# Biharmonic Steklov operator on differential forms

Fida El Chami<EMAIL_ADDRESS>Lebanese University, Faculty of Sciences II,
Department of Mathematics, P.O. Box 90656 Fanar-Matn, Lebanon Nicolas Ginoux
<EMAIL_ADDRESS>Université de Lorraine, CNRS, IECL, F-57000
Metz  Georges Habib<EMAIL_ADDRESS>Lebanese University, Faculty of Sciences
II, Department of Mathematics, P.O. Box 90656 Fanar-Matn, Lebanon Université
de Lorraine, CNRS, IECL, F-57000 Metz  Ola Makhoul<EMAIL_ADDRESS>Lebanese University, Faculty of Sciences II, Department of Mathematics, P.O.
Box 90656 Fanar-Matn, Lebanon

Abstract. We introduce the biharmonic Steklov problem on differential forms by
considering suitable boundary conditions. We characterize its smallest
eigenvalue and prove elementary properties of the spectrum. We obtain various
estimates for the first eigenvalue, some of which involve eigenvalues of other
problems such as the Dirichlet, Neumann, Robin and Steklov ones.
Independently, new inequalities relating the eigenvalues of the latter
problems are proved.
---

Mathematics Subject Classification (2010): 53C21, 58J32, 58C40, 58J50

Keywords: Manifolds with boundary, Steklov operator, biharmonic boundary value
problem, Robin boundary value problem, eigenvalue estimates

##  1 Introduction

Let $(M^{n},g)$ be an $n$-dimensional compact Riemannian manifold with
nonempty smooth boundary $\partial M$. Denote by $\nu$ the inward unit vector
field normal to the boundary and by $\Delta f:=-\mathrm{tr}(\nabla^{2}f)$ the
Laplace operator applied to a smooth function $f$ on $M$. The following fourth
order eigenvalue boundary problem

$\left\\{\begin{array}[]{lll}\Delta^{2}f&=0&\textrm{ on }M\\\ f&=0&\textrm{ on
}\partial M\\\ \Delta f-q\frac{\partial f}{\partial\nu}&=0&\textrm{ on
}\partial M,\end{array}\right.$ (1)

also called biharmonic Steklov problem I, or biharmonic Steklov for
simplicity, was first introduced by Kuttler and Sigillito [16] and Payne [19].
Its physical interpretation in terms of the deformation of an elastic plate
under the action of transversal forces can be found in e.g. [6, p. 316] and
[26, p. 2637]. When $M$ is a bounded domain in ${\mathbb{R}}^{n}$, the
spectrum of this problem has been studied in [6] and proved to be discrete
consisting of positive eigenvalues of finite multiplicities (see also [2] for
the case when the boundary is not smooth). Variational characterizations of
the first eigenvalue $q_{1}$ have been also given in [6] (see also [15]).
Moreover, in the case of a ball, the spectrum has been calculated explicitly.

In the case of a compact $n$-dimensional Riemannian manifold $M$ with smooth
boundary $\partial M$, sharp estimates for the first eigenvalue $q_{1}$ of the
biharmonic Steklov operator are given in [26] and [22]. It is shown that

$q_{1}\leq\frac{{\rm Vol}(\partial M)}{{\rm Vol}(M)}.$ (2)

In [22, Thm. 2], Raulot and Savo proved that, if $M$ is a geodesic ball in a
space form, then (2) is an equality (see also [26, Thm. 1.3]). Under some
assumptions on the Ricci curvature of $M$ and if the mean curvature of the
boundary is bounded below by a positive constant $H_{0}$, then the first
eigenvalue $q_{1}$ also satisfies

$q_{1}\geq nH_{0}.$

Moreover, equality holds if and only if $M$ is isometric to a ball of radius
$\frac{1}{H_{0}}$ in ${\mathbb{R}}^{n}$ (see [26, Thm. 1.2] and [22, Thm. 2]).
For other recent results on the biharmonic Steklov eigenvalue problem, we
refer to [8, 10] and the references listed therein.

On the other hand, recall that a compact Riemannian manifold $M$ with smooth
nonempty boundary $\partial M$ is called a _harmonic domain_ [21, p. 893] if
and only if it supports a (necessarily unique) solution $f$ to the Serrin
boundary value problem

$\left\\{\begin{array}[]{rrl}\Delta f&=1&\textrm{on }M\\\ f&=0&\textrm{on
}\partial M\\\ \partial_{\nu}f&=c&\textrm{on }\partial M\end{array}\right.$
(3)

for some constant $c\in{\mathbb{R}}$. From [25, 27], we know that the only
harmonic domains in ${\mathbb{R}}^{n}$ are the Euclidean balls of radius
$nc=\frac{n\mathrm{Vol}(M)}{\mathrm{Vol}(\partial M)}$. Independently, it is
not difficult to check that a solution to the Serrin problem (3) is an
eigenfunction of (1).

The aim of this paper is first to extend the biharmonic Steklov problem (1) to
the context of differential forms. As we mentioned above, there is a
relationship between problems (1) and (3), hence the idea is to also define
the Serrin problem on differential forms. For this purpose, we assume that the
manifold $M$ carries a non-trivial parallel form and introduce the
generalization of (3) (see (4)). In this case, we show that $M$ is a harmonic
domain if and only if there exists on $M$ a solution to (4). Also in Section
2, we provide a natural extension of problem (1) to the case of differential
forms (problem (5)). Applying the techniques used in [6] by Ferrero, Gazzola
and Weth, we show in Theorem 2.3 that problem (5) has a discrete spectrum
consisting of a countable number of positive eigenvalues of finite
multiplicities. This involves proving the ellipticity of problem (55) (see
Lemma $6.1$ in the appendix). Moreover, we give two variational
characterizations of the first eigenvalue of this problem (Theorem 2.6) which
will be useful to establish inequalities in the sequel.

In Section 3, we obtain different estimates regarding the eigenvalues of
problem (5). As a preliminary step, we prove an interesting property of that
problem, namely its invariance by the Hodge star operator. On the other hand,
under curvature assumptions, using a previous result by Raulot and Savo [22,
Thm. 10], we derive a lower bound (15) for the first eigenvalue $q_{1,p}$,
which generalizes the estimates [26, Thm. 1.2] and [22, Thm. 2]. On the other
hand, when the manifold $M$ supports a non-trivial parallel $p$-form, we show
that

$q_{1,p}\leq\frac{{\rm Vol}(\partial M)}{{\rm Vol}(M)},$

with equality if and only if $M$ is a harmonic domain. Surprisingly, when $M$
is a domain of ${\mathbb{R}}^{n}$, the eigenvalues of the biharmonic Steklov
problem on differential forms are the same as those of the scalar problem,
without taking into consideration their multiplicities. It should be noted
that the same type of result is true for the eigenvalues of the Dirichlet
problem. We end Section 3 with an inequality relating the eigenvalues
corresponding to degrees $p-1$, $p$ and $p+1$ on the sphere. That inequality
is established by first proving a more general result when $M$ is
isometrically immersed in a Euclidean space and by using the variational
characterization (12) of the first eigenvalue. The computations involved in
these results being rather technical, we place the details in the appendix
(Lemma 6.2 and Proposition 6.3) to lighten the text.

In Sections 4 and 5, we establish several bounds concerning the eigenvalues of
various differential operators, again using variational characterizations. For
example, we give estimates involving the eigenvalues of the Robin eigenvalue
problem (38) introduced in [4] and those of Neumann (40), Dirichlet (41),
biharmonic Steklov (5) eigenvalue problems on differential forms (Theorems 4.1
and 5.1). Note that similar results are known in the case of scalar problems
(see [13, Thm. 1.17]). In Theorem 4.2, we also give an estimate, under some
curvature condition along the boundary, for the difference between the first
eigenvalues of the Robin problem on $q$ and $(q-p)$-forms, for some $p$ and
$q$ such that $p\leq q$.
Other inequalities involving eigenvalues of the Steklov problem on forms (see
Section 5) are obtained using properties already established for the
eigenvalues and eigenforms of problem (5).

Acknowledgment: We thank Alberto Ferrero for his skilled guidance in the
theory of elliptic boundary value problems and for important references. The
first, third and fourth named benefited from a grant of the Lebanese
University. The second named author thanks the Agence Universitaire de la
Francophonie (AUF) for its financial support and the Lebanese University for
its warm welcome during his stay. The third named author also thanks the
Humboldt Foundation for its support.

##  2 Biharmonic Steklov operator on differential forms

###  2.1 Serrin problem

In this subsection, we extend the Serrin problem to differential forms. This
extension will motivate us to define the biharmonic Steklov problem on
differential forms.

Recall that the Serrin problem is given by the following [25]: Let
$M\subset{\mathbb{R}}^{n}$ be a bounded domain and let $f$ be a solution to
the problem

$\left\\{\begin{array}[]{lll}\Delta f&=1&\textrm{ on }M\\\ f&=0&\textrm{ on
}\partial M.\\\ \end{array}\right.$

If the inner normal derivative of the function $f$ is a constant $c$, then the
domain $M$ must be a ball of radius $nc$ and the function $f$ has the form
$(n^{2}c^{2}-r^{2})/2n$. Here $c$ is equal to $\frac{{\rm Vol}(M)}{{\rm
Vol}(\partial M)}$. The proof that the Euclidean ball is the unique domain in
${\mathbb{R}}^{n}$ supporting a solution to the Serrin problem was given in
[25, Thm. 1] by using the method of moving planes, which is based on Hopf’s
maximum principle. In [27], H.F. Weinberger suggested an elementary proof
introducing so-called $P$-functions for the Laplacian. Since then, the Serrin
problem has been generalized to several contexts and when the ambient space is
a simply connected space form [14, 18, 3].

A natural question to ask in this set-up is whether the Serrin problem can be
extended to differential forms on a domain in ${\mathbb{R}}^{n}$. For this
purpose, we fix $p\in\\{0,\cdots,n\\}$ and consider, on the set of
differential $p$-forms $\Omega^{p}(M)$, a solution to the system

$\left\\{\begin{array}[]{lll}\Delta\omega&=\omega_{0}&\textrm{ on }M\\\
\omega&=0&\textrm{ on }\partial M,\\\ \end{array}\right.$

where $\omega_{0}$ is a given parallel form on ${\mathbb{R}}^{n}$ assumed to
be of norm $1$. We now set the following question: If the conditions
$\nu\lrcorner d\omega=c\iota^{*}\omega_{0}$ and
$\iota^{*}(\delta\omega)=-c\nu\lrcorner\omega_{0}$ are satisfied on $\partial
M$ for some constant $c$ and where $\nu$ is the inner unit normal vector field
to $\partial M$, can one deduce that the manifold $M$ is a ball of radius
$nc$? Here, $\iota:\partial M\to M$ denotes the inclusion map. Notice that for
$p=0$, the problem that we propose reduces to the usual one on functions. The
answer of this question is given in Proposition 2.1 below.

###### Proposition 2.1

Let $M$ be a compact manifold with smooth nonempty boundary $\partial M$ and
carrying a nontrivial parallel $p$-form $\omega_{0}$.

  1. 1.

If $M$ is a harmonic domain, then for the solution $f$ to the Serrin problem
on $M$ the $p$-form $\omega:=f\cdot\omega_{0}$ is the unique solution to the
boundary value problem

$\left\\{\begin{array}[]{lll}\Delta\omega&=\omega_{0}&\textrm{on }M\\\
\omega_{|_{\partial M}}&=0&\textrm{on }\partial M\\\ \nu\lrcorner
d\omega&=c\iota^{*}\omega_{0}&\textrm{on }\partial M\\\
\iota^{*}(\delta\omega)&=-c\nu\lrcorner\omega_{0}&\textrm{on }\partial
M\end{array}\right.$ (4)

for some constant $c\in{\mathbb{R}}$.

  2. 2.

Conversely, if (4) has a solution $\omega\in\Omega^{p}(M)$, then assuming
w.l.o.g. that $|\omega_{0}|=1$ on $M$, we have that $\omega=f\cdot\omega_{0}$
where $f$ solves (3). As a consequence, $M$ must be a harmonic domain.

Proof. Before proving the proposition, we begin with the following fact. Given
any parallel form $\alpha$ and a smooth function $h$ on $M$, we have that
$\Delta(h\alpha)=(\Delta h)\alpha$. To see this, we first have
$d(h\alpha)=dh\wedge\alpha$ and $\delta(h\alpha)=-dh\lrcorner\alpha$, as
$\alpha$ is parallel. Therefore, if we take $\\{e_{1},\cdots,e_{n}\\}$ a local
orthonormal frame of $TM$, we compute

$\displaystyle\Delta(h\alpha)$ $\displaystyle=$
$\displaystyle\delta(dh\wedge\alpha)-d(dh\lrcorner\alpha)$ $\displaystyle=$
$\displaystyle-\sum_{i=1}^{n}e_{i}\lrcorner(\nabla_{e_{i}}dh\wedge\alpha)-\sum_{i=1}^{n}e_{i}\wedge(\nabla_{e_{i}}dh\lrcorner\alpha)$
$\displaystyle=$ $\displaystyle(\Delta
h)\alpha+\sum_{i=1}^{n}\nabla_{e_{i}}dh\wedge(e_{i}\lrcorner\alpha)-\sum_{i=1}^{n}e_{i}\wedge(\nabla_{e_{i}}dh\lrcorner\alpha)$
$\displaystyle=$ $\displaystyle(\Delta h)\alpha,$

since $\nabla dh$ is a symmetric $2$-tensor field.
As a first consequence, if $f$ solves (3), then for any parallel $p$-form
$\omega_{0}$ on $M$ the $p$-form $\omega:=f\cdot\omega_{0}$ solves
$\Delta\omega=(\Delta f)\cdot\omega_{0}=\omega_{0}$ on $M$ together with
$\omega_{|_{\partial M}}=f_{|_{\partial M}}\cdot\omega_{0}=0$. As for the
other two boundary conditions, note that, if $\omega_{|_{\partial M}}=0$, then
by $\nabla_{X}\omega=0$ for all $X\in T\partial M$ we have that
$d\omega=\nu^{\flat}\wedge\nabla_{\nu}\omega$ and
$\delta\omega=-\nu\lrcorner\,\nabla_{\nu}\omega$. Therefore,

$\nu\lrcorner\,d\omega=\nu\lrcorner(\nu^{\flat}\wedge\nabla_{\nu}\omega)=\nabla_{\nu}\omega-\nu^{\flat}\wedge(\nu\lrcorner\,\nabla_{\nu}\omega)=\iota^{*}\nabla_{\nu}\omega=\partial_{\nu}f\cdot\iota^{*}\omega_{0}=c\iota^{*}\omega_{0}$

and

$\iota^{*}(\delta\omega)=-\iota^{*}(\nu\lrcorner\nabla_{\nu}\omega)=-\nu\lrcorner\nabla_{\nu}\omega=-\partial_{\nu}f\cdot\nu\lrcorner\,\omega_{0}=-c\nu\lrcorner\,\omega_{0},$

so that $\omega$ solves (4). Note that, since the Dirichlet boundary condition
$\omega_{|_{\partial M}}=0$ forces $\ker(\Delta)=\\{0\\}$ [1, Thm. p. 445],
the $p$-form $\omega$ is necessarily the only solution to (4). This proves 1.
Conversely, let $\omega$ solve (4) for some nontrivial parallel $p$-form
$\omega_{0}$. Up to rescaling $\omega_{0}$ we may assume that $|\omega_{0}|=1$
on $M$. We consider the function $f:=\langle\omega,\omega_{0}\rangle$ on $M$.
By the Bochner formula and $\nabla\omega_{0}=0$, we have that

$\Delta
f=\langle\nabla^{*}\nabla\omega,\omega_{0}\rangle=\langle\Delta\omega,\omega_{0}\rangle-\underbrace{\langle\mathfrak{B}^{[p]}\omega,\omega_{0}\rangle}_{0}=|\omega_{0}|^{2}=1.$

Here, we use the fact that the Bochner operator $\mathfrak{B}^{[p]}$ is a
symmetric tensor. Also, it is immediate to see that $f_{|_{\partial M}}=0$
since $\omega_{|_{\partial M}}=0$. Therefore, we deduce that

$\Delta(\omega-f\omega_{0})=\omega_{0}-(\Delta f)\omega_{0}=0,$

on $M$ and $(\omega-f\omega_{0})|_{\partial M}=0$. Hence by triviality of the
Dirichlet kernel, we deduce that $\omega=f\cdot\omega_{0}$ on $M$. In order to
finish the proof, we still have to compute the normal derivative of $f$:

$\displaystyle\partial_{\nu}f$ $\displaystyle=$
$\displaystyle\langle\nabla_{\nu}\omega,\omega_{0}\rangle$ $\displaystyle=$
$\displaystyle\langle\iota^{*}\nabla_{\nu}\omega,\iota^{*}\omega_{0}\rangle+\langle\nu\lrcorner\nabla_{\nu}\omega,\nu\lrcorner\omega_{0}\rangle$
$\displaystyle=$ $\displaystyle\langle\nu\lrcorner
d\omega,\iota^{*}\omega_{0}\rangle-\langle\iota^{*}(\delta\omega),\nu\lrcorner\omega_{0}\rangle$
$\displaystyle=$ $\displaystyle
c\langle\iota^{*}\omega_{0},\iota^{*}\omega_{0}\rangle+c\langle\nu\lrcorner\omega_{0},\nu\lrcorner\omega_{0}\rangle$
$\displaystyle=$ $\displaystyle c|\omega_{0}|^{2}=c.$

Here we used the identities

$\left\\{\begin{array}[]{lll}\nu\lrcorner\nabla_{\nu}\omega&=&\delta^{\partial
M}(\iota^{*}\omega)-\iota^{*}(\delta\omega)-S^{[p-1]}(\nu\lrcorner\omega)+(n-1)H\nu\lrcorner\omega\\\
\iota^{*}\nabla_{\nu}\omega&=&d^{\partial M}(\nu\lrcorner\omega)+\nu\lrcorner
d\omega+S^{[p]}(\iota^{*}\omega).\end{array}\right.$

stated in [20, Lem. 18]. Therefore the function $f$ is a solution to the
Serrin problem on $M$. This concludes the proof of 2 and of Proposition 2.1.
$\square$

###### Remark 2.2

We notice that if we impose that $\iota^{*}\omega_{0}$ is nowhere vanishing
along the boundary, the last boundary condition in (4) can be dropped. Indeed,
with the boundary condition $\nu\lrcorner d\omega=c\iota^{*}\omega_{0}$ and
the explicit form $\omega=f\cdot\omega_{0}$, we compute

$c\iota^{*}\omega_{0}=\nu\lrcorner d\omega=\nu\lrcorner
d(f\omega_{0})=(\partial_{\nu}f)\omega_{0}-df\wedge(\nu\lrcorner\omega_{0})=(\partial_{\nu}f)\iota^{*}\omega_{0},$

from which $\partial_{\nu}f=c$ along $\partial M$ follows. However, the
condition $\iota^{*}\omega_{0}\neq 0$ is not always assured.

The Serrin problem on functions is closely related to the biharmonic Steklov
operator, that is the boundary problem (1). Indeed, as mentioned in the
introduction, on a given compact Riemannian manifold $(M^{n},g)$ (not
necessarily a domain in ${\mathbb{R}}^{n}$), any solution to the Serrin
problem is a solution to (1) with $q=\frac{1}{c}$. Conversely, it was shown in
[21, Thm. 10] that the first positive eigenvalue $q_{1}$ of problem (1) is
bounded from below by the first eigenvalue of the Dirichlet-to-Neumann
operator on $n$-forms (see Section 5 for the definition) and, when equality
occurs, the corresponding eigenfunction $f$ of (1) is a solution to the Serrin
problem. Notice here that, by [6], problem (1) admits a discrete spectrum that
consists of a nondecreasing sequence of positive eigenvalues $(q_{i})_{i}$ of
finite multiplicities.

In order to have a similar situation on differential forms, we come back to
the Serrin problem defined in (4) for domains in ${\mathbb{R}}^{n}$. In fact,
one can easily see that any solution to the Serrin problem (4) gives rise to a
solution to the following boundary problem

$\left\\{\begin{array}[]{lll}\Delta^{2}\omega&=0&\textrm{ on }M\\\
\omega&=0&\textrm{ on }\partial M\\\
\nu\lrcorner\Delta\omega+q\iota^{*}\delta\omega&=0&\textrm{ on }\partial M\\\
\iota^{*}\Delta\omega-q\nu\lrcorner\,d\omega&=0&\textrm{ on }\partial
M,\end{array}\right.$ (5)

with $q=\frac{1}{c}$. The equation $\Delta^{2}\omega=0$ comes from taking the
Laplacian of $\Delta\omega=\omega_{0}$ and using the fact that $\omega_{0}$ is
a parallel form. Note here that, because of $\omega_{|_{\partial M}}=0$, the
last two boundary conditions in (5) are actually equivalent to

$\Delta\omega=q\nabla_{\nu}\omega$

along $\partial M$ since $\nu\lrcorner\,d\omega=\iota^{*}\nabla_{\nu}\omega$
and $\iota^{*}\delta\omega=-\nu\lrcorner\nabla_{\nu}\omega$ as we have seen in
the proof of Proposition 2.1.

It is then natural to study problem (5) for compact Riemannian manifolds with
smooth boundary that are not necessarily domains in ${\mathbb{R}}^{n}$.

###  2.2 Biharmonic Steklov operator

In this section, we will show that the spectrum of problem (5) is discrete and
it consists entirely of eigenvalues of finite multiplicities. We mainly follow
[6].

