# On the structure of the ring of integer-valued entire functions

Bechir Amri
<EMAIL_ADDRESS>
University of Carthage, Department of Mathematics, Faculty of Sciences of
Bizerte, Tunisia.

###### Abstract

The present paper studies structure of the ring of integer-valued entire
functions. We characterize certain classes of prime and maximal ideals and
investigate some of their properties.

##  1 Introduction

The study of algebraic and topological structures of rings of functions has
been a subject of interest for many years. In particular, rings of continuous
functions on a topological spaces and of holomorphic functions have fairly
rich structures, especially in characterization of prime and maximal ideals. A
pioneer papers in this subject are [2, 4, 5].

This paper study the ring $\mathfrak{A}$ of integer-valued entire functions,
that is the set of entire function in the complex plane and takes values in
$\mathbb{Z}$ at all points in $\mathbb{Z}$. It has been inspired and motivated
by the study of the ring of entire functions, presented in the papers [1, 3].
Although they have similar properties, the structure of ring $\mathfrak{A}$ is
quite different: its not a Bezout ring and has more classes of prime and
maximal ideals. However, the ring $\mathfrak{U}$ is not similar in structure
to that of integer-valued polynomials, for example an integer-valued entire
function is not necessarily continuous for p-adic metric, an interesting
argument to classify maximal ideals. The ideals of $\mathfrak{A}$ are herein
classified as in [4], free and fixed, according to the intersection of the
sets of irreducible devisors of its elements is EmptySet or not. We first
establish necessary and sufficiently condition for finitely generated ideal to
be principal. A such condition will be frequently invoked when studying the
problem of identifying prime and maximal ideals.

##  2  Elementary properties of $\mathfrak{A}$

Let us start with the following lemma,

######  Lemma 2.1.

For a given sequence $(a_{k})_{k}$ of complex numbers, there exists an entire
function $f$ such that for all $k\in\mathbb{Z}$, $f(k)=a_{k}$.

This can seen by virtue of Weierstrass theorem or directly by taking the
following function

$f(z)=\sum_{k\in\mathbb{Z}}a_{k}\left(\frac{\sin(\pi(z-\pi))}{\pi(z-k)}\right)^{|k|+[|a_{k}|]}=\sum_{k\in\mathbb{Z}}u_{k}(z)$

where $[x]$ denotes the integer part of a real number $x$. The series is
normally convergent in every compact set. In fact, for $M>0$ and $|z|\leq M$
we have $|\sin\pi z|\leq e^{\pi M}$ and $\pi|z-k|\geq 2e^{\pi M}$, for $|k|$
big enough. Then

$|u_{k}(z)|\leq\frac{|a_{k}|}{2^{|k|+[|a_{k}|]}}\leq\frac{1}{2^{|k|-1}}.$

######  Corollary 2.2.

For a given sequence $(a_{k})_{k}$ of nonzero complex numbers, there exists a
non-vanishing entire function such that $f(k)=a_{k}$, for all
$k\in\mathbb{Z}$.

###### Proof.

As the exponential function is surjective we let $a_{k}=e^{z_{k}}$,
$k\in\mathbb{Z}$. From the above lemma there exists an entire function $f$
such that $f(k)=z_{k}$, for all $k\in\mathbb{Z}$. Hence a desired function is
given by $g(z)=e^{f(z)}$ ∎

In what follows we denote by $\mathfrak{A}$ the ring of integer-valued entire
functions, that is the set of entire functions $f$ which satisfy
$f(\mathbb{Z})\subset\mathbb{Z}$. From the isolated zeros property the ring
$\mathfrak{A}$ is an integral domain. However, $f\in\mathfrak{A}$ is
invertible if and only if it is non-vanishing function with
$f(k)\in\\{-1,1\\}$, $k\in\mathbb{Z}$.

######  Proposition 2.3.

The irreducible elements of $\mathfrak{A}$ are functions $f$ that satisfy one
of the following conditions:
(i) $f$ does not vanish and there exists only one $n\in\mathbb{Z}$ such that
$f(n)$ is a prime integer and $f(k)\in\\{1,-1\\}$ for all $k\in\mathbb{Z}$,
$k\neq n$.
(ii) $f$ vanishes only at one complex number
$a\in\mathbb{C}\setminus\mathbb{Z}$ which is a simple zero and
$f(\mathbb{Z})\subset\\{1,-1\\}$

###### Proof.

It is not hard to verify that a function of $\mathfrak{A}$ that satisfying i)
or ii) is irreducible. Conversely, let $f$ be an irreducible element of
$\mathfrak{A}$. Then for all $n\in\mathbb{Z}$, $f(n)$ is irreducible or
invertible in $\mathbb{Z}$. In fact, suppose that for some $n\in\mathbb{Z}$,
$f(n)=rs$ with $r,s\in\mathbb{Z}$, $r\neq 0$ and $s\notin\\{-1,1\\}$. Let
$g\in\mathfrak{A}$ be a none-vanishing function such that $g(n)=r$ and
$g(k)=1$, for all $k\in\mathbb{Z}$, $k\neq n$. Then $g$ divides $f$ and $f/g$
is not invertible since $f/g(n)=s$, this contradicts the fact that $f$ is
irreducible. Furthermore, we exclude the case where there are two distinct
integers $n_{1},n_{2}\in\mathbb{Z}$ such that $f(n_{1})$ and $f(n_{2})$ are
irreducible integers, since a none-vanishing function $g\in\mathfrak{A}$ such
that $g(n_{1})=f(n_{1})$ and $g(k)=1$, for all $k\in\mathbb{Z}$, $k\neq n_{1}$
divides $f$. Now if $f(\mathbb{Z})\subset\\{-1,1\\}$ then $f$ vanishes on
$\mathbb{C}\setminus\mathbb{Z}$ since otherwise $f$ is invertible. Suppose
that $f(a)=f(b)=0$ for two distinct complex numbers
$a,b\in\mathbb{C}\setminus\mathbb{Z}$. Taking a none vanishing
$h\in\mathfrak{A}$ such that $h(k)=1/(k-a)$ for all $k\in\mathbb{Z}$, then the
function $g(z)=(z-a)h(z)$ divides $f$. Also $g$ divides $f$ in the case
$f(a)=f^{\prime}(a)=0$. Now, by virtue of the above discussions the function
$f$ must satisfy condition i) or ii). ∎

The following corollary is an immediate consequence.

######  Corollary 2.4.

The prime elements of $\mathfrak{A}$ are exactly the irreducible elements.

######  Proposition 2.5.

$\mathfrak{A}$ is not a factorial ring.

###### Proof.

The function $f(z)=z$ has an infinite number of prime divisors, since for
$n\in\mathbb{Z}$ and $p$ a prime divisor of $n$ then the none vanishing
function $g\in\mathfrak{A}$ such that $g(n)=p$ and $g(k)=1$ for $k\neq n$ is a
prime divisor of $f$. ∎

######  Proposition 2.6.

$\mathfrak{A}$ is not a Noetherian ring.

###### Proof.

Consider the the sequence $(H_{n})_{n}$ of $\mathfrak{A}$,

$H_{n}(z)=\prod_{k=n}^{+\infty}\left(1-\frac{\sin^{2}(\pi z)}{k^{2}}\right).$

The ascending chains of ideals $(H_{n})_{n}$ is not stable. ∎

For $f$ is an entire function, we let $Z(f)$ the algebraic set of its zeros (
each zero occurring a number of times equal to its multiplicity )

######  Lemma 2.7.

If $f,g\in\mathfrak{A}$ such that $Z(f)\cap Z(g)=\emptyset$ and
$gcd(f(n),g(n))=1$ for all $n\in\mathbb{Z}$, then there are
$u,v\in\mathfrak{A}$ such that $uf+vg=1$.

###### Proof.

We know from Theorem 9 of [1] that there are entire functions $u_{1}$ and
$v_{1}$ such that $u_{1}f+v_{1}g=1$. On the other hand for each
$n\in\mathbb{Z}$ there exist $a_{n},b_{n}\in\mathbb{Z}$ such that
$a_{n}f(n)+b_{n}g(n)=1$. We let $k$ be an entire function such that

$k(n)=\frac{a_{n}-u_{1}(n)}{g(n)}\;\;\text{if $g(n)\neq 0$ and
}k(n)=\frac{v_{1}(n)-b_{n}}{f(n)}\;\;\text{if $g(n)=0$. }$

Noting here that when both $g(n)$ and $f(n)$ are non-zero we have

$\frac{a_{n}-u_{1}(n)}{g(n)}=\frac{v_{1}(n)-b_{n}}{f(n)}$

Thus putting $u=u_{1}+kg$ $v=v_{1}-kf$. Clearly $uf+vg=1$ and
$u,v\in\mathfrak{A}$, since $u(n)=a_{n}$ and $v(n)=b_{n}$ for all
$n\in\mathbb{Z}$. ∎

######  Lemma 2.8.

If $f,g\in\mathfrak{A}$ such that $Z(f)\cap Z(g)\cap\mathbb{Z}=\emptyset$ then
the ideal $(f,g)=f\mathfrak{A}+g\mathfrak{A}$ is principal.

###### Proof.

Let $d$ be an entire function such that $Z(d)=Z(f)\cap Z(g)$. We assume that
$d(n)=gcd(f(n),g(n))$, for all $n\in\mathbb{Z}$ ( by virtue of corollary 2.2
one can multiply $d$ by a none vanishing entire function $h$ such that
$h(n)=d(n)^{-1}gcd(f(n),g(n)),\;n\in\mathbb{Z})$. Thus $f/d$ and $g/d$ are
elements of $\mathfrak{A}$ that satisfy the hypothesises of Lemma 2.7 and from
which it follows that $f\mathfrak{A}+g\mathfrak{A}=d\mathfrak{A}$. ∎

######  Theorem 2.9.

Let $f_{1},f_{2},...f_{r}\in\mathfrak{A}$. The finitely generated ideal
$I=(f_{1},f_{2},..,f_{r})$ is principal if and only if for each
$n\in\mathbb{Z}$ there exists $j_{n}\in\\{1,2,...,r\\}$ such that
$(f_{j}/f_{j_{n}})(n)\in\mathbb{Q}$ for all $j=1,2,...,r.$

###### Proof.

Let us suppose first that $I$ is principal ideal and put $I=d\mathfrak{A}$ for
some $d\in\mathfrak{A}$. Clearly $Z(d)=\cap_{j=1,2,...r}Z(f_{j})$ and for
$n\in\mathbb{Z}$ there exists $j_{n}$ such that $o(d,n)=o(f_{j_{n}},n)$ (
multiplicity of $n$ ). As $(f_{j_{n}}/d)(n)\in\mathbb{Z}\setminus\\{0\\}$ and
$(f_{j}/d)(n)\in\mathbb{Z}$ for all $j=1,2,...,r$ then
$(fj/f_{j_{n}})(n)\in\mathbb{Q}$. Conversely, suppose first that
$\cap_{j=1,2,...,r}Z(f_{j})=\emptyset$. Then for all $n\in\mathbb{Z}$ the
integers $f_{1}(n),\;f_{2}(n),...,f_{r}(n)$ are not all zero, and hence one
can choose $a_{1,n},a_{2,n},...,a_{r,n}\in\mathbb{Z}$ such that

$a_{1,n}f_{1}(n)+a_{2,n}f_{2}(n)+...+a_{r,n}f_{r}(n)\neq 0.$

Taking $h_{j}\in\mathfrak{A}$, $j=1,2,...,r$ such that $h_{j}(n)=a_{j,n}$, for
all $n\in\mathbb{Z}$ and let $h=h_{1}f_{1}+h_{2}f_{2}+...+,h_{r}f_{r}$. Now as
$I=(h,f_{1},f_{2},...,f_{r})$ then we can apply successively Lemma 2.7 to
conclude that I is principal.

