# The Veronese variety is determinantal; an elementary proof

Rahim Zaare-Nahandi
Rahim Zaare-Nahandi
University of Tehran
<EMAIL_ADDRESS>
###### Abstract.

Let $\nu_{d}:\mathbb{P}^{n}\longrightarrow\mathbb{P}^{N}$ be the Veronese
mapping of degree $d$ where $N={n+d\choose n}-1$. By an elementary proof it is
shown that $\nu_{d}$ is an isomorphism of $\mathbb{P}^{n}$ onto the projective
variety defined by the $2$-minors of a $(n+1)\times{n+d-1\choose n}$ matrix in
the coordinates $[z_{i_{0},\dots,i_{n}}:i_{0}+\dots+i_{n}=d]$.

##  1\. Introduction

The Veronese mapping, supposedly introduced by the Italian mathematician
Giuseppe Veronese (1854-1917), is the map
$\nu_{d}:\mathbb{P}^{n}\longrightarrow\mathbb{P}^{N}$ given by

$[x_{0}:x_{1}:\cdots:x_{n}]\mapsto[x_{0}^{d}:x_{0}^{d-1}x_{1}:x_{0}^{d-1}x_{2}:\cdots:x_{n-1}x_{n}^{d-1}:x_{n}^{d}].$

It is well-known that this map is an embedding, i.e., its image is a
projective variety and the map is an isomorphism between $\mathbb{P}^{n}$ and
the image variety. The image of $\nu_{d}$ is usually called the Veronese
variety of degree $d$. The images of a fixed projective space under Veronese
mappings of different degrees are isomorphic, but their homogenous coordinate
rings are not. I.e., unlike the case of affine varieties, the homogenous
coordinate ring is not a projective invariant. The Veronese embedding allows
one to reduce the study of some problems concerning hypersurfaces of degree
$d$ to the case of hyperplanes [4, 4.4, Example 2]. It is a basic tool in
algebraic geometry (e.g. see the Index of [3] under -Uple embedding and the
Index of [4] under Veronese embedding). It also has applications in other
areas such as Coding Theory [6].

There are several known results in algebraic geometry, the so called
“folklores”, which require difficult or tediously long proofs where most
readers do not bother reproving them. In most text books when discussing the
classical algebraic geometry, a rigorous proof of the fact that the Veronese
variety is indeed a projective variety isomorphic to the initial projective
space, is usually taken for granted. Hartshorne [3, Exercises 2.12 and 3.4]
has left the proof to the reader in a couple of exercises. Smith et. al. [5,
5.1, Proposition] have provided a “proof” which may lead to misunderstanding
by some readers in a first algebraic geometry course (see Remark 2.2).
Shafarevich [4, 4.4, Example 2] has given the main steps of the proof without
much details; the defining equations are presented, it is hinted how certain
open subsets cover the image of the Veronese mapping, and then the inverse
morphism is piecewise defined, whereas, checking most of these steps requires
computations even for mappings of low degree. In a recent paper by Abdolmaleki
and Zaare-Nahandi [1], the generating set of the defining ideal of some
varieties including the Veronese variety is provided as the $2$-minors of a
matrix. In this note, using elementary methods suitable to a first course in
algebraic geometry, it is shown that the presentation of the ”defining ideal”
of the Veronese variety by the $2$-minors of a matrix as provided in [1],
helps in a systematic way, to give a rigorous proof for the same steps stated
by Shafarevich. It needs mentioning that by the term determinantal we mean
represented by minors of a matrix, although in the literature, this term has
also been used for certain varieties of specific type.

##  2\. The construction and the proof of the claim

Construction. Order the set of all monomials of degree $d$ in
$x_{0},\dots,x_{n}$ by the lexicographic order induced by
$x_{0}>x_{1}>\cdots>x_{n}$ . Let $L$ be the matrix where the entries of its
$i$th row are all monomials of degree $d$ in $x_{0},\dots,x_{n}$ with $x_{i}$
as a factor written decreasingly by the induced lexicographic order. Let $M$
be the matrix obtained from $L$ by replacing $x_{0}^{i_{0}}\dots
x_{n}^{i_{n}}$ with $z_{i_{0},\dots,i_{n}}$, the coordinates in
$\mathbb{P}^{N}$. It is clear that both $L$ and $M$ are
$(n+1)\times{n+d-1\choose n}$ matrices.

Example. Consider the embedding

$\nu_{3}:\mathbb{P}^{2}\longrightarrow\mathbb{P}^{9},$
$[x_{0}:x_{1}:x_{2}]\mapsto[x_{0}^{3}:x_{0}^{2}x_{1}:x_{0}^{2}x_{2}:x_{0}x_{1}^{2}:x_{0}x_{1}x_{2}:x_{0}x_{2}^{2}:x_{1}^{3}:x_{1}^{2}x_{2}:x_{1}x_{2}^{2}:x_{2}^{3}].$

Then

$L=\left(\begin{array}[]{cccccc}x_{0}^{3}&x_{0}^{2}x_{1}&x_{0}^{2}x_{2}&x_{0}x_{1}^{2}&x_{0}x_{1}x_{2}&x_{0}x_{2}^{2}\\\
x_{0}^{2}x_{1}&x_{0}x_{1}^{2}&x_{0}x_{1}x_{2}&x_{1}^{3}&x_{1}^{2}x_{2}&x_{1}x_{2}^{2}\\\
x_{0}^{2}x_{2}&x_{0}x_{1}x_{2}&x_{0}x_{2}^{2}&x_{1}^{2}x_{2}&x_{1}x_{2}^{2}&x_{2}^{3}\end{array}\right),$

and

$M=\left(\begin{array}[]{cccccc}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{3,0,0}}&z_{2,1,0}&z_{2,0,1}&z_{1,2,0}&z_{1,1,1}&z_{1,0,2}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{2,1,0}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{1,2,0}}&z_{1,1,1}&z_{0,3,0}&z_{0,2,1}&z_{0,1,2}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{2,0,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{1,1,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{1,0,2}}&z_{0,2,1}&z_{0,1,2}&z_{0,0,3}\end{array}\right).$

######  Theorem 2.1.

Let $\nu_{d}:\mathbb{P}^{n}\longrightarrow\mathbb{P}^{N}$ be the Veronese
mapping of degree $d$. Let $I_{2}(M)$ be the ideal generated by the set of
$2$-minors of $M$ constructed above. Let
$W=\mathbb{V}(I_{2}(M))\subset\mathbb{P}^{N}$ be the projective variety
defined by $I_{2}(M)$. Then $W={\rm Im}(\nu_{d})$ and the Veronese mapping is
an embedding of $\mathbb{P}^{n}$ onto $W$.

###### Proof.

By the construction, $L$ and $M$ may be presented by the following matrices,
respectively:

(1)
$\begin{pmatrix}x_{0}^{d}&x_{0}^{d-1}x_{1}&\cdots&x_{0}^{d-1}x_{n}&x_{0}^{d-2}x_{1}^{2}&\cdots&x_{0}^{d-2}x_{n}^{2}&\cdots&x_{0}x_{n}^{d-1}\\\
x_{0}^{d-1}x_{1}&x_{0}^{d-2}x_{1}^{2}&\cdots&x_{0}^{d-2}x_{1}x_{n}&x_{0}^{d-3}x_{1}^{3}&\cdots&x_{0}^{d-3}x_{1}x_{n}^{2}&\cdots&x_{1}x_{n}^{d-1}\\\
x_{0}^{d-1}x_{2}&x_{0}^{d-2}x_{1}x_{2}&\cdots&x_{0}^{d-2}x_{2}x_{n}&x_{0}^{d-3}x_{1}^{2}x_{2}&\cdots&x_{0}^{d-3}x_{2}x_{n}^{2}&\cdots&x_{2}x_{n}^{d-1}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\
x_{0}^{d-1}x_{n}&x_{0}^{d-2}x_{1}x_{n}&\cdots&x_{0}^{d-2}x_{n}^{2}&\cdots&x_{0}x_{n}^{d-1}&\cdots&x_{n-1}x_{n}^{d-1}&x_{n}^{d}\\\
\end{pmatrix},$

