mathbb{R}^{n})\right)=\binom{n}{k}.$

    * $\bullet$

$\wedge,$ $\lrcorner$, $\langle;\rangle$ and, respectively, $*$ denote the
exterior product, interior product, the scalar product and, respectively, the
Hodge star operator.

    * $\bullet$

For $b\in\Lambda^{k},$ $\operatorname{rank}[b]$ denotes the rank of the
exterior $k$-form $b.$

    * $\bullet$

If $\\{e^{1},\ldots,e^{n}\\}$ is a basis of $\mathbb{R}^{n}$, then,
identifying $\Lambda^{1}$ with $\mathbb{R}^{n},$

$\\{e^{i_{1}}\wedge\cdots\wedge e^{i_{k}}:1\leq i_{1}<\cdots<i_{k}\leq n\\}$

is a basis of $\Lambda^{k}.$

    * $\bullet$

For $E\subseteq\Lambda^{k},$ $\operatorname{span}E$ denotes the subspace
spanned by $E.$

    * $\bullet$

Let $W$ be a subspace of $\Lambda^{k}$. We write $\dim W$ to denote the
dimension of $W$ and $W^{\perp}$ to denote the orthogonal complement of $W.$

    * $\bullet$

For $b\in\Lambda^{k},$ we write, identifying again $\Lambda^{1}$ with
$\mathbb{R}^{n},$

$\mathbb{R}^{n}\wedge b=\Lambda^{1}\wedge b=\\{x\wedge
b:x\in\Lambda^{1}\\}\subseteq\Lambda^{k+1}.$

  * (2)

Let $\Omega\subseteq\mathbb{R}^{n}$ be a bounded open set.

    * $\bullet$

The spaces $C^{1}(\Omega;\Lambda^{k}),$ $W^{1,p}(\Omega;\Lambda^{k})$ and
$W^{1,p}_{0}(\Omega;\Lambda^{k}),$ $1\leq p\leq\infty$ are defined in the
usual way.

    * $\bullet$

For $\eta\in W^{1,p}(\Omega;\Lambda^{k}),$
$\displaystyle\int\limits_{\Omega}\eta\ $ denotes the exterior $k$-form
obtained by integrating componentwise the differential form $\eta$.
Explicitly, for $1\leq i_{1}<\cdots<i_{k}\leq n,$

$\left(\displaystyle\int\limits_{\Omega}\eta\right)_{i_{1}\cdots
i_{k}}=\int\limits_{\Omega}\eta_{i_{1}\cdots i_{k}}.$

    * $\bullet$

For $\eta\in W^{1,p}(\Omega;\Lambda^{k}),$ the exterior derivative $d\eta$
belongs to $L^{p}(\Omega;\Lambda^{k+1})$ and is defined by

$\left(d\eta\right)_{i_{1}\cdots
i_{k+1}}=\displaystyle\sum\limits_{j=1}^{k+1}(-1)^{j+1}\frac{\partial\eta_{i_{1}\cdots
i_{j-1}i_{j+1}\cdots i_{k+1}}}{\partial x_{i_{j}}},$

for $1\leq i_{1}<\cdots<i_{k+1}\leq n.$ If $k=0,$ then
$d\eta\simeq\operatorname{grad}\eta$ . If $k=1,$ then for $1\leq i<j\leq n,$

$(d\eta)_{ij}=\frac{\partial\eta_{j}}{\partial
x_{i}}-\frac{\partial\eta_{i}}{\partial x_{j}}$

i.e., $d\eta\simeq\operatorname{curl}\eta.$

  * (3)

For subsets $C,V\subseteq\Lambda^{k},$

    * $\bullet$

$\operatorname{co}C$ denotes the convex hull of $C;$

    * $\bullet$

$\operatorname{int}_{V}C$ denotes the interior of $C$ with respect to the
topology relative to $V.$

  * (4)

For a convex set $C\subseteq\Lambda^{k},$

    * $\bullet$

$\operatorname{aff}C$ denotes the affine hull of $C$ which is the intersection
of all affine subsets of $\Lambda^{k}$ containing $C;$

    * $\bullet$

$\operatorname{ri}C$ denotes the relative interior of $C$ which is the
interior of $C$ with respect to the topology relative to affine hull of $C.$
Equivalently $\operatorname{ri}C=\operatorname{int}_{\operatorname{aff}C}C$;

    * $\bullet$

$\operatorname{rbd}C$ denotes the relative boundary of $C$ which is
$\overline{C}\setminus\operatorname{ri}C.$

  * (5)

For a set $A\subseteq\mathbb{R}^{n}$ $\operatorname{meas}(A)$ denotes the
Lebesgue measure of $A.$

  * (6)

$\mathbb{R}_{+}$ denotes the set of all non-negative real numbers.

##  3\. Non Existence of Solution: Dimensionality of E

In this section, we will prove a non-existence result, namely, that there is
no $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ satisfying

$\operatorname{curl}\eta\in E,\textrm{ a.e. in
}\Omega,\displaystyle\int\limits_{\Omega}\eta\neq 0,$

when $\dim\operatorname{span}E=n.$ The following lemma plays the main role.

######  Lemma 3.1.

Let $n\in\mathbb{N},$ $n\geq 4$ and let $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be
continuous with $f(0)\neq 0$. Then

$\dim\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}\neq n.$

######  Remark 3.2.

Lemma 3.1 is not true when $n=3.$ Let us define
$f:\mathbb{R}^{3}\to\mathbb{R}^{3}$ by

$f(x):=(e^{1}\otimes e^{1})x+e^{2},\textrm{ for all }x\in\mathbb{R}^{3}.$

Then $\dim\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{3}\\}=3.$ To see
this, we note that $e^{1}\wedge f(e^{1})=e^{1}\wedge e^{2},e^{3}\wedge
f(e^{3})=-e^{2}\wedge e^{3},$ and $(e^{1}+e^{3})\wedge
f(e^{1}+e^{3})=e^{1}\wedge e^{2}-e^{2}\wedge e^{3}-e^{1}\wedge e^{3}.$

###### Proof.

Let us set

$\mathcal{S}:=\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}.$

We prove by contradiction. Let us suppose to the contrary that
$\dim\mathcal{S}=n$ . Note that, $\mathbb{R}^{n}\wedge
f(0)\subseteq\mathcal{S},$ see Proposition $2.2$ of [3]. Furthermore, we can
find $x_{0}\in\mathbb{R}^{n}$ such that $x_{0}\wedge f(x_{0})\wedge f(0)\neq
0.$ Indeed, if this was not the case, we would have