Suppose now $\cap_{j=1,2,...,r}Z(f_{j})\neq\emptyset$ and let $g$ be an entire
function such that $Z(g)=\cap_{j=1,2,...,r}Z(f_{j})$. By hypothesis we set

$\frac{f_{j}}{f_{j_{n}}}(n)=\frac{p_{j,n}}{q_{j,n}},\qquad\text{with}\qquad
gcd(p_{j,n},q_{j,n})=1,\quad n\in\mathbb{Z},\;j=1,2,...,r.$

Let $q_{n}$ be the least common multiple of all $q_{j,n}$, clearly this is a
divisor of $f_{j_{n}}(n)$. Choosing a none vanishing entire function $\alpha$
such that

$\alpha(n)=\frac{f_{j_{n}}}{q_{n}g}(n),\qquad n\in\mathbb{Z}$

and put $D=\alpha g$. Noting here that $(f_{j_{n}}/g)(n)\neq 0$, since
otherwise we would have $(f_{j}/g)(n)=0$ for all $j=1,2,...,r$ which cannot
occur because $Z(g)=\cap_{j=1,2,...,r}Z(f_{j})$. Thus we have
$D\in\mathfrak{A}$ and divides all $f_{j}$, $j=1,2,...,r$. As Now
$\cap_{j=1,2,...,r}Z(f_{j}/D)=\emptyset$ then the ideal
$(f_{1}/D,f_{2}/D,...,f_{r}/D)$ is principal and it is the same for the ideal
$I$. ∎

Taking the following example, $f(z)=z$ and $g(z)=\sin(\pi z)$. Since
$(g/f)(0)=\pi$, then the ideal $(f,g)$ is not principal

######  Corollary 2.10.

An invertible ideal of $\mathfrak{A}$ is principal.

###### Proof.

Suppose that $I$ is an invertible ideal of $\mathfrak{A}$, then it’s inverse
is the $\mathfrak{A}$-module given by
$I^{-1}=\\{h\in\mathfrak{K},hI\subset\mathfrak{A}\\}$, where $\mathfrak{K}$ is
the fraction field of the ring $\mathfrak{A}$. It follows that $I$ and
$I^{-1}$ are finitely generated, let $I=(f_{1},f_{2},...,f_{s})$ and
$I^{-1}=(h_{1},h_{2},...,h_{s})$ with $\sum_{j=1}^{s}h_{j}f_{j}=1$. Thus for
$n\in\mathbb{Z}$ there exists $j_{n}$ such that $(h_{j_{n}}f_{j_{n}})(n)\neq
0$, hence

$\frac{f_{j}}{f_{j_{n}}}(n)=\frac{(h_{j_{n}}f_{j})(n)}{(h_{j_{n}}f_{j_{n}})(n)}\in\mathbb{Q},\qquad
j=1,2,...,s$

and $I$ is principal by Theorem 2.9. ∎

##  3 Maximal and prime ideals

There are two remarkable examples of maximal ideals. The First type is given
by ideals $M_{a}=\\{f\in\mathfrak{A},\;f(a)=0\\}$, for
$a\in\mathbb{C}\setminus\mathbb{Z}$ and the second type is given by ideals
$M_{n,p}=\\{f\in\mathfrak{A},\;f(n)\equiv 0(p)\\}$ for $n\in\mathbb{Z}$ and a
prime number $p$.

######  Definition 3.1.

Let $\Pi(f)$ be the set of the irreducible divisors of $f\in\mathfrak{A}$. For
$\delta\in\Pi(f)$, we define

$o(\delta,f)=\max\\{k;\;\delta^{k}dividesf\\},\qquad o(\delta,0)=\infty.$

and

$m(f)=\sup\\{o(\delta,f),\delta\in\Pi(f)\\}$

Observe that when $m(f)=1$ then all zeros of $f$ are simple and $f(n)$ is a
free integer square for all $n\in\mathbb{Z}$. In addition, if
$\Pi(f)\cap\Pi(g)=\emptyset$ then $Z(f)\cap Z(g)=\emptyset$ and
$gcd(f(n),g(n))=1$ for all $n\in\mathbb{Z}$.

######  Lemma 3.2.

Each (proper) maximal ideal of $\mathfrak{A}$ contains an element $f$ with
$m(f)=1$.

###### Proof.

Let $M$ be a maximal ideal. Let see first that $M$ contains a function not
vanishing on $\mathbb{Z}$. indeed, its obvious that $2$ or $3$ does not
belongs to $M$. If $2\notin M$ (similarly if $3\notin M$ ) and as $A/M$ is a
field then there is $f\in M$ and $g\in\mathfrak{A}$ such that $2g+f=1$ which
asserts that the function $f$ doest not vanish on $\mathbb{Z}$. Now for an
each $f$ we let $h\in\mathfrak{A}$ a none vanishing function such that

$h(n)=\prod_{p/f(n)}p,\qquad n\in\mathbb{Z}$

where the product is over prime integers divisors of $f(n)$ . If $h\in M$ then
$h$ is a desired function and the lemma is proved. Otherwise there exist
$u\in\mathfrak{A}$ and $v\in M$ such that $uh+v=1$. This implies that for
$n\in\mathbb{Z}$, $gcd(h(n),v(n))=1$ and consequently $gcd(f(n),v(n))=1$. Thus
from Lemma 2.1 there exist $\alpha,\;\beta\in\mathfrak{A}$ such that $d=\alpha
f+\beta v\in M$ with $d(n)=1$ for all $n\in\mathbb{Z}$. Now observe that
$Z(d)\neq\emptyset$, since otherwise $d$ is invertible. Then by Weierstrass
theorem one can construct entire function $\mathfrak{d}$ that has the same
zeros as $d$ with multiplicities equal to one and such that
$\mathfrak{d}(n)=1$ for all $n\in\mathbb{Z}$. If $\mathfrak{d}\notin M$ then
there exist $a\in\mathfrak{A}$ and $b\in M$ such that $a\mathfrak{d}+b=1$.
Hence $Z(d)\cap Z(b)=\emptyset$ and from Lemma 2.7 there are
$r,s\in\mathfrak{A}$ such that $rd+sb=1\in M$, this yields a contradiction
since $M\neq\mathfrak{A}$. This achieves the proof of the lemma. ∎

######  Theorem 3.3.

A prime ideal $P$ of $\mathfrak{A}$ is maximal if and only if $P$ contains an
element $f$ with $m(f)=1$.

###### Proof.

Let $M$ be a maximal ideal containing $P$ and let $f\in P$ with $m(f)=1$. We
will prove that $M=P$. In fact, let $h\in M$, there exists $d\in\mathfrak{A}$
such that $(f,h)=(d)$. Then one can write $f=gd$ with $Z(g)\cap
Z(d)=\emptyset$ and $gcd(g(n),d(n))=1$, for all $n\in\mathbb{Z}$. In view of
Lemma 2.1 the function $g$ can not belongs to $M$ and thus $d\in P$, since $P$
is a prime ideal. Therefore $h\in P$. ∎

######  Corollary 3.4.

All finite family of functions of a maximal ideal $M$ of $\mathfrak{A}$ have a
common divisor in $M$.

###### Proof.

Let $f_{1},f_{2},...,f_{r}$ be a functions in $M$ and let $f\in M$ with
$m(f)=1$ ( which does not vanish on $\mathbb{Z}$). By Theorem 2.9 there is
exists $d\in\mathfrak{A}$ such that ideal $(f,f_{1},f_{2},...,f_{r})=(d)$.
Then we must have $d\in M$ and it is a common divisor of
$f_{1},f_{2},...,f_{r}$. ∎

######  Theorem 3.5.

Let $P$ be a prime ideal of $\mathfrak{A}$ such that $\cap_{f\in
P}\Pi(f)\neq\emptyset$.
(i) If $\cap_{f\in P}Z(f)\neq\emptyset$. then there exists $a\in\mathbb{C}$
such that $P=\\{f\in\mathfrak{A},\;f(a)=0\\}$.
(ii) If $\cap_{f\in P}Z(f)=\emptyset$ then there exist $n\in\mathbb{Z}$ and a
prime integer $p$ such that $P=\\{f\in\mathfrak{A},\;f(n)\equiv 0(p)\\}$.

###### Proof.

(i) We let $D=\cap_{f\in P}Z(f)$. We suppose first that
$D\cap\mathbb{Z}=\emptyset$. Let $a\in D$ and $h$ be an none vanishing entire
function such that $h(n)=(n-a)^{-1}$ for all $n\in\mathbb{Z}$. For $f\in P$ we
let $g=((z-a)h)^{k_{0}}$ where $k_{0}=o(f,a)$. Clearly $g\in\mathfrak{A}$ and
divides $f$. Moreover, since $P$ is a prime ideal, $g$ belongs to $P$.
Therefore $d=(z-a)h$ is an element of $P$ which irreducible and $P$ is the
maximal ideal $\\{\mathfrak{f}\in\mathfrak{A},\;\mathfrak{f}(a)=0\\}$. Suppose
now $D\cap\mathbb{Z}=\emptyset$. Let $n\in D\cap\mathbb{Z}$ and $f\in P$
written as $f=(z-n)^{r}\ell$, with $\ell(n)\neq 0$. Given $\alpha$ a none
vanishing entire function such that $\alpha(k)=(k-n)^{r}$, for
$k\in\mathbb{Z}$, $k\neq n$ and $\alpha(n)=1/\ell(n)$. Since ze have
$\alpha\ell\notin P$ and $(z-n)^{r+1}\alpha\ell\in P$, then we conclude that
$P$ contains the polynomial $z-n$. Now for $b\in\mathfrak{A}$ with $b(n)=0$,
we write $b^{2}=(z-n)a$. Since we have $a(k)\in\mathbb{Q}$ for all
$k\in\mathbb{Z}$ then one can find a none vanishing entire function $\beta$
such that $\beta a\in\mathfrak{A}$. Thus we have $\beta b^{2}\in P$ and so
$b\in P$. This proves that
$P=\\{\mathfrak{f}\in\mathfrak{A},\;\mathfrak{f}(n)=0\\}$.
ii) In this case the functions in $P$ have a common irreducible divisor of
type $\delta_{n,p}$. It’s not hard to see that $\delta_{n,p}\in P$ and then
$P=\\{f\in\mathfrak{A},\;f(n)\equiv 0(p)\\}$. ∎

######  Corollary 3.6.

Let $n\in\mathbb{Z}$ and $p$ be a prime integer. The maximal ideal
$M_{n,p}=\\{f\in\mathfrak{A},\;f(n)\equiv 0(mod\;p)\\}$ contains one and only
one proper prime ideal, namely $P_{n}=\\{f\in\mathfrak{A},\;f(n)=0\\}$.
Moreover, the maximal ideals containing $P_{n}$ are all in the form of
$M_{n,p}$.