(2)
$\begin{pmatrix}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d,0,\dots,0}}&z_{d-1,1,\dots,0}&\cdots&z_{d-1,0,\dots,0,1}&z_{d-2,2,\dots,0}&\cdots&z_{d-2,0,\dots,2}&\cdots&z_{1,0,\dots,d-1}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-1,1,\dots,0}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-2,2,\dots,0}}&\cdots&z_{d-2,1,0,\dots,1}&z_{d-3,3,\dots,0}&\cdots&z_{d-3,1,\dots,2}&\cdots&z_{0,1,\dots,d-1}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-1,0,1,\dots,0}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-2,1,1,\dots,0}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&z_{d-2,0,1,\dots,1}&z_{d-3,2,1\dots,0}&\cdots&z_{d-3,0,1,\dots,2}&\cdots&z_{0,0,1,\dots,d-1}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&\cdots&\cdots&\cdots&\cdots&\cdots\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&\cdots&\cdots&\cdots&\cdots\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-1,0,\dots,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-2,1,0,\dots,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{d-2,0,\dots,2}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\cdots}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}z_{1,0,\dots,d-1}}&\cdots&z_{0,\dots,1,d-1}&z_{0,0,\dots,d}\\\
\end{pmatrix}.$

We proceed the proof in the following three steps:

Step 1. We show that if $W_{i}\subset W$, $i=0,1,\dots,n$, is the open subset
of points where the coordinate $z_{0,\dots,0,d,0,\dots,0}$, corresponding to
$x_{i}^{d}$, is nonzero, then $W_{0},\dots,W_{n}$ cover $W$. We prove that if
$W_{i}$’s do not cover $W$, i.e., at some point $Q\in W$ all coordinates
$z_{0,\dots,0,d,0,\dots,0}$, corresponding to $x_{i}^{d}$, $i=0\dots,n$, are
zero, then all coordinates $[z_{i_{0},\dots,i_{n}}:i_{0}+\dots+i_{n}=d]$ are
zero, which is a contradiction. We begin to show that all coordinates on the
first row of $M$ are zero. Let $\Delta$ be the lower triangular portion of the
$(n+1)\times(n+1)$ sub-matrix on the leftmost of $M$ (red colored in $M$ and
the Example). Observe that the entries of $\Delta$ are the same as the first
row of $M$. By the construction of $M$, the order of the first row is
preserved downward and then left to the right in $\Delta$. At first note that
the vanishing of the minor

(3) $\begin{vmatrix}z_{d,0,\dots,0}&z_{d-1,1,\dots,0}\\\
z_{d-1,1,0,\dots,0}&z_{d-2,2,0,\dots,0}\\\ \end{vmatrix}$

implies that $z_{d-1,1,\dots,0}=0$. Assume that the vanishing of $2$-minors of
$M$ implies that all entries on the first row of $M$ preceding the entry
$z_{j_{0},\dots,j_{n}}$, for some $j_{0}$, $1\leq j_{0}\leq d-1$, are zero.
The entry $z_{j_{0},\dots,j_{n}}$ also appears in $\Delta$. Consider the
$2$-minor of $M$

(4) $m=\begin{vmatrix}z_{i_{0},\dots,i_{n}}&z_{j_{0},\dots,j_{n}}\\\
z_{j_{0},\dots,j_{n}}&z_{k_{0},\dots,k_{n}}\\\ \end{vmatrix}$

for appropriate $z_{i_{0},\dots,i_{n}}$ and $z_{k_{0},\dots,k_{n}}$. By the
lexicographic order, $z_{i_{0},\dots,i_{n}}$ appears before
$z_{j_{0},\dots,j_{n}}$ on the first row. Therefore, it is zero by the
assumption. Hence, the vanishing of $m$ implies that
$z_{j_{0},\dots,j_{n}}=0$. The vanishing of the entries on the other rows
follows similarly by changing the order of $z_{i_{0},\dots,i_{n}}$’s in the
Lexicographic order.

Step 2. We define maps $\varphi_{i}:W_{i}\longrightarrow\mathbb{P}^{n}$ for
$i=0,\dots,n$ and show that these maps patch together to define a morphism
$\varphi:W\longrightarrow\mathbb{P}^{n}$. Let
$\varphi_{i}:W_{i}\longrightarrow\mathbb{P}^{n}$ be given by

$\varphi([z_{i_{0},\dots,i_{n}},i_{0}+\dots+i_{n}=d])={\rm the\ column\ of}\
M\ {\rm containing}\ z_{0,\dots,0,d,0,\dots,0},$

where $z_{0,\dots,0,d,0,\dots,0}$ corresponds to $x_{i}^{d}$. We claim that
$\varphi_{i}$ and $\varphi_{j}$ agree on $W_{i}\cap W_{j}$. We do this for the
columns containing $z_{d,0,\dots,0}$ and $z_{0,d,0,\dots,0}$, the columns
corresponding to $x_{0}^{d}$ and $x_{1}^{d}$, respectively. The procedure for
other relevant columns is similar. In fact, the vanishing of the $2$-minors of
the sub-matrix of $M$ consisting of the above two columns implies that these
two columns are proportional, presenting the same point in $\mathbb{P}^{n}$.
Therefore, the maps $\varphi_{i}$ match together to give a morphism
$\varphi:W\longrightarrow\mathbb{P}^{n}$.

Step 3. We show that $\varphi$ is the inverse of $\nu_{d}$. This implies that
${\rm Im}(\nu_{d})=W$ is closed and $\nu_{d}$ in an embedding of
$\mathbb{P}^{n}$ onto ${\rm Im}(\nu_{d})$. The equality
$\varphi\circ\nu_{d}=id_{\mathbb{P}^{n}}$ is straightforward. Let
$[x_{0}:\dots:x_{n}]\in\mathbb{P}^{n}$ where $x_{i}\not=0$ for some $i$. Then
$\nu_{d}([x_{0}:\dots:x_{n}])\in W_{i}$. Thus,
$\varphi(\nu_{d}([x_{0}:\dots:x_{n}]))=\varphi_{i}(\nu_{d}([x_{0}:\dots:x_{n}]))=[x_{0}x_{i}^{d-1}:\dots:x_{i}^{d}:\dots:x_{i}^{d-1}x_{n}]=[x_{0}:\dots:x_{n}]$.
To show $\nu_{d}\circ\varphi=id_{W}$, let
$Q=[z_{i_{0},\dots,i_{n}},i_{0}+\dots+i_{n}=d]\in W_{i}$. We assume that
$i=0$. The argument for other values of $i$ is similar. Thus,

$\nu_{d}\circ\varphi(Q)=\nu_{d}\circ\varphi_{0}(Q)=\nu_{d}([z_{d,0,\dots,0}:z_{d-1,1,0,\dots,0}:\cdots:z_{d-1,0,\dots,0,1}])=$
$=[z_{d,0,\dots,0}^{d}:z_{d,0,\dots,0}^{d-1}z_{d-1,1,0,\dots,0}:\cdots:z_{d-1,0,\dots,0,1}^{d}].$

Given any coordinate $z_{d,0,\dots,0}^{i_{0}}z_{d-1,1,0,\dots,0}^{i_{1}}\cdots
z_{d-1,0,\dots,0,1}^{i_{n}}$ of $\nu_{d}\circ\varphi_{0}(Q)$, using some
$2$-minors of $M$ involving its first row and first column, we show that

$z_{d,0,\dots,0}^{i_{0}}z_{d-1,1,0,\dots,0}^{i_{1}}\cdots
z_{d-1,0,\dots,0,1}^{i_{n}}=z_{d,0,\dots,0}^{d-1}z_{i_{0},\dots,i_{n}}.$

This would mean that $\nu_{d}\circ\varphi=id_{W}$. The claim is already valid
for $i_{0}=d,d-1$. Thus we assume that $i_{0}>d-1$. Let $i_{s}$ be the last
nonzero integer in $(i_{1},\dots,i_{n})$. Let $z_{d-1,0,\dots,0,1,0\dots,0}$
correspond to $x_{0}^{d-1}x_{s}$. Let $i_{s}\geq 2$. Using the minor

(5) $\begin{vmatrix}z_{d,0,\dots,0}&z_{d-1,0,\dots,0,1,0\dots,0}\\\
z_{d-1,0,\dots,0,1,0\dots,0}&z_{d-2,0,\dots,0,2,0,\dots,0}\\\ \end{vmatrix},$

it follows that

$z_{d-1,0,\dots,0,1,0\dots,0}^{i_{s}}=z_{d,0,\dots,0}z_{d-1,0,\dots,0,1,0,\dots,0}^{i_{s}-2}z_{d-2,0,\dots,0,2,0,\dots,0}.$