$x\wedge f(x)\wedge f(0)=0\textrm{ for all }x\in\mathbb{R}^{n}.$

Cartan’s lemma, see Theorem $2.42$ of [7], then guarantees the existence of
$u_{x}\in\mathbb{R}^{n}$ satisfying

$x\wedge f(x)=u_{x}\wedge f(0)\textrm{ for all, }x\in\mathbb{R}^{n},$

which implies that

$\mathcal{S}=\mathbb{R}^{n}\wedge f(0).$

This is a contradiction as $\dim(\mathbb{R}^{n}\wedge f(0))=n-1$ [ Lemma $2.1$
of [3] ], whereas $\dim\mathcal{S}=n.$ Therefore, we indeed have a
$x_{0}\in\mathbb{R}^{n}\setminus\\{0\\}$ such that

$x_{0}\wedge f(x_{0})\wedge f(0)\neq 0.$

Since $f$ is continuous at $x_{0}$, there exists an $\epsilon>0$ such that

$x\wedge f(x)\wedge f(0)\neq 0\textrm{ for all }x\in B_{\epsilon}(x_{0}).$

Let us find a basis $\\{a^{1},\ldots,a^{n}\\}$ of $\mathbb{R}^{n}$ inside
$B_{\epsilon}(x_{0})$. Then

$a^{i}\wedge f(a^{i})\wedge f(0)\neq 0\textrm{ for all }i=1,\ldots,n.$ (3.1)

Let us write

$\mathcal{S}=\left[\mathbb{R}^{n}\wedge
f(0)\right]\oplus\left[\mathbb{R}^{n}\wedge
f(0)\right]^{\perp}=\left[\mathbb{R}^{n}\wedge
f(0)\right]\oplus\operatorname{span}\\{\omega\\},$

where $\omega\in\left[\mathbb{R}^{n}\wedge
f(0)\right]^{\perp}\setminus\\{0\\}.$ For each $i=1,\ldots,n,$ we have

$a^{i}\wedge f(a^{i})=c_{i}\wedge f(0)+\beta_{i}\omega,$ (3.2)

for some $c_{i}\in\mathbb{R}^{n}$ and $\beta_{i}\in\mathbb{R}.$ Note that,
thanks to equation 3.1, we have $\beta_{i}\neq 0,$ for all $i=1,\ldots,n$. It
follows from Equation 3.2 that, for all $i=1,\ldots,n$,

$\beta_{i}\ \omega\wedge f(0)\wedge a_{i}=0.$

Since $\beta_{i}\neq 0$ for every $i=1,\ldots,n$, we have

$\omega\wedge f(0)\wedge a^{i}=0\textrm{ for every }i=1,\ldots,n,$

which implies that $\omega\wedge f(0)=0$ as $\\{a^{1},\ldots,a^{n}\\}$ is a
basis of $\mathbb{R}^{n}$. Using Proposition $2.16$ of [7], we also note that

$\displaystyle\langle f(0)\lrcorner\omega;x\rangle$
$\displaystyle=(-1)^{(1+1)}\langle\omega;f(0)\wedge x\rangle$
$\displaystyle=0,\textrm{ for all }x\in\mathbb{R}^{n},$

as $\omega\in\mathcal{S}\cap\left[\mathbb{R}^{n}\wedge f(0)\right]^{\perp}.$
It follows that $f(0)\lrcorner\omega=0.$ This, combined with $\omega\wedge
f(0)=0$ and Proposition $2.16$ of [7] implies that

$\displaystyle\|f(0)\|^{2}\omega=f(0)\lrcorner\left(f(0)\wedge\omega\right)+f(0)\wedge\left(f(0)\lrcorner\
\omega\right)=0.$

Since $f(0)\neq 0$, we have $\omega=0$, which is a contradiction. Therefore
$\dim\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}\neq n.$ ∎

######  Theorem 3.3.

Let $n\in\mathbb{N}$ with $n\geq 4$, let $\Omega\subseteq\mathbb{R}^{n}$ be
open, bounded and let $E\subseteq\Lambda^{2}(\mathbb{R}^{n})$. Then there is
no $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ satisfying

$\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega\textrm{ and
}\displaystyle\int\limits_{\Omega}\eta\neq 0$ (3.3)

if

  1. (i)

$\dim\operatorname{span}E=n$, and

  2. (ii)

$\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0$ for all
$e\in E$

######  Remark 3.4.

Theorem 3.3 is not true when $n=3,$ see Theorem $4.15$ of [2]. The solution
$\eta$ constructed in the proof of Theorem $4.15$ of [2] has the property that
$\displaystyle\int\limits_{\Omega}\eta\neq 0.$

######  Remark 3.5.

The case $n\leq 3$ has been done completely in [2]. We don’t need to take the
case $\int\limits_{\Omega}\eta\neq 0$ for $n\leq 3.$

###### Proof.

Let

$\mathcal{P}:\Lambda^{2}({\mathbb{R}^{n}})\to\Lambda^{2}(\mathbb{R}^{n})$

be the projection onto the orthogonal complement of $\operatorname{span}E.$
Since $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ extending $\eta$ by
$0$ to $\mathbb{R}^{n}$, it follows that

$\mathcal{P}(\operatorname{curl}\eta)=0\textrm{ a.e. in }\mathbb{R}^{n}.$

Applying the Fourier transform, we obtain

$\mathcal{P}(x\wedge\hat{\eta}(x))=0\textrm{ for all }x\in\mathbb{R}^{n}$

which implies that

$x\wedge\hat{\eta}(x)\in\operatorname{span}E\textrm{ for all
}x\in\mathbb{R}^{n},$

where
$\hat{\eta}(x)=\displaystyle\int\limits_{\mathbb{R}^{n}}\eta(y)\cos(2\pi\langle
x;y\rangle)dy.$ Together with the Proposition $2.2$ of [3] and the above we
can conclude that

$\mathbb{R}^{n}\wedge\hat{\eta}(0)\subseteq\operatorname{span}\\{x\wedge\hat{\eta}(x):x\in\mathbb{R}^{n}\\}\subseteq\operatorname{span}E.$
(3.4)

We will now show that

$\operatorname{span}\\{x\wedge\hat{\eta}(x):x\in\mathbb{R}^{n}\\}=\operatorname{span}E.$

Suppose not, i.e.,
$\operatorname{span}\\{x\wedge\hat{\eta}(x):x\in\mathbb{R}^{n}\\}\subsetneqq\operatorname{span}E$.
Let
$m\in\operatorname{span}\\{x\wedge\hat{\eta}(x):x\in\mathbb{R}^{n}\\}^{\perp}$.
Then