###### Proof.

Let $P$ be a proper prime ideal included in $M_{n,p}$. Suppose that there is
$f\in P$ with $f(n)\neq 0$ and write $f(n)=p^{r}q$ with $gcd(p,q)=1$. Let $h$
be an irreducible element of $\mathfrak{A}$ such that $h(n)=p$ and $h(k)=1$
for $k\in\mathbb{Z}$, $k\neq n$. Then $f=h^{r}g$ with $gcd(g(n),p)=1$ and
$g\notin M$. As $P$ is a prime ideal it follows that $h\in P$ and hence
$P=M_{n,p}$, contradiction. We conclude that
$P=\\{f\in\mathfrak{A},\;f(n)=0\\}$ by using Theorem 3.5.

Let now $N$ be a (proper) maximal ideal that contains $P_{n}$ for some
$n\in\mathbb{Z}$. Let $f\in N$ and $f\notin P_{n}$. We will show that
$f(n)\neq\pm 1$. In fact, suppose that $f(n)=\pm 1$ and choose a none
vanishing entire function $\alpha$ with $\alpha(k)=(k-n)^{-1}$, for all
$k\in\mathbb{Z}$, $k\neq n$ and $\alpha(n)=1$. Put $h=(z-n)\alpha\in P_{n}$.
We have $Z(h)\cap Z(f)=\emptyset$ and $gcd(h(k),f(k))=1$, for all
$k\in\mathbb{Z}$ . But this leads to $N=\mathfrak{A}$, thanks to Lemma 2.7, so
we must have $f(n)\neq\pm 1$. Now decompose into prime factors $f(n)=\pm
p_{1}^{r_{1}}p_{2}^{r_{2}}...p_{\ell}^{r_{\ell}}$ and let
$h_{1},h_{2}...,h_{\ell}$ be irreducible elements of $\mathfrak{A}$ such that
$h_{j}(n)=p_{j}$, $j=1,2,...,\ell$. Thus one can write
$f=h_{1}^{r_{1}}h_{2}^{r_{2}}...h_{\ell}^{r_{\ell}}g$ with $g(n)=1$. As
$g\notin N$, the $N$ contains one of $h_{j}$ and thus
$N=N_{n,p_{j}}\\{f\in\mathfrak{A},\;f(n)\equiv 0(p_{j})\\}$, again thanks to
Lemma 2.1. ∎

######  Theorem 3.7.

Let $P$ be a prime ideal of $\mathfrak{A}$ containing a function which does
not vanish on $\mathbb{Z}$ and let $f\in\mathfrak{A}$ with $m(f)=1$. Then the
ideal $M=(P,f)$ is maximal.

###### Proof.

It’s enough to show that $M$ is a prime ideal. Suppose first $f=uv$ for some
$u,v\in\mathfrak{A}$ and show that $u\in M$ or $v\in M$. In fact, let $g\in P$
that does not vanish on $\mathbb{Z}$. Taking $h\in\mathfrak{A}$ as follows:
the set of zeros of $h$ is the intersection of those of $u$ and $g$ and each
zero has the same order as zero of g and further for $n\in\mathbb{Z}$,

$h(n)=\left\\{\begin{array}[]{ll}\prod_{p/u(n)}p^{v_{p}(g(n))},&\hbox{ if
$u(n)\neq 1$;}\\\ 1,&\hbox{ if $u(n)=1$.}\end{array}\right.$

where the product is taken on the prime integers divisors of $u(n)$ and
$v_{p}(g(n))=\max\\{e;\;p^{e}\;divides\;g(n)\\}$. Now clearly $g$ is written
as $g=hr$ with $(h,v)=(r,u)=\mathfrak{A}$. Hence $u=\alpha f+\beta hu$ and
$v=\theta f+\lambda rv$, for some
$\alpha,\beta,\theta,\lambda\in\mathfrak{A}$. As $P$ is a prime ideal of
$\mathfrak{A}$ thus we have that $h\in P$ or $r\in P$ and consequently $u\in
M$ or $v\in M$.

Now suppose that we have $f_{1}f_{2}\in M$ with

$f_{1}f_{2}=af+p$ (3.1)

for some $a\in\mathfrak{A}$ and $p\in P$. Suppose further that $f_{1}\notin M$
and let $f_{1}=df_{1}^{*}$ and $f=df^{*}$ with
$(f_{1}^{*},f^{*})=\mathfrak{A}$. So, from the above $f^{*}$ must be in $M$.
Now in view of (3.1), $d(f_{1}^{*}f_{2}-af^{*})=p$, then
$f_{1}^{*}f_{2}-af^{*}\in P$ and $f_{1}^{*}f_{2}\in M$. However, if we write
1= $\mu f_{1}^{*}+\nu f^{*}$, some $\mu,\nu\in\mathfrak{A}$, then we obtain
that $f_{2}=\mu f_{2}f_{1}^{*}+\nu f_{2}f^{*}\in M$. This achieves the proof
of the theorem. ∎

######  Corollary 3.8.

Let $P$ be a prime ideal of $\mathfrak{A}$ containing a function that does not
vanish on $\mathbb{Z}$. Then $P$ is included in one and only one maximal ideal
of $\mathfrak{A}$.

###### Proof.

Suppose that we have $M_{1}$ and $M_{2}$ two proper maximal ideals of
$\mathfrak{A}$ containing $P$. Let $f_{1}\in M_{1}$ and $f_{2}\in M_{2}$ with
$m(f_{1})=m(f_{2})=1$. Then from Theorem 3.7, $M_{1}=(P,f_{1})$ and
$M_{2}=(P,f_{2})$. Now let $(d)=(f_{1},f_{2})$ and write $f_{1}=df_{1}^{*}$,
$f_{2}=df_{2}^{*}$ with $(f_{1}^{*},f_{2}^{*})=\mathfrak{A}$. Suppose that
$d\notin M_{1}$ and $d\notin M_{2}$ then we have $M_{1}=(P,f_{1}^{*})$ and
$M_{2}=(P,f_{2}^{*})$. Now for a given $g\in P$ that does not vanish on
$\mathbb{Z}$ we can split $g=g_{1}g_{2}$ with
$(g_{1},f_{1}^{*})=(g_{2},f_{2}^{*})=\mathfrak{A}$. But this yields that
$M_{1}=\mathfrak{A}$ or $M_{2}=\mathfrak{A}$, since we have $g_{1}\in P$ or
$g_{2}\in P$ which contradicts the fact that $M_{1}$ and $M_{2}$ are proper
ideals of $\mathfrak{A}$. Therefore we have $d\in M_{1}$ or $d\in M_{2}$ which
imply that $M_{1}=M_{2}$. ∎

######  Theorem 3.9.

Let $P$ be a prime ideal of $\mathfrak{A}$ containing a function that does not
vanish on $\mathbb{Z}$. The residue class ring $\mathfrak{A}/P$ is a valuation
ring whose the unique maximal ideal $M/P$ is principal.

###### Proof.

We suppose that $P$ is not maximal. Let us denote $\bar{f}$ the class modulo
$P$ of $f\in\mathfrak{A}$. Noting that $\mathfrak{A}/P$ is a valuation ring if
and only if for all $f,g\in\mathfrak{A}$ we have $\bar{f}$ divides $\bar{g}$
or $\bar{g}$ divides $\bar{f}$. Let $M$ be the maximal ideal containing $P$.
As $M$ is the only maximal ideal containing $P$ then for all $f\notin M$ the
ideal $(P,f)=\mathfrak{A}$ and there exists $h\in\mathfrak{A}$ such that
$fh-1\in P$, implies that $\bar{f}$ is a unit in the ring $\mathfrak{A}/P$.
Let now $f,g\in M\setminus P$ and $h\in P$ that does not vanish on
$\mathbb{Z}$. Then if we let $(d)=(f,g,h)$, we must have that $d\notin P$ and
$h/d\in P$, which imply that $f/d$ or $g/d$ does not belong to $M$ and thus is
a unit modulo $P$. Hence we conclude that $\bar{f}$ divides $\bar{g}$ or
$\bar{g}$ divides $\bar{f}$. Further, if we take $f\in M\setminus M^{2}$ then
it is clearly that $M/P$ is generated by $(\bar{f})$. ∎

######  Theorem 3.10.

Let $\mathfrak{E}$ be the ring of entire functions.
(i) The mapping $P\longrightarrow P\cap\mathfrak{A}$ is injective that does
not surjective from the set of prime ideals of $\mathfrak{E}$ into the set of
prime ideals of $\mathfrak{A}$
(ii) If $M$ is a maximal ideal of $\mathfrak{E}$ then $M\cap\mathfrak{A}$ is a
maximal ideal of $\mathfrak{A}$ if and only if $M$ contains a function which
does not vanish on $\mathbb{Z}$.