If $i_{s}-2\geq 1$, then using the $2$-minor

(6) $\begin{vmatrix}z_{d,0,\dots,0}&z_{d-2,0,\dots,0,2,0\dots,0}\\\
z_{d-1,0,\dots,0,1,0\dots,0}&z_{d-3,0,\dots,0,3,0,\dots,0}\\\ \end{vmatrix},$

it follows that

$z_{d-1,0,\dots,0,1,0,\dots,0}^{i_{s}-2}z_{d-2,0,\dots,0,2,0,\dots,0}=z_{d,0,\dots,0}z_{d-1,0,\dots,0,1,0,\dots,0}^{i_{s}-3}z_{d-3,0,\dots,0,3,0,\dots,0}.$

Continuing this procedure, it follows that

$z_{d-1,0,\dots,0,1,0\dots,0}^{i_{s}}=z_{d,0,\dots,0}^{i_{s}-1}z_{d-i_{s},0,\dots,0,i_{s},0,\dots,0}.$

Observe that, since for all $k\geq 1$, $d-i_{s}-k>1$, we are always using some
$2$-minors involving the first row and the first column in the $k$-th
reduction procedure. Now, let $i_{r}$ be the nonzero integer in
$(i_{1},\dots,i_{n})$ right before $i_{s}$. Using some $2$-minors of $M$
involving the first row and the first column similar to the case of $i_{s}$,
we get

$z_{d-1,0,\dots,0,1,0\dots,0}^{i_{r}}z_{d-i_{s},0,\dots,0,i_{s},0\dots,0}=z_{d,0,\dots,0}z_{d-1,0,\dots,0,1,0\dots,0}^{i_{r}-1}z_{d-i_{s}-1,0,\dots,0,1,0,\dots,0,i_{s},0,\dots,0}.$

Continuing this we reach

$z_{d-1,0,\dots,0,1,0,\dots,0}^{i_{r}}z_{d-1,0,\dots,0,1,0,\dots,0}^{i_{s}}=z_{d,0,\dots,0}^{i_{r}+i_{s}-1}z_{d-i_{r}-i_{s},0,\dots,0,i_{r},0,\dots,0,i_{s},0,\dots,0}.$

Observe that, again $d-i_{r}-i_{s}-k>1,$ for $k\geq 1$. Further procedure on
the other nonzero powers of $z_{d-1,0,\dots,0,1,0,\dots,0}$’s corresponding to
$x_{0}^{d-1}x_{t}$ with $t<r$, it follows that

$z_{d,0,\dots,0}^{i_{0}}z_{d-1,1,0,\dots,0}^{i_{1}}\cdots
z_{d-1,0,\dots,0,1}^{i_{n}}=z_{d,0,\dots,0}^{d-1}z_{i_{0},\dots,i_{n}}.$

Since for $k\geq 1$, we have $d-(i_{1}+\cdots+i_{n})-k>1$, in all reduction
procedures, we have been using certain $2$-minors involving the first row and
the first column of $M$.

######  Remark 2.2.

The proof of Theorem 2.1 by Smith et. al. [5, 5.1, Proposition] begins with
defining the inverse map of $\nu_{d}$. In fact, it is necessary to mention
that the image of $\nu_{d}$ is closed in the ambient projective space, namely,
it is a projective variety. To show this, one way is to prove that the image
of $\nu_{d}$ is given by certain equations (such equations are given in the
book only after the two examples following the proof of the Proposition with
no indication of their use in the proof). This property is not valid for maps
of affine varieties. As an example, the image of the hyperbola $xy-1=0$ under
the projection to its $x$-coordinate is not closed in the affine line. It is
in fact a fundamental property of maps of projective varieties that the image
of a projective variety under a projective map is closed. This is the so
called “Fundamental Theorem of Elimination Theory”’ [4, 5.2, Theorem 2] and
[2, Theorem 14.1 and Corollary 14.2]. The proof by Smith et. al. ends with
stating that the composition map
$W\longrightarrow\mathbb{P}^{N}\longrightarrow W$ is the identity map. Again,
to show this, one needs to use the equations proposed for the image of
$\nu_{d}$.

Acknowledgements
The author would like to thank his brother Rashid Zaare-Nahandi for bringing
attention to his joint paper [1] which motivated this note. He thanks Hassan
Haghighi for reading the first manuscript of the paper and for some useful
comments.

## References

  * [1] R. Abdolmaleki and R. Zaare-Nahandi, Toric ideals which are determinantal, arXiv:2106.03900v.
  * [2] D. Eisenbud, Commutative Algebra wit a View Toward Algebraic Geometry, Graduate Texts in Mathematics 150, Springer-Verlag, 1994.
  * [3] R. Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics 52, Springer-Verlag, 1977.
  * [4] I. R. Shafarevich, Basic Algebraic Geometry 1, Varieties in Projective Space, Second Edition, Springer-Verlag, 1994.
  * [5] K.E. Smith, L. Kahanpää, P. Kekäläinen, W. Traves, An Invitation to Algebraic Geometry, Universitext, Springer, 2000.
  * [6] L. Giuzzi and V. Pepe, Families of twisted tensor product codes, Des. Codes Cryptogr. 67, 375–384 (2013).


# On the (growing) gap between Dirichlet and Neumann eigenvalues

Pedro Freitas  Grupo de Física Matemática, Instituto Superior Técnico,
Universidade de Lisboa, Av. Rovisco Pais, 1049-001 Lisboa, Portugal
<EMAIL_ADDRESS>
###### Abstract.

We provide an answer to a question raised by Levine and Weinberger in their
$1986$ paper concerning the difference between Dirichlet and Neumann
eigenvalues of the Laplacian on bounded domains in $\mathbb{R}^{n}$. More
precisely, we show that for a certain class of domains there exists a sequence
$p(k)$ such that $\lambda_{k}\geq\mu_{k+p(k)}$ for sufficiently large $k$.
This sequence, which is given explicitly, grows with $k^{1-1/n}$ as $k$ goes
to infinity, which we conjecture to be optimal, and may be chosen
independently of the domain. We also prove the existence of a sequence, now
not given explicitly and only of order $k^{1-3/n}$ but valid for bounded
Lipschitz domains in $\mathbb{R}^{n}(n\geq 4)$, for which a similar inequality
holds for all $k$. From these results and the analysis of some particular
examples we formulate a conjecture for general Euclidean domains.

###### 2020 Mathematics Subject Classification:

Primary: 35P15 Secondary: 35P20

##  1\. Introduction

We consider the Dirichlet and Neumann eigenvalue problems for the Laplacian on
a bounded domain $\Omega$ in $\mathbb{R}^{n}$ defined by

(1.1) $\left\\{\begin{array}[]{rl}\Delta u+\lambda
u=0,&x\in\Omega\vspace*{2mm}\\\ u=0,&x\in\partial\Omega\end{array}\right.$

and

(1.2) $\left\\{\begin{array}[]{rl}\Delta v+\mu v=0,&x\in\Omega\vspace*{2mm}\\\
\frac{\displaystyle\partial
v}{\displaystyle\partial\nu}=0,&x\in\partial\Omega,\end{array}\right.$

respectively, where $\nu$ denotes the outer unit normal on the boundary of
$\Omega$. We shall denote the corresponding spectra by $\Sigma_{D}$ and
$\Sigma_{N}$, respectively, and write the eigenvalues as

$0<\lambda_{1}\leq\lambda_{2}\leq\dots$

and

$0=\mu_{1}\leq\mu_{2}\leq\dots.$

The study of inequalities of the type $\lambda_{k}\geq\mu_{k+m}$ for all $k$
and some fixed $m$ dates at least as far back as the work of Payne in $1955$,
who showed that $\lambda_{k}\geq\mu_{k+2}$ for planar convex domains with a
sufficiently smooth boundary [P55]. It took about thirty years for this result
to be generalised to higher dimensions by Aviles [A86] and Levine and
Weinberger [LW86] in 1986. Among other results where the curvature of the
boundary plays a key role, it is shown in [LW86] that for smooth bounded
convex domains in $\mathbb{R}^{n}$ we have

(1.3) $\lambda_{k}\geq\mu_{k+n}.$

The other main development in this direction was made by Friedlander in
$1991$, who proved a conjecture of Payne’s (see [L88, P91]), namely, that

(1.4) $\lambda_{k}>\mu_{k+1}$

for all bounded sufficiently smooth domains $\Omega$ in $\mathbb{R}^{n}$ [F91]
– a different proof that extended this result to domains for which an
embedding condition is satisfied was later given by Filonov [F05]; see also
Remark 1.9 in [S08], observing that Filonov’s proof holds for general domains.