$\langle x\wedge\hat{\eta}(x);m\rangle=0\textrm{ for all }x\in\mathbb{R}^{n}.$

Using Plancherel Theorem, this implies that

$\langle\operatorname{curl}\eta(x);m\rangle=0\textrm{ for all
}x\in\mathbb{R}^{n}\textrm{ a.e. }$

and hence

$\langle e;m\rangle=0\textrm{ for all }e\in E.$

This gives us that $m\in(\operatorname{span}E)^{\perp}.$ So,
$\operatorname{span}E\subseteq\operatorname{span}\\{x\wedge\hat{\eta}(x):x\in\mathbb{R}^{n}\\}$.
Therefore

$\dim\operatorname{span}\\{x\wedge\hat{\eta}(x):x\in\mathbb{R}^{n}\\}=n.$

But it can not happen because of lemma 3.1. Thus there does not exist any
solution $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ of the problem
(3.3) if $\dim\operatorname{span}E=n.$ ∎

##  4\. Restrictions on the Curl Set

In this section, we will see that $\dim\operatorname{span}E\leq n$ if we add
one constraint on $E$ that $\operatorname{rank}[e-f]\leq 2$ for any $e,f\in
E$, where $E\subseteq\Lambda^{2}(\mathbb{R}^{n})$ and $n\geq 4$. In lemma 4.1,
we will prove it for $n=4$ and in lemma 4.6, we will do it for $n\geq 5.$ Let
us notice here one thing that between two statements
‘$\operatorname{rank}[e-f]\leq 2$ for any $e,f\in E$’ and ‘$e\wedge f=0$ for
all $e,f\in E$’, the later one will always imply the first one but the
converse may not be true. We have given one example in remark (ii)$(ii)$ in
this respect. Finally, we will establish Theorem 4.8. Let us first prove the
lemma below.

######  Lemma 4.1.

Let $E\subseteq\Lambda^{2}(\mathbb{R}^{4})$ be such that
$\operatorname{rank}[e-f]\leq 2$ for all $e,f\in E.$ Then
$\dim\operatorname{span}E\leq 4.$

###### Proof.

Let $V:=\Lambda^{2}(\mathbb{R}^{4}).$ Let us define a bilinear map $B:V\times
V\to\mathbb{R}$ by

$B(u,v):=c(u\wedge v)\textrm{ for all }u,v\in V,$

where $c(u\wedge v)\in\mathbb{R}$ is such that

$u\wedge v=c(u\wedge v)e^{1}\wedge e^{2}\wedge e^{3}\wedge e^{4}.$

Clearly, $B$ is symmetric and non-degenerate. For any subspace $F\subseteq V,$
let us define

$\tilde{F}:=\\{v\in V:B(v,f)=0\textrm{ for all }f\in F\\}.$

As $B$ is non-degenerate, the map

$B_{1}:V\to V^{*},\textrm{ defined by }v\mapsto B(v,.)$

is one-one and hence onto. Therefore, the map

$B_{2}:V\to F^{*},\textrm{ defined by }v\mapsto B(v,.)$

is also onto because it can be written as $B_{2}=\psi\circ B_{1},$ where
$\psi:V^{*}\to F^{*}.$ Hence

$\dim V=\dim\ker B_{2}+\dim F^{*}.$

Clearly, $\ker B_{2}=\tilde{F}$ and $\dim F^{*}=\dim F$. So

$\dim F+\dim\tilde{F}=\dim V.$ (4.1)

Now suppose that $F\subseteq V$ is an isotropic subspace, i.e., $B|_{F\times
F}=0.$ In other words, $F\subseteq\tilde{F}.$ In this case, we can say from
(4.1) that

$\displaystyle\dim V$ $\displaystyle=\dim F+\dim\tilde{F}$
$\displaystyle\geq\dim F+\dim F,$

i.e., $\dim F\leq\frac{\dim V}{2}=3.$ Therefore, if $F\subseteq V$ is any
isotropic subspace of $V,$ then $\dim F\leq 3.$

Now suppose that $E\subseteq V$ is such that for any $e,f\in E,$ there exist
$x,y\in\mathbb{R}^{4}$ such that $e-f=x\wedge y,$ i.e.,
$\operatorname{rank}[e-f]\leq 2.$ We will show that
$\dim\operatorname{span}E\leq 4.$

In contrary, let us suppose that $E$ contains five linearly independent
elements $\zeta^{0},\zeta^{1},\zeta^{2},\zeta^{3}$, $\zeta^{4}$ and let

$\xi^{i}:=\zeta^{i}-\zeta^{0}\textrm{ for }i=1,2,3,4.$

As every $\xi^{i}$ has rank less than or equals to $2$,

$\xi^{i}\wedge\xi^{i}=0\textrm{ for all }i=1,2,3,4\ \ \textrm{ [see
proposition 2.37(iii) of \cite[cite]{[\@@bibref{}{csato2011pullback}{}{}]}]}$

i.e.,

$B(\xi^{i},\xi^{i})=0\textrm{ for all }i=1,2,3,4.$

Now $\xi^{i}-\xi^{j}=\zeta^{i}-\zeta^{j}$ and
$(\zeta^{i}-\zeta^{j})\wedge(\zeta^{i}-\zeta^{j})=0$ so
$(\xi^{i}-\xi^{j})\wedge(\xi^{i}-\xi^{j})=0,$ i.e.,
$B(\xi^{i}-\xi^{j},\xi^{i}-\xi^{j})=0$ for $i,j\in\\{1,2,3,4\\}.$ This implies
that

$\xi^{i}\wedge\xi^{j}=0\textrm{ for all }i,j\in\\{1,2,3,4\\}.$

If we take

$F=\\{\xi^{i}:i=1,2,3,4\\}\textrm{ and }F^{\prime}:=\operatorname{span}F$

then

$B|_{F^{\prime}\times F^{\prime}}=0,\textrm{ i.e., }F^{\prime}\textrm{ is an
isotropic subspace of }V$

and $\dim F^{\prime}=4.$ It contradicts that $\dim F^{\prime}\leq 3.$
Therefore, $E$ can not contain $5$ linearly independent elements. Hence
$\dim\operatorname{span}E\leq 4.$ ∎

We will state two trivial lemmas below, the proofs of which are
straightforward. We will use these lemmas 4.2 and 4.3 in the proofs of lemmas
4.4 and 4.6.