###### Proof.

(i) Suppose $P_{1}$ and $P_{2}$ are distinct prime ideals of $\mathfrak{E}$
and there is $u\in P_{1}$ and $u\notin P_{2}$. Let $v\in\mathfrak{E}$ be a
none vanishing entire function such that $uv\in\mathfrak{A}$. Then $uv\in
P_{1}\cap\mathfrak{A}$ and $uv\notin P_{2}\cap\mathfrak{A}$ and so
$P_{1}\cap\mathfrak{A}\neq P_{2}\cap\mathfrak{A}$. For the surjectivity we
note first that if $P_{1}\cap\mathfrak{A}\subset P_{2}\cap\mathfrak{A}$ then
$P_{1}\subset P_{2}$. In fact, let $g\in P_{1}$ and $h\in\mathfrak{E}$ a none
vanishing function such that $hg\in\mathfrak{A}$. Then we have that $hg\in
P_{2}\cap\mathfrak{A}$ and $g\in P_{2}$, since $h\notin P_{2}$. Consider now
the prime ideal of $\mathfrak{A}$, $M=\\{f\in\mathfrak{A},\;f(0)\equiv
0(2)\\}$ and suppose that $M=P\cap\mathfrak{A}$ where $P$ is a prime ideal of
$\mathfrak{E}$. However $I=\\{f\in\mathfrak{E},\;f(0)=0\\}$ is a maximal ideal
of $\mathfrak{E}$ and $I\cap\mathfrak{A}\subset P\cap\mathfrak{A}$ which
implies that $I\subset P$ and then $I=P$ which cannot hold.
(ii) Let $M$ be a (proper) maximal ideal of $\mathfrak{E}$. We will affirm
that $\mathfrak{A}/\mathfrak{A}\cap M$ is a field. Let $h\in M$ assumed not
vanish on $\mathbb{Z}$ and let $f\in\mathfrak{A}$, $f\notin M$. There exists
$g\in\mathfrak{E}$ such that $gf-1\in M$. Let $\kappa\in\mathfrak{E}$ such
that

$\kappa(n)=\frac{g(n)}{h(n)},\quad n\in\mathbb{Z}.$

Putting $L=\kappa h\in M$ and $H=g-L\in\mathfrak{A}$. Thus we have
$Hf-1=gf-1-Lf\in M$ and so $f$ is invertible in $\mathfrak{A}/\mathfrak{A}\cap
M$. To prove the converse, assume that $M\cap\mathfrak{A}$ is a maximal ideal
of $\mathfrak{A}$. Since the constant function $2$ does not belong to
$M\cap\mathfrak{A}$, then there is $d\in\mathfrak{A}$ such that $2d-1\in
M\cap\mathfrak{A}$ and this function does not vanish on $\mathbb{Z}$. ∎

######  Theorem 3.11.

The ring $\mathfrak{A}$ is integrally closed.

###### Proof.

Let us observe that the rings $\mathfrak{A}$ and $\mathfrak{E}$ have the same
fractions field, namely the field of the meromorphic functions, since for
$f,g\in\mathfrak{E}$ we can write

$\frac{f}{g}(z)=\frac{f(z)\sin(\pi z)}{g(z)\sin(\pi z)}.$

Now, $\mathfrak{A}$ is integrally closed can be easily deduced from fact that
$\mathfrak{E}$ and $\mathbb{Z}$ are integrally closed ring ( gcd-domains ). ∎

######  Theorem 3.12.

The ring $\mathfrak{A}$ is nearly analytic, namely that
i) if $M$ is a maximal ideal of $\mathfrak{A}$ then its powers $M^{n}$ for
$n=1,2,...$ are distinct,
ii) $P^{*}=\cap_{n=1}^{\infty}M^{n}$ is a prime ideal of $\mathfrak{A}$ and is
the largest non-maximal prime ideal contained in M.

###### Proof.

i) We proceed as in [3]. First we note that if $g\in M^{n}$ for some positive
integer $n$ then $g=ud^{n}$, where $u\in\mathfrak{A}$ and $d\in M$. This can
be follow by recurrence on $n$. In fact, let $g\in M^{n+1}=MM^{n}$, which can
be written as a finite sum of the form $\sum_{i}a_{i}b_{i}^{n}$ with
$a_{i},b_{i}\in M$. Choose a function $h\in M$ that does not vanish on
$\mathbb{Z}$ and let $(d)=(a_{1},a_{2},,,b_{1},b_{2},...,h)$, it follows that
$d\in M$ and hence $g$ takes the form $g=ud^{n+1}$. Now using this fact we
deduce that for $f\in M$ with $m(f)=1$ we have $f^{n}\in M^{n}\backslash
M^{n+1}$, for all positive integer $n$.
ii) Suppose we have $f_{1}f_{2}\in P^{*}$. If $f_{1}\notin M$ (or $f_{2}\notin
M)$ then there are $\alpha\in\mathfrak{A}$ and $m\in M$ such that $\alpha
f_{1}+m=1$ and one can write $f_{2}=f_{1}f_{2}+mf_{2}$, from which $f_{2}\in
P^{*}$. If both $f_{1},f_{2}\in M\setminus P^{*}$ then there exist a positive
integers $r,s$ such that $f_{1}\in M^{r}\setminus M^{r+1}$ and $f_{2}\in
M^{s}\setminus M^{s+1}$, and one write $f_{1}=u(f_{1}^{*})^{r}$ and
$f_{2}=v(f_{2}^{*})^{s}$ where $u,v\in\mathfrak{A}\setminus M$ and
$f_{1}^{*},f_{2}^{*}\in M\setminus M^{2}$. On the other hand , in view of the
corollary 3.4 if we let $d\in M\setminus M^{2}$ be a common divisor of
$f_{1}^{*}$ and $f_{2}^{*}$ then we must have $f_{1}f_{2}=wd^{r+s}$ where
$w\in\mathfrak{A}\setminus M$. Now as $f_{1}f_{2}\in P^{*}$, its written as
$f_{1}f_{2}=ab^{r+s+1}$ with $b\in M$. Letting $\delta\in M$ be a common
divisor of $d$ and $b$, we have
$f_{1}f_{2}=\alpha\delta^{r+s}=\beta\delta^{r+s+1}$ with
$\alpha\in\mathfrak{A}\setminus M$, but this does not carry out since
$\mathfrak{A}$ is an integral domain, and thus $f_{1}$ or $f_{2}$ must be in
$P^{*}$. Hence we conclude that $P^{*}$ is a prime ideal.

In order to prove the second part of the theorem, we let $P$ be a non-maximal
prime ideal contained in $M$ and $f\in M$ with $m(f)=1$. Let $g\in P$ and
$d\in M$ be a common divisor of $f$ and $g$, where we write $f=df^{*}$ and
$g=dg^{*}$. Since $m(d)=1$ then $d\notin P$ and thus $g^{*}\in P$. Hence we
conclude that $MP=P$ and this fact lead to the inclusion of $P$ in $P^{*}$. ∎

In the next we let $M$ be a proper maximal ideal of $\mathfrak{A}$. Following
[3], for $f,g\in M$ we have $f\geq g$ if there is $e\in M$ such that

$\Pi(e)\subset\Pi(f)\cap\Pi(g)\quad\text{and}\quad o(f,\delta)\geq
o(g,\delta),\qquad\text{ for all }\;\delta\in\Pi(e).$

We denote $f\gg g$ if $f\geq g^{N}$ all positive integers $N$.

######  Remark 3.13.

The function $e$ can be taken with $m(e)=1$. Indeed, let $h\in M$ with
$m(h)=1$ and $d\in M$ such that $(e,h)=(d)$. Then $\Pi(d)\subset\Pi(e)$ and
$m(d)=1$.

######  Lemma 3.14.

If $f,g\in M$ then $f\geq g$ or $g\geq f$.

###### Proof.

Let $h\in M$ with $m(h)=1$ and $d\in M$ such that $(f,g,h)=(d)$. Clearly f
$\\{\delta\in\Pi(d);\;o(f,\delta)\geq o(g,\delta)\\}=\emptyset$ then $g\geq
f$. Otherwise, let $v\in\mathfrak{U}$ be a divisor of $d$ such that
$\Pi(v)=\\{\delta\in\Pi(d);\;o(f,\delta)\geq o(g,\delta)\\}$ and
$o(v,\delta)=o(d,\delta)$ for all $\delta\in\Pi(v)$. If $v\in M$ then clearly
$f\geq g$. If $v\notin M$ then $d/v\in M$ and we must have $g\geq f$. ∎

######  Lemma 3.15.

If $f_{1},f_{2},g\in M$ such that $f_{1}f_{2}\geq g^{2N}$ for some positive
integer $N$, then we have $f_{1}\geq g^{N}$ or $f_{2}\geq g^{N}$.

###### Proof.

Let $e\in M$ with $m(e)=1$, such that $\Pi(e)\subset\Pi(f_{1}f_{2})\cap\Pi(g)$
and $o(f_{1}f_{2},\delta)\geq o(g^{2N},\delta)$, for all $\delta\in\Pi(e).$
Assume that $A=\\{\delta\in\Pi(e);\;o(f_{1},\delta)\geq
o(g^{N},\delta)\\}\neq\emptyset$ and let $d\in\mathfrak{U}$ be a divisor of
$e$, so that $\Pi(d)=A$. Therefore, when $d\in M$ we have that $f_{1}\geq
g^{N}$, otherwise we have $e/d\in M$ and $f_{2}\geq g^{N}$, noting here that
$\Pi(e/d)=\\{\delta\in\Pi(e);\;o(f_{2},\delta)>o(g^{N},\delta)\\}$. This
completes the proof and the lemma ∎

######  Theorem 3.16.

Let $\Omega$ a subset of $M$ and

$P_{\Omega}=\\{f\in M;\;f\gg g\;\text{for all}\;g\in\Omega\\}.$

Then $P_{\Omega}$ is a prime ideal.

###### Proof.

We prove first that $P_{g}$, $g\in M$ is an ideal and its prime. Let $N$ be a
positive integer. For $f_{1},f_{2}\in P_{g}$, there exist $e_{1},e_{2}\in M$
such that we have the following,

$\Pi(e_{1})\subset\Pi(f_{1})\cap\Pi(g),\;\Pi(e_{2})\subset\Pi(f_{2})\cap\Pi(g),\;\;o(f_{1},\delta_{1})\geq
o(g^{N},\delta_{1}),$ $o(f_{2},\delta_{2})\geq o(g^{N},\delta_{2})$

for all $\delta_{1}\in\Pi(e_{1})$ and $\delta_{2}\in\Pi(e_{2})$. Let $h\in M$
be a function that does not vanish on $\mathbb{Z}$ and let $d\in M$ such that
$(d)=(e_{1},e_{2},h)$. Clearly

$\Pi(d)\subset\Pi(f_{1})\cap\Pi(f_{2})\cap\Pi(g)\subset\Pi(f_{1}+f_{2})\cap\Pi(g)$

and for $\delta\in\Pi(d),$

$o(f_{1}+f_{2},\delta)\geq\min(o(f_{1},\delta),o(f_{2},\delta))\geq
o(g^{N},\delta)$

Hence we have that $f_{1}+f_{2}\in P_{g}$. Similarly one can see that for
$f_{1}\in P_{g}$ and $f_{2}\in\mathfrak{A}$ the product $f_{1}f_{2}\in P_{g}.$
We thus conclude that $P_{g}$ is an ideal.