In this paper we shall consider two aspects related to inequalities between
Dirichlet and Neumann eigenvalues. The first of these is motivated by a
question posed at the end of [LW86], asking whether inequality (1.3) may be
replaced by a better inequality of the form

(1.5) $\lambda_{k}>\mu_{\phi(n,k)}$

for convex $n-$dimensional domains. See also the comments in the second to
last paragraph on page $44$ of [M91], referring to the behaviour for large
$k$. To address this question, we consider the two-term Weyl asymptotics for
eigenvalues of problems (1.1) and (1.2), namely,

(1.6) $\lambda_{k}=c_{0}k^{2/n}+c_{1}k^{1/n}+{\rm o}\left(k^{1/n}\right)$

and

$\mu_{k}=c_{0}k^{2/n}-c_{1}k^{1/n}+{\rm o}\left(k^{1/n}\right),$

as $k\to\infty$. Here

$\begin{array}[]{lcl}c_{0}=\frac{\displaystyle
4\pi^{2}}{\displaystyle\left(\omega_{n}\left|\Omega\right|\right)^{2/n}}&\mbox{
and }&c_{1}=\frac{\displaystyle
2\pi^{2}\omega_{n-1}\left|\partial\Omega\right|}{\displaystyle
n\left(\omega_{n}\left|\Omega\right|\right)^{1+1/n}}\end{array}$

where $\omega_{n}$ denotes the volume of the unit ball in $\mathbb{R}^{n}$
and, with a slight abuse of notation, $|\Omega|$ and $|\partial\Omega|$ denote
the $n-$ and $(n-1)-$volume of $\Omega$ and $\partial\Omega$, respectively.
The above asymptotics hold under a non-periodicity condition on the set of the
billiard orbits defined on $\Omega$, namely, that the set of such orbits which
are periodic has measure zero – see [SV97] for precise definitions and
statements. This yields that for domains for which these two-term asymptotics
are valid, such as convex domains with an analytic boundary or convex
polyhedra, the difference between Dirichlet and Neumann eigenvalues satisfies

$\lambda_{k}-\mu_{k}=2c_{1}k^{1/n}+{\rm o}\left(k^{1/n}\right),$

growing to infinity with $k$ and thus suggesting that it might be possible to
determine an increasing sequence of natural numbers $p=p(k)$ for which
$\lambda_{k}\geq\mu_{k+p(k)}$. As far as we are aware, all existing results
for differences between Dirichlet and Neumann eigenvalues pertain to a fixed
gap between the corresponding Dirichlet and Neumann indexes. One of the aims
of this paper is thus to provide a first answer to Levine and Weinberger’s
question and show that better inequalities of the form (1.5) are indeed
possible in dimensions two and higher. More precisely, we shall prove that
there exists an index function $\phi(n,k)=k+p(k)$ with the sequence $p(k)$ of
order $k^{1-1/n}$ and independent of the domain, such that
$\lambda_{k}\geq\mu_{k+p(k)}$ for all sufficiently large values of $k$ – see
Theorem 2.1 for the precise formulation of this result.

For general Lipschitz domains and in dimensions higher than three, we show
that there exists a sequence $p(k)$ such that $\lambda_{k}\geq\mu_{k+p(k)}$
for all $k$, as a direct consequence of a result of Safarov and Filonov’s for
the difference between the counting functions for Dirichlet and Neumann
eigenvalues. However, now $p$ depends on an unspecified constant and its
asymptotic behaviour is slightly worse than that of the sequence above – see
Theorem 2.2 and Remark 2.4 for a discussion of the two results.

The second aspect we consider concerning this type of inequality turns up as a
consequence of the proof of Theorem 2.1. Together with the above two-term
asymptotic expansions, it turns out that another key ingredient appearing in
the proof of Theorem 2.1 is the Euclidean geometric isoperimetric inequality,
namely,

(1.7) $\left|\partial\Omega\right|\geq
n\left|\Omega\right|^{1-1/n}\omega_{n}^{1/n}.$

This points in the direction that convexity might not be a crucial condition,
or that at least it might be possible to hope for a result of this type for
all bounded domains. A role of the isoperimetric inequality is to determine
the form of the asymptotic behaviour of $p$ as $k$ goes to infinity. This
connection was also noted in [CMS19] with respect to the number of Neumann
eigenvalues which are smaller than or equal to the first Dirichlet eigenvalue.
We note further that the proof of the results for rectangles given in Section
3.1, now valid for all $k$, also makes the corresponding relation between the
perimeter and area of a rectangle appear explicitly. In fact one of the
results we obtain is of the form

$\lambda_{k}\geq\mu_{k+\left\lfloor P\sqrt{\frac{\displaystyle
k}{\displaystyle\pi A}}\right\rfloor}\;\;\mbox{ for all }k\in\mathbb{N},$

where $\left\lfloor\cdot\right\rfloor$ denotes the floor function, with $P$
and $A$ the perimeter and area of the rectangle, respectively. This points in
the direction that the larger the isoperimetric constant associated with the
rectangle, the larger the gap between the indexes of the corresponding
Dirichlet and Neumann eigenvalues appearing in the inequalities. It is
possible to do variations on these inequalities, either emphasising the
dependence on the side-lengths of the rectangle or on obtaining inequalities
which are independent of these – see Section 3.1 for other results.

On the other hand, it is known that inequalities of this type between
Dirichlet and Neumann eigenvalues in some non-Euclidean settings such as
spheres, may not hold and may, in fact, be reversed – see the results and
discussions in [AL97], [L88] and [M91]; in particular, the second and third
papers both refer back to a result that may be found in [C84], namely, that
for geodesic disks on $\mathbb{S}^{n}$ whose radius is strictly between
$\pi/2$ and $\pi$, we have $\lambda_{1}<\mu_{2}$. Note that for these disks
the non-periodicity condition is not satisfied, again suggesting this to be a
relevant condition. We also have that the isoperimetric inequality satisfied
by domains on $\mathbb{S}^{n}$ is, of course, not the same as that in
$n-$Euclidean space but, as we shall see in Section 3.3, this by itself will
not be an obstacle for inequalities analogous to those in Euclidean space to
hold.

Based on the combination of results obtained we believe that the two key
ingredients mentioned above and which appear in the proof of Theorem 2.1 are
decisive factors for a result of this type to hold. As such, we formulate the
following conjecture for general Euclidean bounded domains – note that the
non-periodicity condition is conjectured to hold for general Euclidean
domains.

######  Conjecture 1.

Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$. Then its Dirichlet and
Neumann eigenvalues satisfy the inequalities

$\lambda_{k}(\Omega)\geq\mu_{k+p(k)}(\Omega),\mbox{ for all }k\in\mathbb{N},$

where $p(k)=\left\lfloor\frac{\displaystyle n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}k^{1-1/n}\right\rfloor$. Furthermore, the power $1-1/n$ is
optimal.

In essence, this is stating that Theorem 2.1 holds for all positive integer
$k$, and that the power in $k$ given there cannot be improved.

It is also possible to formulate other conjectures with a sequence $p$
depending on the isoperimetric constant of the given domain. As an example,
consider the following conjecture which, for simplicity, is only stated in the
planar case. This is now essentially the result for rectangles in Theorem 3.1,
already mentioned above.

######  Conjecture 2.

Let $\Omega$ be a bounded planar domain with perimeter $P$ and area $A$. Then
its Dirichlet and Neumann eigenvalues satisfy the inequalities
$\lambda_{k}\geq\mu_{k+\left\lfloor P\sqrt{\frac{\displaystyle
k}{\displaystyle\pi A}}\right\rfloor}\;\;\mbox{ for all }k\in\mathbb{N}.$

######  Remark 1.1.

As has been pointed out in [BLP09], the claim made in [LW86] that for the
annular sector given by

$D=\left\\{(r,\theta):1<r<2\wedge 0<\theta<3\pi/2\right\\}$

we have $\mu_{3}>\lambda_{1}$ is incorrect – note that if this were not the
case, it would immediately prove both conjectures wrong. As is well known,
both the Dirichlet and Neumann eigenvalues of $D$ may be obtained by
separation of variables and then solving the resulting equations involving the
Bessel functions $J_{2k/3}$, $J_{2k/3\pm 1}$, $Y_{2k/3}$ and $Y_{2k/3\pm 1}$,
$k\in\mathbb{N}_{0}$. The first Dirichlet eigenvalue is given by
$\lambda_{1}(D)\approx 9.96001$, while the corresponding Neumann eigenvalues
are (approximately) given by

$\left\\{0,\,0.204718,\,0.811126,\,1.79721,\,3.13054,\,4.77455,\,6.69575,\,8.86914,\,10.2181,\,10.4649\right\\}.$

We thus see that the first Neumann eigenvalue to be larger than
$\lambda_{1}(D)$ is $\mu_{9}(D)$. We further note that this behaviour is not
very different from what one has for a rectangle with side lengths $1$ and
$9\pi/4$, corresponding to a rectangle with the same perimeter and area as the
annular sector. In this case $\lambda_{1}=\pi^{2}+16/81\approx 10.0671$, and
the first ten Neumann eigenvalues are (approximately) given by

$\left\\{0.,0.197531,0.790123,1.77778,3.16049,4.93827,7.11111,9.67901,9.8696,10.0671\right\\},$

showing that $\mu_{10}$ is now the first Neumann eigenvalue to equal
$\lambda_{1}$.