First, note that on a compact Riemannian manifold $(M^{n},g)$ with smooth
boundary $\partial M$, we have the following integration by parts, which is
valid for any $\omega,\omega^{\prime}\in\Omega^{p}(M)$:

$\displaystyle\int_{M}\Big{(}\langle\Delta\omega,\omega^{\prime}\rangle-\langle\omega,\Delta\omega^{\prime}\rangle\Big{)}\,d\mu_{g}$
(6) $\displaystyle=$ $\displaystyle\int_{\partial M}\Big{(}\langle\nu\lrcorner
d\omega,\iota^{*}\omega^{\prime}\rangle-\langle\iota^{*}\omega,\nu\lrcorner
d\omega^{\prime}\rangle+\langle\nu\lrcorner\omega,\iota^{*}\delta\omega^{\prime}\rangle-\langle\iota^{*}\delta\omega,\nu\lrcorner\omega^{\prime}\rangle\Big{)}\,d\mu_{g}.$

Thus, replacing $\omega$ by $\Delta\omega$ in (6), we obtain:

$\displaystyle\int_{M}\langle\Delta^{2}\omega,\omega^{\prime}\rangle\,d\mu_{g}$
$\displaystyle=$
$\displaystyle\int_{M}\langle\Delta\omega,\Delta\omega^{\prime}\rangle\,d\mu_{g}$
(7) $\displaystyle+$ $\displaystyle\int_{\partial M}\Big{(}\langle\nu\lrcorner
d\Delta\omega,\iota^{*}\omega^{\prime}\rangle-\langle\iota^{*}\Delta\omega,\nu\lrcorner
d\omega^{\prime}\rangle+\langle\nu\lrcorner\Delta\omega,\iota^{*}\delta\omega^{\prime}\rangle-\langle\iota^{*}\delta\Delta\omega,\nu\lrcorner\omega^{\prime}\rangle\Big{)}\,d\mu_{g}.$

The main result of this section is the following:

###### Theorem 2.3

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary
$\partial M$ and let $\nu$ be the inward unit vector field normal to the
boundary. Then the boundary problem

$\left\\{\begin{array}[]{lll}\Delta^{2}\omega&=0&\textrm{ on }M\\\
\omega&=0&\textrm{ on }\partial M\\\
\nu\lrcorner\Delta\omega+q\iota^{*}\delta\omega&=0&\textrm{ on }\partial M\\\
\iota^{*}\Delta\omega-q\nu\lrcorner\,d\omega&=0&\textrm{ on }\partial
M,\end{array}\right.$

on $p$-forms, has a discrete spectrum consisting of an unbounded monotonously
nondecreasing sequence of positive eigenvalues of finite multiplicities
$(q_{j,p})_{j\geq 1}$.

Proof. As in [6, Eq. (1.7)], we let

$Z:=\left\\{\omega\in\Omega^{p}(M)\,|\,\Delta^{2}\omega=0\;\textrm{on}\;M\textrm{
and }\omega_{|_{\partial M}}=0\right\\}.$

We define the following Hermitian sesquilinear forms on $Z$: for all
$\omega,\omega^{\prime}\in\Omega^{p}(M)$,

$\left(\omega,\omega^{\prime}\right)_{V}:=\int_{M}\langle\Delta\omega,\Delta\omega^{\prime}\rangle\,d\mu_{g}\;\textrm{and}\;\left(\omega,\omega^{\prime}\right)_{W}:=\int_{\partial
M}\langle\nu\lrcorner d\omega,\nu\lrcorner
d\omega^{\prime}\rangle\,d\mu_{g}+\int_{\partial
M}\langle\iota^{*}\delta\omega,\iota^{*}\delta\omega^{\prime}\rangle\,d\mu_{g}.$

We split the proof of Theorem 2.3 into the following lemmas.

###### Lemma 2.4

For $\omega\in Z$, we have

$\int_{M}|\Delta\omega|^{2}\,d\mu_{g}+\int_{\partial
M}\Big{(}\langle\nu\lrcorner\Delta\omega,\iota^{*}\delta\omega\rangle-\langle\iota^{*}\Delta\omega,\nu\lrcorner
d\omega\rangle\Big{)}\,d\mu_{g}=0.$ (8)

The forms $(\cdot\,,\cdot)_{V}$ and $(\cdot\,,\cdot)_{W}$ are positive
definite on $Z$. Moreover, there exists a positive constant $C$ such that
$\|\cdot\|_{W}\leq C\cdot\|\cdot\|_{V}$ on $Z$. As a consequence, if we denote
by $V$ (resp. $W$) the completion of $Z$ w.r.t. $\|\cdot\|_{V}$ (resp.
$\|\cdot\|_{W}$) as Hilbert spaces, then there is a natural bounded linear map
$I_{1}\colon V\to W$ extending the identity map $\mathrm{Id}_{Z}$.

Proof. To prove (8) for $\omega\in Z$, we replace $\omega^{\prime}$ in (7) by
$\omega$ and use the fact that $\Delta^{2}\omega=0$ and
$\iota^{*}\omega=\nu\lrcorner\omega=0$. Since, if $\Delta\omega=0$ on $M$ and
$\omega_{|_{\partial M}}=0$, then $\omega=0$ (see e.g. [1, Thm. p. 445], the
sesquilinear form $(\cdot\,,\cdot)_{V}$ is positive definite. For
$(\cdot\,,\cdot)_{W}$, positive definiteness is a consequence of (8). In fact,
if $\left(\omega,\omega\right)_{W}=0$, then $\nu\lrcorner d\omega=0$ and
$\iota^{*}\delta\omega=0$ on $\partial M$ and therefore, from equation (8),
$\Delta\omega=0$ on $M$, from which $\omega=0$ on $M$ follows again by [1,
Thm. p. 445] since $\omega_{|_{\partial M}}=0$. We now show, as in [6, Sec.
2], the existence of a positive constant $C$ such that $\|\cdot\|_{W}\leq
C\cdot\|\cdot\|_{V}$ on $Z$. First, both $\|\cdot\|_{V}$ and
$\|\cdot\|_{H^{2}(M)}$ are equivalent on $Z$. To see this, we have for any
$\omega\in Z$

$\|\omega\|_{V}=\|\Delta\omega\|_{L^{2}(M)}\leq C\cdot\|\omega\|_{H^{2}(M)}$

for some constant $C$ depending only on $M$ and $n$. On the other hand, by the
elliptic estimates and using the fact that, given any $f\in L^{2}(M)$, there
exists a _unique_ weak solution $\omega$ to the boundary value problem
$\Delta\omega=f$ on $M$ with $\omega_{|_{\partial M}}=0$, we have, for any
$\omega\in Z$ that $\|\omega\|_{L^{2}(M)}\leq
C\cdot\|\Delta\omega\|_{L^{2}(M)}$ for some constant $C$, so that

$\|\omega\|_{H^{2}(M)}\leq
C\cdot\left(\|\Delta\omega\|_{L^{2}(M)}+\|\omega\|_{L^{2}(M)}\right)\leq
C\cdot\|\Delta\omega\|_{L^{2}(M)}=C\cdot\|\omega\|_{V}$

for some positive constant that we also denote by $C$ and which again depends
only on $M$ and $n$, see e.g. [5, Thm. 4 in Sec. 6.3]. Therefore, both
$\|\cdot\|_{V}$ and $\|\cdot\|_{H^{2}(M)}$ are equivalent on $Z$. Finally,
using the fact that both $d$ and $\delta$ are first-order linear differential
operators, we estimate, for any $\omega\in Z$,

$\displaystyle\|\omega\|_{W}^{2}$ $\displaystyle=$
$\displaystyle\|\nu\lrcorner d\omega\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial M)}^{2}$ $\displaystyle\leq$
$\displaystyle C\cdot\|\omega\|_{H^{1}(\partial M)}^{2}$ $\displaystyle\leq$
$\displaystyle C\cdot\|\omega\|_{H^{2}(M)}^{2}$ $\displaystyle\leq$
$\displaystyle C\cdot\|\omega\|_{V}^{2}$

for some positive constant that we also denote by $C$, which again depends
only on $M$ and $n$. Here we have also used the boundedness of the trace map
$T\colon H^{2}(M)\to H^{1}(\partial M)$. $\square$

Next we consider the linear operator $K\colon V\to V$ defined by

$K:=D_{V}^{-1}\circ{}^{t}\\!I_{1}\circ D_{W}\circ I_{1},$

where $D_{V}\colon V\to V^{\prime}$ and $D_{W}\colon W\to W^{\prime}$ are the
natural duality isomorphisms, i.e.
$D_{V}(\omega):=\left(\cdot,\omega\right)_{V}$ and
$D_{W}(\omega):=\left(\cdot,\omega\right)_{W}$ for every $\omega$ in $V$ resp.
$W$. As usual, ${}^{t}\\!I_{1}(\theta):=\theta\circ I_{1}\in V^{\prime}$ for
every $\theta\in W^{\prime}$. Actually $K$ can be defined via the identity

$\left(K\omega,\omega^{\prime}\right)_{V}=\left(I_{1}\omega,I_{1}\omega^{\prime}\right)_{W}\left(=\left(\omega,\omega^{\prime}\right)_{W}\right)$

for all $\omega,\omega^{\prime}\in V$. By definition, the operator $K$ is
self-adjoint and positive semi-definite. We need now to prove the following.

###### Lemma 2.5

The map $I_{1}$, defined in the previous lemma, is compact and injective.
Therefore $K$ is also compact and injective.

Proof. Let $I_{3}\colon V\to L^{2}(\partial M)\oplus L^{2}(\partial M)$ be the
composition of the following linear maps:

$\begin{array}[]{ccccc}V&\longrightarrow&H^{\frac{1}{2}}(\partial M)\oplus
H^{\frac{1}{2}}(\partial M)&\longrightarrow&L^{2}(\partial M)\oplus
L^{2}(\partial M)\\\ \omega&\longmapsto&(\nu\lrcorner
d\omega,\iota^{*}\delta\omega)&\longmapsto&(\nu\lrcorner
d\omega,\iota^{*}\delta\omega).\end{array}$

Note that $I_{3}$ is well-defined since the trace operator maps $H^{1}(M)$
into (and onto) $H^{\frac{1}{2}}(\partial M)$. Moreover, since the inclusion
map $H^{\frac{1}{2}}(\partial M)\to L^{2}(\partial M)$ is compact by the
Rellich-Kondrachov theorem, so is $I_{3}$. Now
$(Z,\left(\cdot\,,\cdot\right)_{W})\to L^{2}(\partial M)\oplus L^{2}(\partial
M)$, $\omega\mapsto(\nu\lrcorner d\omega,\iota^{*}\delta\omega)$, is a linear
isometry, therefore it extends to a linear isometry $I_{2}\colon W\to
L^{2}(\partial M)\oplus L^{2}(\partial M)$ with $I_{3}=I_{2}\circ I_{1}$.
Since $I_{3}$ is compact and $I_{2}$ is a linear isometry, $I_{1}$ must be
compact.
We now prove that $I_{3}$ is injective, so that $I_{1}$ must be injective as
well. First, we show the inclusion

$V\subset\left\\{\omega\in H^{2}(M)\cap
H_{0}^{1}(M)\,|\,\Delta^{2}\omega=0\textrm{ weakly on }M\right\\},$

where the concept of a weak solution is defined by the following: given $f\in
L^{2}(M)$, a weak solution $\omega$ to $\Delta^{2}\omega=f$ on $M$ is a form
$\omega\in H^{2}\cap H_{0}^{1}(M):=H^{2}(M)\cap H_{0}^{1}(M)$ with

$\left(\Delta\omega,\Delta\omega^{\prime}\right)_{L^{2}(M)}=\left(f,\omega^{\prime}\right)_{L^{2}(M)}\;\forall\,\omega^{\prime}\in
H_{0}^{2}(M),$ (9)

where $H_{0}^{2}(M):=\left\\{\omega\in H^{2}(M)\,|\,\omega_{|_{\partial
M}}=0\textrm{ and }(\nabla\omega)_{|_{\partial M}}=0\right\\}$. Note that the
condition $(\nabla\omega)_{|_{\partial M}}=0$ can be replaced by
$\nabla_{\nu}\omega=0$ along $\partial M$ because of $\omega_{|_{\partial
M}}=0$. Namely $V\subset H^{2}(M)$ already holds because of the equivalence of
$\|\cdot\|_{V}$ and $\|\cdot\|_{H^{2}(M)}$ on $Z$. Moreover $V\subset
H_{0}^{1}(M)$ holds as well because of the continuous inclusion map
$H^{2}(M)\subset H^{1}(M)$ and the continuity of the trace operator
$H^{1}(M)\to L^{2}(\partial M)$. Thus $V\subset H^{2}\cap H_{0}^{1}(M)$.
Furthermore, if $\omega\in V$ is given, then there exists a sequence
$(\omega_{m})_{m}$ in $Z$ with
$\|\omega_{m}-\omega\|_{V}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}0$. Because of $V\subset H^{2}(M)$
and the equivalence of $\|\cdot\|_{V}$ and $\|\cdot\|_{H^{2}(M)}$, the
sequence $(\omega_{m})_{m}$ goes to $\omega$ in $H^{2}(M)$ and hence
$(\Delta\omega_{m})_{m}$ goes to $\Delta\omega$ in $L^{2}(M)$ . But since, as
a consequence of (7), we have, for all $m\in\mathbb{N}$ and
$\omega^{\prime}\in\left\\{\omega\in\Omega^{p}(M)\,|\,\omega_{|_{\partial
M}}=0\textrm{ and }(\nabla\omega)_{|_{\partial M}}=0\right\\}$,

$0=\left(\Delta\omega_{m},\Delta\omega^{\prime}\right)_{L^{2}(M)},$

we can deduce that
$\left(\Delta\omega,\Delta\omega^{\prime}\right)_{L^{2}(M)}=0$ for all
$\omega^{\prime}$ as above and therefore for all $\omega^{\prime}\in
H_{0}^{2}(M)$. This shows that $\omega\in H^{2}\cap H_{0}^{1}(M)$ satisfies
$\Delta^{2}\omega=0$ weakly on $M$ and therefore the inclusion is proved. Now,
we come back to the injectivity of $I_{3}$. Consider $\omega\in V$ such that
$I_{3}(\omega)=0$, that is $\nu\lrcorner d\omega=0$ and
$\iota^{*}\delta\omega=0$ in $L^{2}(\partial M)$. Then both $d\omega$ and
$\delta\omega$ vanish along $\partial M$ because of $\iota^{*}\omega=0$,
$\nu\lrcorner\omega=0$ and the identities $[d,\iota^{*}]=0$ and
$\\{\delta,\nu\lrcorner\\}=0$. Again, because of $\omega_{|_{\partial M}}=0$,
we have $(d\omega)_{|_{\partial M}}=\nu^{\flat}\wedge\nabla_{\nu}\omega$ and
$(\delta\omega)_{|_{\partial M}}=-\nu\lrcorner\nabla_{\nu}\omega$, so that
$\nu^{\flat}\wedge\nabla_{\nu}\omega=0$ and $\nu\lrcorner\nabla_{\nu}\omega=0$
along $\partial M$, from which $\nabla_{\nu}\omega=0$ on $\partial M$ follows.
This shows that $\omega\in H_{0}^{2}(M)$. Taking $\omega^{\prime}=\omega$ in
(9), we deduce that $\Delta\omega=0$ and therefore $\omega=0$ on $M$. This
shows $I_{3}$ and hence $I_{1}$ to be injective. $\square$

We end the proof of Theorem 2.3. Since $K$ is compact, self-adjoint and
positive definite in the Hilbert space $V$, there is a countable Hilbert
o.n.b. $(\omega_{i})_{i\geq 1}$ of $V$ for which a monotonously nonincreasing
positive real sequence $(\mu_{i})_{i\geq 1}$ going to $0$ exists such that
$K\omega_{i}=\mu_{i}\omega_{i}$ for all $i\geq 1$. We want to show that, for
each $i\geq 1$, the eigenform $\omega_{i}$ lies in $Z$ and satisfies
$\mu_{i}\iota^{*}\Delta\omega_{i}=\nu\lrcorner d\omega_{i}$ as well as
$\mu_{i}\nu\lrcorner\Delta\omega_{i}=-\iota^{*}\delta\omega_{i}$ along
$\partial M$. Hence, for each $i$, the form $\omega_{i}$ becomes a smooth
eigenform for problem (5) associated with the eigenvalue
$q_{i,p}=\frac{1}{\mu_{i}}$ which is of finite multiplicity, since $\mu_{i}$
is.
For this purpose, fix $i\geq 1$. Since $\omega_{i}\in V$, we already know that
$\Delta^{2}\omega_{i}=0$ holds weakly on $M$ with $\omega_{i}{}_{|_{\partial
M}}=0$. It remains to show that $\omega_{i}$ is smooth and satisfies
$\iota^{*}\Delta\omega_{i}=\frac{1}{\mu_{i}}\nu\lrcorner d\omega_{i}$ as well
as $\nu\lrcorner\Delta\omega_{i}=-\frac{1}{\mu_{i}}\iota^{*}\delta\omega_{i}$
along $\partial M$. By definition, for every $\omega\in Z$,

$\displaystyle\mu_{i}\left(\Delta\omega_{i},\Delta\omega\right)_{L^{2}(M)}$
$\displaystyle=$ $\displaystyle\mu_{i}\left(\omega_{i},\omega\right)_{V}$ (10)
$\displaystyle=$ $\displaystyle\left(K\omega_{i},\omega\right)_{V}$
$\displaystyle=$ $\displaystyle\left(\omega_{i},\omega\right)_{W}$
$\displaystyle=$ $\displaystyle\left(\nu\lrcorner d\omega_{i},\nu\lrcorner
d\omega\right)_{L^{2}(\partial
M)}+\left(\iota^{*}\delta\omega_{i},\iota^{*}\delta\omega\right)_{L^{2}(\partial
M)}.$

But by (7), we have, still for every $\omega\in Z$,

$\displaystyle\left(\Delta\omega_{i},\Delta\omega\right)_{L^{2}(M)}$
$\displaystyle=$
$\displaystyle\left(\underbrace{\Delta^{2}\omega_{i}}_{0},\omega\right)_{L^{2}(M)}-\left(\nu\lrcorner
d\Delta\omega_{i},\underbrace{\iota^{*}\omega}_{0}\right)_{L^{2}(\partial
M)}+\left(\iota^{*}\Delta\omega_{i},\nu\lrcorner
d\omega\right)_{L^{2}(\partial M)}$ (11)
$\displaystyle-\left(\nu\lrcorner\Delta\omega_{i},\iota^{*}\delta\omega\right)_{L^{2}(\partial
M)}+\left(\iota^{*}\delta(\Delta\omega_{i}),\underbrace{\nu\lrcorner\omega}_{0}\right)_{L^{2}(\partial
M)}$ $\displaystyle=$
$\displaystyle\left(\iota^{*}\Delta\omega_{i},\nu\lrcorner
d\omega\right)_{L^{2}(\partial
M)}-\left(\nu\lrcorner\Delta\omega_{i},\iota^{*}\delta\omega\right)_{L^{2}(\partial
M)}.$

Here we see both $\iota^{*}\Delta\omega_{i}$ and
$\nu\lrcorner\Delta\omega_{i}$ as elements in $H^{-\frac{1}{2}}(\partial M)$.
Comparing (10) and (11), we deduce that

$\left(\iota^{*}\Delta\omega_{i}-\frac{1}{\mu_{i}}\nu\lrcorner
d\omega_{i},\nu\lrcorner d\omega\right)_{L^{2}(\partial
M)}-\left(\nu\lrcorner\Delta\omega_{i}+\frac{1}{\mu_{i}}\iota^{*}\delta\omega_{i},\iota^{*}\delta\omega\right)_{L^{2}(\partial
M)}=0$

for all $\omega\in Z$. Note that the map $Z\to\Omega^{p}(\partial
M)\oplus\Omega^{p-1}(\partial M)$, $\omega\mapsto(\nu\lrcorner
d\omega,\iota^{*}\delta\omega)$ is continuous w.r.t. $\|\cdot\|_{V}$ and
$\|\cdot\|_{L^{2}(\partial M)}$ and is injective since it is the restriction
to $Z$ of the map $I_{3}$ from above. Now, Lemma $6.1$ in the appendix shows
that this map is onto. Hence, it follows that
$\iota^{*}\Delta\omega_{i}-\frac{1}{\mu_{i}}\nu\lrcorner d\omega_{i}=0$ and
$\nu\lrcorner\Delta\omega_{i}+\frac{1}{\mu_{i}}\iota^{*}\delta\omega_{i}=0$.
Therefore, $\omega_{i}$ is an eigenform for (5) associated to the eigenvalue
$q_{i,p}=\frac{1}{\mu_{i}}$. The smoothness of $\omega_{i}$ follows from the
fact that both boundary conditions
$\iota^{*}\Delta\omega_{i}-\frac{1}{\mu_{i}}\nu\lrcorner d\omega_{i}=0$ and
$\nu\lrcorner\Delta\omega_{i}+\frac{1}{\mu_{i}}\iota^{*}\delta\omega_{i}=0$
together with $\omega_{i}{}_{|_{\partial M}}=0$ define elliptic boundary
conditions for $\Delta^{2}$.