######  Lemma 4.2.

Let $n\in\mathbb{N}$ and $n\geq 3.$ Let $\\{\omega,\omega^{\prime}\\}$ be a
linearly independent subset of $\Lambda^{2}(\mathbb{R}^{n})$ such that
$\operatorname{rank}[\omega]=2$, $\operatorname{rank}[\omega^{\prime}]=2$ and
$\omega\wedge\omega^{\prime}=0$. Then
$\dim[\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}]=1.$ Also
$\dim[\ker\\{\omega\\}\cap\ker\\{\omega^{\prime}\\}]=n-3.$

###### Proof.

As $\operatorname{rank}[\omega]=\operatorname{rank}[\omega^{\prime}]=2,$ let
us suppose that

$\omega=x\wedge y\textrm{ and }\omega^{\prime}=x^{\prime}\wedge
y^{\prime}\textrm{ where }x,y,x^{\prime},y^{\prime}\in\mathbb{R}^{n}.$

Again since $\omega\wedge\omega^{\prime}=0$, the set
$\\{x,y,x^{\prime},y^{\prime}\\}$ is linearly dependent [see Theorem
$2.3$,[7]]. This gives us that

$\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}\neq\\{0\\},$

because
$\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}=\operatorname{span}\\{x,y\\}\cap\operatorname{span}\\{x^{\prime},y^{\prime}\\}=\\{0\\}$
implies that $\\{x,y,x^{\prime},y^{\prime}\\}$ is linearly independent, which
is a contradiction. Hence

$\dim[\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}]\geq 1.$

Now if $\dim[\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}]=2$
then clearly,

$\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}=\ker\\{\omega\\}^{\perp}=\ker\\{\omega^{\prime}\\}^{\perp},$

since $\dim\ker\\{\omega\\}^{\perp}=\dim\ker\\{\omega^{\prime}\\}^{\perp}=2.$
This implies that

$\operatorname{span}\\{x,y\\}=\operatorname{span}\\{x^{\prime},y^{\prime}\\},\textrm{
i.e., }\\{\omega,\omega^{\prime}\\}\textrm{ is linearly dependent, }$

which is a contradiction. Therefore
$\dim[\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}]=1.$

For the second part,

$\displaystyle\dim[\ker\\{\omega\\}\cap\ker\\{\omega^{\prime}\\}]$
$\displaystyle=n-\dim[\\{\ker\\{\omega\\}\cap\ker\\{\omega^{\prime}\\}\\}^{\perp}]$
$\displaystyle=n-[\dim\ker\\{\omega\\}^{\perp}+\ker\\{\omega^{\prime}\\}^{\perp}]$
$\displaystyle=n-[\dim\ker\\{\omega\\}^{\perp}+\dim\ker\\{\omega^{\prime}\\}^{\perp}-\dim\\{\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}\\}]$
$\displaystyle=n-[2+2-1],\ \ \textrm{ as
}\dim\\{\ker\\{\omega\\}^{\perp}\cap\ker\\{\omega^{\prime}\\}^{\perp}\\}=1$
$\displaystyle=n-3.$

Therefore $\dim[\ker\\{\omega\\}\cap\ker\\{\omega^{\prime}\\}]=n-3.$ ∎

######  Lemma 4.3.

Let $n\in\mathbb{N}$ and $b\in\mathbb{R}^{n}\setminus\\{0\\}.$ Let
$\\{\omega_{1},\omega_{2},\ldots,\omega_{m}\\}\subseteq\mathbb{R}^{n}\wedge b$
be a linearly independent subset, where $m\in\mathbb{N},m\leq n-1$. If
$\omega_{i}=x_{i}\wedge b$ for some $x_{i}\in\mathbb{R}^{n},i=1,\ldots,m$ then
$\\{b,x_{1},\ldots,x_{m}\\}$ is linearly independent.

###### Proof.

Let $\alpha b+\alpha_{1}x_{1}+\cdots+\alpha_{m}x_{m}=0$, where
$\alpha,\alpha_{1},\ldots,\alpha_{m}\in\mathbb{R}.$ Then

$\displaystyle 0=0\wedge b$ $\displaystyle=(\alpha
b+\alpha_{1}x_{1}+\cdots+\alpha_{m}x_{m})\wedge b$
$\displaystyle=\alpha_{1}x_{1}\wedge b+\cdots+\alpha_{m}x_{m}\wedge b$
$\displaystyle=\alpha_{1}\omega_{1}+\cdots+\alpha_{m}\omega_{m}.$

As $\\{\omega_{1},\ldots,\omega_{m}\\}$ is linearly independent,
$\alpha_{i}=0$ for $i=1,\ldots,m.$ So $\alpha b=0$ and this gives $\alpha=0.$
Therefore $\\{b,x_{1},\ldots,x_{m}\\}$ is linearly independent. ∎

Now let us consider $n\geq 5.$ Using this lemma 4.4 below we will prove
another main lemma 4.6 of this section which states that
$\dim\operatorname{span}E\leq n$ if $\operatorname{rank}[e-f]\leq 2$ for any
$e,f\in E.$

######  Lemma 4.4.

Let $E\subseteq\Lambda^{2}(\mathbb{R}^{n})\setminus\\{0\\},\ n\geq 5$. Let
$\dim\operatorname{span}E=n-1$ and $\omega\wedge\omega^{\prime}=0$ for any
$\omega,\omega^{\prime}\in E$, then

$\displaystyle\bigcap_{\omega\in E}\ker\\{\omega\\}^{\perp}\neq\\{0\\}.$

###### Proof.

Let $\\{\omega_{1},\omega_{2},\ldots,\omega_{n-1}\\}$ be a basis of
$\operatorname{span}E$ and

$A_{i}:=\ker\\{\omega_{i}\\},\ i=1,2,\ldots,n-1.$

Now two cases may arise:

Case 1: In this case, let there exists $1\leq i<j<k\leq n-1$ such that

$\dim(A_{i}^{\perp}\cap A_{j}^{\perp}\cap A_{k}^{\perp})=1.$

Without loss of generality, let

$\dim(A_{1}^{\perp}\cap A_{2}^{\perp}\cap A_{3}^{\perp})=1.$

If $\displaystyle\bigcap\limits_{i=1}^{4}A_{i}^{\perp}=\\{0\\},\ $ then
together with $\dim(A_{1}^{\perp}\cap A_{2}^{\perp}\cap A_{3}^{\perp})=1$ and
$\omega_{4}\wedge\omega_{4}=0,$ we can say that

$(A_{1}^{\perp}\cap A_{2}^{\perp}\cap
A_{3}^{\perp})=\operatorname{span}\\{b\\}$

and

$\omega_{4}=\alpha\wedge\beta\textrm{ with
}\dim\operatorname{span}\\{\alpha,\beta,b\\}=3\textrm{ for some
}\alpha,\beta,b\in\mathbb{R}^{n}\setminus\\{0\\}.$

Now let

$\omega_{1}=x\wedge b,\omega_{2}=y\wedge b,\omega_{3}=z\wedge b\textrm{ for
some }x,y,z\in\mathbb{R}^{n}\setminus\\{0\\},$

the existence of $x,y,z$ follows from Cartan’s lemma, see Theorem $2.42$ of
[7]. As $\\{\omega_{1},\omega_{2},\omega_{3}\\}$ is linearly independent, it
follows from lemma 4.3 that