##  2\. General results for Euclidean domains

In this section we provide what might be called proof-of-concept results, the
first of which requiring the non-periodicity condition to hold, and valid only
for sufficiently large enough values of the indexes.

######  Theorem 2.1.

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded domain satisfying the non-
periodicity condition and which is not a ball. Then there exists
$k^{*}=k^{*}(\Omega)$ such that

$\lambda_{k}(\Omega)\geq\mu_{k+p}(\Omega),\mbox{ for all }k\geq k^{*}$

where $p=\left\lfloor\frac{\displaystyle n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}k^{1-1/n}\right\rfloor$.

###### Proof.

From

$\lambda_{k}=c_{0}k^{2/n}+c_{1}k^{1/n}+r_{D}(k)$

and

$\mu_{k}=c_{0}k^{2/n}-c_{1}k^{1/n}+r_{N}(k)$

we obtain

$\begin{array}[]{lll}\mu_{k+p}&=&c_{0}(k+p)^{2/n}-c_{1}(k+p)^{1/n}+r_{N}(k)\vspace*{2mm}\\\
&=&c_{0}k^{2/n}\left(1+\frac{\displaystyle p}{\displaystyle
k}\right)^{2/n}-c_{1}k^{1/n}\left(1+\frac{\displaystyle p}{\displaystyle
k}\right)^{1/n}+r_{N}(k)\vspace*{2mm}\\\
&=&c_{0}k^{2/n}+c_{1}k^{1/n}+r_{D}(k)\vspace*{2mm}\\\ &&\hskip
14.22636pt+c_{0}k^{2/n}\left[\left(1+\frac{\displaystyle p}{\displaystyle
k}\right)^{2/n}-1\right]-c_{1}k^{1/n}\left[1+\left(1+\frac{\displaystyle
p}{\displaystyle k}\right)^{1/n}\right]+r_{N}(k)-r_{D}(k)\vspace*{2mm}\\\
&=&\lambda_{k}+c_{0}k^{2/n}\left[\left(1+\frac{\displaystyle p}{\displaystyle
k}\right)^{2/n}-1\right]-c_{1}k^{1/n}\left[1+\left(1+\frac{\displaystyle
p}{\displaystyle k}\right)^{1/n}\right]+r_{N}(k)-r_{D}(k).\end{array}$

Hence

$\lambda_{k}-\mu_{k+p}=r_{D}(k)-r_{N}(k)+\underbrace{c_{1}k^{1/n}\left[1+\left(1+\frac{\displaystyle
p}{\displaystyle
k}\right)^{1/n}\right]-c_{0}k^{2/n}\left[\left(1+\frac{\displaystyle
p}{\displaystyle k}\right)^{2/n}-1\right]}_{\text{g(n,k,p)}}.$

The remainder of the proof is divided into two parts. We shall first derive a
condition for the term $g(n,k,p)$ on the right to be positive, and then show
that this is satisfied by the expression for $p$ given above. We then show
that this gives a term of order $k^{1/n}$, and is thus larger than the
difference $r_{D}(k)-r_{K}(k)={\rm o}(k^{1/n})$ for sufficiently large $k$.

Letting $x=(k+p)^{1/n}$ we may write $g$ as

$\displaystyle g(n,k,p)$
$\displaystyle=c_{1}\left(k^{1/n}+x\right)-c_{0}\left(x^{2}-k^{2/n}\right)\vspace*{2mm}$
(2.1)
$\displaystyle=-c_{0}x^{2}+c_{1}x+\left(c_{0}k^{2/n}+c_{1}k^{1/n}\right).$

This will be positive if

$\begin{array}[]{lll}\left(0\leq\right)x&\leq&\frac{\displaystyle
c_{1}+\sqrt{c_{1}^{2}+4k^{1/n}\left(c_{0}k^{1/n}+c_{1}\right)c_{0}}}{\displaystyle
2c_{0}}\vspace*{2mm}\\\ &=&\frac{\displaystyle c_{1}}{\displaystyle
c_{0}}+k^{1/n}.\end{array}$

We thus have that $p$ must satisfy

$\begin{array}[]{lll}p&\leq&\left(\frac{\displaystyle c_{1}}{\displaystyle
c_{0}}+k^{1/n}\right)^{n}-k\vspace*{2mm}\\\
&=&\left[\frac{\displaystyle\omega_{n-1}\left|\partial\Omega\right|}{\displaystyle
2n\left(\omega_{n}\left|\Omega\right|\right)^{1-1/n}}+k^{1/n}\right]^{n}-k\end{array}$

From the Euclidean geometric isoperimetric inequality (1.7) we have that the
right-hand side above satisfies

$\begin{array}[]{lll}\left[\frac{\displaystyle\omega_{n-1}\left|\partial\Omega\right|}{\displaystyle
2n\left(\omega_{n}\left|\Omega\right|\right)^{1-1/n}}+k^{1/n}\right]^{n}-k&\geq&\left(\frac{\displaystyle\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-1/(2n)}}+k^{1/n}\right)^{n}-k\vspace*{2mm}\\\
&=&k\left[\left(1+\frac{\displaystyle\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}\times\frac{\displaystyle 1}{\displaystyle
k^{1/n}}\right)^{n}-1\right]\vspace*{2mm}\\\ &>&k\left(1+\frac{\displaystyle
n\omega_{n-1}}{\displaystyle 2\omega_{n}^{1-2/n}}\times\frac{\displaystyle
1}{\displaystyle k^{1/n}}-1\right)\vspace*{2mm}\\\ &=&\frac{\displaystyle
n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}\times{k^{1-1/n}},\end{array}$

where the strict inequality comes from applying Bernoulli’s inequality with
$n$ greater than or equal to two. From this it follows that if we take $p$ to
satisfy

(2.2) $p=\left\lfloor\frac{\displaystyle n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}k^{1-1/n}\right\rfloor$

then $g$ is a strictly positive function. It remains to prove that the
resulting term when $p$ takes on this value is of order $k^{1/n}$. We first
note that from (2.1) it follows that $g$ will be strictly decreasing in $p$
for sufficiently large $k$. Thus, showing that this will be of order $k^{1/n}$
for a value of $p$ larger than that given by (2.2) will imply the desired
result. From

$p=\left\lfloor\frac{\displaystyle n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}k^{1-1/n}\right\rfloor\leq\frac{\displaystyle
n\omega_{n-1}}{\displaystyle 2\omega_{n}^{1-2/n}}k^{1-1/n}\leq
n\frac{\displaystyle c_{1}}{\displaystyle c_{0}}k^{1-1/n}$

we see that it is enough to prove the asymptotic behaviour for
$g\left(n,k,n\frac{\displaystyle c_{1}}{\displaystyle c_{0}}k^{1-1/n}\right)$.
Writing

$g\left(n,k,n\frac{\displaystyle c_{1}}{\displaystyle
c_{0}}k^{1-1/n}\right)=c_{1}k^{1/n}\left[1+\left(1+\frac{\displaystyle\alpha}{\displaystyle
k^{1/n}}\right)^{1/n}\right]-c_{0}k^{2/n}\left[\left(1+\frac{\displaystyle\alpha}{\displaystyle
k^{1/n}}\right)^{2/n}-1\right]$

with $\alpha=\frac{\displaystyle n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-1/n}}$, we then have

$\begin{array}[]{lll}g\left(n,k,n\frac{\displaystyle c_{1}}{\displaystyle
c_{0}}k^{1-1/n}\right)&\approx&k^{1/n}\left[c_{1}+c_{1}\left(1+\frac{\displaystyle\alpha}{\displaystyle
nk^{1/n}}+\dots\right)-c_{0}k^{1/n}\left(1+\frac{\displaystyle
2\alpha}{\displaystyle nk^{1/n}}+\dots-1\right)\right]\vspace*{2mm}\\\
&=&k^{1/n}\left(2c_{1}+\frac{\displaystyle\alpha}{\displaystyle
nk^{1/n}}-2\frac{\displaystyle\alpha c_{0}}{\displaystyle
n}+\dots\right)\vspace*{2mm}\\\ &=&2\left(c_{1}-\frac{\displaystyle\alpha
c_{0}}{\displaystyle n}\right)k^{1/n}+{\rm O}(1),\end{array}$

as $k$ goes to infinity. Since $\alpha<\frac{\displaystyle
nc_{1}}{\displaystyle c_{0}}$, provided $\Omega$ is not a ball, the
coefficient affecting the leading term $k^{1/n}$ is strictly positive, proving
the result. ∎

######  Remark 2.1.

We believe the exclusion of the ball from the result to be a purely technical
matter.