In order to finish the proof of Theorem 2.3, it remains to show that there is
a one-to-one correspondence between solutions of (5) and eigenforms of $K$. We
have already shown that every eigenform $\omega$ of $K$, associated to some
eigenvalue $\mu>0$, satisfies (5) with $q=\frac{1}{\mu}$. Conversely, if
$q\in{\mathbb{R}}$ is given for which a nontrivial solution $\omega$ to (5)
exists, then by (7), we have, for every $\omega^{\prime}\in Z$,

$\left(\Delta\omega,\Delta\omega^{\prime}\right)_{L^{2}(M)}=q\cdot\left(\left(\nu\lrcorner
d\omega,\nu\lrcorner d\omega^{\prime}\right)_{L^{2}(\partial
M)}+\left(\iota^{*}\delta\omega,\iota^{*}\delta\omega^{\prime}\right)_{L^{2}(\partial
M)}\right),$

that is, seeing both $\omega,\omega^{\prime}$ as elements of $V$,

$\left(\omega,\omega^{\prime}\right)_{V}=q\cdot\left(\omega,\omega^{\prime}\right)_{W}.$

Note that necessarily $q>0$ holds, otherwise $\omega=0$ would follow. By
definition of $K$, we then have
$\left(\omega,\omega^{\prime}\right)_{V}=q\cdot\left(K\omega,\omega^{\prime}\right)_{V}$
for all $\omega^{\prime}\in Z$ and hence in $V$, therefore
$K\omega=\frac{1}{q}\omega$. This shows $\omega$ to be an eigenform of $K$
associated to the eigenvalue $\mu=\frac{1}{q}$. This shows the correspondence
to be one-to-one. This concludes the proof. $\square$

In the following, we give a characterization for the first eigenvalue
$q_{1,p}$ on $p$-forms. This will be used later in order to get estimates for
the eigenvalues.

###### Theorem 2.6

The first eigenvalue $q_{1,p}$ of the boundary problem (5) is characterized by

$\displaystyle q_{1,p}$ $\displaystyle=$
$\displaystyle\inf\left\\{\frac{\|\Delta\omega\|_{L^{2}(M)}^{2}}{\|\nu\lrcorner
d\omega\|_{L^{2}(\partial M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial
M)}^{2}}\,|\,\omega\in\Omega^{p}(M),\,\omega_{|_{\partial M}}=0\textrm{ and
}\nabla_{\nu}\omega\neq 0\right\\}$ (12) $\displaystyle=$
$\displaystyle\inf\left\\{\frac{\|\omega\|_{L^{2}(\partial
M)}^{2}}{\|\omega\|_{L^{2}(M)}^{2}}\,|\,\omega\in\Omega^{p}(M)\setminus\\{0\\},\,\;\Delta\omega=0\textrm{
on }M\right\\}.$ (13)

Both infima are indeed minima, (12) is attained by an eigenform of (5),
associated to $q_{1,p}$ and (13) is attained by $\Delta\omega$, where $\omega$
is an eigenform of (5), associated to $q_{1,p}$.

Proof. As mentioned above, it follows from (8) that, given any nonzero
eigenform $\omega$ associated to a positive eigenvalue $q$ of (5), we have

$\int_{M}|\Delta\omega|^{2}\,d\mu_{g}=q\int_{\partial M}|\nu\lrcorner
d\omega|^{2}\,d\mu_{g}+q\int_{\partial
M}|\iota^{*}\delta\omega|^{2}\,d\mu_{g},$

so that $q_{1,p}\leq\frac{\|\Delta\omega\|_{L^{2}(M)}^{2}}{\|\nu\lrcorner
d\omega\|_{L^{2}(\partial M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial
M)}^{2}}$ for every such eigenform, with equality for $\omega$ associated to
$q_{1,p}$ of course. More generally, for every $\omega\in V$, we can write
$\omega=\sum_{i}\left(\omega,\omega_{i}\right)_{V}\cdot\omega_{i}$ because
$(\omega_{i})_{i}$ is a Hilbert orthonormal basis of $V$. From this, we can
express

$\|\Delta\omega\|_{L^{2}(M)}^{2}=\|\omega\|_{V}^{2}=\sum_{i}\left|\left(\omega,\omega_{i}\right)_{V}\right|^{2}$

on the one hand, and

$\displaystyle\|\nu\lrcorner d\omega\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial M)}^{2}$ $\displaystyle=$
$\displaystyle\|\omega\|_{W}^{2}$ $\displaystyle=$
$\displaystyle\left(K\omega,\omega\right)_{V}$ $\displaystyle=$
$\displaystyle\sum_{i}\frac{1}{q_{i,p}}\left|\left(\omega,\omega_{i}\right)_{V}\right|^{2}$
$\displaystyle\leq$
$\displaystyle\frac{1}{q_{1,p}}\sum_{i}\left|\left(\omega,\omega_{i}\right)_{V}\right|^{2}$
$\displaystyle\leq$
$\displaystyle\frac{1}{q_{1,p}}\|\Delta\omega\|_{L^{2}(M)}^{2}$

on the other hand, therefore

$q_{1,p}=\inf\left\\{\frac{\|\Delta\omega\|_{L^{2}(M)}^{2}}{\|\nu\lrcorner
d\omega\|_{L^{2}(\partial M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial
M)}^{2}}\,|\,\omega\in V\setminus\\{0\\}\right\\}.$

Now, we will show that this infimum can be taken over all $\omega\in(H^{2}\cap
H_{0}^{1})(M)\setminus\\{0\\}$, as well as over all smooth forms vanishing and
whose normal derivative does not vanish identically along $\partial M$. Recall
that,

$H_{0}^{2}(M)=\left\\{\omega\in(H^{2}\cap
H_{0}^{1})(M)\,|\,\iota^{*}\delta\omega=0\textrm{ and }\nu\lrcorner
d\omega=0\right\\},$

since, as we noticed above, if $\omega\in(H^{2}\cap H_{0}^{1})(M)$ is such
that both $\iota^{*}\delta\omega$ and $\nu\lrcorner d\omega$ vanish along
$\partial M$, then so does $\nabla_{\nu}\omega$. As in [6, Thm. 1.2], we have
the following lemma:

###### Lemma 2.7

The inner product $\left(\cdot\,,\cdot\right)_{V}$ is well defined on
$(H^{2}\cap H_{0}^{1})(M)$ and we have the following
$\left(\cdot\,,\cdot\right)_{V}$-orthogonal splitting:

$H^{2}\cap H_{0}^{1}(M)=V\oplus H_{0}^{2}(M).$

Proof. By its definition, $\left(\cdot\,,\cdot\right)_{V}$ is well defined on
$(H^{2}\cap H_{0}^{1})(M)$. Furthermore, (7) already implies that, for all
$(\omega,\omega^{\prime})\in Z\times H^{2}\cap H_{0}^{1}(M)$,

$\left(\omega,\omega^{\prime}\right)_{V}=\left(\iota^{*}\Delta\omega,\nu\lrcorner
d\omega^{\prime}\right)_{L^{2}(\partial
M)}-\left(\nu\lrcorner\Delta\omega,\iota^{*}\delta\omega^{\prime}\right)_{L^{2}(\partial
M)},$ (14)

so that $\left(\omega,\omega^{\prime}\right)_{V}=0$ as soon as
$\omega^{\prime}\in H_{0}^{2}(M)$. This shows that $H_{0}^{2}(M)\subset
Z^{\perp}=V^{\perp}$. Conversely, let $\omega^{\prime}\in V^{\perp}\subset
H^{2}\cap H_{0}^{1}(M)$. Then $\left(\omega_{i},\omega^{\prime}\right)_{V}=0$
for all $i$. By (14), this is equivalent to $\left(\nu\lrcorner
d\omega_{i},\nu\lrcorner d\omega^{\prime}\right)_{L^{2}(\partial
M)}+\left(\iota^{*}\delta\omega_{i},\iota^{*}\delta\omega^{\prime}\right)_{L^{2}(\partial
M)}=0$ for all $i$ by
$\iota^{*}\Delta\omega_{i}=\frac{1}{\mu_{i}}\nu\lrcorner\,d\omega_{i}$ and
$\nu\lrcorner\Delta\omega_{i}=-\frac{1}{\mu_{i}}\iota^{*}(\delta\omega_{i})$.
Since the map $Z\to\Omega^{p}(\partial M)\oplus\Omega^{p-1}(\partial M)$,
$\omega\mapsto(\nu\lrcorner d\omega,\iota^{*}\delta\omega)$ is bounded (w.r.t.
$\|\cdot\|_{V}$ and $\|\cdot\|_{L^{2}(\partial M)}$), onto and the
$\omega_{i}$’s span a dense subspace of $V$, we obtain that
$\left(\omega_{1},\nu\lrcorner d\omega^{\prime}\right)_{L^{2}(\partial M)}=0$
as well as $\left(\omega_{2},\nu\lrcorner
d\omega^{\prime}\right)_{L^{2}(\partial M)}=0$ for all
$(\omega_{1},\omega_{2})\in\Omega^{p}(\partial M)\oplus\Omega^{p-1}(\partial
M)$ and therefore $\nu\lrcorner d\omega^{\prime}=0$ and
$\iota^{*}\delta\omega^{\prime}=0$ hold along $\partial M$. This shows that
$\omega^{\prime}\in H_{0}^{2}(M)$. On the whole, $H_{0}^{2}(M)=V^{\perp}$ and
the orthogonal splitting is proved. $\square$

It remains to notice that, for any $\omega\in\Omega^{p}(M)$ with
$\omega_{|_{\partial M}}=0$, we can split
$\left(\cdot\,,\cdot\right)_{V}$-orthogonally
$\omega=\omega_{V}+\hat{\omega}$, where $\omega_{V}\in V$ and $\hat{\omega}\in
H_{0}^{2}(M)$. Then

$\|\Delta\omega\|_{L^{2}(M)}^{2}=\|\omega\|_{V}^{2}=\|\omega_{V}\|_{V}^{2}+\|\hat{\omega}\|_{V}^{2}$

on the one hand, and

$\|\nu\lrcorner d\omega\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial M)}^{2}=\|\nu\lrcorner
d\omega_{V}\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\omega_{V}\|_{L^{2}(\partial M)}^{2}$

on the other hand, so that

$\displaystyle\|\Delta\omega\|_{L^{2}(M)}^{2}$ $\displaystyle=$
$\displaystyle\|\Delta\omega_{V}\|_{L^{2}(M)}^{2}+\|\Delta\hat{\omega}\|_{L^{2}(M)}^{2}$
$\displaystyle\geq$ $\displaystyle\|\Delta\omega_{V}\|_{L^{2}(M)}^{2}$
$\displaystyle\geq$ $\displaystyle q_{1,p}\cdot\left(\|\nu\lrcorner
d\omega_{V}\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\omega_{V}\|_{L^{2}(\partial M)}^{2}\right)$
$\displaystyle=$ $\displaystyle q_{1,p}\cdot\left(\|\nu\lrcorner
d\omega\|_{L^{2}(\partial M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial
M)}^{2}\right),$

which proves (12). Furthermore, the r.h.s. of (12) is actually a minimum
attained exactly by those eigenforms of the biharmonic Steklov problem that
are associated to the smallest positive eigenvalue $q_{1,p}$.

We now prove the following lemma:

###### Lemma 2.8

The infimum in (13), that we denote by $q_{1,p}^{\prime}$, is a positive
minimum.

Proof. To prove this result, we apply the same argument as in [6, p. 318].
Namely the standard Rellich-Kondrachov compactness theorem ensures the natural
inclusion map $H^{\frac{1}{2}}(\partial M)\to L^{2}(\partial M)$ to be
compact. Hence its transpose map $L^{2}(\partial M)\to
H^{-\frac{1}{2}}(\partial M)$ is compact, as a straightforward consequence.
Moreover, the harmonic extension from $\partial M$ to $M$ defines a bounded
linear map $H^{-\frac{1}{2}}(\partial M)\to L^{2}(M)$, see e.g. [17, Thm. 6.6
chap. 2]. Therefore the composition $L^{2}(\partial M)\to
H^{-\frac{1}{2}}(\partial M)\to L^{2}(M)$ of both maps defines a compact
linear map $E\colon L^{2}(\partial M)\to L^{2}(M)$, which already shows that
$q_{1,p}^{\prime}$ to be positive using only the boundedness of the map.
Furthermore, because the image by $E$ of the unit sphere
$\mathbb{S}:=\left\\{\omega\in L^{2}(\partial
M)\,,\,\|\omega\|_{L^{2}(\partial M)}=1\right\\}$ of $L^{2}(\partial M)$ is
relatively compact in $L^{2}(M)$, there exists
$\tilde{\omega}\in\overline{E(\mathbb{S})}$ such that
$\|\tilde{\omega}\|_{L^{2}(M)}=\sup\left\\{\|E\omega\|_{L^{2}(M)}\,|\,\omega\in\mathbb{S}\right\\}$.
By definition of the closure, there exists a sequence $(\omega_{m})_{m}$ in
$\mathbb{S}$ such that $E\omega_{m}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\tilde{\omega}$ in $L^{2}(M)$. But
then $\Delta(E\omega_{m})\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\Delta\tilde{\omega}$ in
$H^{-2}(M)=(H_{0}^{2}(M))^{\prime}$, so that necessarily
$\Delta\tilde{\omega}=0$ holds in $H^{-2}(M)$. Note that, because
$\Delta\tilde{\omega}=0$ and $\Delta(E\omega_{m})=0$ for all $m$, we can also
claim that $\Delta(E\omega_{m})\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\Delta\tilde{\omega}$ in $L^{2}(M)$.
By Gårding’s inequality and since both $E\omega_{m}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\tilde{\omega}$ and
$\Delta(E\omega_{m})\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\Delta\tilde{\omega}$ in $L^{2}(M)$,
we have that $E\omega_{m}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\tilde{\omega}$ in $H^{2}(M)$. As a
consequence, because of $\nu\lrcorner\,E\omega_{m}=0$ along $\partial M$ and
$\nu\lrcorner\,E\omega_{m}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\nu\lrcorner\,\tilde{\omega}$ in
$L^{2}(\partial M)$, necessarily $\nu\lrcorner\,\tilde{\omega}=0$ holds along
$\partial M$. Now again $E\omega_{m}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\tilde{\omega}$ in $H^{2}(M)$ also
implies $\omega_{m}=(E\omega_{m})_{|_{\partial M}}\mathrel{\mathop{\kern
0.0pt\longrightarrow}\limits_{m\to\infty}}\tilde{\omega}_{|_{\partial M}}$ in
$L^{2}(\partial M)$ (actually also in $H^{1}(\partial M)$), so that
$\|\tilde{\omega}_{|_{\partial M}}\|_{L^{2}(\partial M)}=1$. This shows that
$\tilde{\omega}\in E(\mathbb{S})$ and hence
$\|\tilde{\omega}\|_{L^{2}(M)}=\max\left\\{\|E\omega\|_{L^{2}(M)}\,|\,\omega\in\mathbb{S}\right\\}$
satisfies $\|\tilde{\omega}\|_{L^{2}(M)}=\frac{1}{q_{1,p}^{\prime}}$. This
shows the positive r.h.s. $q_{1,p}^{\prime}$ of (13) to be a minimum.
$\square$

In order to finish the proof of Theorem 2.6, we want to show that
$q_{1,p}=q_{1,p}^{\prime}$. Pick any eigenform $\omega$ associated to
$q_{1,p}$. Up to rescaling $\omega$, we may assume that $\|\nu\lrcorner
d\omega\|_{L^{2}(\partial M)}^{2}+\|\iota^{*}\delta\omega\|_{L^{2}(\partial
M)}^{2}=1$. Let $\hat{\omega}\in\Omega^{p}(M)$ be the unique solution to
$\Delta\hat{\omega}=0$ on $M$ with $\iota^{*}\hat{\omega}_{|_{\partial
M}}=-q_{1,p}\nu\lrcorner d\omega$ a well as
$\nu\lrcorner\hat{\omega}=q_{1,p}\iota^{*}\delta\omega$ along $\partial M$. By
(6),

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\int_{M}\langle\Delta\hat{\omega},\omega\rangle\,d\mu_{g}$
$\displaystyle=$
$\displaystyle\int_{M}\langle\hat{\omega},\Delta\omega\rangle\,d\mu_{g}+q_{1,p}\int_{\partial
M}(|\nu\lrcorner d\omega|^{2}+|\iota^{*}\delta\omega|^{2})\,d\mu_{g}$
$\displaystyle=$
$\displaystyle\int_{M}\langle\hat{\omega},\Delta\omega\rangle\,d\mu_{g}+q_{1,p},$

so that Cauchy-Schwarz inequality leads to
$q_{1,p}\leq\|\Delta\omega\|_{L^{2}(M)}\cdot\|\hat{\omega}\|_{L^{2}(M)}$.
Using $q_{1,p}=\|\Delta\omega\|_{L^{2}(M)}^{2}$, we obtain
$q_{1,p}\leq\|\hat{\omega}\|_{L^{2}(M)}^{2}$. Therefore,

$\frac{\|\hat{\omega}\|_{L^{2}(\partial
M)}^{2}}{\|\hat{\omega}\|_{L^{2}(M)}^{2}}=\frac{q_{1,p}^{2}}{\|\hat{\omega}\|_{L^{2}(M)}^{2}}\leq
q_{1,p},$

from which $q_{1,p}^{\prime}\leq q_{1,p}$ follows. Conversely, if
$\omega\in\Omega^{p}(M)\setminus\\{0\\}$ satisfies $\Delta\omega=0$ on $M$ and
$\|\omega\|_{L^{2}(\partial
M)}^{2}=q_{1,p}^{\prime}\|\omega\|_{L^{2}(M)}^{2}$, then let $\hat{\omega}$ be
the solution to $\Delta\hat{\omega}=\omega$ on $M$ with
$\hat{\omega}_{|_{\partial M}}=0$. Then, again by (6), $\omega_{|_{\partial
M}}=\iota^{*}\omega+\nu^{\flat}\wedge(\nu\lrcorner\,\omega)$ and Cauchy-
Schwarz inequality, we have

$\displaystyle\left\|\omega\right\|_{L^{2}(M)}^{2}$ $\displaystyle=$
$\displaystyle\left(\Delta\hat{\omega},\omega\right)_{L^{2}(M)}$
$\displaystyle=$
$\displaystyle\left(\hat{\omega},\underbrace{\Delta\omega}_{0}\right)_{L^{2}(M)}+\left(\nu\lrcorner
d\hat{\omega},\iota^{*}\omega\right)_{L^{2}(\partial
M)}-\left(\iota^{*}\delta\hat{\omega},\nu\lrcorner\omega\right)_{L^{2}(\partial
M)}$ $\displaystyle=$ $\displaystyle\left(\nu\lrcorner
d\hat{\omega},\omega\right)_{L^{2}(\partial
M)}-\left(\nu^{\flat}\wedge\iota^{*}\delta\hat{\omega},\omega\right)_{L^{2}(\partial
M)}$ $\displaystyle=$
$\displaystyle\left(\nu\lrcorner\,d\hat{\omega}-\nu^{\flat}\wedge\iota^{*}\delta\hat{\omega},\omega\right)_{L^{2}(\partial
M)}$ $\displaystyle\leq$
$\displaystyle\left\|\nu\lrcorner\,d\hat{\omega}-\nu^{\flat}\wedge\iota^{*}\delta\hat{\omega}\right\|_{L^{2}(\partial
M)}\cdot\left\|\omega\right\|_{L^{2}(\partial M)}.$

But since
$\left(\nu\lrcorner\,d\hat{\omega},\nu^{\flat}\wedge\iota^{*}\delta\hat{\omega}\right)_{L^{2}(\partial
M)}=0$, we have

$\left\|\nu\lrcorner\,d\hat{\omega}-\nu^{\flat}\wedge\iota^{*}\delta\hat{\omega}\right\|_{L^{2}(\partial
M)}^{2}=\left\|\nu\lrcorner\,d\hat{\omega}\right\|_{L^{2}(\partial
M)}^{2}+\left\|\nu^{\flat}\wedge\iota^{*}\delta\hat{\omega}\right\|_{L^{2}(\partial
M)}^{2}=\left\|\nu\lrcorner\,d\hat{\omega}\right\|_{L^{2}(\partial
M)}^{2}+\left\|\iota^{*}\delta\hat{\omega}\right\|_{L^{2}(\partial M)}^{2},$

so that $\left\|\omega\right\|_{L^{2}(M)}^{2}\leq\left(\left\|\nu\lrcorner
d\hat{\omega}\right\|_{L^{2}(\partial
M)}^{2}+\left\|\iota^{*}\delta\hat{\omega}\right\|_{L^{2}(\partial
M)}^{2}\right)^{\frac{1}{2}}\cdot\left\|\omega\right\|_{L^{2}(\partial M)}$.
Therefore,

$\displaystyle\frac{\|\Delta\hat{\omega}\|_{L^{2}(M)}^{2}}{\|\nu\lrcorner
d\hat{\omega}\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\hat{\omega}\|_{L^{2}(\partial M)}^{2}}$
$\displaystyle=$ $\displaystyle\frac{\|\omega\|_{L^{2}(M)}^{2}}{\|\nu\lrcorner
d\hat{\omega}\|_{L^{2}(\partial
M)}^{2}+\|\iota^{*}\delta\hat{\omega}\|_{L^{2}(\partial M)}^{2}}$
$\displaystyle\leq$
$\displaystyle\frac{\|\omega\|_{L^{2}(M)}^{2}\cdot\left\|\omega\right\|_{L^{2}(\partial
M)}^{2}}{\|\omega\|_{L^{2}(M)}^{4}}$ $\displaystyle=$
$\displaystyle\frac{\left\|\omega\right\|_{L^{2}(\partial
M)}^{2}}{\|\omega\|_{L^{2}(M)}^{2}}$ $\displaystyle=$ $\displaystyle
q_{1,p}^{\prime},$

from which $q_{1,p}^{\prime}\geq q_{1,p}$ follows. On the whole, we deduce
that $q_{1,p}=q_{1,p}^{\prime}$, as we claimed. Moreover, the $p$-form
$\hat{\omega}$ defined above, because it now minimizes $q_{1,p}$, must be an
eigenform of the biharmonic Steklov operator associated to the eigenvalue
$q_{1,p}$. As a consequence, $\omega=\Delta\hat{\omega}$ where $\hat{\omega}$
is an eigenform of the biharmonic Steklov operator associated to the
eigenvalue $q_{1,p}$. $\square$

##  3 Eigenvalues of the biharmonic Steklov operator

In this section, we will establish some eigenvalue estimates for the first
eigenvalue of the biharmonic Steklov operator defined in the previous section.