Now, we show that $P_{g}$ is prime. Assume that $f_{1}f_{2}\in P_{g}$, for
$f_{1},f_{2}\in\mathfrak{A}$. In particular $f_{1}f_{2}\in M$ and then $f_{1}$
or $f_{2}$ must be in $M$. If first both $f_{1}$ and $f_{2}$ belong to $M$
then by Lemma 3.15 one has

$f_{1}\geq g^{N}\quad\text{or}\quad f_{2}\geq g^{N}$

for all positive integers $N$. As at least one of the sets $\\{N;\;f_{1}\geq
g^{N}\\}$ and $\\{N;\;f_{2}\geq g^{N}\\}$ has infinitely many positive
integers $N$ then we must have $f_{1}\in P_{g}$ or $f_{2}\in P_{g}$.
Otherwise, suppose that $f_{1}\notin M$. Let $N$ be a positive integer and
$e\in M$ with $m(e)=1$ (see remark 3.13) and such that
$\Pi(e)\subset\Pi(f_{1}f_{2})\cap\Pi(g)$ and

$o(f_{1}f_{2},\delta)\geq o(g^{N},\delta),\quad\text{for all
}\;\delta\in\Pi(e).$

Putting $e=e_{1}e_{2}$ with $\Pi(e_{1})\cap\Pi(f_{1})=\emptyset$ and
$\Pi(e_{2})\subset\Pi(f_{1})$. Clearly we cannot have $e_{2}\in M$, since in
case $M$ contains an element $\omega$ so that
$\Pi(e_{2})\cap\Pi(\omega)=\emptyset$. Hence $e_{1}\in M$ and as
$\Pi(e_{1})\subset\Pi(e)\subset\Pi(f_{1}f_{2})\cap\Pi(g)$ and

$o(f_{2},\delta)=o(f_{1}f_{2},\delta)\geq
o(g^{N},\delta),\quad\delta\in\Pi(e_{1})$

then $f_{2}\geq g^{N}$. Since $N$ is arbitrary then $f_{2}\gg g$.

Now we come back to the ideal $P_{\Omega}=\cap_{g\in\Omega}P_{g}$ and prove
that is prime. If we let $f_{1},f_{2}\in M\setminus P_{\Omega}$ then there are
$g_{1},g_{2}\in M$ such that $f_{1}\notin P_{g_{1}}$ and $f_{2}\notin
P_{g_{2}}$. But in view of Lemma 3.14 we have $P_{g_{1}}\subset P_{g_{2}}$ or
$P_{g_{2}}\subset P_{g_{1}}$ which implies that $f_{1}.f_{2}\notin P_{g_{1}}$
or $f_{1}.f_{2}\notin P_{g_{2}}$ and hence $f_{1}.f_{2}\notin P_{\Omega}$.
This completes the proof of Theorem 3.16.

######  Example 3.17.

Let $p$ be a prime integer. Consider the ideal of functions $f\in\mathfrak{A}$
such the set $\\{n\in\mathbb{Z},\;gcd(f(n),p)=1\\}$ is finite. Let $M$ be a
maximal ideal contains $I$. For a nonnegative integer $k$ taken $f_{k}\in M$
such that

$f_{k}(n)=\left\\{\begin{array}[]{ll}1,&\hbox{ if $|n|\leq k$;}\\\
p^{n^{k}},&\hbox{ if $|n|>k$;.}\end{array}\right.$

Then we have $f_{k+1}\gg f_{k}$ and $P_{f_{k}}\varsubsetneq P_{f_{k+1}}$. The
Ring $\mathfrak{A}$ has infinite Krull dimension.

∎

## References

  * [1] O. Helmer, Divisibility properties of integral functions, Duke Math. J. 6(1940), 345-356.
  * [2] M. Henriksen, On the ideal structure of the ring of entire functions, Pacific J. Math.,II (1952), 179-184.
  * [3] M. Henriksen, On the prime ideals of the ring of entire functions, ibid. 3 (1953), 711–720
  * [4] E. Hewitt, Rings of real-valued continuous functions. I, Trans. Amer. Math. Soc. vol. 64 (1948) pp. 54-99.
  * [5] L. Gillman and M. Jerison, Rings of continuous functions, New York, Van Nostrand, 1960\.
  * [6] R. S. Pierce, Rings of integer-valued continuous functions, Trans. Amer. Math. Soc. 100 (1961), 371-394.


#  On geometrical origin of Kodama vector

Shunichiro Kinoshita<EMAIL_ADDRESS>Department of
Physics, College of Humanities and Sciences, Nihon University, Sakurajosui,
Setagaya-ku, Tokyo 156-8550, Japan

###### Abstract

It has been known that warped product spacetimes such as spherically symmetric
ones admit the Kodama vector. This vector provides a locally conserved current
made by contraction of the Einstein tensor, even though there is no Killing
vector. In addition, a quasilocal mass, Birkhoff’s theorem and various
properties are closely related to the Kodama vector. Recently, it is shown
that the notion of the Kodama vector can be extended to three-dimensional
axisymmetric spacetimes even if the spacetimes are not warped product. This
implies that warped product may not be a necessary condition for a spacetime
to admit the Kodama vector. We show properties of the Kodama vector originate
from the conformal Killing-Yano two-form. In particular, the well-known
spacetimes that admit the Kodama vector have a closed conformal Killing-Yano
two-form. Furthermore, we show the Kodama vector provides local conserved
currents for each order of the Lovelock tensor as well as the Einstein tensor.

##  I Introduction and summary

The Kodama vector, which was at first found in four-dimensional spherically
symmetric spacetimes [1], provides a locally conserved current for the
Einstein tensor even in spacetimes without Killing vectors such as dynamical
spacetimes. Since this vector $K^{a}$ satisfies $G^{ab}\nabla_{a}K_{b}=0$ for
the Einstein tensor $G_{ab}$, a current $J^{a}\equiv G^{ab}K_{b}$ is locally
conserved, i.e., $\nabla_{a}J^{a}=0$. If $K^{a}$ is timelike, this current can
be interpreted as an appropriate energy current with assuming the Einstein
equation and its associated charge yields the so-called Misner-Sharp
quasilocal mass [2, 3]. This notion has been generalized to higher dimensions
straightforwardly. It is worth noting that spherical symmetry is not essential
for a spacetime to admit the Kodama vector but warped product with two-
dimensional base space plays an important role. Moreover, it is known that the
Kodama vector is closely related to Birkhoff’s theorem (see [4], for example).
This theorem states that all spherically symmetric solutions of the Einstein
equation in vacuum must be static. It can be rephrased in terms of the Kodama
vector as follows. The warped product spacetimes, including spherically
symmetric spacetimes, admit the Kodama vector. If the spacetime is Einstein
manifold, then the Kodama vector becomes the Killing vector.

Recently, it is shown that in three-dimensional axisymmetric spacetimes even
for non-warped product spacetimes such as rotating ones, the notion of the
Kodama vector can be extended [5, 6]. This vector can provide a local
conserved current and quasilocal mass taking into account of angular momentum,
as in the cases of warped product spacetimes. This fact suggests that warped
product does not seem to be necessary for a spacetime to admit the Kodama
vector.

In this paper we show properties of the Kodama vector geometrically originate
from conformal Killing-Yano (CKY) two-form. Various conserved currents and
charges associated with (conformal) Killing tensors and (conformal) Killing-
Yano forms have been reported in the literature [7, 8, 9, 10, 11, 14, 15, 12,
13]. What we emphasis here is that the Kodama vector is the so-called
associated vector with a CKY two-form. In particular, all the well-known
spacetimes admitting the Kodama vector have closed conformal Killing-Yano
(CCKY) two-forms, which belong to a subclass of CKY two-forms.

Furthermore, we show that the associated vector of the CKY two-form can yield
conserved currents not only for the Einstein tensor but also for each order of
the Lovelock tensor [16, 17]. This means that the Kodama vector provides a
locally conserved energy current in Lovelock gravity, which has been partially
proved and conjectured for symmetric spacetimes such as spherically symmetric
one in [18, 19]. (In warped product spacetimes of a two-dimensional base and
an Einstein space, the Kodama vector and the Misner-Sharp quasilocal mass were
studied in Ref. [20].)

This paper is organized as follows. In Sec. II we present definitions and some
basic properties of CKY two-forms. We reveal the relation between the Kodama
vector and the associated vector of a CKY two-form. In Sec. III we exhibit
some explicit examples of the known Kodama vectors in terms of CKY two-forms.

##  II conformal Killing-Yano two-form and Kodama vector

###  II.1 CKY two-form and conserved current for Einstein tensor

We consider that a $D$-dimensional spacetime with the metric $g_{ab}$ admits a
conformal Killing-Yano (CKY) two-form, $h_{ab}$. The conformal Killing-Yano
two-form [21, 22] (also, see [23] and references therein) satisfies

$\nabla_{c}h_{ab}=g_{ca}K_{b}-g_{cb}K_{a}+L_{abc},\quad
K_{a}\equiv-\frac{1}{D-1}\nabla^{b}h_{ab},\quad
L_{abc}\equiv\nabla_{[a}h_{bc]},$ (1)

where the vector field $K_{a}$ is the so-called associated vector of $h_{ab}$.
If $L_{abc}=0$, $h_{ab}$ reduces to a closed conformal Killing-Yano (CCKY)
two-form. In this case, a Hodge dual of $h_{ab}$ yields a Killing-Yano
$(D-2)$-form $f_{a_{1}\cdots a_{D-2}}$, which satisfies
$\nabla_{a}f_{b_{1}\cdots b_{D-2}}=\nabla_{[a}f_{b_{1}\cdots b_{D-2}]}$.

Covariant derivative of the associated vector $K_{a}$ is

$\displaystyle\nabla_{a}K_{b}$
$\displaystyle=-\frac{1}{D-1}\nabla_{a}\nabla^{c}h_{bc}=-\frac{1}{D-1}\left(\nabla^{c}\nabla_{a}h_{bc}+R_{a}{}^{c}{}_{b}{}^{d}h_{dc}+R_{a}{}^{c}{}_{c}{}^{d}h_{bd}\right)$
(2)
$\displaystyle=\frac{1}{D-1}\nabla_{a}K_{b}+\frac{1}{D-1}R_{acbd}h^{cd}+\frac{1}{D-1}R_{a}{}^{c}h_{bc}-\frac{1}{D-1}\nabla^{c}L_{abc}.$

This can be rewritten as

$\displaystyle\nabla_{a}K_{b}$
$\displaystyle=\frac{1}{2(D-2)}R_{abcd}h^{cd}+\frac{1}{D-2}R_{a}{}^{c}h_{bc}-\frac{1}{D-2}\nabla^{c}L_{abc},$
(3)

where we have used the first Bianchi identity $R_{abcd}+R_{acdb}+R_{adbc}=0$.

It turns out that a symmetric part of Eq. (3) is given by

$\nabla_{(a}K_{b)}=\frac{1}{D-2}R_{(a}{}^{c}h_{b)c}.$ (4)

The trace yields

$\nabla_{a}K^{a}=\frac{1}{D-2}R^{ac}h_{ac}=0,$ (5)

implying that the vector field $K_{a}$ is divergence free. For the Einstein
tensor $G_{ab}$, we obtain

$G^{ab}\nabla_{a}K_{b}=\frac{1}{D-2}R^{ab}R_{a}{}^{c}h_{bc}=0.$ (6)

Thus, the associated vector $K^{a}$ for a conformal Killing-Yano two-form
$h_{ab}$ provides the same properties as Kodama vectors and $G_{ab}K^{b}$
becomes a locally conserved current.111These properties have been pointed out
in Refs. [10, 14, 13], where $G_{ab}K^{b}$ is referred to as “Einstein
current.” We note that if the spacetime is an Einstein space, i.e.,
$R_{ab}=\lambda g_{ab}$, then Eq. (4) leads to the Killing equation
$\nabla_{a}K_{b}+\nabla_{b}K_{a}=0$ [22]. This implies a version of Birkhoff’s
theorem that the Kodama vector becomes a Killing vector in vacuum with a
cosmological constant. In four dimensions, the relation between CKY two-form
and Birkhoff’s theorem was discussed [24].