######  Remark 2.2.

We have

$\frac{\displaystyle n\omega_{n-1}}{\displaystyle
2\omega_{n}^{1-2/n}}=\frac{\sqrt{\pi}4^{\frac{1}{n}-1}n\left(n\Gamma\left(\frac{n}{2}\right)\right)^{\frac{n-2}{n}}}{\Gamma\left(\frac{n+1}{2}\right)}\approx
e\sqrt{\frac{\displaystyle\pi}{\displaystyle 2}n}+{\rm o}(\sqrt{n})\mbox{ as
}n\to\infty.$

This means that for large $n$ we cannot expect this sequence $p$ to be optimal
when $k$ is one, at least for convex domains for which we know that (1.3)
holds.

######  Remark 2.3.

Since the asymptotic behaviour of eigenvalues of the Laplace operator with
Robin boundary conditions follows the same behaviour as the two-term
asymptotics for the Neumann problem, the above result also holds for Robin
boundary conditions of the form $\partial u/\partial\nu+\beta u=0$ with
positive $\beta$. However, in this case and since for any given integer $m$,
by making $\beta$ large enough, we can make the first $m$ eigenvalues of the
Robin spectrum as close to the corresponding first $m$ Dirichlet eigenvalues
as we want, we cannot expect any such set of inequalities to be valid for all
$k$ without imposing any further restrictions. When $\beta$ is allowed to be
negative, then it was shown in [GM09] that Friedlander’s inequalities (1.4)
continue to hold, and that this also extends to more general (nonlocal) Robin
boundary conditions.

Using the result of Safarov and Filonov’s for the difference
$N_{N}(\lambda)-N_{D}(\lambda)$ mentioned in the Introduction [SF10], we may
prove a result for all $k$ and general Lipschitz domains. The price to pay is
that this does not provide an explicit constant and the asymptotic growth in
$k$ is weaker.

######  Theorem 2.2.

Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ ($n\geq 4$) with a
Lipschitz boundary. Then there exists a positive constant $C_{\Omega}$ such
that

$\lambda_{k}\geq\mu_{k+\left\lfloor C_{\Omega}k^{1-3/n}\right\rfloor}$

for all positive integer $k$.

######  Remark 2.4.

This result has the advantage that it does hold for all $k$ and general
Lipschitz domains. On the other hand, and apart from requiring $n$ greater
than or equal to four to provide relevant information, the constant
$C_{\Omega}$ is not explicit and the exponent $1-3/n$ is smaller than the
corresponding $1-1/n$ exponent in Theorem 2.1. The latter exponent corresponds
to the conjecture mentioned in Remark 4.3 in [SF10].

###### Proof.

Define the Dirichlet and Neumann counting functions by

(2.3)
$\begin{array}[]{lll}N_{D}(\lambda)=\\#\left\\{\lambda_{k}\in\Sigma_{D}:\lambda_{k}<\lambda\right\\}&\mbox{
and
}&N_{N}(\lambda)=\\#\left\\{\mu_{k}\in\Sigma_{N}:\lambda_{k}<\lambda\right\\}\end{array}.$

We now start from a consequence of Theorem 4.1 in [SF10], namely equation
(4.3) in that paper that states that for Lipschitz domains in $\mathbb{R}^{n}$
there exists a constant $C=C(\Omega)$ such that these functions satisfy

$N_{N}(\lambda)-N_{D}(\lambda)\geq C(\Omega)\lambda^{(n-3)/2}$

for all positive values of $\lambda$. Take
$\lambda\in(\lambda_{k},\lambda_{k+1})$ for some $k$. Then $N_{D}(\lambda)=k$
and we have from [LY83] that $\lambda$ satisfies

$\lambda>\lambda_{k}\geq\frac{\displaystyle n}{\displaystyle
n+2}c_{0}k^{2/n},$

where $c_{0}$ is the same constant as in the first term in the Weyl
asymptotics (1.6). Hence

$N_{N}(\lambda)\geq k+C(\Omega)\left(\frac{\displaystyle n}{\displaystyle
n+2}c_{0}\right)^{(n-3)/2}k^{(n-3)/n}=k+C_{\Omega}k^{1-3/n},$

for some constant $C_{\Omega}$. We thus conclude that

$\lambda\geq\mu_{N_{N}(\lambda)}\geq\mu_{k+C_{\Omega}k^{1-3/n}}$

and since we may take $\lambda$ arbitrarily close to $\lambda_{k}$ we obtain
the desired result. ∎

##  3\. Two–dimensional examples

The purpose of this section is to explore further examples illustrating the
type of results that may be expected to hold. For simplicity we have
restricted ourselves to the planar case.

###  3.1. Rectangles

We begin by giving two different results for rectangles to illustrate what may
(and may not) be expected for general domains. Similar results may be obtained
for higher dimensions, essentially in the same way but with the calculations
becoming more involved.

######  Theorem 3.1.

For any rectangle and all positive integer $k$ we have
$\lambda_{k}\geq\mu_{k+\left\lfloor P\sqrt{\frac{\displaystyle
k}{\displaystyle\pi A}}\right\rfloor}.$

###### Proof.

The proof is similar to that of Theorem 2.4. Let $R$ be a rectangle with side
lengths $a$ and $b$. The Dirichlet and Neumann eigenvalues of $R$ are given by

$\lambda_{k}=\pi^{2}\left(\frac{\displaystyle q^{2}}{\displaystyle
a^{2}}+\frac{\displaystyle r^{2}}{\displaystyle
b^{2}}\right),q,r\in\mathbb{N}$

and

$\mu_{k}=\pi^{2}\left(\frac{\displaystyle q^{2}}{\displaystyle
a^{2}}+\frac{\displaystyle r^{2}}{\displaystyle
b^{2}}\right),q,r\in\mathbb{N}_{0}.$

As usual we associate this with an integer lattice counting problem on the
plane $qr$. With $N_{D}$ and $N_{N}$ the counting functions defined by (2.3)
we have that the difference between these two functions is now given precisely
by the number of points on the positive $q$ and $r$ axes plus one
(corresponding to the zero Neumann eigenvalue). More precisely,

$N_{N}(\lambda)-N_{D}(\lambda)=1+\left\lfloor\frac{\displaystyle
a\sqrt{\lambda}}{\displaystyle\pi}\right\rfloor+\left\lfloor\frac{\displaystyle
b\sqrt{\lambda}}{\displaystyle\pi}\right\rfloor.$

We now proceed exactly as before, except that since rectangles satisfy Pólya’s
conjecture we have the stronger inequality $\lambda_{k}\geq 4\pi k/(ab)$
yielding

(3.1) $\displaystyle N_{N}(\lambda)$ $\displaystyle\geq k+1+\left\lfloor
2\sqrt{\frac{\displaystyle ak}{\displaystyle b\pi}}\right\rfloor+\left\lfloor
2\sqrt{\frac{\displaystyle bk}{\displaystyle a\pi}}\right\rfloor\vspace*{2mm}$
$\displaystyle=k+\left\lfloor 2\left(\frac{\displaystyle
a+b}{\displaystyle\sqrt{ab}}\right)\sqrt{\frac{\displaystyle
k}{\displaystyle\pi}}\right\rfloor\vspace*{2mm}$ $\displaystyle=k+\left\lfloor
P\sqrt{\frac{\displaystyle k}{\displaystyle\pi A}}\right\rfloor.$

The remaining part of the proof now follows as before. ∎

Following in a similar path, but then considering some exceptional cases
separately, it is possible to derive a result independent of the dimensions of
the rectangle.