As before, we will consider a compact Riemannian manifold $(M^{n},g)$ with
smooth boundary $\partial M$. Notice first the following fact:

###### Lemma 3.1

The biharmonic Steklov operator is preserved by the Hodge star operator
$*_{M}$ on $M$.

Proof. We only need to check that the last two boundary conditions in (5) are
preserved. For this purpose, using the equality
$\iota^{*}(*_{M}\alpha)=*_{\partial M}(\nu\lrcorner\alpha)$ for any form
$\alpha$, we compute, for any solution $\omega$ of degree $p$ to problem (5),

$\displaystyle\nu\lrcorner\Delta(*_{M}\omega)$ $\displaystyle=$
$\displaystyle\nu\lrcorner(*_{M}\Delta\omega)$ $\displaystyle=$
$\displaystyle(-1)^{p}*_{\partial M}\iota^{*}(\Delta\omega)$ $\displaystyle=$
$\displaystyle(-1)^{p}q*_{\partial M}(\nu\lrcorner d\omega)$ $\displaystyle=$
$\displaystyle(-1)^{p}q\iota^{*}(*_{M}d\omega)$ $\displaystyle=$
$\displaystyle-q\iota^{*}(\delta*_{M}\omega).$

In the last equality, we use the fact that $*_{M}d=(-1)^{p-1}\delta*_{M}$ on
$p$-forms. For the other boundary condition, we have

$\iota^{*}(\Delta*_{M}\omega)=\iota^{*}(*_{M}\Delta\omega)=*_{\partial
M}(\nu\lrcorner\Delta\omega)=-q*_{\partial
M}(\iota^{*}\delta\omega)=(-1)^{p}q\nu\lrcorner*_{M}\delta\omega=q\nu\lrcorner
d(*_{M}\omega).$

Also here we use the fact that $d*_{M}=(-1)^{p}*_{M}\delta$ on $p$-forms. This
finishes the proof. $\square$

###### Remark 3.2

As a direct consequence of the invariance of the biharmonic Steklov operator
by the Hodge star operator is that $q_{i,p}=q_{i,n-p}$ for any $i\geq 1$ and
$p\leq n$.

In the following, we recall the estimate stated by S. Raulot and A. Savo in
[22] for subharmonic functions that we will use in order to get a lower bound
of the first eigenvalue $q_{1,p}$. Let $(M^{n},g)$ be a compact Riemannian
manifold with smooth boundary such that the Ricci curvature of $M$ satisfies
${\rm Ric}_{M}\geq(n-1)K$ and the mean curvature of the boundary satisfies
$H\geq H_{0}$ for some real numbers $K$ and $H_{0}$. Let $R$ be the inner
radius of the manifold $M$, that is

$R={\rm max}\\{{\rm dist}(x,\partial M)|\,x\in M\\},$

and $\Theta(r)=(s^{\prime}_{K}(r)-H_{0}s_{K}(r))^{n-1}$ for all $r$, where the
function $s_{K}$ is being given by

$s_{K}(r):=\left\\{\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\frac{1}{\sqrt{K}}\,\sin(r\sqrt{K})&&\text{if}\,\,K>0,\\\ \vskip 6.0pt
plus 2.0pt minus 2.0ptr&&\text{if}\,\,K=0,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\frac{1}{\sqrt{|K|}}\,\sinh(r\sqrt{|K|})&&\text{if}\,\,K<0.\end{array}\right.$

It was shown in [22, Prop. 14] (see also [12, Thm. A]) that the function
$\Theta$ is smooth and positive on $[0,R[$ and $\Theta(R)=0$ when $M$ is a
geodesic ball in $M_{K}$, the space form of sectional curvature $K$. The
following result was proved in [22]:

###### Theorem 3.3

[22, Thm. 10] Let $(M^{n},g)$ be a compact Riemannian manifold with smooth
boundary. Assume that the Ricci curvature of $M$ satisfies ${\rm
Ric}_{M}\geq(n-1)K$ and the mean curvature $H\geq H_{0}$ for some real numbers
$K$ and $H_{0}$. If $h$ is a non-trivial, nonnegative subharmonic function on
$M$ (i.e. $\Delta h\leq 0$ on $M$), then

$\frac{\int_{\partial
M}hd\mu_{g}}{\int_{M}hd\mu_{g}}\geq\frac{1}{\int_{0}^{R}\Theta(r)dr}.$

Using this result and the Bochner formula
$\Delta=\nabla^{*}\nabla+\mathfrak{B}^{[p]}$ on $p$-forms, we prove the
following:

###### Theorem 3.4

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary. Assume
that the Ricci curvature of $M$ satisfies ${\rm Ric}_{M}\geq(n-1)K$ and the
mean curvature $H\geq H_{0}$ for some real numbers $K$ and $H_{0}$. Assume
also that the Bochner operator $\mathfrak{B}^{[p]}$ is nonnegative for some
$p$. Then, the inequality

$q_{1,p}\geq\frac{1}{\int_{0}^{R}\Theta(r)\,dr}$ (15)

holds.

Proof. Applying the Bochner formula to $\Delta\omega$ where $\omega$ is a
$p$-eigenform of the biharmonic Steklov operator associated with $q_{1,p}$, we
get after taking the pointwise scalar product with $\Delta\omega$ that

$0=\langle\Delta^{2}\omega,\Delta\omega\rangle=|\nabla\Delta\omega|^{2}+\frac{1}{2}\Delta(|\Delta\omega|^{2})+\langle\mathfrak{B}^{[p]}\Delta\omega,\Delta\omega\rangle.$

Since $\mathfrak{B}^{[p]}$ is nonnegative, we deduce that
$\Delta(|\Delta\omega|^{2})$ is nonpositive or equivalently the function
$h:=|\Delta\omega|^{2}$ is subharmonic. Therefore, by the previous theorem, we
can say that

$\frac{\int_{\partial
M}|\Delta\omega|^{2}d\mu_{g}}{\int_{M}|\Delta\omega|^{2}d\mu_{g}}\geq\frac{1}{\int_{0}^{R}\Theta(r)dr}.$

Now, Characterization (13) gives the result and finishes the proof of the
theorem. $\square$

###### Remark 3.5

Depending on the sign of $K$ and $H_{0}$, we can estimate explicitly
$\int_{0}^{R}\Theta(r)dr$ in terms of $R$ and $H_{0}$, as in [22, Thms. 12 &
13]. Therefore, one can deduce several estimates for $q_{1,p}$ in terms of $R$
and $H_{0}$.

We will now provide an estimate for the first eigenvalue of problem (5) on
manifolds carrying parallel forms and study the limiting case of the estimate.
Recall that a harmonic domain is a compact Riemannian manifold $(M^{n},g)$
with smooth boundary $\partial M$ supporting a solution to the Serrin problem
(3). We have the following result:

###### Theorem 3.6

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary. Assume
that $M$ supports a non-trivial parallel $p$-form $\omega_{0}$ for some
$p=0,\ldots,n$. Then

$q_{1,p}\leq\frac{{\rm Vol}(\partial M)}{{\rm Vol}(M)}.$ (16)

Moreover, if equality holds in (16), then $f\cdot\omega_{0}$ is an eigenform
associated to $q_{1,p}$, where $f$ is the solution of (3) and therefore $M$
must be a harmonic domain (and hence a Euclidean ball if
$M\subset{\mathbb{R}}^{n}$).
Conversely, if $M$ is a harmonic domain, then $\frac{{\rm Vol}(\partial
M)}{{\rm Vol}(M)}$ is an eigenvalue of the biharmonic Steklov problem (5).

Proof. As $\omega_{0}$ is a parallel form, we can assume that
$|\omega_{0}|=1$. By using the variational characterization (13), we obtain
that

$q_{1,p}\leq\dfrac{\|\omega_{0}\|_{L^{2}(\partial
M)}^{2}}{\|\omega_{0}\|_{L^{2}(M)}^{2}}=\dfrac{{\rm Vol}(\partial M)}{{\rm
Vol}(M)}.$

If equality occurs in (16), then $\omega_{0}=\Delta\omega$ for some eigenform
$\omega$ associated with $q_{1,p}$ by Theorem 2.6. Now Proposition 2.1 implies
that $M$ carries a solution $f$ to the Serrin problem (3) and that
$\omega=f\cdot\omega_{0}$. To check the converse, we take a function $f$
solution to the Serrin problem (3), then using again Proposition 2.1 the
$p$-form $\omega:=f\cdot\omega_{0}$ is an eigenform of (5) associated with the
eigenvalue $q=\frac{1}{c}$, where $c=\dfrac{{\rm Vol}(M)}{{\rm Vol}(\partial
M)}$. $\square$

###### Remark 3.7

If a compact Riemannian manifold $M$ with smooth boundary carries a nontrivial
_harmonic form of constant length_ $\omega_{0}$, then (16) remains valid.
Moreover, if (16) is an equality, then there still exists an eigenform
$\omega$ to the biharmonic Steklov operator on $p$-forms such that
$\Delta\omega=\omega_{0}$, nevertheless it is no more true in general that $M$
must be a harmonic domain and that $\omega=f\cdot\omega_{0}$ for some solution
$f$ to the scalar Serrin problem.

Next we compare the first eigenvalues of the biharmonic Steklov operator for
successive degrees, when the manifold $M$ is a domain in ${\mathbb{R}}^{n}$ or
$\mathbb{S}^{n}$. We first notice that, if $f$ is any eigenfunction to the
scalar biharmonic Steklov problem (1), then for any parallel $p$-form
$\omega_{0}$ on $M$ the form $f\cdot\omega_{0}$ is an eigenform to the
biharmonic Steklov problems on $p$-forms and associated to the _same
eigenvalue_. Therefore, for every eigenvalue $q$ of the scalar biharmonic
Steklov operator, we have an embedding

$\ker(\mathrm{BS}_{0}-q)\otimes\mathrm{P}_{p}\hookrightarrow\ker(\mathrm{BS}_{p}-q),$

where $\ker(\mathrm{BS}_{j}-q)$ denotes the eigenspace for the biharmonic
Steklov operator on $j$-forms and associated to the eigenvalue $q$, and
$\mathrm{P}_{p}$ denotes the space of parallel $p$-forms on $M$. When
$M\subset{\mathbb{R}}^{n}$, then conversely for any
$\omega\in\ker(\mathrm{BS}_{p}-q)$, there exists a parallel $p$-form
$\omega_{0}$ on $M$ with $|\omega_{0}|=1$ and
$\langle\omega,\omega_{0}\rangle\neq 0$ (non identically vanishing) on $M$.
But then $f$ can be easily shown to lie in $\ker(\mathrm{BS}_{0}-q)$. This
shows that, when $M\subset{\mathbb{R}}^{n}$, both $0$\- and $p$-biharmonic
Steklov eigenvalues coincide, their multiplicities being ignored.

In what follows, we assume that $(M^{n},g)$ is isometrically immersed into the
Euclidean space ${\mathbb{R}}^{n+m}$. For any given smooth normal vector field
$N$ to $M$, we denote by $I\\!I_{N}$ the associated Weingarten map, that is,
the endomorphism field of $TM$ defined by

$\langle I\\!I_{N}(X),Y\rangle=\langle N,I\\!I(X,Y)\rangle$

for all $X,Y$ tangent to $M$, where $I\\!I$ is the second fundamental form of
the immersion. Recall that any endomorphism $A$ of $TM$ can be extended to the
set of differential $p$-forms on $M$ as follows: For any $p$-form $\omega$ on
$M$, we define

$A^{[p]}\omega(X_{1},\cdots,X_{p})=\sum_{i=1}^{p}\omega(X_{1},\cdots,A(X_{i}),\cdots,X_{p}),$
(17)

for all $X_{1},\cdots,X_{p}$ vector fields in $TM$. In particular, this
applies to $I\\!I_{N}$ for all $N\in T^{\perp}M$. The following lemma is
technical but will be useful for the comparison.

###### Lemma 3.8

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary
$\partial M$. Assume that $M$ is isometrically immersed into the Euclidean
space ${\mathbb{R}}^{n+m}$. Let $\omega$ be any $p$-eigenform of the
biharmonic Steklov operator. Then we have

$\displaystyle pq_{1,p-1}\int_{\partial M}(|\nu\lrcorner
d\omega|^{2}+|\iota^{*}(\delta\omega)|^{2})d\mu_{g}\leq\sum_{i=1}^{n+m}\int_{M}|-2I\\!I_{n\widetilde{H}}(\partial_{x_{i}}^{T})\lrcorner\omega+2\sum_{a=1}^{m}I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner
I\\!I_{f_{a}}^{[p]}\omega$
$\displaystyle-2\sum_{s=1}^{n}(I\\!I_{\partial_{x_{i}}^{\perp}}e_{s})\lrcorner(\nabla_{e_{s}}\omega)-n\left(d(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)\right)\lrcorner\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega|^{2}d\mu_{g}.$

Here $\\{e_{1},\cdots,e_{n}\\}$ and $\\{f_{1},\cdots,f_{m}\\}$ are
respectively orthonormal bases of $TM$ and $T^{\perp}M$ and $\widetilde{H}$ is
the mean curvature field of the immersion.

Proof. Let $\omega$ be any eigenform of the biharmonic Steklov problem
associated with $q_{1,p}$. For each $i=1,\cdots,n+m$, the unit parallel vector
field $\partial_{x_{i}}$ on ${\mathbb{R}}^{n+m}$ splits into
$\partial_{x_{i}}=(\partial_{x_{i}})^{T}+(\partial_{x_{i}})^{\perp}$ where
$(\partial_{x_{i}})^{T}$ is the tangent part in $TM$ and
$(\partial_{x_{i}})^{\perp}$ is the orthogonal one in $T^{\perp}M$. We
consider the $(p-1)$-form $(\partial_{x_{i}})^{T}\lrcorner\omega$ on $M$ which
clearly vanishes on $\partial M$. By applying to it the variational
characterization (12), we get, for each $i$,

$q_{1,p-1}\int_{\partial M}\left(|\nu\lrcorner
d((\partial_{x_{i}})^{T}\lrcorner\omega)|^{2}+|\iota^{*}\delta((\partial_{x_{i}})^{T}\lrcorner\omega)|^{2}\right)d\mu_{g}\leq\int_{M}|\Delta((\partial_{x_{i}})^{T}\lrcorner\omega)|^{2}d\mu_{g}.$
(19)

Now we want to sum over $i=1,\cdots,n+m$. We first begin with the l.h.s.
Recall the Cartan formula:
$\mathcal{L}_{X}\omega=d(X\lrcorner\omega)+X\lrcorner d\omega$, for any vector
field $X$ on $M$. Using this formula, we have for each $i$,

$\displaystyle d(\partial_{x_{i}}^{T}\lrcorner\omega)$ $\displaystyle=$
$\displaystyle\mathcal{L}_{\partial_{x_{i}}^{T}}\omega-\partial_{x_{i}}^{T}\lrcorner
d\omega$ (20) $\displaystyle=$
$\displaystyle\nabla_{\partial_{x_{i}}^{T}}\omega+I\\!I_{\partial_{x_{i}}^{\perp}}^{[p]}\omega-\partial_{x_{i}}^{T}\lrcorner
d\omega.$

In the last equality, we used the splitting of the Lie derivative in terms of
the connection as follows:
$\mathcal{L}_{X^{T}}\omega=\nabla_{X^{T}}\omega+I\\!I_{X^{\perp}}^{[p]}\omega$,
for a parallel vector field $X\in{\mathbb{R}}^{n+m}$, proved in [9, Eq. (4.3)
p. 337]. Since $\omega=0$ on $\partial M$, we have that for any $X\in
T\partial M$ [20, Eq. (23)]

$\nu\lrcorner\nabla_{X}\omega=\nabla_{X}^{\partial
M}(\nu\lrcorner\omega)+S(X)\lrcorner(\iota^{*}\omega)=0.$

Here $S$ denotes the second fundamental form of the boundary. Therefore, we
deduce that

$\displaystyle\nu\lrcorner d((\partial_{x_{i}})^{T}\lrcorner\omega)$
$\displaystyle=$
$\displaystyle\nu\lrcorner\nabla_{\partial_{x_{i}}^{T}}\omega+\partial_{x_{i}}^{T}\lrcorner(\nu\lrcorner
d\omega)$ (21) $\displaystyle=$ $\displaystyle
g((\partial_{x_{i}})^{T},\nu)\nu\lrcorner\nabla_{\nu}\omega+\partial_{x_{i}}^{T}\lrcorner(\nu\lrcorner
d\omega)$ $\displaystyle=$
$\displaystyle-g(\partial_{x_{i}},\nu)\iota^{*}(\delta\omega)+\partial_{x_{i}}^{T}\lrcorner(\nu\lrcorner
d\omega).$

In the last equality, we use the identity [20, Lem. 18]

$\nu\lrcorner\nabla_{\nu}\omega=\delta^{\partial
M}(\iota^{*}\omega)-\iota^{*}(\delta\omega)-S^{[p-1]}(\nu\lrcorner\omega)+(n-1)H\nu\lrcorner\omega=-\iota^{*}(\delta\omega).$

Independently, by a straightforward computation, we check that, for any
$p$-form $\alpha$,

$\sum_{i=1}^{n+m}\partial_{x_{i}}^{T}\wedge(\partial_{x_{i}}^{T}\lrcorner\alpha)=\sum_{i=1}^{n}e_{i}\wedge(e_{i}\lrcorner\alpha)=p\alpha.$

As a consequence, we obtain for any $p$-forms $\alpha$ and $\beta$ on $M$

$\sum_{i=1}^{n+m}\langle\partial_{x_{i}}^{T}\lrcorner\alpha,\partial_{x_{i}}^{T}\lrcorner\beta\rangle=p\langle\alpha,\beta\rangle.$
(22)

We take the norm of (21) and sum over $i$ to get

$\displaystyle\sum_{i=1}^{n+m}|\nu\lrcorner
d((\partial_{x_{i}})^{T}\lrcorner\omega)|^{2}$
$\displaystyle\mathrel{\mathop{\kern
0.0pt=}\limits^{\eqref{eq:scalinnerpro}}}$
$\displaystyle|\iota^{*}(\delta\omega)|^{2}+p|\nu\lrcorner
d\omega|^{2}-2\sum_{i=1}^{n+m}g(\partial_{x_{i}},\nu)\langle\iota^{*}(\delta\omega),\partial_{x_{i}}^{T}\lrcorner(\nu\lrcorner
d\omega)\rangle$ (23) $\displaystyle=$
$\displaystyle|\iota^{*}\delta\omega|^{2}+p|\nu\lrcorner
d\omega|^{2}-2\langle\iota^{*}\delta\omega,\nu^{T}\lrcorner(\nu\lrcorner
d\omega)\rangle$ $\displaystyle=$
$\displaystyle|\iota^{*}\delta\omega|^{2}+p|\nu\lrcorner d\omega|^{2}.$

Here we notice that by the Gauß formula and the fact that $\partial_{x_{i}}$
is parallel in ${\mathbb{R}}^{n+m}$, we have
$\nabla_{X}\partial_{x_{i}}^{T}=I\\!I_{\partial_{x_{i}}^{\perp}}(X)$, so that

$\displaystyle\delta(\partial_{x_{i}}^{T}\lrcorner\omega)$ $\displaystyle=$
$\displaystyle-\sum_{j=1}^{n}e_{j}\lrcorner\,\nabla_{e_{j}}(\partial_{x_{i}}^{T}\lrcorner\omega)$
$\displaystyle=$
$\displaystyle-\sum_{j=1}^{n}e_{j}\lrcorner\,(\nabla_{e_{j}}\partial_{x_{i}}^{T}\lrcorner\,\omega+\partial_{x_{i}}^{T}\lrcorner\,\nabla_{e_{j}}\omega)$
$\displaystyle=$
$\displaystyle-\sum_{j=1}^{n}e_{j}\lrcorner\,I\\!I_{\partial_{x_{i}}^{\perp}}(e_{j})\lrcorner\,\omega+\sum_{j=1}^{n}\partial_{x_{i}}^{T}\lrcorner\,(e_{j}\lrcorner\,\nabla_{e_{j}}\omega)$
$\displaystyle=$ $\displaystyle-\sum_{j,k=1}^{n}\langle
I\\!I_{\partial_{x_{i}}^{\perp}}(e_{j}),e_{k}\rangle
e_{j}\lrcorner\,e_{k}\lrcorner\,\omega-\partial_{x_{i}}^{T}\lrcorner\,\delta\omega$
$\displaystyle=$ $\displaystyle-\partial_{x_{i}}^{T}\lrcorner\,\delta\omega$

because of the expression $\langle
I\\!I_{\partial_{x_{i}}^{\perp}}(e_{j}),e_{k}\rangle=\langle
I\\!I(e_{j},e_{k}),\partial_{x_{i}}^{\perp}\rangle$ being symmetric in $j,k$.
Thus
$\delta(\partial_{x_{i}}^{T}\lrcorner\omega)=-\partial_{x_{i}}^{T}\lrcorner\delta\omega$
for any $i$. Using also the fact that
$\nu\lrcorner\,\delta(\partial_{x_{i}}^{T}\lrcorner\omega)=-\delta^{\partial
M}(\nu\lrcorner\,\partial_{x_{i}}^{T}\lrcorner\omega)=0$, we get that