We can rewrite $G_{ab}K^{b}$ as

$\displaystyle G_{ab}K^{b}$
$\displaystyle=\frac{1}{2(D-3)}\nabla^{b}\left(R_{abcd}h^{cd}+4R_{[a}{}^{c}h_{b]c}+Rh_{ab}\right)$
(7)
$\displaystyle=\nabla^{b}\left[\frac{1}{2(D-3)}W_{abcd}h^{cd}+\frac{2}{D-2}R_{[a}{}^{c}h_{b]c}+\frac{D}{2(D-1)(D-2)}Rh_{ab}\right]$
$\displaystyle=\frac{1}{2(D-2)}C_{abc}h^{bc}+\nabla^{b}\left[\frac{2}{D-2}R_{[a}{}^{c}h_{b]c}+\frac{D}{2(D-1)(D-2)}Rh_{ab}\right],$

where $W_{abcd}$ denotes the Weyl curvature tensor and the Cotton tensor
$C_{abc}$ is defined as

$C_{abc}\equiv
2\nabla_{[c}R_{b]a}-\frac{1}{D-1}g_{a[b}\nabla_{c]}R=\frac{D-2}{D-3}\nabla^{d}W_{adbc}.$
(8)

Since $G_{ab}K^{b}$ is given by a divergence of two-form “potential” in Eq.
(7), we can explicitly see this current is locally conserved. It is worth
noting that the expressions in the first and second lines of (7) are valid in
$D>3$ dimensions, because both $\mathcal{P}_{abcd}\equiv
R_{abcd}-2R_{a[c}g_{d]b}+2R_{b[c}g_{d]a}+Rg_{a[c}g_{d]b}$222This rank-$4$
tensor is divergence free and its indices have the same symmetries of the
Riemann tensor, which can be also written as
$\delta_{abb_{1}b_{2}}^{cda_{1}a_{2}}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}=4\mathcal{P}_{ab}{}^{cd}$
by using the generalized Kronecker delta symbol. This type of tensor has been
used in Ref. [32], for example.  and $W_{abcd}$ are identically zero in three
dimensions. However, that in the last line is valid even in $D=3$ dimensions.
We note that $C_{abc}h^{bc}$ is a so-called Cotton current in Ref. [11].333A
conserved current for the Cotton tensor was discussed in Ref. [25], also.

In a specific case, if $h_{ab}$ is a Killing-Yano tensor, then the “potential”
two-form field $R_{abcd}h^{cd}+4R_{[a}{}^{c}h_{b]c}+Rh_{ab}$ itself can be
conserved. This is referred to as the Yano current [8]. It is equivalent to
the fact that the associated vector for the Killing-Yano tensor will vanish in
Eq. (7).

###  II.2 Generalization to Lovelock tensor

By using the fact that the Kodama vector is provided by a CKY $2$-form, we can
prove the Kodama vector yields conserved currents for each order of the
Lovelock tensor as well as the Einstein tensor.

The $n$th order Lovelock tensor ($0<n<D/2$) in $D$ dimensions [17, 16] (also,
see [26] and references therein) is given by

$G^{(n)}{}^{a}{}_{b}\equiv-\frac{1}{2^{n+1}}\delta^{aa_{1}\cdots
a_{2n}}_{bb_{1}\cdots b_{2n}}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}},$ (9)

which reduces to the Einstein tensor for $n=1$. Note that symbol
$\delta^{a_{1}\cdots a_{k}}_{b_{1}\cdots b_{k}}$ is the generalized Kronecker
delta symbol, defined by

$\delta^{a_{1}\cdots a_{k}}_{b_{1}\cdots b_{k}}=k!g^{a_{1}}_{[b_{1}}\cdots
g^{a_{k}}_{b_{k}]}=-\frac{1}{(D-k)!}\epsilon^{a_{1}\cdots a_{k}c_{k+1}\cdots
c_{D}}\epsilon_{b_{1}\cdots b_{k}c_{k+1}\cdots c_{D}},$ (10)

where $\epsilon_{a_{1}\cdots a_{D}}$ denotes the totally-antisymmetric
$D$-dimensional volume form.

We introduce the following two-form field consisting of a CKY $2$-form
$h_{ab}$ and $n$ powers of the Riemann tensors,

$F^{(n)}{}_{ab}\equiv\delta^{cda_{1}\cdots a_{2n}}_{abb_{1}\cdots
b_{2n}}h_{cd}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}.$ (11)

It turns out that

$\displaystyle\nabla^{b}F^{(n)}{}_{ab}$ $\displaystyle=\delta^{cda_{1}\cdots
a_{2n}}_{abb_{1}\cdots
b_{2n}}\nabla^{b}h_{cd}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$ (12)
$\displaystyle\quad+\delta^{cda_{1}\cdots a_{2n}}_{abb_{1}\cdots
b_{2n}}h_{cd}\sum_{k=1}^{n}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots\nabla^{b}R_{a_{2k-1}a_{2k}}{}^{b_{2k-1}b_{2k}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$ $\displaystyle=\delta^{cda_{1}\cdots
a_{2n}}_{abb_{1}\cdots
b_{2n}}(g^{b}{}_{c}K_{d}-g^{b}{}_{d}K_{c}+L^{b}{}_{cd})R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$
$\displaystyle=-2(D-2n-1)\delta^{da_{1}\cdots a_{2n}}_{ab_{1}\cdots
b_{2n}}K_{d}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$
$\displaystyle=2^{n+2}(D-2n-1)G^{(n)}{}_{ad}K^{d}.$

The second equality follows from the second Bianchi identity,
$\nabla_{[a}R_{bc]de}=0$, and the third equality does from the first Bianchi
identity. Since $F_{ab}$ is anti-symmetric, $G^{(n)}{}_{ad}K^{d}$ is
divergence free. Hence, we have also a local conserved current for the $n$th
order Lovelock tensor as

$J^{(n)}{}^{a}\equiv
G^{(n)}{}^{a}{}_{b}K^{b}=\frac{1}{2^{n+2}(D-2n-1)}\nabla_{b}F^{(n)ab}.$ (13)

Note that, for $n=1$, the previous result for the Einstein tensor is obviously
reproduced. On arbitrary spacelike hypersurfaces $\Sigma$ with a common
boundary $\partial\Sigma$, by using Stokes’ theorem, we have a conserved
charge written in the boundary integral. An $n$th order quasilocal charge
becomes

$Q^{(n)}[\partial\Sigma]=\int_{\Sigma}J^{(n)}{}^{a}d\Sigma_{a}=\frac{1}{2^{n+2}(D-2n-1)}\oint_{\partial\Sigma}F^{(n)ab}dS_{ab}.$
(14)

We note that the potential two-form field (11) seems to be very similar to a
part of the Killing-Lovelock potential [27, 28] to define improved Komar
integrals in Lovelock theory. The $n$th Killing-Lovelock potential for the
$n$th order Lovelock term, however, consists of ($n-1$) powers of the Riemann
tensor. On the other hand, in Ref [15], the authors introduced a two-form
field with the same powers of the Riemann tensor as (11) for Killing-Yano two-
forms but not for conformal Killing-Yano two-forms. In that case, the two-form
field itself is conserved.

##  III Applications to known examples

###  III.1 Warped-product spacetime

It is known that warped-product spacetimes with two-dimensional base possess
the Kodama vector field. We revisit the known results for the Kodama vector in
terms of CKY two-forms (also see appendix D in [29]).

We consider that the metric of a $D$-dimensional warped-product spacetime,
$\mathcal{B}\times_{r}\mathcal{F}$, is given by

$g_{ab}dx^{a}dx^{b}=\gamma_{\mu\nu}(y)dy^{\mu}dy^{\nu}+r(y)^{2}\omega_{IJ}(\sigma)d\sigma^{I}d\sigma^{J},$
(15)

where $\gamma_{\mu\nu}$ and $\omega_{IJ}$ denote metrics on the two-
dimensional base space $\mathcal{B}$ and the $(D-2)$-dimensional fiber
$\mathcal{F}$, respectively. The positive function $r(y)$ is a warp factor
depending only on the coordinates on the base space, $\\{y^{\mu}\\}$. On
$\mathcal{F}$ the metric $\omega_{IJ}$ itself becomes a rank-two Killing
tensor and the associated $(D-2)$-dimensional volume form is a Killing-Yano
$(D-2)$-form. It follows from the lifting theorem in [30] that we can lift it
to a Killing-Yano $(D-2)$-form on the whole spacetime. As a result, we find
that this spacetime admits a CCKY $2$-form given by

$\frac{1}{2}h_{ab}dx^{a}\wedge
dx^{b}\equiv\frac{r}{2}\,{}^{(\gamma)}\epsilon_{ab}dx^{a}\wedge
dx^{b}=r\sqrt{-\gamma}dy^{0}\wedge dy^{1},$ (16)

where ${}^{(\gamma)}\epsilon_{ab}$ is the $2$-dimensional volume form
associated with the metric $\gamma_{\mu\nu}$. Note that this is equivalent to
the Hodge dual $f=*h$ being the Killing-Yano ($D-2$)-form.