######  Theorem 3.2.

For any rectangle and all positive integer $k$ the corresponding eigenvalues
satisfy

$\lambda_{k}\geq\mu_{k+\left\lfloor 2\sqrt{k}\right\rfloor+1}.$

###### Proof.

For simplicity we shall fix the area of the rectangle to be one, and take side
lengths to equal $b=1/a\leq 1\leq a$. As in the proof of Theorem 3.1 we have

$\begin{array}[]{lll}N_{N}(\lambda)-N_{D}(\lambda)&=&1+\left\lfloor\frac{\displaystyle
a\sqrt{\lambda}}{\displaystyle\pi}\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{\lambda}}{\displaystyle
a\pi}\right\rfloor\vspace*{2mm}\\\ &\geq&1+\left\lfloor
2a\sqrt{\frac{\displaystyle
k}{\displaystyle\pi}}\right\rfloor+\left\lfloor\frac{\displaystyle
2}{\displaystyle a}\sqrt{\frac{\displaystyle
k}{\displaystyle\pi}}\right\rfloor.\end{array}$

We want to prove that the functions $g_{k}:[1,+\infty)\to\mathbb{Z}$ defined
by

$g_{k}(a)=\left\lfloor 2a\sqrt{\frac{\displaystyle
k}{\displaystyle\pi}}\right\rfloor+\left\lfloor\frac{\displaystyle
2}{\displaystyle a}\sqrt{\frac{\displaystyle
k}{\displaystyle\pi}}\right\rfloor-\left\lfloor 2\sqrt{k}\right\rfloor$

are non-negative for all $k\in\mathbb{N}$. We shall break the proof into
several cases.

First we observe that if $a\geq\sqrt{\pi}$ we have $\left\lfloor
2a\sqrt{\frac{\displaystyle k}{\displaystyle\pi}}\right\rfloor\geq\left\lfloor
2\sqrt{k}\right\rfloor$ and $g_{k}(a)\geq 0$.

Using $\lfloor x\rfloor+\lfloor y\rfloor+1\geq\lfloor x+y\rfloor$ we obtain

$\begin{array}[]{lll}g_{k}(a)&\geq&\left\lfloor 2\left(a+\frac{\displaystyle
1}{\displaystyle a}\right)\sqrt{\frac{\displaystyle
k}{\displaystyle\pi}}\right\rfloor-\left\lfloor
2\sqrt{k}\right\rfloor-1\vspace*{2mm}\\\ &\geq&\left\lfloor\frac{\displaystyle
4}{\displaystyle\sqrt{\pi}}\sqrt{k}\right\rfloor-\left\lfloor
2\sqrt{k}\right\rfloor-1\vspace*{2mm}\\\ &=&\left\lfloor
2\sqrt{k}+\left(\frac{\displaystyle
4}{\displaystyle\sqrt{\pi}}-2\right)\sqrt{k}\right\rfloor-\left\lfloor
2\sqrt{k}\right\rfloor-1\end{array}$

and so the results holds if $\left(\frac{\displaystyle
4}{\displaystyle\sqrt{\pi}}-2\right)\sqrt{k}\geq 1$, which is equivalent to
$k\geq\frac{\displaystyle\pi}{\displaystyle\left(4-2\sqrt{\pi}\right)^{2}}\approx
15.2.$

It remains thus to prove the result for $k=1,\dots,15$ and $1\leq
a<\sqrt{\pi}$. If we analyse the original function $g_{k}$ for
$a\in[1,\sqrt{\pi})$ and $k=1,2,\dots,15$ we see that non-negativity fails for
$k=1,3,4,7,9$ – see Figure 1; a rigorous analysis may be carried out for each
case, but since this is similar to what is presented below for the five
exceptional values of $k$, we skip those calculations here.

Figure 1. Graphs of $g_{k}(a)$ for $a\in[1,\sqrt{\pi})$ and $k=1,\dots,15$.

In order to prove the result in these cases, we will use a stronger inequality
than Pólya’s, namely, we shall estimate each of these eigenvalues by their
optimal values under an area restriction. This problem was considered in
[AF13], and the optimal values $a_{k}^{*}$ and $\lambda_{k}^{*}$ for the
values of $k$ mentioned above are given in Table 1.

$k$ | 1 | 3 | 4 | 7 | 9
---|---|---|---|---|---
$\left(a_{k}^{*}\right)^{4}$ | $1$ | $1$ | $\frac{\displaystyle 8}{\displaystyle 3}$ | $\frac{\displaystyle 7}{\displaystyle 3}$ | $\frac{\displaystyle 3}{\displaystyle 2}$
$\lambda_{k}^{*}$ | $2\pi^{2}$ | $5\pi^{2}$ | $\frac{\displaystyle 35\pi^{2}}{\displaystyle 2\sqrt{6}}$ | $\frac{\displaystyle 55\pi^{2}}{\displaystyle\sqrt{21}}$ | $\frac{\displaystyle 35\pi^{2}}{\displaystyle\sqrt{6}}$
Table 1. Optimal values $a_{k}^{*}$ and $\lambda_{k}^{*}$ for $k=1,3,4,7,9$
(taken from [AF13]).

Using each of these values as a lower bound for the corresponding eigenvalue,
it is possible to obtain stronger bounds for the difference
$N_{N}(\lambda)-N_{D}(\lambda)=N_{N}(\lambda)-k$.

  1. $k=1$:

We have $\lambda_{1}\geq 2\pi^{2}$ which, when replaced in the original
expression for $N_{N}$ with $N_{D}=1$, yields $N_{N}(\lambda)\geq
1+1+\left\lfloor\sqrt{2}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{2}}{\displaystyle
a}\right\rfloor$. Possible transition points occur when $a$ equals either
$m\sqrt{2}$ or $\sqrt{2}/m$ for positive integer $m$, these being the only
points that need to be checked, and only when yielding values between $1$ and
$\sqrt{\pi}$. Since

$\left\lfloor\sqrt{2}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{2}}{\displaystyle
a}\right\rfloor=\left\\{\begin{array}[]{ll}1+1=2,&1\leq
a<\sqrt{2}\vspace*{2mm}\\\ 0+2=2,&\sqrt{2}\leq a<3/\sqrt{2}\vspace*{2mm}\\\
\geq 3,&3/\sqrt{2}\leq a\end{array}\right.$

we obtain that $N_{N}(\lambda)\geq 1+1+2=4$. Note that $3/\sqrt{2}>\sqrt{\pi}$
and so we do not actually need the last line in the above calculations.

  2. $k=3$:

Now $\lambda_{3}\geq 5\pi^{2}$ and $N_{N}(\lambda)\geq
3+1+\left\lfloor\sqrt{5}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{5}}{\displaystyle
a}\right\rfloor$. Since $3+1+\lfloor 2\sqrt{3}\rfloor=7$, we need to show that
$\left\lfloor\sqrt{5}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{5}}{\displaystyle
a}\right\rfloor\geq 3$, which follows from

$\left\lfloor\sqrt{5}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{5}}{\displaystyle
a}\right\rfloor=\left\\{\begin{array}[]{ll}2+2=4,&1\leq
a<\sqrt{5}/2\vspace*{2mm}\\\ 1+2=3,&\sqrt{5}/2\leq
a<3/\sqrt{5}\vspace*{2mm}\\\ 1+3=4,&3/\sqrt{5}\leq
a<4/\sqrt{5}\vspace*{2mm}\\\ 1+4=5,&4/\sqrt{5}\leq a<\sqrt{5}\vspace*{2mm}\\\
\geq 5,&a\geq\sqrt{5}\end{array}.\right.$

Again, since $4/\sqrt{5}\approx 1.789>1.772\approx\sqrt{\pi}$ we only need the
first three lines. From now on, we will only present the calculations for
$a<\sqrt{\pi}$.

  3. $k=4$:

$\lambda_{4}\geq\frac{\displaystyle 35\pi^{2}}{\displaystyle 2\sqrt{6}}$ and
we need to show that
$\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle\sqrt{2}\;6^{1/4}}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle\sqrt{2}\;6^{1/4}a}\right\rfloor\geq
4$.