$\sum_{i=1}^{n+m}|\iota^{*}\delta(\partial_{x_{i}}^{T}\lrcorner\omega)|^{2}=\sum_{i=1}^{n+m}|\delta(\partial_{x_{i}}^{T}\lrcorner\omega)|^{2}\mathrel{\mathop{\kern
0.0pt=}\limits^{\eqref{eq:scalinnerpro}}}(p-1)|\iota^{*}\delta\omega|^{2}.$
(24)

Hence adding Equations (23) and (24) allows to find the l.h.s. of Inequality
(3.8). We are now going to estimate the term
$\displaystyle\sum_{i=1}^{n+m}|\Delta(\partial_{x_{i}}^{T}\lrcorner\omega)|^{2}$
in (19). Taking the divergence of (20) and the differential of the identity
$\delta(\partial_{x_{i}}^{T}\lrcorner\omega)=-\partial_{x_{i}}^{T}\lrcorner\delta\omega$
along with the Cartan formula and the decomposition of the Lie derivative as
in (20), we get that

$\Delta(\partial_{x_{i}}^{T}\lrcorner\omega)=[\delta,\nabla_{\partial_{x_{i}}^{T}}](\omega)+\delta(I\\!I_{\partial_{x_{i}}^{\perp}}^{[p]}\omega)-I\\!I^{[p-1]}_{\partial_{x_{i}}^{\perp}}(\delta\omega)+\partial_{x_{i}}^{T}\lrcorner\Delta\omega.$
(25)

In the following, we will compute each term of (25) separately. First, take an
orthonormal frame $\\{e_{1},\cdots,e_{n}\\}$ of $TM$ such that $\nabla
e_{i}=0$ at some point. Then we have, for any vector field $X$ on $M$,

$\displaystyle[\delta,\nabla_{X}](\omega)$ $\displaystyle=$
$\displaystyle\delta(\nabla_{X}\omega)-\nabla_{X}\delta\omega$ (26)
$\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner\nabla_{e_{s}}\nabla_{X}\omega-\nabla_{X}\delta\omega$
$\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner\left(R(e_{s},X)\omega+\nabla_{X}\nabla_{e_{s}}\omega+\nabla_{[e_{s},X]}\omega\right)-\nabla_{X}\delta\omega$
$\displaystyle=$ $\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},X)\omega+\nabla_{X}\delta\omega-\sum_{s=1}^{n}e_{s}\lrcorner\nabla_{\nabla_{e_{s}}X}\omega-\nabla_{X}\delta\omega$
$\displaystyle=$ $\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},X)\omega-\sum_{s=1}^{n}\
e_{s}\lrcorner\nabla_{\nabla_{e_{s}}X}\omega.$

Now we use the fact that for any tensor field $A$, any vector field $X$ and
any $p$-form $\alpha$ on $M$, $\nabla_{X}A^{[p]}=(\nabla_{X}A)^{[p]}$ and
$A^{[p-1]}(X\lrcorner\alpha)=X\lrcorner A^{[p]}\alpha-A(X)\lrcorner\alpha$,
which both can be proved by a straightforward computation. Thus for any $N\in
T^{\perp}M$, we can write

$\displaystyle\delta(I\\!I^{[p]}_{N}\omega)-I\\!I^{[p-1]}_{N}(\delta\omega)$
$\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner\nabla_{e_{s}}(I\\!I^{[p]}_{N}\omega)-I\\!I^{[p-1]}_{N}(\delta\omega)$
(27) $\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner(\nabla_{e_{s}}I\\!I_{N})^{[p]}(\omega)-\sum_{s=1}^{n}e_{s}\lrcorner
I\\!I_{N}^{[p]}(\nabla_{e_{s}}\omega)-I\\!I^{[p-1]}_{N}(\delta\omega)$
$\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{N})^{[p-1]}(e_{s}\lrcorner\omega)-\sum_{s=1}^{n}((\nabla_{e_{s}}I\\!I_{N})(e_{s}))\lrcorner\omega-\sum_{s=1}^{n}I\\!I^{[p-1]}_{N}(e_{s}\lrcorner\nabla_{e_{s}}\omega)$
$\displaystyle-\sum_{s=1}^{n}(I\\!I_{N}e_{s})\lrcorner(\nabla_{e_{s}}\omega)-I\\!I^{[p-1]}_{N}(\delta\omega)$
$\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{N})^{[p-1]}(e_{s}\lrcorner\omega)+\delta(I\\!I_{N})\lrcorner\omega-\sum_{s=1}^{n}(I\\!I_{N}e_{s})\lrcorner(\nabla_{e_{s}}\omega).$

By taking $X=\partial_{x_{i}}^{T}$ in (26) and $N=\partial_{x_{i}}^{\perp}$ in
(27), Equation (25) reduces after using the fact that
$\nabla\partial_{x_{i}}^{T}=I\\!I_{\partial_{x_{i}}^{\perp}}$ as a consequence
of the parallelism of the vector field $\partial_{x_{i}}$, to

$\displaystyle\Delta(\partial_{x_{i}}^{T}\lrcorner\omega)$ $\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},\partial_{x_{i}}^{T})\omega-2\sum_{s=1}^{n}(I\\!I_{\partial_{x_{i}}^{\perp}}e_{s})\lrcorner(\nabla_{e_{s}}\omega)$
(28)
$\displaystyle-\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})^{[p-1]}(e_{s}\lrcorner\omega)+\delta(I\\!I_{\partial_{x_{i}}^{\perp}})\lrcorner\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega.$

We proceed in the computation of (28). Let us compute the term
$\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})^{[p-1]}(e_{s}\lrcorner\omega)$.
Using the fact that
$A^{[p]}=\sum_{l=1}^{n}e_{l}\wedge\left(A(e_{l})\lrcorner\right)$ for any
symmetric tensor $A$, we compute with the help of Equation (56) in the
appendix

$\displaystyle\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})^{[p-1]}(e_{s}\lrcorner\omega)$
$\displaystyle=$
$\displaystyle\sum_{l,s=1}^{n}e_{l}\wedge\left((\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})(e_{l})\lrcorner
e_{s}\lrcorner\omega\right)$ (29) $\displaystyle=$
$\displaystyle\sum_{l,s=1}^{n}e_{l}\wedge\left((\nabla_{e_{l}}I\\!I_{\partial_{x_{i}}^{\perp}})(e_{s})\lrcorner
e_{s}\lrcorner\omega\right)-\sum_{l,s=1}^{n}e_{l}\wedge\left(I\\!I_{(\nabla^{{\mathbb{R}}^{n+m}}_{e_{l}}\partial_{x_{i}}^{\perp})^{\perp}})(e_{s})\lrcorner
e_{s}\lrcorner\omega\right)$
$\displaystyle+\sum_{l,s=1}^{n}e_{l}\wedge\left(I\\!I_{(\nabla^{{\mathbb{R}}^{n+m}}_{e_{s}}\partial_{x_{i}}^{\perp})^{\perp}})(e_{l})\lrcorner
e_{s}\lrcorner\omega\right).$

The first two sums vanish identically, since $\sum_{s}A(e_{s})\lrcorner
e_{s}\lrcorner\omega=0$ for any symmetric endomorphism $A$. Hence with the
help of Equation (58), Equality (29) reduces to

$\displaystyle\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})^{[p-1]}(e_{s}\lrcorner\omega)$
$\displaystyle=$
$\displaystyle-\sum_{l,s=1}^{n}e_{l}\wedge\left(I\\!I_{I\\!I(e_{s},\partial_{x_{i}}^{T})}(e_{l})\lrcorner
e_{s}\lrcorner\omega\right)$ $\displaystyle=$
$\displaystyle-\sum_{l,s=1}^{n}\sum_{a=1}^{m}\langle
I\\!I(e_{s},\partial_{x_{i}}^{T}),f_{a}\rangle
e_{l}\wedge\left(I\\!I_{f_{a}}(e_{l})\lrcorner e_{s}\lrcorner\omega\right)$
$\displaystyle=$
$\displaystyle-\sum_{l=1}^{n}\sum_{a=1}^{m}e_{l}\wedge\left(I\\!I_{f_{a}}(e_{l})\lrcorner
I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner\omega\right)=-\sum_{a=1}^{m}I\\!I_{f_{a}}^{[p-1]}(I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner\omega)$

where $\\{f_{1},\cdots,f_{m}\\}$ is an orthonormal frame of $T^{\perp}M$. In
the last equality, we used again the expression
$A^{[p]}=\sum_{l}e_{l}\wedge(A(e_{l})\lrcorner)$ for any symmetric
endomorphism $A$. Hence after replacing Equations (57), proved in the
appendix, and (3) into Equation (28), we finally get

$\displaystyle\Delta(\partial_{x_{i}}^{T}\lrcorner\omega)$ $\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},\partial_{x_{i}}^{T})\omega-2\sum_{s=1}^{n}(I\\!I_{\partial_{x_{i}}^{\perp}}e_{s})\lrcorner(\nabla_{e_{s}}\omega)$
(31)
$\displaystyle+\sum_{a=1}^{m}I\\!I_{f_{a}}^{[p-1]}(I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner\omega)+\left(\sum_{a=1}^{m}I\\!I^{2}_{f_{a}}(\partial_{x_{i}}^{T})-nd(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)-nI\\!I_{\widetilde{H}}(\partial_{x_{i}}^{T})\right)\lrcorner\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega$
$\displaystyle=$ $\displaystyle-\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},\partial_{x_{i}}^{T})\omega-2\sum_{s=1}^{n}(I\\!I_{\partial_{x_{i}}^{\perp}}e_{s})\lrcorner(\nabla_{e_{s}}\omega)$
$\displaystyle+\sum_{a=1}^{m}I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner
I\\!I_{f_{a}}^{[p]}\omega+\left(-nd(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)-nI\\!I_{\widetilde{H}}(\partial_{x_{i}}^{T})\right)\lrcorner\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega$
$\displaystyle=$
$\displaystyle-I\\!I_{n\widetilde{H}}(\partial_{x_{i}}^{T})\lrcorner\omega+\sum_{a=1}^{m}I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner
I\\!I_{f_{a}}^{[p]}\omega-2\sum_{s=1}^{n}(I\\!I_{\partial_{x_{i}}^{\perp}}e_{s})\lrcorner(\nabla_{e_{s}}\omega)$
$\displaystyle+\sum_{a=1}^{m}I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner
I\\!I_{f_{a}}^{[p]}\omega+\left(-nd(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)-nI\\!I_{\widetilde{H}}(\partial_{x_{i}}^{T})\right)\lrcorner\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega$
$\displaystyle=$
$\displaystyle-2I\\!I_{n\widetilde{H}}(\partial_{x_{i}}^{T})\lrcorner\omega+2\sum_{a=1}^{m}I\\!I_{f_{a}}(\partial_{x_{i}}^{T})\lrcorner
I\\!I_{f_{a}}^{[p]}\omega-2\sum_{s=1}^{n}(I\\!I_{\partial_{x_{i}}^{\perp}}e_{s})\lrcorner(\nabla_{e_{s}}\omega)-n\big{(}d\big{(}\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)\big{)}\lrcorner\omega$
$\displaystyle+\partial_{x_{i}}^{T}\lrcorner\Delta\omega.$

In the second equality, we used again the relation
$A^{[p-1]}(X\lrcorner\alpha)=X\lrcorner A^{[p]}\alpha-A(X)\lrcorner\alpha$ for
any $p$-form $\alpha$ and $X\in TM$, and in the third equality, we used
Proposition 6.3 in the appendix. Equation (31), along with (23) and (24),
gives the result, by using Inequality (19). $\square$

In general, it is difficult to control all the terms in Inequality (3.8) in
order to compare $q_{1,p-1}$ and $q_{1,p}$. Therefore, we shall restrict
ourselves to the case when $M$ is a domain in $\mathbb{S}^{n}$. We have the
following result:

###### Theorem 3.9

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary. If $M$
is a domain in $\mathbb{S}^{n}$, then we have that

$q_{1,p-1}+(n-p)q_{1,p+1}<C_{p,n}q_{1,p},$ (32)

where $C_{p,n}$ is some constant that depends on $p$ and $n$ and whose
explicit expression is given in (37).

Proof. We consider the isometric immersion
$M\subset\mathbb{S}^{n}\hookrightarrow{\mathbb{R}}^{n+1}$. In this case, we
have that $m=1$, the orthonormal basis of $T^{\perp}M$ reduces to the inward
unit vector field $\tilde{\nu}=-\sum_{i=1}^{n+1}x_{i}\partial_{x_{i}}$, the
second fundamental form is given by $I\\!I_{\widetilde{\nu}}={\rm Id}$ and
$\widetilde{H}=\tilde{\nu}$. Therefore, Inequality (3.8) becomes

$\displaystyle pq_{1,p-1}\int_{\partial M}(|\nu\lrcorner
d\omega|^{2}+|\iota^{*}(\delta\omega)|^{2})d\mu_{g}$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{n+1}\int_{M}\Big{|}(2p-2n)\partial_{x_{i}}^{T}\lrcorner\omega+2\langle\partial_{x_{i}}^{\perp},\widetilde{\nu}\rangle\delta\omega-
nd(\langle\widetilde{\nu},\partial_{x_{i}}^{\perp}\rangle)\lrcorner\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega\Big{|}^{2}d\mu_{g}.$

Now, an elementary computation shows that
$d(\langle\widetilde{\nu},\partial_{x_{i}}^{\perp}\rangle)=-\partial_{x_{i}}^{T}$.
Therefore, the above inequality reduces to

$\displaystyle pq_{1,p-1}\int_{\partial M}(|\nu\lrcorner
d\omega|^{2}+|\iota^{*}(\delta\omega)|^{2})d\mu_{g}$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{n+1}\int_{M}|(2p-n)\partial_{x_{i}}^{T}\lrcorner\omega+2\langle\partial_{x_{i}}^{\perp},\widetilde{\nu}\rangle\delta\omega+\partial_{x_{i}}^{T}\lrcorner\Delta\omega|^{2}d\mu_{g}$
(33) $\displaystyle=$
$\displaystyle\int_{M}\Big{(}(2p-n)^{2}p|\omega|^{2}+4|\delta\omega|^{2}+p|\Delta\omega|^{2}$
$\displaystyle\phantom{\int_{M}\Big{(}}+2p(2p-n)\langle\omega,\Delta\omega\rangle\Big{)}d\mu_{g}$
$\displaystyle=$
$\displaystyle\int_{M}(4|\delta\omega|^{2}+p|\Delta\omega+(2p-n)\omega|^{2})d\mu_{g}.$

Here, we use Identity (22) and that

$\sum_{i=1}^{n+1}\langle\partial_{x_{i}}^{\perp},\widetilde{\nu}\rangle\langle\partial_{x_{i}}^{T}\lrcorner\omega,\delta\omega\rangle=\sum_{i=1}^{n+1}\langle\partial_{x_{i}},\widetilde{\nu}\rangle\langle\partial_{x_{i}}^{T}\lrcorner\omega,\delta\omega\rangle=\langle\widetilde{\nu}^{T}\lrcorner\omega,\delta\omega\rangle=0.$

The same argument applies to
$\sum_{i=1}^{n+1}\langle\partial_{x_{i}}^{\perp},\widetilde{\nu}\rangle\langle\delta\omega,\partial_{x_{i}}^{T}\lrcorner\Delta\omega\rangle=0$.
Since Inequality (33) is true for any $p$-eigenform $\omega$, we apply it to
the $(n-p)$-eigenform $*_{M}\omega$ to get

$(n-p)q_{1,n-p-1}\int_{\partial M}(|\iota^{*}(\delta\omega)|^{2}+|\nu\lrcorner
d\omega|^{2})d\mu_{g}\leq\int_{M}(4|d\omega|^{2}+(n-p)|\Delta\omega+(2n-2p-n)\omega|^{2})d\mu_{g}.$
(34)

Summing inequalities (33) and (34) and using the fact that
$q_{1,n-p-1}=q_{1,p+1}$ yield the following:

$\displaystyle(pq_{1,p-1}+(n-p)q_{1,p+1})\int_{\partial M}(|\nu\lrcorner
d\omega|^{2}+|\iota^{*}(\delta\omega)|^{2})d\mu_{g}$ (35) $\displaystyle\leq$
$\displaystyle\int_{M}(4|d\omega|^{2}+4|\delta\omega|^{2}+p|\Delta\omega+(2p-n)\omega|^{2}+(n-p)|\Delta\omega+(n-2p)\omega|^{2})d\mu_{g}$
$\displaystyle=$
$\displaystyle\int_{M}(4\langle\Delta\omega,\omega\rangle+p|\Delta\omega+(2p-n)\omega|^{2}+(n-p)|\Delta\omega+(n-2p)\omega|^{2})d\mu_{g}.$

Here, we used the fact that $\int_{M}\langle\Delta\omega,\omega\rangle
d\mu_{g}=\int_{M}(|d\omega|^{2}+|\delta\omega|^{2})d\mu_{g}$ as a consequence
of the boundary condition $\omega_{|_{\partial M}}=0$. Now, the Bochner
formula applied to the eigenform $\omega$, with the help of the pointwise
inequality
$|\nabla\alpha|^{2}\geq\frac{1}{p+1}|d\alpha|^{2}+\frac{1}{n-p+1}|\delta\alpha|^{2}$
which is true for any $p$-form $\alpha$ [7, Lemme 6.8], gives that

$\displaystyle\int_{M}\langle\Delta\omega,\omega\rangle d\mu_{g}$
$\displaystyle=$
$\displaystyle\int_{M}|\nabla\omega|^{2}d\mu_{g}+\frac{1}{2}\int_{M}\Delta(|\omega|^{2})d\mu_{g}+\int_{M}\langle\mathfrak{B}^{[p]}\omega,\omega\rangle
d\mu_{g}$ $\displaystyle\geq$
$\displaystyle\int_{M}\frac{1}{a(p,n)}(|d\omega|^{2}+|\delta\omega|^{2})d\mu_{g}+\int_{\partial{M}}\langle\nabla_{\nu}\omega,\omega\rangle
d\mu_{g}+\int_{M}p(n-p)|\omega|^{2}d\mu_{g}$ $\displaystyle=$
$\displaystyle\frac{1}{a(p,n)}\int_{M}\langle\Delta\omega,\omega\rangle
d\mu_{g}+\int_{M}p(n-p)|\omega|^{2}d\mu_{g}.$

Here, we have set $a(p,n)={\rm max}(p+1,n-p+1)$ and used the fact that the
Bochner operator $\mathfrak{B}^{[p]}$ is equal to $p(n-p)\mathrm{Id}$ on the
round sphere $\mathbb{S}^{n}$, see e.g. [7, Cor. 2.6] and [7, Rem. 6.15].
Thus, we deduce that

$p(n-p)\|\omega\|^{2}_{L^{2}(M)}\leq\left(1-\frac{1}{a(p,n)}\right)\int_{M}\langle\Delta\omega,\omega\rangle
d\mu_{g}\leq\left(1-\frac{1}{a(p,n)}\right)\|\Delta\omega\|_{L^{2}(M)}\|\omega\|_{L^{2}(M)},$

so that

$\|\omega\|_{L^{2}(M)}\leq\frac{a(p,n)-1}{p(n-p)a(p,n)}\|\Delta\omega\|_{L^{2}(M)}.$
(36)

Coming back to Inequality (35) and using again the Cauchy-Schwarz inequality
as well as the estimate $\|\alpha+\beta\|^{2}_{L^{2}(M)}\leq
2(\|\alpha\|^{2}_{L^{2}(M)}+\|\beta\|^{2}_{L^{2}(M)})$, we obtain

$\displaystyle(pq_{1,p-1}+(n-p)q_{1,p+1})\int_{\partial M}(|\nu\lrcorner
d\omega|^{2}+|\iota^{*}(\delta\omega)|^{2})d\mu_{g}$ $\displaystyle\leq$
$\displaystyle
4\|\Delta\omega\|_{L^{2}(M)}\|\omega\|_{L^{2}(M)}+2p\|\Delta\omega\|^{2}_{L^{2}(M)}$
$\displaystyle+2p(2p-n)^{2}\|\omega\|_{L^{2}(M)}^{2}+2(n-p)\|\Delta\omega\|^{2}_{L^{2}(M)}$
$\displaystyle+2(n-p)(n-2p)^{2}\|\omega\|^{2}_{L^{2}(M)}$
$\displaystyle\mathrel{\mathop{\kern 0.0pt\leq}\limits^{\eqref{eq:caushw}}}$
$\displaystyle C(p,n)\|\Delta\omega\|^{2}_{L^{2}(M)},$

where $C(p,n)$ is the constant given by

$\displaystyle C(p,n)$ $\displaystyle=$ $\displaystyle
4\frac{a(p,n)-1}{p(n-p)a(p,n)}+2n+(2p(2p-n)^{2}+2(n-p)(n-2p)^{2}))(\frac{a(p,n)-1}{p(n-p)a(p,n)})^{2}$
(37) $\displaystyle=$ $\displaystyle
4\frac{a(p,n)-1}{p(n-p)a(p,n)}+2n+2n(2p-n)^{2}(\frac{a(p,n)-1}{p(n-p)a(p,n)})^{2}.$

Finally Characterization (12) allows to get the result. Notice that, if (32)
were an equality, then by the limiting case in the Cauchy-Schwarz inequality,
the form $\Delta\omega$ would be parallel to $\omega$. But by
$\Delta^{2}\omega=0$, this would imply that $\Delta\omega=0$. Because of
$\omega_{|_{\partial M}}=0$, we would deduce from [1] that $\omega=0$ on $M$,
which is a contradiction. Therefore, (32) always remains strict. $\square$

##  4 Robin vs. Dirichlet and Neumann eigenvalue problems

In this section, we will establish estimates for the Robin eigenvalue problem
on differential forms defined in [4]. We mainly generalize some results in
[13] to differential forms. For this purpose, we recall the Robin problem on
forms. Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary
$\partial M$. Fix a positive real number $\tau$. Then the boundary value
problem [4]

$\left\\{\begin{array}[]{lll}\Delta\omega&=\lambda\omega&\textrm{ on }M\\\
\iota^{*}(\nu\lrcorner\,d\omega-\tau\omega)&=0&\textrm{ on }\partial M\\\
\nu\lrcorner\,\omega&=0&\textrm{ on }\partial M\end{array}\right.$ (38)

is elliptic and self-adjoint. It admits an increasing unbounded sequence of
positive real eigenvalues with finite multiplicities

$\lambda_{1,p}(\tau)\leq\lambda_{2,p}(\tau)\leq\cdots$

The first eigenvalue $\lambda_{1,p}(\tau)$ of the Robin boundary problem (38)
can be characterized as follows:

$\lambda_{1,p}(\tau)={\rm
inf}\left\\{\frac{\displaystyle\int_{M}\left(|d\omega|^{2}+|\delta\omega|^{2}\right)d\mu_{g}+\tau\int_{\partial
M}|\iota^{*}\omega|^{2}d\mu_{g}}{\displaystyle\int_{M}|\omega|^{2}d\mu_{g}}\right\\},$
(39)

where $\omega$ runs over all non-identically vanishing $p$-forms on $M$ such
that $\nu\lrcorner\omega=0$. When the parameter $\tau$ tends to $0$, the Robin
problem (38) reduces to the Neumann boundary problem, that is

$\left\\{\begin{array}[]{lll}\Delta\omega&=\lambda\omega&\textrm{ on }M\\\
\nu\lrcorner\,d\omega&=0&\textrm{ on }\partial M\\\
\nu\lrcorner\,\omega&=0&\textrm{ on }\partial M.\end{array}\right.$ (40)

Notice that the first eigenvalue $\lambda_{1,p}^{N}$ of (40) is nonnegative
and the kernel of the operator (40) is isomorphic to the so-called absolute de
Rham cohomology, which is defined by

$H_{A}^{p}(M)=\\{\omega\in\Omega^{p}(M)|\,\,d\omega=\delta\omega=0\,\,\text{on}\,\,M\,\,\,\text{and}\,\,\,\nu\lrcorner\omega=0\,\,\text{on}\,\,\partial
M\\}.$

When $\tau\to\infty$, the Robin problem (38) reduces to the Dirichlet boundary
problem

$\left\\{\begin{array}[]{lll}\Delta\omega&=\lambda\omega&\textrm{ on }M\\\
\omega&=0&\textrm{ on }\partial M.\\\ \end{array}\right.$ (41)

By [1], the first eigenvalue $\lambda_{1,p}^{D}$ of problem (41) is positive.
We also have the estimate [4, Prop. 5.4]

$\lambda_{1,p}^{N}\leq\lambda_{1,p}(\tau)\leq\lambda_{1,p}^{D}.$

In the following, we will establish another estimates for
$\lambda_{1,p}(\tau)$ in terms of the Neumann and Dirichlet ones. We have:

###### Theorem 4.1

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary. We have
the following estimates for the first eigenvalue of the Robin boundary
problem:

  1. 1.