The associated vector with this CCKY $2$-form yields the Kodama vector as
follows:

$\nabla_{a}h^{ab}=\frac{1}{\sqrt{-g}}\partial_{a}\left(\sqrt{-g}r{}^{(\gamma)}\epsilon^{ab}\right)=\frac{1}{r^{D-2}\sqrt{-\gamma}\sqrt{\omega}}\partial_{a}\left(r^{D-1}\sqrt{-\gamma}\sqrt{\omega}{}^{(\gamma)}\epsilon^{ab}\right)=(D-1){}^{(\gamma)}\epsilon^{ab}\nabla_{a}r,$
(17)

where the conventional Kodama vector is given by
$K^{a}=-{}^{(\gamma)}\epsilon^{ab}\nabla_{b}r$. In fact, the warp factor $r$
is given by a “norm” of the CCKY two-form $h$ [or the KY ($D-2$)-form $f$] as

$r^{2}=-\frac{1}{2}h_{ab}h^{ab}.$ (18)

For the Einstein tensor, the components on the two-dimensional base space are

$G_{\mu\nu}=-\frac{D-2}{r}\bar{\nabla}_{\mu}\bar{\nabla}_{\nu}r+\left[\frac{(D-2)(D-3)}{2r^{2}}\bar{\nabla}^{\lambda}r\bar{\nabla}_{\lambda}r+\frac{D-2}{r}\bar{\nabla}_{\lambda}\bar{\nabla}^{\lambda}r-\frac{1}{2r^{2}}{}^{(\omega)}R\right]\gamma_{\mu\nu},$
(19)

where $\bar{\nabla}_{\mu}$ denotes the covariant derivative associated with
$\gamma_{\mu\nu}$ and ${}^{(\omega)}R$ is the scalar curvature of the
($D-2$)-dimensional metric $\omega_{IJ}$. For a conserved current
$G_{ab}K^{b}$, we have

$\displaystyle G_{ab}K^{b}$
$\displaystyle=-\frac{1}{r^{D-2}}{}^{(\gamma)}\epsilon_{ab}\nabla^{b}\left[(D-2)\frac{r^{D-3}}{2}\nabla^{c}r\nabla_{c}r-\frac{r^{D-3}}{2(D-3)}{}^{(\omega)}R\right]$
(20) $\displaystyle=\frac{1}{r^{D-1}}h_{ab}\nabla^{b}m,$

where a mass function can be defined by

$m=\frac{D-2}{2}r^{D-3}\left[K^{a}K_{a}+\frac{{}^{(\omega)}R}{(D-2)(D-3)}\right].$
(21)

Since $K^{a}$ is divergence free, the Kodama vector itself becomes a conserved
current for the metric tensor $g_{ab}$. By definition, a charge associated
with this current is given by

$K_{a}=-\frac{1}{r}h_{ab}\nabla^{b}r=-\frac{1}{(D-1)r^{D-1}}h_{ab}\nabla^{b}r^{D-1}.$
(22)

If we consider the Einstein equation with a cosmological constant term,
$G_{ab}+\Lambda g_{ab}=T_{ab}$, the Misner-Sharp quasilocal mass

$m_{\text{MS}}=\frac{D-2}{2}r^{D-3}\left[-\frac{2\Lambda}{(D-1)(D-2)}r^{2}+K^{a}K_{a}+\frac{{}^{(\omega)}R}{(D-2)(D-3)}\right]$
(23)

is obtained by combining two conserved charges, including only the
contribution of matter without a cosmological constant. It is built from the
CCKY two-form and the Ricci scalar on the fiber $\mathcal{F}$.

###  III.2 Three-dimensional spacetime

In three dimensions one can consider that spacetimes are not warped product
but axisymmetric, such as a rotating spacetime with angular momentum. In this
case the Kodama vector can be defined and it provides conserved current and
charge [5, 6].

Let us suppose $\psi_{a}$ is a Killing vector satisfying

$\nabla_{a}\psi_{b}+\nabla_{b}\psi_{a}=0.$ (24)

The Hodge dual of it provides a CCKY two-form given by

$h_{ab}\equiv\epsilon_{abc}\psi^{c}.$ (25)

Note that we can directly confirm

$\displaystyle\nabla_{c}h_{ab}$
$\displaystyle=\epsilon_{abd}\nabla_{c}\psi^{d}$ (26)
$\displaystyle=g_{ac}K_{b}-g_{bc}K_{a},$

where the associated vector is given by

$K_{a}\equiv-\frac{1}{2}\nabla^{b}h_{ab}=-\frac{1}{2}\epsilon_{abc}\nabla^{b}\psi^{c}.$
(27)

This is the extended Kodama vector, which has been introduced in [5, 6]. Note
that $\nabla_{a}\psi_{b}=\epsilon_{abc}K^{c}$. We have

$\displaystyle\nabla_{a}K_{b}$
$\displaystyle=-\frac{1}{2}\epsilon_{b}{}^{cd}\nabla_{a}\nabla_{c}\psi_{d}$
(28) $\displaystyle=\frac{1}{2}\epsilon_{b}{}^{cd}R_{cda}{}^{e}\psi_{e}$
$\displaystyle=-\epsilon_{acd}G_{b}{}^{d}\psi^{c}=h_{ac}G_{b}{}^{c},$

which yields $G^{ab}\nabla_{a}K_{b}=0$. A straightforward calculation shows

$\displaystyle\nabla_{a}(K^{b}K_{b})$ $\displaystyle=2K^{b}\nabla_{a}K_{b}$
(29)
$\displaystyle=-2\epsilon_{acd}\psi^{c}G^{d}{}_{b}K^{b}=2h_{ac}G^{c}{}_{b}K^{b},$
$\displaystyle\nabla_{a}(\psi^{b}\psi_{b})$
$\displaystyle=2\psi^{b}\nabla_{a}\psi_{b}$ (30)
$\displaystyle=2\epsilon_{abc}\psi^{b}K^{c}=-2h_{ab}K^{b},$

and

$\displaystyle\nabla_{a}(\psi^{b}K_{b})$
$\displaystyle=\psi^{b}\nabla_{a}K_{b}+K_{b}\nabla_{a}\psi^{b}$ (31)
$\displaystyle=-\epsilon_{acd}\psi^{c}G_{b}{}^{d}\psi^{b}=h_{ac}G^{c}{}_{b}\psi^{b}.$

This implies that the above scalar quantities $K^{a}K_{a}$, $\psi^{a}\psi_{a}$
and $\psi^{a}K_{a}$ are conserved charges associated with conserved currents
$G^{a}{}_{b}K^{b}$, $K^{a}$ and $G^{a}{}_{b}\psi^{b}$, respectively.

If we assume that $\psi^{a}$ is an axial Killing vector and the Einstein
equation $G_{ab}+\Lambda g_{ab}=T_{ab}$ is satisfied, the following scalar
functions

$\displaystyle m$
$\displaystyle\equiv\frac{1}{2}(-\Lambda\psi^{a}\psi_{a}+K^{a}K_{a}),$ (32)
$\displaystyle j$ $\displaystyle\equiv-\psi^{a}K_{a},$

can be interpreted as a Misner-Sharp quasilocal mass and Komar angular-
momentum in three-dimensional axisymmetric spacetimes.

###  III.3 Generalized Misner-Sharp mass in Lovelock gravity

In this subsection we consider $D$-dimensional warped product spacetime (15),
again. For simplicity, we focus on the cases in which the metric $\omega_{IJ}$
on the ($D-2$)-dimensional subspace $\mathcal{F}$ is maximally symmetric,
i.e., ${}^{(\omega)}R=(D-2)(D-3)k$. The real constant $k$ denotes a curvature
scale on the ($D-2$)-dimensional subspace.

Now, because the whole spacetime is warped product, components of two-form
potential only on the two-dimensional base should contribute to the conserved
charge by integrating the conserved current for the $n$th Lovelock tensor. The
CCKY two-form of Eq. (16), $h_{ab}$, is proportional to the volume form of the
two-dimensional base space. We have

$\displaystyle F^{(n)}_{ab}h^{ab}$ $\displaystyle=\delta^{cda_{1}\cdots
a_{2n}}_{abb_{1}\cdots b_{2n}}h^{ab}h_{cd}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$ (33)
$\displaystyle=-4r^{2}\delta^{I_{1}\cdots I_{2n}}_{J_{1}\cdots
J_{2n}}R_{I_{1}I_{2}}{}^{J_{1}J_{2}}\cdots
R_{I_{2n-1}I_{2n}}{}^{J_{2n-1}J_{2n}}$
$\displaystyle=-\frac{(D-2)!2^{n+2}}{(D-2n-2)!r^{2n-2}}(k+K^{a}K_{a})^{n},$

where ($D-2$)-dimensional components of the Riemann curvature tensor are given
by

$R_{IJ}{}^{KL}=\frac{k+K^{a}K_{a}}{r^{2}}\delta^{KL}_{IJ}.$ (34)

Note that $\delta^{KL}_{IJ}$ denotes the generalized Kronecker delta symbol on
the $(D-2)$ dimensions, we have used the formulae $\epsilon_{abc_{1}\cdots
c_{D-2}}h^{ab}=-2r^{D-1}{}^{(\omega)}\epsilon_{c_{1}\cdots c_{D-2}}$ and
$\delta^{I_{1}I_{2}\cdots I_{2n-1}I_{2n}}_{J_{1}J_{2}\cdots
J_{2n-1}J_{2n}}\delta^{J_{1}J_{2}}_{I_{1}I_{2}}\cdots\delta^{J_{2n-1}J_{2n}}_{I_{2n-1}I_{2n}}=2^{n}(D-2)!/(D-2n-2)!$.
As the result, a conserved current and a quasilocal charge for $n$th Lovelock
tensor are

$G^{(n)a}{}_{b}K^{b}=\nabla_{b}\left(\frac{m^{(n)}}{r^{D-1}}h^{ab}\right),$
(35)

where

$m^{(n)}\equiv\frac{(D-2)!}{2(D-2n-1)!}r^{D-2n-1}(k+K^{a}K_{a})^{n}.$ (36)

Since, for each order of the Lovelock tensor, each current and each charge are
conserved, linear combinations of these quantities should be conserved. Hence,
according to the field equations, they can reproduce the generalized Misner-
Sharp quasilocal mass in Lovelock gravity, which has been proposed in Refs.
[18, 19].

We note that, when the ($D-2$)-dimensional subspace is described by Einstein
spaces as well as maximally symmetric spaces, the Misner-Sharp quasilocal mass
was provided in Ref. [20]. In that case, the quasilocal mass contains the Weyl
curvature of the ($D-2$)-dimensional Einstein space. (More generally, it
comprises the sum of every order of Lovelock terms for the ($D-2$)-dimensional
subspace, as shown in appendix A.)

##  IV discussion

In this paper, we have shown that the associated vector of a conformal
Killing-Yano two-form is the origin of the Kodama vector. In spacetimes
admitting a CKY two-form, each order of the Lovelock tensors as well as the
Einstein tensor contracted with the Kodama vector yields a locally conserved
current. This fact results from purely geometrical properties of CKY forms
without the field equations in gravitational theories. Physical
interpretations of the conserved current such as an energy current should be
provided through the field equations. The Kodama vectors that has been known
in the literature arise from closed CKY two-forms. We expect that various
arguments based on the Kodama vector can be extended to spacetimes admitting
CKY two-forms as well as closed ones. Unfortunately, little is known about
ansatz of non-trivial spacetimes admitting a CKY two-form such that its
associated vector is not Killing vector. If a spacetime admits a CCKY two-
form, we can obtain the spacetime admitting the CKY two-form by conformal
transformation.

For each order of the Lovelock tensor, including the Einstein tensor and
metric tensor (i.e. cosmological constant term), each current provided by the
Kodama vector can be individually conserved. This means there are individual,
conserved charges associated with each current. It is expected that in terms
of these charges we can obtain thermodynamic relations such as Smarr formula
and the first law [19, 31]. In particular, this nature may play a significant
role in extracting the contribution of a cosmological constant from a
definition of energy [32, 33].