$\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle\sqrt{2}\;6^{1/4}}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle\sqrt{2}\;6^{1/4}a}\right\rfloor=\left\\{\begin{array}[]{ll}4,&1\leq
a<\frac{3\sqrt{2}\;6^{1/4}}{\sqrt{35}}\vspace*{2mm}\\\
5,&\frac{3\sqrt{2}\;6^{1/4}}{\sqrt{35}}\leq
a<\frac{\sqrt{35}}{2\sqrt{2}\;6^{1/4}}\vspace*{2mm}\\\
4,&\frac{\sqrt{35}}{2\sqrt{2}\;6^{1/4}}\leq
a<\frac{4\sqrt{2}\;6^{1/4}}{\sqrt{35}}\vspace*{2mm}\\\
5,&\frac{4\sqrt{2}\;6^{1/4}}{\sqrt{35}}\leq a<\sqrt{\pi}\end{array}.\right.$

  4. $k=7$:

$\lambda_{7}\geq\frac{\displaystyle 55\pi^{2}}{\displaystyle\sqrt{21}}$ and we
need to show that $\left\lfloor\frac{\displaystyle\sqrt{55}}{\displaystyle
21^{1/4}}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{55}}{\displaystyle
21^{1/4}a}\right\rfloor\geq 5$.

$\left\lfloor\frac{\displaystyle\sqrt{55}}{\displaystyle
21^{1/4}}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{55}}{\displaystyle
21^{1/4}a}\right\rfloor=\left\\{\begin{array}[]{ll}6,&1\leq a<\frac{5\times
21^{1/4}}{\sqrt{55}}\vspace*{2mm}\\\ 7,&\frac{5\times 21^{1/4}}{\sqrt{55}}\leq
a<\sqrt{\pi}\end{array}.\right.$

  5. $k=9$:

$\lambda_{9}\geq\frac{\displaystyle 35\pi^{2}}{\displaystyle\sqrt{6}}$ and we
need to show that $\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle
6^{1/4}}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle
6^{1/4}a}\right\rfloor\geq 6$.

$\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle
6^{1/4}}a\right\rfloor+\left\lfloor\frac{\displaystyle\sqrt{35}}{\displaystyle
6^{1/4}a}\right\rfloor=\left\\{\begin{array}[]{ll}6,&1\leq a<\frac{4\times
6^{1/4}}{\sqrt{35}}\vspace*{2mm}\\\ 7,&\frac{4\times 6^{1/4}}{\sqrt{35}}\leq
a<\frac{\sqrt{35}}{3\times 6^{1/4}}\vspace*{2mm}\\\
6,&\frac{\sqrt{35}}{3\times 6^{1/4}}\leq a<\frac{5\times
6^{1/4}}{\sqrt{35}}\vspace*{2mm}\\\ 7,&\frac{5\times 6^{1/4}}{\sqrt{35}}\leq
a<\frac{6\times 6^{1/4}}{\sqrt{35}}\vspace*{2mm}\\\ 8,&\frac{6\times
6^{1/4}}{\sqrt{35}}\leq a<\sqrt{\pi}\vspace*{2mm}\\\ \end{array}.\right.$

This concludes the proof. ∎

Note that while for long thin rectangles Theorem 3.1 gives a stronger result,
Theorem 3.2 provides a better value for some values of $k$. In particular,
while in general using the former result we can only ensure that for
$\lambda_{1}\geq\mu_{3}$, the later result yields $\lambda_{1}\geq\mu_{4}$
directly. While it would still be possible to recover this last inequality for
the square, for instance, from (3.1) in the proof of Theorem 3.1, to prove it
for general rectangles would require an analysis similar to what was carried
out in the proof of Theorem 3.2. In Figure 2 we show the first $1,000$ values
for the differences $\lambda_{k}-\mu_{k+p(k)}$ for the square with the
sequences $p(k)$ given by (3.1), and Theorems 3.1 and 3.2.

Figure 2. Graphs of the differences of the eigenvalues of the square for (from
left to right) $p(k)=1+\left\lfloor 2\sqrt{\frac{\displaystyle
ak}{\displaystyle b\pi}}\right\rfloor+\left\lfloor 2\sqrt{\frac{\displaystyle
bk}{\displaystyle a\pi}}\right\rfloor$, $p(k)=\left\lfloor P\sqrt{k/(\pi
A)}\right\rfloor$ and $p(k)=\left\lfloor 2\sqrt{k}\right\rfloor+1$.

###  3.2. General planar domains

The application of Bernoulli’s inequality in the proof of Theorem 2.1 limits
the powers of $n$ appearing in that result. By considering the two-dimensional
case we can actually easily do that calculation explicitly with all the terms
and recover the same inequality as in Theorem 3.2, except now only for
sufficiently large $k$. In fact, this result cannot hold for the first
Dirichlet eigenvalue of the disk as $\lambda_{1}(D)<\mu_{4}(D)$, and hence it
cannot hold for the first eigenvalue of domains sufficiently close to the disk
either (in an appropriate sense).

######  Theorem 3.3.

Let $\Omega$ be a bounded planar domain satisfying the non-periodicity
condition and which is not a disk. Then there exists $k^{*}=k^{*}(\Omega)$
such that

$\lambda_{k}\geq\mu_{k+\left\lfloor\sqrt{2k}\right\rfloor+1},\mbox{ for all
}k\geq k^{*}.$

###### Proof.

The proof proceeds in the same way as that of Theorem 2.1 to obtain

$\lambda_{k}-\mu_{k+p}=r_{D}(k)-r_{N}(k+p)+\frac{\displaystyle
2\sqrt{\pi}}{\displaystyle
A}\underbrace{\left[\left(\sqrt{k+p}+\sqrt{k}\right)\frac{\displaystyle
P}{\displaystyle\sqrt{A}}-2p\sqrt{\pi}\right]}_{\text{g(k,p)}},$

where $P$ and $A$ denote the perimeter and area of $\Omega$, respectively. The
function $g(k,p)$ will be non-negative if

$(0<)\;p\leq\frac{\displaystyle P^{2}}{\displaystyle 4\pi
A}+\frac{\displaystyle P}{\displaystyle\sqrt{\pi A}}\sqrt{k}.$

From the two-dimensional isoperimetric inequality we have that the right-hand
side above satisfies

$\frac{\displaystyle P^{2}}{\displaystyle 4\pi A}+\frac{\displaystyle
P}{\displaystyle\sqrt{\pi A}}\sqrt{k}\geq 1+2\sqrt{k}$

and we will thus take $p=\left\lfloor 2\sqrt{k}\right\rfloor+1\leq
2\sqrt{k}+1$, concluding that $g\left(k,\left\lfloor
2\sqrt{k}\right\rfloor+1\right)\geq 0$. Finally, note that we may write

$g(k,2\sqrt{k}+1)=\left(\frac{\displaystyle
P}{\displaystyle\sqrt{A}}-2\sqrt{\pi}\right)\left(2\sqrt{k}+1\right)$

showing that, except possibly for the disk, $g\left(k,\left\lfloor
2\sqrt{k}\right\rfloor+1\right)$ is of order $k^{1/2}$. ∎

Note that the actual form of the isoperimetric inequality, at least in the
proof above, is important only to determine the form of the expression for
$p$. Once this has been done, the dominant term will always be of order
$\sqrt{k}$, except for domains giving equality in the isoperimetric
inequality.

###  3.3. The sphere $\mathbb{S}^{2}$

Proceeding in the same way as above for planar domains, it is possible to
derive a similar result for domains on the sphere that also satisfy the non-
periodicity condition. The resulting expressions are now more involved,
mirroring the version of the isoperimetric inequality on $\mathbb{S}^{2}$,
namely, [Le51]

$P^{2}\geq 4\pi A-A^{2}.$

######  Theorem 3.4.

Let $\Omega\subsetneq\mathbb{S}^{2}$ de a domain satisfying the non-
periodicity condition and which is not a geodesic disk. Then there exists
$k^{*}=k^{*}(\Omega)$ such that

$\lambda_{k}\geq\mu_{k+p(k)},\mbox{ for all }k\geq k^{*},$

where $p$ is given by

$p(k)=\left\lfloor 1-\frac{\displaystyle A}{\displaystyle
4\pi}+2\sqrt{1-\frac{\displaystyle A}{\displaystyle
4\pi}}\sqrt{k}\right\rfloor.$

###### Proof.

The proof proceeds in the same way as that for Theorem 3.3. ∎

As mentioned in the Introduction, geodesic disks on $\mathbb{S}^{2}$ with
radius larger than or equal to $\pi/2$ do not satisfy the non-periodicity
condition. On the other hand, those with radius smaller than $\pi/2$ do, and
again we would expect these to satisfy the above inequality.

## Acknowledgements

It is a pleasure to acknowledge several exchanges with Mark Ashbaugh
concerning this problem and, in particular, for having mentioned [L88] and
other relevant articles. We are also indebted to Pedro Antunes for a numerical
check on the eigenvalues of the annular sector example. This work was
partially supported by the Fundação para a Ciência e a Tecnologia (Portugal)
through project UIDB/00208/2020.

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