Assume that the absolute de Rham cohomology $H_{A}^{p}(M)$ does not vanish. We
denote by $\omega_{D}$ an eigenform of the Dirichlet boundary problem
associated to $\lambda^{D}_{1,p}$ and let $\omega_{0}$ be the orthogonal
projection of $\omega_{D}$ on the space $H_{A}^{p}(M)$, assumed to be nonzero.
Then

$\dfrac{1}{\lambda_{1,p}(\tau)}\geq\dfrac{1}{\lambda^{D}_{1,p}}+\dfrac{\|\omega_{0}\|^{4}_{L^{2}(M)}}{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}.$

  2. 2.

Assume that the first eigenvalue $\lambda^{N}_{1,p}$ of the Neumann boundary
problem is positive, then

$\dfrac{1}{\lambda_{1,p}(\tau)}\geq\dfrac{1}{\lambda^{N}_{1,p}}-\dfrac{\tau\alpha_{N}(\lambda^{D}_{1,p}-\lambda^{N}_{1,p})}{\lambda^{N}_{1,p}(\tau\alpha_{N}\lambda^{D}_{1,p}+\lambda^{N}_{1,p}(\lambda^{D}_{1,p}-\lambda^{N}_{1,p}))},$

where $\alpha_{N}=\dfrac{\|\omega_{N}\|^{2}_{L^{2}(\partial
M)}}{\|\omega_{N}\|^{2}_{L^{2}(M)}}$ and $\omega_{N}$ is being an eigenform of
the Neumann boundary problem associated to $\lambda^{N}_{1,p}$.

Proof. We begin with the proof of the first point. Let $\omega_{D}$ be a
$p$-eigenform associated to the first eigenvalue $\lambda_{1,p}^{D}$ that is
assumed to be of $L^{2}$-norm equal to $1$. Let $\omega_{0}$ be the orthogonal
projection of $\omega_{D}$ to $H_{A}^{p}(M)$. For any real number $t$, we
consider the $p$-form

$\omega_{t}=\omega_{D}+t\,\omega_{0}.$

Clearly, we have that $\nu\lrcorner\,\omega_{t}=0$. Therefore, by the
characterization of the first eigenvalue $\lambda_{1,p}(\tau)$, we have that

$\lambda_{1,p}(\tau)\leq\dfrac{\displaystyle\int_{M}\left(|d\omega_{t}|^{2}+|\delta\omega_{t}|^{2}\right)d\mu_{g}+\tau\int_{\partial
M}|\omega_{t}|^{2}d\mu_{g}}{\displaystyle\int_{M}|\omega_{t}|^{2}d\mu_{g}}.$

By the definition of the form $\omega_{0}$, we have that

$\|d\omega_{t}\|^{2}_{L^{2}(M)}+\|\delta\omega_{t}\|^{2}_{L^{2}(M)}=\lambda_{1,p}^{D},\,\|\omega_{t}\|^{2}_{L^{2}(\partial
M)}=t^{2}\|\omega_{0}\|^{2}_{L^{2}(\partial M)}.$

Also, we have that

$\|\omega_{t}\|^{2}_{L^{2}(M)}=1+t^{2}\|\omega_{0}\|^{2}_{L^{2}(M)}+2t\|\omega_{0}\|^{2}_{L^{2}(M)}.$

The last term comes from the fact that $\omega_{0}$ is the orthogonal
projection of $\omega_{D}$. Thus by plugging in the above inequality, we get
that

$\lambda_{1,p}(\tau)\leq\dfrac{\lambda^{D}_{1,p}+t^{2}\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{1+t^{2}\|\omega_{0}\|^{2}_{L^{2}(M)}+2t\|\omega_{0}\|^{2}_{L^{2}(M)}}.$

Now, we take the inverse of this last inequality, then add and subtract the
term $\dfrac{t^{2}\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}$ in the numerator to find that

$\dfrac{1}{\lambda_{1,p}(\tau)}\geq\dfrac{1}{\lambda^{D}_{1,p}}+\dfrac{t^{2}\left(\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}\right)+2t\|\omega_{0}\|^{2}_{L^{2}(M)}}{\lambda^{D}_{1,p}+t^{2}\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}.$

Since this is true for any real number $t$, then we deduce that
$\dfrac{1}{\lambda_{1,p}(\tau)}\geq\dfrac{1}{\lambda^{D}_{1,p}}+\mathop{\rm
sup}\limits_{{\mathbb{R}}}(f)$ where $f$ is the function given by

$f(t)=\dfrac{t^{2}\left(\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}\right)+2t\|\omega_{0}\|^{2}_{L^{2}(M)}}{\lambda^{D}_{1,p}+t^{2}\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}=\dfrac{At^{2}+Bt}{Ct^{2}+D}$

with

$\displaystyle
A=\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}},$ $\displaystyle B=2\|\omega_{0}\|^{2}_{L^{2}(M)},$
$\displaystyle C=\tau\|\omega_{0}\|^{2}_{L^{2}(\partial M)},$ $\displaystyle
D=\lambda^{D}_{1,p}.$

It is easy to check that the supremum of $f$ is attained at
$t_{0}=\dfrac{AD+\sqrt{A^{2}D^{2}+B^{2}CD}}{BC}$ which is equal to

$\sup_{\mathbb{R}}f=f(t_{0})=\dfrac{t_{0}(At_{0}+B)}{Ct_{0}^{2}+D}=\dfrac{t_{0}(At_{0}+B)}{\dfrac{2ADt_{0}}{B}+D+D}=\dfrac{Bt_{0}}{2D}=\dfrac{A+\sqrt{A^{2}+\dfrac{B^{2}C}{D}}}{2C}.$

Now, by replacing $A,B,C$ and $D$ by their values, we estimate

$\displaystyle A^{2}+\dfrac{B^{2}C}{D}$ $\displaystyle=$
$\displaystyle\left(\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}\right)^{2}+\dfrac{4\|\omega_{0}\|^{4}_{L^{2}(M)}\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}$ $\displaystyle=$
$\displaystyle\left(\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}-2\|\omega_{0}\|^{4}_{L^{2}(M)}\right)^{2}-4\|\omega_{0}\|^{8}_{L^{2}(M)}+4\|\omega_{0}\|^{6}_{L^{2}(M)}$
$\displaystyle\geq$
$\displaystyle\left(\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}-2\|\omega_{0}\|^{4}_{L^{2}(M)}\right)^{2},$

since $\|\omega_{0}\|^{2}_{L^{2}(M)}\leq\|\omega_{D}\|^{2}_{L^{2}(M)}=1$. Then

$\displaystyle\sup_{\mathbb{R}}f$ $\displaystyle\geq$
$\displaystyle\dfrac{\left(\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}\right)+\left|\,\|\omega_{0}\|^{2}_{L^{2}(M)}-\dfrac{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}{\lambda^{D}_{1,p}}-2\|\omega_{0}\|^{4}_{L^{2}(M)}\,\right|}{2\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}$ $\displaystyle\geq$
$\displaystyle\dfrac{\|\omega_{0}\|^{4}_{L^{2}(M)}}{\tau\|\omega_{0}\|^{2}_{L^{2}(\partial
M)}}.$

This shows the required estimate. To prove the second inequality, let
$\omega_{D}$ (resp. $\omega_{N}$) an eigenform of the Dirichlet (resp.
Neumann) boundary problem associated to $\lambda^{D}_{1,p}$ (resp.
$\lambda^{N}_{1,p}$) such that
$\|\omega_{D}\|_{L^{2}(M)}=\|\omega_{N}\|_{L^{2}(M)}=1$. For any nonnegative
number $s$, we consider the $p$-form $\omega_{s}=s\omega_{D}+\omega_{N}$,
which clearly satisfies $\nu\lrcorner\,\omega_{s}=0$. In order to use the
Rayleigh inequality for the eigenvalue $\lambda_{1,p}(\tau)$, we compute

$\displaystyle\int_{M}(|d\omega_{s}|^{2}+|\delta\omega_{s}|^{2})d\mu_{g}$
$\displaystyle=$ $\displaystyle
s^{2}\int_{M}(|d\omega_{D}|^{2}+|\delta\omega_{D}|^{2})d\mu_{g}+\int_{M}(|d\omega_{N}|^{2}+|\delta\omega_{N}|^{2})d\mu_{g}$
$\displaystyle+2s\int_{M}\langle d\omega_{D},d\omega_{N}\rangle
d\mu_{g}+2s\int_{M}\langle\delta\omega_{D},\delta\omega_{N}\rangle d\mu_{g}$
$\displaystyle=$ $\displaystyle
s^{2}\lambda^{D}_{1,p}+\lambda^{N}_{1,p}+2s\lambda^{N}_{1,p}(\omega_{D},\omega_{N})_{L^{2}(M)}.$

In the last equality, we use the Stokes formula. Also, we have

$\|\omega_{s}\|_{L^{2}(\partial M)}^{2}=\|\omega_{N}\|_{L^{2}(\partial
M)}^{2}\,\,\,\text{and}\,\,\,\|\omega_{s}\|_{L^{2}(M)}^{2}=s^{2}+1+2s(\omega_{D},\omega_{N})_{L^{2}(M)}.$

Therefore, after replacing we get that

$\lambda_{1,p}(\tau)\leq\dfrac{s^{2}\lambda^{D}_{1,p}+\lambda^{N}_{1,p}+2s\lambda^{N}_{1,p}(\omega_{D},\omega_{N})_{L^{2}(M)}+\tau\|\omega_{N}\|_{L^{2}(\partial
M)}^{2}}{s^{2}+1+2s(\omega_{D},\omega_{N})_{L^{2}(M)}}.$

As we did before, we take the inverse of this last inequality, then add and
subtract the term
$\frac{s^{2}\lambda_{1,p}^{D}+\tau\|\omega\|^{2}_{L^{2}(\partial
M)}}{\lambda^{N}_{1,p}}$ in the numerator to get that

$\dfrac{1}{\lambda_{1,p}(\tau)}\geq\dfrac{1}{\lambda^{N}_{1,p}}+\dfrac{s^{2}\left(1-\dfrac{\lambda^{D}_{1,p}}{\lambda^{N}_{1,p}}\right)-\dfrac{\tau\alpha_{N}}{\lambda^{N}_{1,p}}}{s^{2}\lambda^{D}_{1,p}+\lambda^{N}_{1,p}+2s\lambda^{N}_{1,p}+\tau\alpha_{N}}.$

Here, we also use the fact that $|(\omega_{D},\omega_{N})_{L^{2}(M)}|\leq 1$
by the Cauchy-Schwarz inequality and the fact that $s\geq 0$. In order to get
the lower bound, we need to compute the supremum of the function $g$ which is
given by

$g(s)=\dfrac{As^{2}+B}{Cs^{2}+Ds+E},$

with

$\displaystyle A=1-\dfrac{\lambda^{D}_{1,p}}{\lambda^{N}_{1,p}},$
$\displaystyle B=-\dfrac{\tau\alpha_{N}}{\lambda^{N}_{1,p}},$ $\displaystyle
C=\lambda^{D}_{1,p},$ $\displaystyle D=2\lambda^{N}_{1,p},$ $\displaystyle
E=\lambda^{N}_{1,p}+\tau\alpha_{N}.$

A direct computation shows that the supremum of the function $g$ is attained
at $s_{2}=\dfrac{\tau\alpha_{N}}{\lambda^{D}_{1,p}-\lambda^{N}_{1,p}}$ and
thus

${\rm
sup}_{{\mathbb{R}}}(g)=g(s_{2})=-\dfrac{\tau\alpha_{N}(\lambda^{D}_{1,p}-\lambda^{N}_{1,p})}{\lambda^{N}_{1,p}(\tau\alpha_{N}\lambda^{D}_{1,p}+\lambda^{N}_{1,p}(\lambda^{D}_{1,p}-\lambda^{N}_{1,p}))}.$

This finishes the proof of the theorem. $\square$

In the following result, we will give a gap inequality between the eigenvalues
of the Robin Laplacian under some curvature conditions. Let $(M^{n},g)$ be a
Riemannian manifold with smooth boundary and let
$\eta_{1}(x),\eta_{2}(x),\cdots,\eta_{n-1}(x)$ be the principal curvatures at
a point $x$ of the boundary (i.e. eigenvalues of the second fundamental form
of the Weingarten tensor $S$). We assume that
$\eta_{1}(x)\leq\eta_{2}(x)\leq\cdots\leq\eta_{n-1}(x)$. For any integer
$p\in\\{1,\cdots,n-1\\}$, the $p$-curvatures $\sigma_{p}(x)$ are defined as
$\sigma_{p}(x)=\eta_{1}(x)+\cdots+\eta_{p}(x)$. One can easily check that for
any integer $p$ and $q$ with $p\leq q$, we have that
$\frac{\sigma_{p}(x)}{p}\leq\frac{\sigma_{q}(x)}{q}$ with equality if and only
if $\eta_{1}(x)=\eta_{2}(x)=\cdots=\eta_{q}(x)$. Hence, we deduce that
$H\geq\frac{\sigma_{p}(x)}{p}$ for any integer $p\in\\{1,\cdots,n-1\\}$. In
the next theorem, we set

$\sigma_{p}=\mathop{\rm inf}\limits_{x\in\partial M}\sigma_{p}(x)$

We state the result which generalizes [4, Thm. 5.8].

###### Theorem 4.2

Let $M$ be a compact domain in ${\mathbb{R}}^{n}$. Fix an integer number
$q\in\\{1,\cdots,n-1\\}$ and Let $\omega$ be a $q$-eigenform of the Robin
Laplacian. If $\sigma_{p}>0$ for some $p\leq q$, then we have

$\frac{\|\omega\|^{2}_{L^{2}(\partial
M)}}{\|\omega\|^{2}_{L^{2}(M)}}\leq\frac{1}{\sigma_{p}}(\lambda_{1,q}-\lambda_{1,q-p}).$

Proof. We mainly follow the same computations as in [4, Thm. 5.8]. Let
$\omega$ be a $q$-eigenform of the Robin Laplacian and, for any $p\leq q$,
consider the $(q-p)$-form
$\phi_{i_{1},\cdots,i_{p}}:=\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\omega$,
for $i_{k}=1,\cdots,n$ with $k=1,\cdots,p$. Clearly, we have that
$\nu\lrcorner\phi_{i_{1},\cdots,i_{p}}=0$. Hence by the characterization (39)
of the first eigenvalue, we get that

$\lambda_{1,q-p}(\tau)\int_{M}|\phi_{i_{1},\cdots,i_{p}}|^{2}d\mu_{g}\leq\int_{M}(|d\phi_{i_{1},\cdots,i_{p}}|^{2}+|\delta\phi_{i_{1},\cdots,i_{p}}|^{2})d\mu_{g}+\tau\int_{\partial
M}|\phi_{i_{1},\cdots,i_{p}}|^{2}d\mu_{g}.$ (42)

Next we sum over $i_{1},\cdots,i_{p}$. We begin with the l.h.s. Applying
successively (22), we have

$\displaystyle\sum_{i_{1},\cdots,i_{p}}|\phi_{i_{1},\cdots,i_{p}}|^{2}$
$\displaystyle=$
$\displaystyle\sum_{i_{1},\cdots,i_{p}}|\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\omega|^{2}$
(43) $\displaystyle=$ $\displaystyle(q-(p-1))\cdots
q\cdot|\omega|^{2}=\frac{q!}{(q-p)!}|\omega|^{2}.$

For the r.h.s., we first compute

$\displaystyle\sum_{i_{1},\cdots,i_{p}}|\delta\phi_{i_{1},\cdots,i_{p}}|^{2}$
$\displaystyle=$
$\displaystyle\sum_{i_{1},\cdots,i_{p}}|\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\delta\omega|^{2}\mathrel{\mathop{\kern
0.0pt=}\limits^{\eqref{eq:scalinnerpro}}}\frac{(q-1)!}{(q-1-p)!}|\delta\omega|^{2}.$
(44)

In order to compute the term
$\sum_{i_{1},\cdots,i_{p}}|d\phi_{i_{1},\cdots,i_{p}}|^{2}$, we proceed as in
Equation (20). Using repeatedly the identity
$d(X\lrcorner\alpha)=\nabla_{X}\alpha-X\lrcorner d\alpha$, true for any
parallel vector field $X$, we get

$\displaystyle
d\phi_{i_{1},\cdots,i_{p}}=(-1)^{p}\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner
d\omega+\sum_{l=1}^{p}(-1)^{l+1}\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega.$

Thus, we find that the sum
$\sum_{i_{1},\cdots,i_{p}}|d\phi_{i_{1},\cdots,i_{p}}|^{2}$ is equal to

$\displaystyle\sum_{i_{1},\cdots,i_{p}}|\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner
d\omega|^{2}+\sum_{i_{1},\cdots,i_{p}}|\sum_{l=1}^{p}(-1)^{l+1}\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega|^{2}$
$\displaystyle+2(-1)^{p}\sum_{i_{1},\cdots,i_{p}}\sum_{l=1}^{p}(-1)^{l+1}\langle\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner
d\omega,\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega\rangle$
$\displaystyle\mathrel{\mathop{\kern
0.0pt=}\limits^{\eqref{eq:scalinnerpro}}}$
$\displaystyle\frac{(q+1)!}{(q+1-p)!}|d\omega|^{2}+\sum_{i_{1},\cdots,i_{p}}\sum_{l}\langle\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega,\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega\rangle$
$\displaystyle-2\sum_{i_{1},\cdots,i_{p}}\sum_{l<s}(-1)^{l+s}\langle\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega,\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{s}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{s}}}}\omega\rangle$
$\displaystyle-2\sum_{i_{1},\cdots,i_{p}}\sum_{l=1}^{p}\langle\partial_{x_{i_{1}}}\lrcorner\cdots\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner(\partial_{x_{i_{l}}}\lrcorner
d\omega),\partial_{x_{i_{1}}}\lrcorner\cdots\lrcorner\widehat{\partial_{x_{i_{l}}}}\lrcorner\cdots\lrcorner\partial_{x_{i_{p}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega\rangle$
$\displaystyle=$
$\displaystyle\frac{(q+1)!}{(q+1-p)!}|d\omega|^{2}+\frac{pq!}{(q-p+1)!}|\nabla\omega|^{2}$
$\displaystyle-2\sum_{i_{1},\cdots,i_{p}}\sum_{l<s}(q-1)-(p-3))\cdots(q-1)\langle\partial_{x_{i_{s}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega,\partial_{x_{i_{l}}}\lrcorner\nabla_{\partial_{x_{i_{s}}}}\omega\rangle-2\frac{pq!}{(q-p+1)!}|d\omega|^{2}.$