$\\{x,y,z,b\\}\textrm{ is linearly independent}.$

If possible, let all of $\\{\alpha,\beta,b,x\\},\\{\alpha,\beta,b,y\\}$ and
$\\{\alpha,\beta,b,z\\}$ are linearly dependent. Then we see that

$x,y,z\in\operatorname{span}\\{\alpha,\beta,b\\}$

and therefore

$\operatorname{span}\\{x,y,z\\}=\operatorname{span}\\{\alpha,\beta,b\\}.$

This gives

$b\in\operatorname{span}\\{x,y,z\\},$

which is a contradiction because $\\{x,y,z,b\\}$ is linearly independent. Now
without loss of generality let $\\{\alpha,\beta,x,b\\}$ is linearly
independent then clearly

$\omega_{4}\wedge\omega_{1}\neq 0$

which contradicts our hypothesis that

$\omega_{i}\wedge\omega_{j}=0\textrm{ for all }1\leq i<j\leq 4.$

Therefore $\displaystyle\bigcap\limits_{i=1}^{4}A_{i}^{\perp}\neq\\{0\\}$ and
hence we can say that

$\bigcap\limits_{i=1}^{4}\ker\\{\omega_{i}\\}^{\perp}=\bigcap\limits_{i=1}^{3}\ker\\{\omega_{i}\\}^{\perp},$
(4.2)

as $\dim\bigcap\limits_{i=1}^{3}\ker\\{\omega_{i}\\}^{\perp}=1$. By similar
argument, we can again show that

$\bigcap\limits_{i=1,i\neq
4}^{5}\ker\\{\omega_{i}\\}^{\perp}=\bigcap\limits_{i=1}^{3}\ker\\{\omega_{i}\\}^{\perp}.$
(4.3)

From equations $(\ref{25/2/2022 1})$ and $(\ref{25/2/2022 2})$ we can say that

$\displaystyle\bigcap\limits_{i=1}^{5}\ker\\{\omega_{i}\\}^{\perp}$
$\displaystyle=\left[\bigcap\limits_{i=1}^{4}\ker\\{\omega_{i}\\}^{\perp}\right]\bigcap\left[\bigcap\limits_{i=1,i\neq
2}^{5}\ker\\{\omega_{i}\\}^{\perp}\right]$
$\displaystyle=\bigcap\limits_{i=1}^{3}\ker\\{\omega_{i}\\}^{\perp}.$ (4.4)

Thus we can assert that

$\bigcap\limits_{i=1}^{3}A_{i}^{\perp}=A_{1}^{\perp}\cap A_{2}^{\perp}\cap
A_{3}^{\perp}\cap A_{k}^{\perp}\ \textrm{ for all }\
k\in\\{4,5,\ldots,n-1\\}.$

Therefore

$\displaystyle\bigcap\limits_{i=1}^{n-1}A_{i}^{\perp}$
$\displaystyle=\bigcap\limits_{i=4}^{n-1}\left(A_{1}^{\perp}\cap
A_{2}^{\perp}\cap A_{3}^{\perp}\cap A_{i}^{\perp}\right)$ (4.5)
$\displaystyle=A_{1}^{\perp}\cap A_{2}^{\perp}\cap A_{3}^{\perp}.$ (4.6)

This implies that

$\dim\left(\bigcap\limits_{i=1}^{n-1}A_{i}^{\perp}\right)=1.$

Therefore

$\bigcap\limits_{i=1}^{n-1}A_{i}^{\perp}=\operatorname{span}\\{b\\}\textrm{
for some }\ b\in\mathbb{R}^{n}\setminus\\{0\\}$

and hence

$\operatorname{span}E=\operatorname{span}\\{\omega_{1},\omega_{2},\ldots,\omega_{n-1}\\}=\mathbb{R}^{n}\wedge
b.$

Case 2: In this case we let

$A_{i}^{\perp}\cap A_{j}^{\perp}\cap A_{k}^{\perp}=\\{0\\}\textrm{ for any }\
1\leq i<j<k\leq n-1.$

As $n\geq 5$, there exists $l\in\\{1,2,\ldots,n-1\\}\setminus\\{i,j,k\\}$ such
that

$\dim A_{l}^{\perp}=2\textrm{ and }\ A_{i}^{\perp}\cap A_{j}^{\perp}\cap
A_{k}^{\perp}\subseteq A_{l}^{\perp}.$

This gives us $\dim(A_{l}^{\perp}+(A_{i}^{\perp}\cap A_{j}^{\perp}\cap
A_{k}^{\perp}))=2$ and so

$\dim(A_{l}\cap(A_{i}+A_{j}+A_{k}))=n-2.$ (4.7)

Because of $A_{i}^{\perp}\cap A_{j}^{\perp}\cap A_{k}^{\perp}=\\{0\\}$,

$A_{i}^{\perp}\cap A_{j}^{\perp}\cap A_{k}^{\perp}\cap A_{l}^{\perp}=\\{0\\}$

and it follows that

$\dim(A_{i}+A_{j}+A_{k}+A_{l})=n.$

Now

$\displaystyle\dim(A_{i}+A_{j}+A_{k}+A_{l})$ $\displaystyle=\dim A_{i}+\dim
A_{j}+\dim A_{k}+\dim A_{l}$ $\displaystyle\ \ -\dim(A_{j}\cap
A_{k})-\dim(A_{j}\cap A_{l})-\dim(A_{k}\cap A_{l})$ $\displaystyle\ \
+\dim(A_{j}\cap A_{k}\cap A_{l})-\dim(A_{i}\cap(A_{j}+A_{k}+A_{l}))$
$\displaystyle=(n-2)\times 4-(n-3)\times 3+\dim(A_{j}\cap A_{k}\cap
A_{l})-(n-2).$ $\displaystyle\ \ \hskip 28.45274pt[\textrm{using lemma
\ref{ilem} and equation \ref{25/2/2022 3}}]$ $\displaystyle=\dim(A_{j}\cap
A_{k}\cap A_{l}).$

That is, $\dim(A_{j}\cap A_{k}\cap A_{l})=n$. But $n\geq 5$, so
$\dim(A_{j}\cap A_{k}\cap A_{l})\geq 5$. Also

$A_{j}\cap A_{k}\cap A_{l}\subseteq A_{j}\cap A_{k}$

and $\dim(A_{j}\cap A_{k})=n-3$ by lemma 4.2. So $\dim(A_{i}\cap A_{j}\cap
A_{k})$ can not be equal to $n$ for $n\geq 5.$ Thus we are getting
contradiction and it follows that case 2 can not happen. ∎