The conserved currents associated with Killing vectors and Kodama vectors have
a similar structure [27, 28] built from the following quantities:
$\mathcal{P}^{(n)}{}_{ab}{}^{cd}\equiv\delta^{cda_{1}\cdots
a_{2n}}_{abb_{1}\cdots b_{2n}}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$, which are crucial to the Euler-
Lagrange equations in Lovelock gravity [34, 35, 26]. Because a Killing vector
$\xi^{a}$ is divergence free, we obtain a $2$-form potential $\omega_{ab}$
such that $\xi^{a}=\nabla_{b}\omega^{ab}$. A part of Komar-type potential is
given by $\mathcal{P}^{(n)}_{abcd}\omega^{cd}$. On the other hand, a Kodama
vector is provided by a CKY $2$-form $h_{ab}$ as
$K^{a}=-\nabla_{b}h^{ab}/(D-1)$ and the potential is given by
$\mathcal{P}^{(n)}_{abcd}h^{cd}$. It is fascinating to explore the relation
between these conserved current.

A primitive proof of Birkhoff’s theorem based on the CKY two-form can apply to
only vacuum with a cosmological constant not but electrovac spacetimes,
because it rely on the fact that the spacetime is described by Einstein
metric. However, the condition that the spacetime is described by Einstein
metric is only a sufficient condition for the Kodama vector to be a Killing
vector. The fact that Birkhoff’s theorem holds for a wider class of spacetimes
even in Lovelock gravity [36, 37, 38] implies the proof can be improved. For
example, extending the argument to generalized CKYs or CKYs with torsion [39,
40] may be interesting.

###### Acknowledgements.

I would like to thank T. Houri, M. Nozawa, and N. Dadhich for valuable
comments. This work was supported in part by JSPS KAKENHI Grant No.
JP21H05186.

##  Appendix A Curvature tensors on warped product spacetimes

In this appendix, we summarize useful relations in terms of the curvature
tensor on the $D$-dimensional warped product spacetimes,
$\mathcal{B}\times_{r}\mathcal{F}$, described by the metric (15).

The nonvanishing components of the Riemann tensor are given by

$\displaystyle R_{\mu\nu\alpha\beta}$
$\displaystyle={}^{(\gamma)}R\gamma_{\mu[\alpha}\gamma_{\beta]\nu},$ (37)
$\displaystyle R_{\mu I\nu J}$
$\displaystyle=-\omega_{IJ}r\bar{\nabla}_{\mu}\bar{\nabla}_{\nu}r,$
$\displaystyle R_{IJKL}$
$\displaystyle=r^{2}\left[{}^{(\omega)}R_{IJKL}+2K^{a}K_{a}\omega_{I[K}\omega_{L]J}\right],$

where ${}^{(\omega)}R_{IJKL}$ denotes the Riemann tensor with respect to the
metric $\omega_{IJ}$ on the fiber $\mathcal{F}$, and ${}^{(\gamma)}R$ and
$\bar{\nabla}_{\mu}$ denote, respectively, the Ricci scalar and the covariant
derivative with respect to the metric $\gamma_{\mu\nu}$ on the base
$\mathcal{B}$.

The nonvanishing components of the Ricci tensor are

$\displaystyle R_{\mu\nu}$
$\displaystyle=\frac{{}^{(\gamma)}R}{2}\gamma_{\mu\nu}-\frac{D-2}{r}\bar{\nabla}_{\mu}\bar{\nabla}_{\nu}r,$
(38) $\displaystyle R_{IJ}$
$\displaystyle={}^{(\omega)}R_{IJ}+\left[(D-3)K^{a}K_{a}-r\bar{\nabla}^{\mu}\bar{\nabla}_{\mu}r\right]\omega_{IJ},$

and the Ricci scalar is

$R={}^{(\gamma)}R-\frac{2(D-2)}{r}\bar{\nabla}^{\mu}\bar{\nabla}_{\mu}r+\frac{1}{r^{2}}\left[{}^{(\omega)}R+(D-2)(D-3)K^{a}K_{a}\right].$
(39)

The contraction of $n$ powers of the Riemann tensor with the generalized
Kronecker delta on ($D-2$)-space is given by

$\displaystyle\delta^{I_{1}\cdots I_{2n}}_{J_{1}\cdots
J_{2n}}R_{I_{1}I_{2}}{}^{J_{1}J_{2}}\cdots
R_{I_{2n-1}I_{2n}}{}^{J_{2n-1}J_{2n}}$
$\displaystyle=\frac{1}{r^{2n}}\delta^{I_{1}\cdots I_{2n}}_{J_{1}\cdots
J_{2n}}\left[{}^{(\omega)}R_{I_{1}I_{2}}{}^{J_{1}J_{2}}+K^{a}K_{a}\delta_{I_{1}I_{2}}^{J_{1}J_{2}}\right]\cdots$
(40)
$\displaystyle=\frac{2^{n}}{r^{2n}}\sum_{l=0}^{n}{}_{n}C_{l}\frac{(D-2-2n+2l)!}{(D-2-2n)!}(K^{a}K_{a})^{l}\mathcal{R}^{(n-l)}_{\omega},$

where

$\mathcal{R}_{\omega}^{(k)}\equiv\frac{1}{2^{k}}\delta^{I_{1}\cdots
I_{2k}}_{J_{1}\cdots
J_{2k}}{}^{(\omega)}R_{I_{1}I_{2}}{}^{J_{1}J_{2}}\cdots{}^{(\omega)}R_{I_{2k-1}I_{2k}}{}^{J_{2k-1}J_{2k}}.$
(41)

Note that we have used

$\delta^{I_{1}\cdots I_{2n}}_{J_{1}\cdots
J_{2n}}\delta^{I_{1}I_{2}}_{J_{1}J_{2}}\cdots\delta^{I_{2l-1}I_{2l}}_{J_{2l-1}J_{2l}}=\frac{2^{l}(D-2-2n+2l)!}{(D-2-2n)!}\delta^{I_{1}\cdots
I_{2(n-l)}}_{J_{1}\cdots J_{2(n-l)}}\qquad(n\geq l).$ (42)

If $\omega_{IJ}$ on the fiber $\mathcal{F}$ is a ($D-2$)-dimensional Einstein
metric, i.e, ${}^{(\omega)}R_{IJ}=k(D-3)\omega_{IJ}$, then we have
${}^{(\omega)}R_{IJ}{}^{KL}={}^{(\omega)}W_{IJ}{}^{KL}+k\delta_{IJ}^{KL}$,
where ${}^{(\omega)}W_{IJKL}$ is the Weyl tensor with respect to
$\omega_{IJ}$. Equation (40) reduces to

$\delta^{I_{1}\cdots I_{2n}}_{J_{1}\cdots
J_{2n}}R_{I_{1}I_{2}}{}^{J_{1}J_{2}}\cdots
R_{I_{2n-1}I_{2n}}{}^{J_{2n-1}J_{2n}}=\frac{2^{n}}{r^{2n}}\sum_{l=0}^{n}{}_{n}C_{l}\frac{(D-2-2n+2l)!}{(D-2-2n)!}(k+K^{a}K_{a})^{l}\mathcal{W}^{(n-l)}_{\omega},$
(43)

where $\mathcal{W}_{\omega}^{(k)}$ has been obtained by replacing the Riemann
tensor ${}^{(\omega)}R_{IJ}{}^{KL}$ with the Weyl tensor
${}^{(\omega)}W_{IJ}{}^{KL}$ in Eq. (41).

##  Appendix B Formulae for curvature polynomials

In this appendix, we summarize basic properties of curvature polynomials in
$D$-dimensions. (See, for example, Refs. [26, 34, 35])

The $n$th order Lovelock scalar and Lovelock-Ricci tensor, which are
respectively analogous to the Ricci scalar and the Ricci tensor for $n=1$, are

$\displaystyle\mathcal{R}^{(n)}$
$\displaystyle\equiv\frac{1}{2^{n}}\delta^{a_{1}a_{2}\cdots
a_{2n}}_{b_{1}b_{2}\cdots b_{2n}}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}},$ (44)
$\displaystyle\mathcal{R}^{(n)a}{}_{b}$
$\displaystyle\equiv\frac{n}{2^{n}}\delta^{a_{1}a_{2}\cdots
a_{2n}}_{bb_{2}\cdots b_{2n}}R_{a_{1}a_{2}}{}^{ab_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}.$ (45)

The $n$th order Lovelock tensor, which is the analog of the Einstein tensor,
is given by

$\displaystyle G^{(n)a}{}_{b}$
$\displaystyle\equiv-\frac{1}{2^{n+1}}\delta^{aa_{1}a_{2}\cdots
a_{2n}}_{bb_{1}b_{2}\cdots b_{2n}}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}$ (46)
$\displaystyle=\mathcal{R}^{(n)a}{}_{b}-\frac{1}{2}\mathcal{R}^{(n)}g^{a}{}_{b},$

where the last equality is easily verified by using the following formula:

$\delta^{aa_{1}a_{2}\cdots a_{2n}}_{bb_{1}b_{2}\cdots
b_{2n}}=g^{a}_{b}\delta^{a_{1}a_{2}\cdots a_{2n}}_{b_{1}b_{2}\cdots
b_{2n}}-\sum^{2n}_{k=1}g^{a}_{b_{k}}\delta^{a_{1}a_{2}\cdots a_{k}\cdots
a_{2n}}_{b_{1}b_{2}\cdots b\cdots b_{2n}}.$ (47)

An $n$th order rank-$4$ tensor that consists of $n$ powers of the Riemann
tensor is given by

$\mathcal{P}^{(n)}{}_{ab}{}^{cd}\equiv\delta^{cda_{1}a_{2}\cdots
a_{2n}}_{abb_{1}b_{2}\cdots b_{2n}}R_{a_{1}a_{2}}{}^{b_{1}b_{2}}\cdots
R_{a_{2n-1}a_{2n}}{}^{b_{2n-1}b_{2n}}.$ (48)

Its indices have the same properties as those of the Riemann tensor:

$\mathcal{P}^{(n)}{}_{abcd}=-\mathcal{P}^{(n)}{}_{bacd}=-\mathcal{P}^{(n)}{}_{abdc},\quad\mathcal{P}^{(n)}{}_{abcd}=\mathcal{P}^{(n)}{}_{cdab},\quad\mathcal{P}^{(n)}{}_{[abc]d}=0,$
(49)

and, in addition, it is divergence free for each index:

$\nabla^{a}\mathcal{P}^{(n)}{}_{abcd}=0.$ (50)

This tensor has various useful properties as follows. The contraction yields

$\mathcal{P}^{(n)}{}_{ac}{}^{bc}=-2^{n+1}(D-2n-1)G^{(n)}{}_{a}{}^{b}.$ (51)

Furthermore, we have

$\mathcal{R}^{(n)}=\frac{1}{2^{n}}\mathcal{P}^{(n-1)}{}_{abcd}R^{abcd},\quad\mathcal{R}^{(n)a}{}_{b}=\frac{n}{2^{n}}\mathcal{P}^{(n-1)}{}_{bcde}R^{acde}.$
(52)

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