In this lenghty computation, we used the fact that
$\sum_{i}\langle\partial_{x_{i}}\lrcorner\alpha,\partial_{x_{i}}\lrcorner\beta\rangle=p\langle\alpha,\beta\rangle$
for any $p$-forms $\alpha$ and $\beta$. We also make use of the formula
$d=\sum_{i}\partial_{x_{i}}\wedge\nabla_{\partial_{x_{i}}}$. Now, one can
easily check by using the expression of $d$ and $\nabla$ that the sum term in
the above equality is equal to

$\sum_{i_{1},\cdots,i_{p}}\sum_{l<s}\langle\partial_{x_{i_{s}}}\lrcorner\nabla_{\partial_{x_{i_{l}}}}\omega,\partial_{x_{i_{l}}}\lrcorner\nabla_{\partial_{x_{i_{s}}}}\omega\rangle=\begin{pmatrix}p\\\
2\end{pmatrix}(|\nabla\omega|^{2}-|d\omega|^{2}).$

Hence, after simplifying, we find that

$\sum_{i_{1},\cdots,i_{p}}|d\phi_{i_{1},\cdots,i_{p}}|^{2}=\frac{p(q-1)!}{(q-p)!}|\nabla\omega|^{2}+\frac{(q-1)!}{(q-p-1)!}|d\omega|^{2}.$
(45)

Plugging Equalities (43), (44) and (45) into Inequality (42) yields after
simplifying by $\frac{(q-1)!}{(q-p)!}$

$\displaystyle\lambda_{1,q-p}(\tau)q\int_{M}|\omega|^{2}d\mu_{g}$
$\displaystyle\leq$ $\displaystyle
p\int_{M}|\nabla\omega|^{2}d\mu_{g}+(q-p)\int_{M}(|d\omega|^{2}+|\delta\omega|^{2})d\mu_{g}$
(46) $\displaystyle+\tau q\int_{\partial M}|\omega|^{2}d\mu_{g}.$

Now the Bochner formula $\Delta=\nabla^{*}\nabla+\mathfrak{B}^{[q]}$ applied
to the form $\omega$ gives after integrating that

$\lambda_{1,q}(\tau)\int_{M}|\omega|^{2}d\mu_{g}=\int_{M}\langle\Delta\omega,\omega\rangle
d\mu_{g}=\int_{M}|\nabla\omega|^{2}d\mu_{g}+\int_{\partial
M}\langle\nabla_{\nu}\omega,\omega\rangle d\mu_{g}.$

By using $\langle\nabla_{\nu}\omega,\omega\rangle=\langle
S^{[q]}(\iota^{*}\omega),\iota^{*}\omega\rangle+\tau|\omega|^{2}$ as proved in
[9], we get that

$\int_{M}|\nabla\omega|^{2}d\mu_{g}=\lambda_{1,q}(\tau)\int_{M}|\omega|^{2}d\mu_{g}-\int_{\partial
M}(\langle
S^{[q]}(\iota^{*}\omega),\iota^{*}\omega\rangle+\tau|\omega|^{2})d\mu_{g}.$

Plugging this last equality into Inequality (46), we get after using the
pointwise inequality $\langle
S^{[q]}(\iota^{*}\omega),\iota^{*}\omega\rangle\geq\sigma_{q}|\omega|^{2}$
that

$\displaystyle\lambda_{1,q-p}(\tau)q\int_{M}|\omega|^{2}d\mu_{g}$
$\displaystyle\leq$ $\displaystyle
p\lambda_{1,q}(\tau)\int_{M}|\omega|^{2}d\mu_{g}-p\int_{\partial
M}(\sigma_{q}+\tau)|\omega|^{2}d\mu_{g}$
$\displaystyle+(q-p)\left(\lambda_{1,q}(\tau)\int_{M}|\omega|^{2}d\mu_{g}-\tau\int_{\partial
M}|\omega|^{2}d\mu_{g}\right)+\tau q\int_{\partial M}|\omega|^{2}d\mu_{g}$
$\displaystyle=$ $\displaystyle
q\lambda_{1,q}(\tau)\int_{M}|\omega|^{2}d\mu_{g}-p\sigma_{q}\int_{\partial
M}|\omega|^{2}d\mu_{g}$ $\displaystyle\leq$ $\displaystyle
q\lambda_{1,q}(\tau)\int_{M}|\omega|^{2}d\mu_{g}-q\sigma_{p}\int_{\partial
M}|\omega|^{2}d\mu_{g}.$

In the last inequality, we used the fact that
$\frac{\sigma_{p}}{p}\leq\frac{\sigma_{q}}{q}$. This finishes the proof of the
theorem. $\square$

##  5 Robin and Steklov operators vs. biharmonic Steklov

In this section, we relate the eigenvalues of the Robin problem to those of
the biharmonic Steklov operator. This extends the result stated in [13, Thm.
1.17] for functions.

###### Theorem 5.1

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary. We have
for any $\tau>0$ the estimate

$\dfrac{1}{\lambda_{1,p}(\tau)}\leq\dfrac{1}{\lambda^{D}_{1,p}}+\dfrac{1}{\tau
q_{1,p}},$

where $\lambda^{D}_{1,p}$ is the first eigenvalue of the Dirichlet boundary
problem (41).

Proof. We mainly follow the computations done in [13, Thm. 1.17]. Let $\omega$
be an eigenform of the Robin boundary problem (38) associated to
$\lambda_{1,p}(\tau)$. We denote by $\omega_{1}$ a solution to the problem

$\left\\{\begin{array}[]{lll}\Delta\omega_{1}&=0&\textrm{ on }M\\\
\iota^{*}\omega_{1}&=\iota^{*}\omega&\textrm{ on }\partial M\\\
\nu\lrcorner\,\omega_{1}&=0&\textrm{ on }\partial M.\end{array}\right.$ (47)

Notice that such a problem admits a unique solution by [24]. Now, let us
consider the $p$-form $\omega_{2}:=\omega-\omega_{1}$. It clearly satisfies

$\left\\{\begin{array}[]{lll}\Delta\omega_{2}&=\Delta\omega&\textrm{ on }M\\\
\omega_{2}&=0&\textrm{ on }\partial M.\end{array}\right.$ (48)

By using the triangle inequality, the characterization (13) and the one of the
first eigenvalue $\lambda_{1,p}^{D}$ of the Dirichlet problem (corresponding
to $\tau\to\infty$ in (39)), we have

$\displaystyle\|\omega\|_{L^{2}(M)}$ $\displaystyle\leq$
$\displaystyle\|\omega_{1}\|_{L^{2}(M)}+\|\omega_{2}\|_{L^{2}(M)}$ (49)
$\displaystyle\leq$
$\displaystyle\sqrt{q_{1,p}^{-1}}\|\omega_{1}\|_{L^{2}(\partial
M)}+\sqrt{(\lambda_{1,p}^{D})^{-1}}\left(\|d\omega_{2}\|^{2}_{L^{2}(M)}+\|\delta\omega_{2}\|^{2}_{L^{2}(M)}\right)^{\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\sqrt{q_{1,p}^{-1}}\|\omega\|_{L^{2}(\partial
M)}+\sqrt{(\lambda_{1,p}^{D})^{-1}}\left(\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}\right)^{\frac{1}{2}}.$

Indeed, $\|\omega_{1}\|_{L^{2}(\partial M)}=\|\omega\|_{L^{2}(\partial M)}$
since $\omega_{2}=0$ on $\partial M$ and

$\displaystyle\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}$
$\displaystyle=$
$\displaystyle\|d\omega_{1}\|^{2}_{L^{2}(M)}+\|d\omega_{2}\|^{2}_{L^{2}(M)}+2(d\omega_{1},d\omega_{2})_{L^{2}(M)}$
$\displaystyle+\|\delta\omega_{1}\|^{2}_{L^{2}(M)}+\|\delta\omega_{2}\|^{2}_{L^{2}(M)}+2(\delta\omega_{1},\delta\omega_{2})_{L^{2}(M)}$
$\displaystyle=$
$\displaystyle\|d\omega_{1}\|^{2}_{L^{2}(M)}+\|\delta\omega_{1}\|^{2}_{L^{2}(M)}+\|d\omega_{2}\|^{2}_{L^{2}(M)}+\|\delta\omega_{2}\|^{2}_{L^{2}(M)}$
$\displaystyle+2\int_{M}\langle\delta d\omega_{1},\omega_{2}\rangle
d\mu_{g}-2\int_{\partial
M}\langle\nu\lrcorner\,\omega_{1},\iota^{*}\omega_{2}\rangle d\mu_{g}$
$\displaystyle+2\int_{M}\langle d\delta\omega_{1},\omega_{2}\rangle
d\mu_{g}+2\int_{\partial
M}\langle\iota^{*}\delta\omega_{1},\nu\lrcorner\,\omega_{2}\rangle d\mu_{g}$
$\displaystyle=$
$\displaystyle\|d\omega_{1}\|^{2}_{L^{2}(M)}+\|\delta\omega_{1}\|^{2}_{L^{2}(M)}+\|d\omega_{2}\|^{2}_{L^{2}(M)}+\|\delta\omega_{2}\|^{2}_{L^{2}(M)}$
$\displaystyle\geq$
$\displaystyle\|d\omega_{2}\|^{2}_{L^{2}(M)}+\|\delta\omega_{2}\|^{2}_{L^{2}(M)}.$

Now, we square both sides of Inequality (49) to write

$\displaystyle\|\omega\|^{2}_{L^{2}(M)}$ $\displaystyle\leq$ $\displaystyle
q_{1,p}^{-1}\|\omega\|^{2}_{L^{2}(\partial
M)}+(\lambda_{1,p}^{D})^{-1}\left(\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}\right)$
$\displaystyle+2\left(q_{1,p}^{-1}(\lambda_{1,p}^{D})^{-1}\|\omega\|^{2}_{L^{2}(\partial
M)}\left(\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}\right)\right)^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle q_{1,p}^{-1}\|\omega\|^{2}_{L^{2}(\partial
M)}+(\lambda_{1,p}^{D})^{-1}\left(\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}\right)$
$\displaystyle\tau^{-1}q_{1,p}^{-1}\left(\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}\right)+\tau(\lambda_{1,p}^{D})^{-1}\|\omega\|^{2}_{L^{2}(\partial
M)}$ $\displaystyle=$
$\displaystyle(\tau^{-1}q_{1,p}^{-1}+(\lambda_{1,p}^{D})^{-1})\left(\|d\omega\|^{2}_{L^{2}(M)}+\|\delta\omega\|^{2}_{L^{2}(M)}+\tau\|\omega\|^{2}_{L^{2}(\partial
M)}\right).$

In the second above inequality, we use the fact that
$2\sqrt{ab}\leq\frac{a}{\tau}+\tau b$ for any real positive $\tau$. The
characterization (39) allows to deduce the estimate. $\square$

Now, we come back to the Serrin problem on forms. We will use the existence of
solution to this problem in harmonic domains carrying parallel forms to get an
estimate for the eigenvalues of the absolute Dirichlet-to-Neumann operator
introduced in [21] (see also [11]). Recall the definition of this operator.
Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary
$\partial M$. Let $p\in\\{0,\cdots,n-1\\}$. Given any $p$-form $\omega$ on
$\partial M$, there exists a unique $p$-form $\hat{\omega}$ on $M$ such that
[24]

$\left\\{\begin{array}[]{lll}\Delta\hat{\omega}&=0&\textrm{ on }M\\\
\iota^{*}\hat{\omega}&=\omega&\textrm{ on }\partial M\\\
\nu\lrcorner\omega&=0&\textrm{ on }\partial M.\\\ \end{array}\right.$ (50)

The form $\hat{\omega}$ is usually called the harmonic tangential extension of
$\omega$. The Dirichlet-to-Neumann operator is then defined as
$T^{[p]}:\Lambda^{p}(\partial M)\to\Lambda^{p}(\partial M)$,
$\omega\mapsto-\nu\lrcorner d\hat{\omega}$. When $p=0$, this operator reduces
to the classical Dirichlet-to-Neumann operator on functions, known as Steklov
operator. It is shown in [21] that $T^{[p]}$ is an elliptic self-adjoint
pseudo-differential operator with discrete spectrum consisting of eigenvalues

$0\leq\nu_{1,p}(M)\leq\nu_{2,p}(M)\leq\cdots.$

The kernel of this operator is isomorphic to the absolute de Rham cohomology
$H_{A}^{p}(M)$ introduced in Section 4. The dual problem to (50) (w.r.t. the
Hodge star operator) is called the relative Dirichlet-to-Neumann operator and
is defined by $T^{[p]}_{D}=(-1)^{p(n-1-p)}*_{\partial M}T^{[n-1-p]}*_{\partial
M}$. If $\nu_{1,p}^{D}(M)$ is the first eigenvalue of $T^{[p]}_{D}$, we have

$\nu_{1,p}^{D}(M)=\nu_{1,n-1-p}(M).$

Also, we have the following characterization [22] for the first eigenvalue
$\nu_{1,p}^{D}(M)$:

$\nu_{1,p}^{D}(M)={\rm
inf}\left\\{\frac{\displaystyle\int_{M}(|d\phi|^{2}+|\delta\phi|^{2})d\mu_{g}}{\displaystyle\int_{\partial
M}|\phi|^{2}d\mu_{g}}|\,\,\phi\in\Omega^{p+1}(M),\iota^{*}\phi=0\right\\}.$
(51)

###### Theorem 5.2

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary. Assume
that $M$ is a harmonic domain and carries a parallel $p$-form for some
$p=1,\cdots,n-1$. If moreover $\sigma_{p}>0$ or $\sigma_{n-p}>0$, then

${\rm min}(\nu_{1,p-1}(M),\nu_{1,n-1-p}(M))\leq\frac{{\rm Vol}(\partial
M)}{{\rm Vol}(M)}.$

Proof. Let $\omega$ be any $p$-eigenform of the biharmonic Steklov operator
associated to some eigenvalue, say $q$. We let the $(p+1)$-form
$\phi:=d\omega$. Clearly, we have that $\iota^{*}\phi=0$. Hence, by
Characterization (51), we get that

$\nu_{1,p}^{D}(M)\displaystyle\int_{\partial M}|\nu\lrcorner
d\omega|^{2}d\mu_{g}\leq\displaystyle\int_{M}|\delta d\omega|^{2}d\mu_{g}.$
(52)

Now, applying the same characterization (51) to $d(*_{M}\omega)$, since
$*_{M}\omega$ is also an eigenform of the biharmonic Steklov operator, yields
the inequality

$\nu_{1,n-p}^{D}(M)\displaystyle\int_{\partial
M}|\iota^{*}\delta\omega|^{2}d\mu_{g}\leq\displaystyle\int_{M}|d\delta\omega|^{2}d\mu_{g}.$
(53)

Summing Inequalities (52) and (53) yields

${\rm min}(\nu_{1,p}^{D}(M),\nu_{1,n-p}^{D}(M))\int_{\partial M}(|\nu\lrcorner
d\omega|^{2}+|\iota^{*}\delta\omega|^{2})d\mu_{g}\leq\int_{M}(|\delta
d\omega|^{2}+|d\delta\omega|^{2})d\mu_{g}.$

Now a direct computation using the Stokes formula and the boundary conditions
on $\omega$ gives that

$\int_{M}|\Delta\omega|^{2}d\mu_{g}=\int_{M}(|d\delta\omega|^{2}+|\delta
d\omega|^{2})d\mu_{g}-\frac{2}{q^{2}}\int_{\partial
M}\langle\nu\lrcorner\Delta\omega,\delta^{\partial
M}(\iota^{*}\Delta\omega)\rangle d\mu_{g}.$

Hence after plugging this equality into the above inequality, we deduce that

${\rm min}(\nu_{1,p}^{D}(M),\nu_{1,n-p}^{D}(M))\leq
q+\frac{2}{q^{2}}\frac{\displaystyle\int_{\partial
M}\langle\nu\lrcorner\Delta\omega,\delta^{\partial
M}(\iota^{*}\Delta\omega)\rangle d\mu_{g}}{\displaystyle\int_{\partial
M}(|\nu\lrcorner d\omega|^{2}+|\iota^{*}\delta\omega|^{2})d\mu_{g}}.$ (54)

Notice that this inequality is true for any eigenform $\omega$ of the
biharmonic Steklov operator. From Theorem (3.6), we know that when $M$ is a
harmonic domain carrying a parallel $p$-form $\omega_{0}$, the form
$\omega=f\cdot\omega_{0}$ is an eigenform associated to the eigenvalue
$q=\frac{{\rm Vol}(\partial M)}{{\rm Vol}(M)}$. Hence, we will apply
Inequality (54) to the particular form $\omega=f\cdot\omega_{0}$. For this
purpose, we will check the sign of the integral. Assume first that
$\sigma_{n-p}>0$. We estimate

$\displaystyle\int_{\partial
M}\langle\nu\lrcorner\Delta\omega,\delta^{\partial
M}(\iota^{*}\Delta\omega)\rangle d\mu_{g}$ $\displaystyle=$
$\displaystyle\int_{\partial M}\langle\nu\lrcorner\omega_{0},\delta^{\partial
M}(\iota^{*}\omega_{0})\rangle d\mu_{g}$ $\displaystyle=$
$\displaystyle\int_{\partial
M}\langle\nu\lrcorner\omega_{0},(S^{[p-1]}-(n-1)H)\nu\lrcorner\omega_{0})\rangle
d\mu_{g}$ $\displaystyle\leq$ $\displaystyle\int_{\partial
M}\left((\sigma_{n-1}-\sigma_{n-p})|\nu\lrcorner\omega_{0}|^{2}-\sigma_{n-1}|\nu\lrcorner\omega_{0}|^{2}\right)d\mu_{g}$
$\displaystyle\leq$ $\displaystyle 0.$

In the second equality, we use the identity [20, Lem. 18]

$\delta^{\partial
M}(\iota^{*}\omega_{0})=\nu\lrcorner\nabla_{\nu}\omega_{0}+\iota^{*}(\delta\omega_{0})+S^{[p-1]}(\nu\lrcorner\omega_{0})-(n-1)H\nu\lrcorner\omega_{0}.$

Also, we use the pointwise estimate $\langle
S^{[p]}\alpha,\alpha\rangle\leq(\sigma_{n-1}-\sigma_{n-p-1})|\alpha|^{2}$ for
any $p$-form $\alpha$. Hence, we deduce that

${\rm min}(\nu_{1,p}^{D}(M),\nu_{1,n-p}^{D}(M))\leq\frac{{\rm Vol}(\partial
M)}{{\rm Vol}(M)}.$

Finally, the fact that $\nu_{1,p}^{D}(M)=\nu_{1,n-1-p}(M)$ and
$\nu_{1,n-p}^{D}(M)=\nu_{1,p-1}(M)$ finishes the proof of the statement when
$\sigma_{n-p}>0$. If $\sigma_{p}>0$, then replacing $p$ by $n-p$ and by
invariance of parallel forms under the Hodge star operator on $M$, we obtain
the same inequality. This concludes the proof. $\square$

###### Remark 5.3

Notice that a similar estimate has been established in [21, Cor. 15] for
compact manifolds carrying parallel forms with the assumption
$H_{A}^{p}(M)=H^{p}(M)=0$. The inequality is

$\nu_{1,n-1-p}(M)+\nu_{1,p-1}(M)\leq\frac{{\rm Vol}(\partial M)}{{\rm
Vol}(M)}.$

##  6 Appendix

###### Lemma 6.1

Let $(M^{n},g)$ be a compact Riemannian manifold with smooth boundary
$\partial M$ and let $\nu$ be the inward unit normal vector field to the
boundary. Consider the following boundary value problem

$\left\\{\begin{array}[]{lll}\Delta^{2}\omega&=f&\textrm{ on }M\\\
B_{1}\omega&=\omega_{1}&\textrm{ on }\partial M\\\
B_{2}\omega&=\omega_{2}&\textrm{ on }\partial M\\\
B_{3}\omega&=\omega_{3}&\textrm{ on }\partial M\end{array}\right.$ (55)

for given $f\in\Omega^{p}(M)$,
$\omega_{1}\in\Gamma(\Lambda^{p}T^{*}M_{|_{\partial M}})$,
$\omega_{2}\in\Omega^{p-1}(\partial M)$ and $\omega_{3}\in\Omega^{p}(\partial
M)$ and where if $E_{1}:=\Lambda^{p}T^{*}M_{|_{\partial M}}$,
$E_{2}:=\Lambda^{p-1}T^{*}\partial M$ and $E_{3}:=\Lambda^{p}T^{*}\partial M$,
$B_{1}\colon\Omega^{p}(M)\to\Gamma(E_{1})$ such that
$B_{1}\omega:=\omega_{|_{\partial M}}$,
$B_{2}\colon\Omega^{p}(M)\to\Gamma(E_{2})$ and
$B_{3}\colon\Omega^{p}(M)\to\Gamma(E_{3})$ are either:

  1. 1.

$B_{2}=\iota^{*}\delta\omega$ and $B_{3}=\nu\lrcorner d\omega$. In this case,
$\eqref{eq:bvporder4}$ is elliptic in the sense of Lopatinskiĭ-Shapiro (see
Definition $1.6.1$ in [24]), self-adjoint and its kernel is reduced to
$\\{0\\}$. Or

  2. 2.

$B_{2}\omega:=\nu\lrcorner\Delta\omega+q\iota^{*}(\delta\omega)$ and
$B_{3}\omega:=\iota^{*}\Delta\omega-q\nu\lrcorner\,d\omega$ for some real
constant $q$. In this case, problem $\eqref{eq:bvporder4}$ is elliptic in the
sense of Lopatinskiĭ-Shapiro.