######  Remark 4.5.

  * (i)

We can not apply the proof of the above lemma 4.4 for
$\Lambda^{2}(\mathbb{R}^{4})$, i.e., if $E\subset\Lambda^{2}(\mathbb{R}^{4})$
such that $\dim\operatorname{span}E=3$ and $\omega\wedge\omega^{\prime}=0$ for
any $\omega,\omega^{\prime}\in E$ then it may not happen that
$\operatorname{span}E=\mathbb{R}^{4}\wedge b$ for some
$b\in\mathbb{R}^{4}\setminus\\{0\\}$ because if we take the set $E$ as
$\\{e^{1}\wedge e^{2},e^{1}\wedge e^{3},e^{2}\wedge e^{3}\\}$ then $E$ can not
be written as a subset of $\mathbb{R}^{4}\wedge b$ for any
$b\in\mathbb{R}^{4}\setminus\\{0\\}.$ Importantly, case $2$ of the above lemma
is true for this example.

  * (ii)

The aforementioned lemma is not true if we replace the case
$\omega\wedge\omega^{\prime}=0$ for all $\omega,\omega^{\prime}\in E$ with
$\operatorname{rank}[\omega-\omega^{\prime}]\leq 2$ for any
$\omega,\omega^{\prime}\in E.$ For example, if we take $E$ as

$E=\\{e^{2}\wedge e^{3},e^{2}\wedge e^{3}+e^{1}\wedge e^{2},e^{2}\wedge
e^{3}+e^{1}\wedge e^{3},\ldots,e^{2}\wedge e^{3}+e^{1}\wedge e^{n-1}\\}$

for $n\in\mathbb{N}$ and $n\geq 5,$ then $\dim\operatorname{span}E=n-1$ but
$\operatorname{span}E$ can not be written as $\mathbb{R}^{n}\wedge b$ for any
$b\in\mathbb{R}^{n}.$ For $n=4,$ we can simply take the set $\\{e^{2}\wedge
e^{3},e^{2}\wedge e^{3}+e^{1}\wedge e^{2},e^{2}\wedge e^{3}+e^{1}\wedge
e^{4}\\}$.

We will use the following lemma in Theorem 4.8.

######  Lemma 4.6.

Let $n\in\mathbb{N},n\geq 5$ and $E\subset\Lambda^{2}(\mathbb{R}^{n})$ be such
that $\operatorname{rank}[e-f]\leq 2$ for any $e,f\in E.$ Then
$\dim\operatorname{span}E\leq n.$

######  Remark 4.7.

For $n=4,$ we have done separate proof of this lemma in 4.1 because we will
use lemma 4.4 in the proof below which may not hold for $n=4.$

###### Proof.

We will show that $\dim\operatorname{span}E\leq n$ if we take
$\operatorname{rank}[e-f]\leq 2$ for any $e,f\in E.$ Let us suppose that $E$
contains $n+1$ linearly independent elements
$\omega_{0},\omega_{1},\ldots,\omega_{n}.$ Let

$\psi_{i}:=\omega_{i}-\omega_{0},\ i=1,\ldots,n.$

The set $\\{\psi_{i}:i=1,\ldots,n\\}$ is linearly independent set with

$\psi_{i}\wedge\psi_{j}=0\textrm{ for any }i,j.$

Indeed, $\psi_{i}-\psi_{j}=\omega_{i}-\omega_{j}$ and
$(\omega_{i}-\omega_{j})\wedge(\omega_{i}-\omega_{j})=0$, i.e.,
$(\psi_{i}-\psi_{j})\wedge(\psi_{i}-\psi_{j})=0$. Therefore from lemma 4.4 we
can say that there exists $b\in\mathbb{R}^{n}\setminus\\{0\\}$ such that

$\operatorname{span}\\{\psi_{1},\ldots,\psi_{n-1}\\}=\mathbb{R}^{n}\wedge b.$

Similarly there exists $b^{\prime}\in\mathbb{R}^{n}\setminus\\{0\\}$ such that

$\operatorname{span}\\{\psi_{2},\ldots,\psi_{n}\\}=\mathbb{R}^{n}\wedge
b^{\prime}.$

As $n\geq 5$, we can say from lemma 4.2 that

$\ker\\{\psi_{2}\\}^{\perp}\cap\ker\\{\psi_{3}\\}^{\perp}=\displaystyle\bigcap_{i=1}^{n-1}\ker\\{\psi_{i}\\}^{\perp}$

and

$\ker\\{\psi_{2}\\}^{\perp}\cap\ker\\{\psi_{3}\\}^{\perp}=\displaystyle\bigcap_{i=2}^{n}\ker\\{\psi_{i}\\}^{\perp}$

as $\dim[\ker\\{\psi_{2}\\}^{\perp}\cap\ker\\{\psi_{3}\\}^{\perp}]=1.$
Therefore

$\mathbb{R}^{n}\wedge
b=\operatorname{span}\\{\psi_{1},\ldots,\psi_{n-1}\\}=\mathbb{R}^{n}\wedge
b^{\prime}=\operatorname{span}\\{\psi_{2},\ldots,\psi_{n}\\},$

which is a contradiction because $\\{\psi_{1},\ldots,\psi_{n}\\}$ is a
linearly independent set.

Hence $\dim\operatorname{span}E\leq n.$ ∎

Let us prove the main theorem on differential inclusions of this section using
the previous lemma 4.6 and Theorem 3.3.

######  Theorem 4.8.

Let $n\in\mathbb{N},n\geq 5$ and $E\subseteq\Lambda^{2}(\mathbb{R}^{n})$ be
such that

$\omega\wedge\omega^{\prime}=0\textrm{ for all }\ \omega,\omega^{\prime}\in
E.$

Let $\Omega\subseteq\mathbb{R}^{n}$ be an open, bounded set. Then there exists
$\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ such that

$\displaystyle\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega$
$\displaystyle\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0\textrm{
for all }e\in E$ (4.8) $\displaystyle\textrm{ and
}\int\limits_{\Omega}\eta\neq 0,$

if and only if $0\in\operatorname{rico}E$ and $\dim\operatorname{span}E=n-1.$

###### Proof.

If there exists $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ such that

$\displaystyle\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega$
$\displaystyle\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0\textrm{
for all }e\in E$ (4.9) $\displaystyle\textrm{ and
}\int\limits_{\Omega}\eta\neq 0,$

then from Theorem $2.5$ of [3] we can say that

$\dim\operatorname{span}E\geq n-1.$

As $E$ has the property that $\omega\wedge\omega^{\prime}=0$ for any
$\omega,\omega^{\prime}\in E,$ it implies that

$\operatorname{rank}[\omega-\omega^{\prime}]\leq 2\textrm{ for any
}\omega,\omega^{\prime}\in E.$

Hence from lemma 4.6 it follows that

$\dim\operatorname{span}E\leq n.$

Now for $\dim\operatorname{span}E=n,$ there does not exist any solution
$\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ of the problem (4) which
directly follows from Theorem 3.3. Hence $\dim\operatorname{span}E=n-1.$ Using
lemma $2.4$ of [3], it holds that $0\in\operatorname{rico}E.$

Conversely, if $\dim\operatorname{span}E=n-1,$ then using lemma 4.4 there
exists $b\in\mathbb{R}^{n}\setminus\\{0\\}$ such that
$\operatorname{span}E=\mathbb{R}^{n}\wedge b$. Now a solution $\eta$ exists
satisfying (4.8) from corollary $3.9$ of [3].

∎

######  Remark 4.9.

For $n=4,$ the necessary part of the above theorem is true and it follows from
Theorem 3.3 and lemma 4.1. But the converse part is not true because
$\operatorname{span}E$ may not be written as $\mathbb{R}^{4}\wedge b$ for some
$b\in\mathbb{R}^{4}\setminus\\{0\\}$ always.

##  5\. Existence of Solution: Dimension 2n-3

In this section, we will see one existence result of solution $\eta\in
W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ at dimension $(2n-3)$ for the
following differential inclusion problem

$\displaystyle\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega,$
$\displaystyle\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\eta(x)=e\\}>0\textrm{
for all }e\in E.$

Let $n\in\mathbb{N},n\geq 4.$ Let
$E\subseteq\Lambda^{2}(\mathbb{R}^{n})\setminus\\{0\\}$ and
$\dim\operatorname{span}E=2n-3$ where $E=\\{e^{1}\wedge e^{2},e^{1}\wedge
e^{3},\ldots,e^{1}\wedge e^{n},-e^{1}\wedge e^{2},-e^{1}\wedge
e^{3},\ldots,-e^{1}\wedge e^{n},e^{2}\wedge e^{3},e^{2}\wedge
e^{4},\ldots,e^{2}\wedge e^{n},-e^{2}\wedge e^{3},-e^{2}\wedge
e^{4},\ldots,-e^{2}\wedge e^{n}\\}.$ Clearly,
$\operatorname{span}E=\mathbb{R}^{n}\wedge e^{1}+\mathbb{R}^{n}\wedge e^{2}.$
Let us write $E=E_{1}\cup E_{2}$, where

$E_{1}=\\{e^{1}\wedge e^{2},e^{1}\wedge e^{3},\ldots,e^{1}\wedge
e^{n},-e^{1}\wedge e^{2},-e^{1}\wedge e^{3},\ldots,-e^{1}\wedge e^{n}\\}$

and

$E_{2}=\\{e^{2}\wedge e^{1},e^{2}\wedge e^{3},\ldots,e^{2}\wedge
e^{n},-e^{2}\wedge e^{1},-e^{2}\wedge e^{3},\ldots,-e^{2}\wedge e^{n}\\}.$

Then, $\operatorname{span}E_{1}=\mathbb{R}^{n}\wedge e^{1}$,
$\operatorname{span}E_{2}=\mathbb{R}^{n}\wedge e^{2}$ and
$0\in\operatorname{rico}E_{1}\cap\operatorname{rico}E_{2}$.

Step-1: Let $G=I_{1}\times I_{2}\times\cdots\times I_{n}$ be open unit cube in
$\mathbb{R}^{n}$, where $I_{i}=(0,1)$ for each $i=1,\ldots,n.$ Let us divide
the domain $G$ into two parts as follows:

$G_{1}=\left(0,\frac{1}{2}\right)\times I_{2}\times\cdots\times I_{n}$

and

$G_{2}=\left(\frac{1}{2},1\right)\times I_{2}\times\cdots\times I_{n}.$

Clearly, $G_{1},G_{2}$ are open, bounded sets in $\mathbb{R}^{n}.$ Then there
exist $\overline{\eta_{1}}\in W_{0}^{1,\infty}(G_{1};\mathbb{R}^{n})$ and
$\overline{\eta_{2}}\in W_{0}^{1,\infty}(G_{2};\mathbb{R}^{n})$ such that

$\operatorname{curl}\overline{\eta_{1}}\in E_{1}\textrm{ a.e. in }G_{1}$

and

$\operatorname{curl}\overline{\eta_{2}}\in E_{2}\textrm{ a.e. in }G_{2}.$

Let us define a mapping $\overline{\eta}\in
W_{0}^{1,\infty}(G;\mathbb{R}^{n})$ by

$\overline{\eta}(x)=\begin{cases}&\overline{\eta_{1}}(x),\ \ \textrm{if }x\in
G_{1}\textrm{ a.e. }\\\ &\overline{\eta_{2}}(x),\ \ \textrm{if }x\in
G_{2}\textrm{ a.e. }\end{cases}$

Then

$\operatorname{curl}\overline{\eta}\in E\textrm{ a.e. in }G,$

i.e., there exists $\overline{\eta}\in W_{0}^{1,\infty}(G;\mathbb{R}^{n})$
such that

$\displaystyle\operatorname{curl}\overline{\eta}\in E\textrm{ a.e. in }G,$
$\displaystyle\operatorname{meas}\\{x\in\Omega:\operatorname{curl}\overline{\eta}(x)=e\\}>0\textrm{
for all }e\in E,$

where $\dim\operatorname{span}E=2n-3$ and
$\operatorname{span}E=\mathbb{R}^{n}\wedge e^{1}+\mathbb{R}^{n}\wedge e^{2}.$

Step 2: Let $\Omega\subseteq\mathbb{R}^{n}$ be an open, bounded set. Using
Vitali’s covering theorem, there exists a sequence
$\\{G_{k}:k\in\mathbb{N}\\}$, where $G_{k}$’s are translated and dilated sets
of $\overline{G}$ (closure of $G$) and $G$ is as defined in step-$1$ above
such that

$\displaystyle G_{k}\subseteq\Omega\textrm{ for each }k\in\mathbb{N},$
$\displaystyle G_{h}\cap G_{k}=\emptyset\textrm{ for all
}h,k\in\mathbb{N},h\neq k,$
$\displaystyle\operatorname{meas}\left(\Omega\setminus\bigcup_{k\in\mathbb{N}}G_{k}\right)=0.$

Let $G_{k}:=a_{k}+t_{k}\overline{G}$, where $a_{k}\in\mathbb{R}^{n}$ and
$t_{k}\in\mathbb{R}\setminus\\{0\\}$. Let us define a map $\eta\in
W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ by

$\eta(z)=\begin{cases}\overline{\eta}\left(\frac{z-a_{k}}{t_{k}}\right),\ \
\textrm{ if }z\in G_{k}\\\ 0,\ \ \textrm{ if
}z\in\Omega\setminus\displaystyle\bigcup_{k\in\mathbb{N}}G_{k}.\end{cases}$

Clearly, $\eta\in W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ and

$\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega.$

Thus we are getting a solution $\eta\in
W_{0}^{1,\infty}(\Omega;\mathbb{R}^{n})$ for the following differential
inclusion problem

$\displaystyle\operatorname{curl}\eta\in E\textrm{ a.e. in }\Omega,$
$\displaystyle\operatorname{meas}\\{z\in\Omega:\operatorname{curl}\eta(z)=e\\}>0\textrm{
for all }e\in E.$

##  6\. The Case of a k-form

In this section, we will generalize lemma 3.1 to exterior $k$-form.