Proof. Given any $v\in T_{x}^{*}\partial M\setminus\\{0\\}$ for a fixed
$x\in\partial M$, we consider the space

$\mathcal{M}_{v}^{+}:=\left\\{\textrm{bounded solutions }y=y(t)\textrm{ on
}{\mathbb{R}}_{+}\textrm{ to the ODE
}\sigma_{\Delta^{2}}((-iv,\partial_{t}))y=0\right\\}.$

A direct computation shows that

$\mathcal{M}_{v}^{+}=\mathrm{Span}\left(e^{-|v|t}(at+b)\cdot\omega_{0}\,|\,a,b\in{\mathbb{R}},\,\omega_{0}\in\Lambda^{p}T_{x}^{*}M_{|_{\partial
M}}\right),$

which is hence a space of dimension $N:=2\left(\begin{array}[]{c}n\\\
p\end{array}\right)$. We look at the pointwise map

$\displaystyle\mathcal{M}_{v}^{+}$ $\displaystyle\longrightarrow$
$\displaystyle\bigoplus_{j=1}^{3}E_{j}$ $\displaystyle y$
$\displaystyle\longmapsto$
$\displaystyle\left(\sigma_{B_{1}}((-iv,\partial_{t}))y,\sigma_{B_{2}}((-iv,\partial_{t}))y,\sigma_{B_{3}}((-iv,\partial_{t}))y\right)(0)$

which we want to show to be an isomorphism. Note already that space dimensions
are equal on both sides. Since
$\sigma_{B_{1}}((-iv,\partial_{t}))=\mathrm{Id}$,
$\sigma_{B_{2}}((-iv,\partial_{t}))=-\partial_{t}\cdot\nu\lrcorner\cdot+iv\lrcorner\iota^{*}\cdot$
and $\sigma_{B_{3}}((-iv,\partial
t))=\partial_{t}\cdot\iota^{*}+iv\wedge(\nu\lrcorner\cdot)$, we obtain that,
for any fixed $\omega_{0}\in\Lambda^{p}T_{x}^{*}M$, the element
$e^{-|v|t}\cdot\omega_{0}$ of $\mathcal{M}_{v}^{+}$ (corresponding to $a=0$
and $b=1$) is sent to
$(\omega_{0},|v|\nu\lrcorner\omega_{0}+iv\lrcorner\iota^{*}\omega_{0},-|v|\iota^{*}\omega_{0}+iv\wedge(\nu\lrcorner\omega_{0}))$;
and that the element $te^{-|v|t}\cdot\omega_{0}$ of $\mathcal{M}_{v}^{+}$
(corresponding to $a=1$ and $b=0$) is sent to
$(0,-\nu\lrcorner\omega_{0},\iota^{*}\omega_{0})$. Choosing a basis
$(\omega_{0}^{(1)},\ldots,\omega_{0}^{(N)})$ of $\Lambda^{p}T_{x}^{*}M$, the
basis
$\left(e^{-|v|t}\cdot\omega_{0}^{(1)},\ldots,e^{-|v|t}\cdot\omega_{0}^{(N)},te^{-|v|t}\cdot\omega_{0}^{(1)},\ldots,te^{-|v|t}\cdot\omega_{0}^{(N)}\right)$
of $\mathcal{M}_{v}^{+}$ will therefore be sent to a basis of
$\bigoplus_{j=1}^{3}E_{j}$. This shows the map
$\mathcal{M}_{v}^{+}\longrightarrow\bigoplus_{j=1}^{3}E_{j}$ to be an
isomorphism. Therefore (55) is elliptic.
Using (6), it is easy to see that (55) is also self-adjoint. Moreover, the
kernel of (55) is reduced to $\\{0\\}$: namely, if $\omega\in\Omega^{p}(M)$
solves (55) with $f=0$ as well as $\omega_{1}=\omega_{2}=\omega_{3}=0$, then
(7) implies that $\|\Delta\omega\|_{L^{2}(M)}=0$, from which $\omega=0$ on $M$
follows using $\omega_{|_{\partial M}}=0$.
As a consequence, fixing $f=0$ as well as $\omega_{1}=0$, for any given
$(\omega_{2},\omega_{3})\in\Omega^{p-1}(\partial M)\oplus\Omega^{p}(\partial
M)$, there exists a unique $\omega\in\Omega^{p}(M)$ solving (55). In
particular, $\omega\in Z$. This shows the map $Z\to\Omega^{p}(\partial
M)\oplus\Omega^{p-1}(\partial M)$, $\omega\mapsto(\nu\lrcorner
d\omega,\iota^{*}\delta\omega)$, to be onto. This proves $1.$
Changing the boundary operators $B_{2}$ and $B_{3}$ via
$B_{2}\omega:=\nu\lrcorner\Delta\omega+q\iota^{*}(\delta\omega)$ and
$B_{3}\omega:=\iota^{*}\Delta\omega-q\nu\lrcorner\,d\omega$ for some real
contant $q$ (which actually plays no role since it is only involved in the
first-order-terms of the b.c. and not in their principal symbols), we still
get elliptic boundary conditions for $\Delta^{2}$: for any $v\in
T_{x}^{*}\partial M\setminus\\{0\\}$, the pointwise map
$\mathcal{M}_{v}^{+}\longrightarrow\bigoplus_{j=1}^{3}E_{j}$ from above sends
$e^{-|v|t}\cdot\omega_{0}$ to $(\omega_{0},0,0)$ and sends
$te^{-|v|t}\cdot\omega_{0}$ to
$(0,2|v|\nu\lrcorner\omega_{0},2|v|\iota^{*}\omega_{0})$, which shows that map
to be an isomorphism. Therefore $B_{1},B_{2},B_{3}$ define elliptic boundary
conditions for $\Delta^{2}$, this proves $2.$ $\square$

###### Lemma 6.2

Let $M^{n}\hookrightarrow{\mathbb{R}}^{n+m}$ be an isometric immersion and let
$I\\!I$ be the second fundamental form of the immersion. For all $X,Y\in TM$
and $N\in T^{\perp}M$, we have

$(\nabla_{X}I\\!I_{N})(Y)=(\nabla_{Y}I\\!I_{N})(X)-I\\!I_{\nabla_{Y}^{\perp}N}(X)+I\\!I_{\nabla_{X}^{\perp}N}(Y),$
(56)

where $\nabla_{X}^{\perp}N:=(\nabla_{X}^{{\mathbb{R}}^{n+m}}N)^{\perp}$
defines the normal connection on $T^{\perp}M$. As a consequence, by writing
$\partial_{x_{i}}=\partial_{x_{i}}^{T}+\partial_{x_{i}}^{\perp}$ for all
$i=1,\cdots,n+m$, the divergence of the endomorphism
$I\\!I_{\partial_{x_{i}}^{\perp}}$ is equal to

$\delta(I\\!I_{\partial_{x_{i}}^{\perp}})=-nd(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)-nI\\!I_{\widetilde{H}}(\partial_{x_{i}}^{T})+\sum_{a=1}^{m}I\\!I^{2}_{f_{a}}(\partial_{x_{i}}^{T}).$
(57)

Here $\\{f_{1},\cdots,f_{m}\\}$ is a local orthonormal frame of $T^{\perp}M$
and $\widetilde{H}$ is the mean curvature field of the immersion.

Proof. Let $X,Y,Z$ be vector fields in $TM$ that we assume to be parallel at
some point in $M$ and let $N\in T^{\perp}M$. We compute

$\displaystyle(\nabla_{X}I\\!I_{N})(Y,Z)$ $\displaystyle=$ $\displaystyle
X(I\\!I_{N}(Y,Z))\quad$ $\displaystyle=$ $\displaystyle X\langle
I\\!I(Y,Z),N\rangle$ $\displaystyle=$
$\displaystyle\langle\nabla_{X}^{{\mathbb{R}}^{n+m}}I\\!I(Y,Z),N\rangle+\langle
I\\!I(Y,Z),\nabla_{X}^{{\mathbb{R}}^{n+m}}N\rangle$ $\displaystyle=$
$\displaystyle\langle\nabla_{Y}^{{\mathbb{R}}^{n+m}}I\\!I(X,Z),N\rangle+\langle
I\\!I(Y,Z),\nabla_{X}^{{\mathbb{R}}^{n+m}}N\rangle$ $\displaystyle=$
$\displaystyle Y(I\\!I_{N}(X,Z))-\langle
I\\!I(X,Z),\nabla_{Y}^{{\mathbb{R}}^{n+m}}N\rangle+\langle
I\\!I(Y,Z),\nabla_{X}^{{\mathbb{R}}^{n+m}}N\rangle$ $\displaystyle=$
$\displaystyle(\nabla_{Y}I\\!I_{N})(X,Z)-I\\!I_{\nabla_{Y}^{\perp}N}(X,Z)+I\\!I_{\nabla_{X}^{\perp}N}(Y,Z).$

In the fourth equality, we use the Codazzi equation for submanifolds in
${\mathbb{R}}^{n+m}$. Hence we get Equality (56). To find Equation (57), we
decompose $\partial_{x_{i}}=\partial_{x_{i}}^{T}+\partial_{x_{i}}^{\perp}$ for
all $i=1,\cdots,n+m$. Then from the parallelism of the vector
$\partial_{x_{i}}$ and the Gauss formula, we get that

$\nabla^{{\mathbb{R}}^{n+m}}_{X}\partial_{x_{i}}^{\perp}=-\nabla^{{\mathbb{R}}^{n+m}}_{X}\partial_{x_{i}}^{T}=-\nabla_{X}\partial_{x_{i}}^{T}-I\\!I(X,\partial_{x_{i}}^{T}),$

where $\nabla$ is the Levi-Civita connection on $TM$. Thus, we deduce that

$\nabla_{X}^{\perp}\partial_{x_{i}}^{\perp}=(\nabla^{{\mathbb{R}}^{n+m}}_{X}\partial_{x_{i}}^{\perp})^{\perp}=-I\\!I(X,\partial_{x_{i}}^{T}).$
(58)

The divergence of the endomorphism $I\\!I_{\partial_{x_{i}}^{\perp}}$ can be
computed using Equation (56) with $N=\partial_{x_{i}}^{\perp}$. For $X\in TM$,
we have

$\displaystyle\delta(I\\!I_{\partial_{x_{i}}^{\perp}})(X)$ $\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})(e_{s},X)$
$\displaystyle=$
$\displaystyle-\sum_{s=1}^{n}(\nabla_{e_{s}}I\\!I_{\partial_{x_{i}}^{\perp}})(X,e_{s})$
$\displaystyle\mathrel{\mathop{\kern 0.0pt=}\limits^{\eqref{eq:delta}}}$
$\displaystyle-\sum_{s}(\nabla_{X}I\\!I_{\partial_{x_{i}}^{\perp}})(e_{s},e_{s})-\sum_{s=1}^{n}I\\!I_{I\\!I(X,\partial_{x_{i}}^{T})}(e_{s},e_{s})+\sum_{s=1}^{n}I\\!I_{I\\!I(e_{s},\partial_{x_{i}}^{T})}(X,e_{s})$
$\displaystyle=$ $\displaystyle-
nX(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)-nI\\!I_{\widetilde{H}}(X,\partial_{x_{i}}^{T})+\sum_{a=1}^{m}\sum_{s=1}^{n}\langle
I\\!I(e_{s},\partial_{x_{i}}^{T}),f_{a}\rangle\langle
I\\!I(X,e_{s}),f_{a}\rangle$ $\displaystyle=$ $\displaystyle-
nX(\langle\widetilde{H},\partial_{x_{i}}^{\perp}\rangle)-nI\\!I_{\widetilde{H}}(\partial_{x_{i}}^{T},X)+\sum_{a=1}^{m}g(I\\!I_{f_{a}}(\partial_{x_{i}}^{T}),I\\!I_{f_{a}}(X)).$

This ends the proof of the lemma. $\square$

###### Proposition 6.3

For any $p$-form $\omega$, the curvature term $\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},X)\omega$ is equal to

$\sum_{s=1}^{n}e_{s}\lrcorner
R(e_{s},X)\omega=-\sum_{a=1}^{m}I\\!I_{f_{a}}(X)\lrcorner
I\\!I_{f_{a}}^{[p]}\omega+I\\!I_{n\widetilde{H}}(X)\lrcorner\omega,$

where $\\{e_{1},\cdots,e_{n}\\}$ and $\\{f_{1},\cdots,f_{m}\\}$ are
respectively orthonormal basis of $TM$ and $T^{\perp}M$.

Proof. In order to compute the curvature term, we use the Gauß equation.
Indeed, for any $X,Y,Z,T\in TM$ we have

$R(X,Y,Z,T)=-\langle I\\!I(X,Z),I\\!I(Y,T)\rangle+\langle
I\\!I(X,T),I\\!I(Y,Z)\rangle,$

which can be equivalently written as

$R(X,Y)Z=-\sum_{a=1}^{m}g(I\\!I_{f_{a}}(X),Z)I\\!I_{f_{a}}(Y)+\sum_{a=1}^{m}g(I\\!I_{f_{a}}(Y),Z)I\\!I_{f_{a}}(X).$

Now, due to linearity, we can consider that $\omega$ of the form
$e_{i_{1}}\wedge\cdots\wedge e_{i_{p}}$. Then, we compute

$\displaystyle\sum_{s=1}^{n}e_{s}\lrcorner R(e_{s},X)\omega$ $\displaystyle=$
$\displaystyle\sum_{s=1}^{n}\sum_{j=1}^{p}e_{s}\lrcorner(e_{i_{1}}\wedge\dots\wedge
R(e_{s},X)e_{i_{j}}\wedge\dots\wedge e_{i_{p}})$ $\displaystyle=$
$\displaystyle-\sum_{a=1}^{m}\sum_{s=1}^{n}\sum_{j=1}^{p}g(I\\!I_{f_{a}}(e_{s}),e_{i_{j}})e_{s}\lrcorner(e_{i_{1}}\wedge\dots\wedge\underbrace{I\\!I_{f_{a}}(X)}_{j-th}\wedge\dots\wedge
e_{i_{p}})$
$\displaystyle+\sum_{a=1}^{m}\sum_{s=1}^{n}\sum_{j=1}^{p}g(I\\!I_{f_{a}}(X),e_{i_{j}})e_{s}\lrcorner(e_{i_{1}}\wedge\dots\wedge\underbrace{I\\!I_{f_{a}}(e_{s})}_{j-th}\wedge\dots\wedge
e_{i_{p}})$ $\displaystyle=$
$\displaystyle-\sum_{a=1}^{m}\sum_{s=1}^{n}\,\sum_{\begin{subarray}{c}k,j=1\\\
k\neq
j\end{subarray}}^{p}(-1)^{k+1}g(I\\!I_{f_{a}}(e_{s}),e_{i_{j}})\delta_{si_{k}}e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(X)\wedge\dots\wedge e_{i_{p}}$
$\displaystyle-\sum_{a=1}^{m}\sum_{s=1}^{n}\sum_{j=1}^{p}(-1)^{j+1}g(I\\!I_{f_{a}}(e_{s}),e_{i_{j}})g(I\\!I_{f_{a}}(X),e_{s})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge
e_{i_{p}}$
$\displaystyle+\sum_{a=1}^{m}\sum_{s=1}^{n}\,\sum_{\begin{subarray}{c}k,j=1\\\
k\neq
j\end{subarray}}^{p}(-1)^{k+1}g(I\\!I_{f_{a}}(X),e_{i_{j}})\delta_{si_{k}}e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{s})\wedge\dots\wedge e_{i_{p}}$
$\displaystyle+\sum_{a=1}^{m}\sum_{s=1}^{n}\sum_{j=1}^{p}(-1)^{j+1}g(I\\!I_{f_{a}}(X),e_{i_{j}})g(I\\!I_{f_{a}}(e_{s}),e_{s})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge
e_{i_{p}}.$

By using the symmetry of the second fundamental form, the above computation
reduces to

$\displaystyle\sum_{s=1}^{n}e_{s}\lrcorner R(e_{s},X)\omega$ $\displaystyle=$
$\displaystyle-\sum_{a=1}^{m}\,\sum_{\begin{subarray}{c}k,j=1\\\ k\neq
j\end{subarray}}^{p}(-1)^{k+1}g(I\\!I_{f_{a}}(e_{i_{k}}),e_{i_{j}})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(X)\wedge\dots\wedge e_{i_{p}}$
$\displaystyle-\sum_{a=1}^{m}\sum_{j=1}^{p}(-1)^{j+1}g(I\\!I_{f_{a}}^{2}(X),e_{i_{j}})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge
e_{i_{p}}$ $\displaystyle+\sum_{a=1}^{m}\,\sum_{\begin{subarray}{c}k,j=1\\\
k\neq
j\end{subarray}}^{p}(-1)^{k+1}g(I\\!I_{f_{a}}(X),e_{i_{j}})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\dots\wedge e_{i_{p}}$
$\displaystyle+\sum_{a=1}^{m}\sum_{j=1}^{p}(-1)^{j+1}g(I\\!I_{n\widetilde{H}}(X),e_{i_{j}})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge
e_{i_{p}}.$

Now, we prove that the first sum vanishes. Indeed, by decomposing with respect
to $k<j$ and $k>j$, it is equal to

$\displaystyle\sum_{k<j}(-1)^{k+1}g(I\\!I_{f_{a}}(e_{i_{k}}),e_{i_{j}})e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(X)\wedge\dots\wedge e_{i_{p}}$
$\displaystyle+\sum_{k>j}(-1)^{k+1}g(I\\!I_{f_{a}}(e_{i_{k}}),e_{i_{j}})e_{i_{1}}\wedge\cdots\wedge
I\\!I_{f_{a}}(X)\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\dots\wedge
e_{i_{p}}$ $\displaystyle=$
$\displaystyle\sum_{k<j}(-1)^{k+j-1}g(I\\!I_{f_{a}}(e_{i_{k}}),e_{i_{j}})I\\!I_{f_{a}}(X)\wedge
e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\dots\wedge
e_{i_{p}}$
$\displaystyle+\sum_{k>j}(-1)^{k+j}g(I\\!I_{f_{a}}(e_{i_{k}}),e_{i_{j}})I\\!I_{f_{a}}(X)\wedge
e_{i_{1}}\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge{\widehat{e}_{i_{k}}}\wedge\dots\wedge
e_{i_{p}}$ $\displaystyle=$ $\displaystyle 0.$

In the same way, we prove that the third sum is equal to
$-\sum_{a=1}^{m}I\\!I_{f_{a}}^{[p-1]}(I\\!I_{f_{a}}(X)\lrcorner\omega)$.
Indeed, it is equal to

$\displaystyle\sum_{k<j}(-1)^{k+1}g(I\\!I_{f_{a}}(X),e_{i_{j}})(e_{i_{1}}\wedge\cdots\wedge\widehat{e_{i_{k}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\cdots\wedge e_{i_{p}})$
$\displaystyle+\sum_{k>j}(-1)^{k+1}g(I\\!I_{f_{a}}(X),e_{i_{j}})(e_{i_{1}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\cdots\wedge\widehat{e_{i_{k}}}\wedge\cdots\wedge
e_{i_{p}})$ $\displaystyle=$
$\displaystyle\sum_{k<j}(-1)^{k+1+j-k-1}g(I\\!I_{f_{a}}(X),e_{i_{j}})(e_{i_{1}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge
e_{i_{p}})$
$\displaystyle+\sum_{k>j}(-1)^{k+1+k-j-1}g(I\\!I_{f_{a}}(X),e_{i_{j}})(e_{i_{1}}\wedge\cdots\wedge\widehat{e_{i_{j}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\cdots\wedge e_{i_{p}})$ $\displaystyle=$
$\displaystyle\sum_{k<j}(-1)^{j}g(I\\!I_{f_{a}}(X),e_{i_{j}})(e_{i_{1}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\cdots\wedge{\widehat{e}_{i_{j}}}\wedge\cdots\wedge
e_{i_{p}})$
$\displaystyle+\sum_{k>j}(-1)^{j}g(I\\!I_{f_{a}}(X),e_{i_{j}})(e_{i_{1}}\wedge\cdots\wedge\widehat{e_{i_{j}}}\wedge\cdots\wedge
I\\!I_{f_{a}}(e_{i_{k}})\wedge\cdots\wedge e_{i_{p}})$ $\displaystyle=$
$\displaystyle-I\\!I_{f_{a}}^{[p-1]}(I\\!I_{f_{a}}(X)\lrcorner\omega).$

In the last equality, we use the formula $A^{[p]}(X_{1}\wedge\cdots\wedge
X_{p})=\sum_{i=1}^{p}X_{1}\wedge\cdots\wedge A(X_{i})\wedge\cdots\wedge X_{p}$
for any vector fields $X_{1},\cdots X_{p}$ in $TM$. Therefore, we deduce that

$\displaystyle\sum_{s=1}^{n}e_{s}\lrcorner R(e_{s},X)\omega$ $\displaystyle=$
$\displaystyle-\sum_{a=1}^{m}I\\!I_{f_{a}}^{2}(X)\lrcorner\omega-\sum_{a=1}^{m}I\\!I_{f_{a}}^{[p-1]}(I\\!I_{f_{a}}(X)\lrcorner\omega)+I\\!I_{n\widetilde{H}}(X)\lrcorner\omega$
$\displaystyle=$ $\displaystyle-\sum_{a=1}^{m}I\\!I_{f_{a}}(X)\lrcorner
I\\!I_{f_{a}}^{[p]}\omega+I\\!I_{n\widetilde{H}}(X)\lrcorner\omega.$

In the last equality, we used the identity
$A^{[p-1]}(X\lrcorner\alpha)=X\lrcorner A^{[p]}\alpha-A(X)\lrcorner\alpha$.
$\square$

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