######  Theorem 6.1.

Let $n\in\mathbb{N},$ $1\leq k\leq n-3.$ Suppose
$f:\mathbb{R}^{n}\to\Lambda^{k}(\mathbb{R}^{n})$ be continuous such that
$f(0)\neq 0$ and $f(0)$ is $k$-divisible, i.e., $f(0)=c^{1}\wedge\ldots\wedge
c^{k}$ for some $c^{i}\in\mathbb{R}^{n}\setminus\\{0\\}$ for $i=1,\ldots,k.$
Then

$\dim\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}\neq n-k+1.$

###### Proof.

Let

$\mathcal{S}:=\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}.$

Let us suppose to the contrary that $\dim\mathcal{S}=n-k+1.$ We know that

$\mathbb{R}^{n}\wedge f(0)\subseteq\mathcal{S},$ (6.1)

from proposition $2.2$ of [3]. Therefore

$\mathcal{S}=\left(\mathbb{R}^{n}\wedge
f(0)\right)\oplus\operatorname{span}(\omega),$

for some $\omega\in\Lambda^{k+1}(\mathbb{R}^{n})\setminus\\{0\\}$. Using
proposition $2.16$ of [7], we can write

$\displaystyle\langle f(0)\lrcorner\omega;x\rangle$
$\displaystyle=(-1)^{k+1}\langle\omega;f(0)\wedge x\rangle$
$\displaystyle=0,\textrm{ because }\omega\in\left[\mathbb{R}^{n}\wedge
f(0)\right]^{\perp}\textrm{ for all }x\in\mathbb{R}^{n}.$

Therefore

$f(0)\lrcorner\ \omega=0.$ (6.2)

Let us choose $x^{0}\in\mathbb{R}^{n}$ such that $x^{0}\wedge
f(x^{0})\in\Lambda^{k+1}\setminus(\mathbb{R}^{n}\wedge f(0))$. Such an $x^{0}$
exists because $\mathbb{R}^{n}\wedge f(0)\subsetneqq\mathcal{S}$. Since
$\Lambda^{k+1}\setminus(\mathbb{R}^{n}\wedge f(0))$ is open, it follows from
the continuity of $f$ that, for some $\epsilon>0$

$x\wedge f(x)\in\Lambda^{k+1}\setminus(\mathbb{R}^{n}\wedge f(0)),$

for all $x\in B(x^{0},\epsilon).$

Let us choose a basis $\\{a^{1},\ldots,a^{n}\\}$ of $\mathbb{R}^{n}$ within
$B(x^{0},\epsilon)$. Then for each $j\in\\{1,2,\ldots,n\\}$, we find
$\beta^{j}\in\mathbb{R}$ and $b^{j}\in\mathbb{R}^{n}$ such that

$a^{j}\wedge f(a^{j})=b^{j}\wedge f(0)+\beta^{j}\omega.$

Note that, $\beta^{j}\neq 0\textrm{ for all }j\in\\{1,2,\ldots,n\\}.$

We have $f(0)=c^{1}\wedge\cdots\wedge c^{k}$. Let $1\leq r\leq k$ be fixed.
Then for all $j\in\\{1,\ldots,n\\}$,

$\beta^{j}(\omega\wedge c^{r}\wedge a^{j})=0$

implies that

$(\omega\wedge c^{r})\wedge a^{j}=0\textrm{ for all }j=1,\ldots,n.$

This gives us

$\omega\wedge c^{r}=0\textrm{ for all }r\in\\{1,\ldots,k\\}.$

Therefore, we have

$\omega=(c^{1}\wedge\ldots\wedge c^{k})\wedge\omega^{\prime}$

for some $\omega^{\prime}\in\mathbb{R}^{n},$ where
$c^{i}\lrcorner\omega^{\prime}=0$ for all $i=1,\ldots,k.$

Since $(c^{1}\wedge\ldots\wedge c^{k})\lrcorner\omega=0$ [from equation
(6.2)], we have $\omega^{\prime}=0$.

Hence $\omega=0$, a contradiction. This proves the theorem. ∎

######  Remark 6.2.

For $k=n-2$ and $n\geq 4$, there exists a continuous function such that the
above Theorem 6.1 fails. For example, let us take
$f:\mathbb{R}^{n}\to\Lambda^{n-2}(\mathbb{R}^{n})$ defined by

$f(x_{1},\ldots,x_{n})=(x_{1}+1)e^{1}\wedge\cdots\wedge e^{n-2}+x_{2}\
e^{1}\wedge\cdots\wedge e^{n-3}\wedge e^{n-1},$

for all $x=(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}$. Then

$\displaystyle x\wedge f(x)$
$\displaystyle=(x_{1}e^{1}+\cdots+x_{n}e^{n})\wedge f(x)$
$\displaystyle=\left((-1)^{n-3}x_{n-2}x_{2}+(-1)^{n-2}x_{n-1}(x_{1}+1)\right)e^{1}\wedge\cdots\wedge
e^{n-1}$ $\displaystyle\ +(-1)^{n-2}x_{n}(x_{1}+1)e^{1}\wedge\cdots\wedge
e^{n-2}\wedge e^{n}$ $\displaystyle\
+(-1)^{n-2}x_{n}x_{2}e^{1}\wedge\cdots\wedge e^{n-3}\wedge e^{n-1}\wedge
e^{n}.$

Clearly, $\dim\operatorname{span}\\{x\wedge f(x):x\in\mathbb{R}^{n}\\}$ has
dimension $n-(n-2)+1=3.$

For $k=n-1,$ the above Theorem 6.1 is trivially true.

## Funding

The research of N. Nesha was supported by CSIR Senior Research Fellowship
[File No:09/921(0309)/2020-EMR-I].

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