# Strong $u$-invariant and Period-Index bound for complete ultrametric fields

Shilpi Mandal  Department of Mathematics
Emory University
400 Dowman Drive NE
Atlanta, GA 30322, USA<EMAIL_ADDRESS>
###### Abstract.

Let $k$ be a complete ultrametric field with
$\text{dim}_{\mathbb{Q}}{\sqrt{\mid k^{*}\mid}}=n$ finite, with residue field
$\tilde{k}$, and char$(\tilde{k})\neq 2$. We prove that $u(k)\leq
2^{n}u(\tilde{k})$. Let $u_{s}(k)$ be the strong $u$-invariant of $k$, then we
further show that $u_{s}(k)\leq 2^{n}u_{s}(\tilde{k})$. Let $l$ be a prime
such that the gcd(char$(\tilde{k}),l$) $=1$. If the
$\text{Br}_{l}\text{dim}(\tilde{k})=d$, then we also show that
$\text{Br}_{l}\text{dim}(k)\leq d+n$. Let $C$ be a curve over $k$ and
$F=k(C)$. Then we show that for any quadratic form over $F$ with dimension
$>2^{n+1}u_{s}(\tilde{k})$ is isotropic over $F$. We further show that if
$\text{Br}_{l}\text{dim}(\tilde{k})\leq d$ and
$\text{Br}_{l}\text{dim}(\tilde{k}(T))\leq d+1$, then
$\text{Br}_{l}\text{dim}(F)\leq d+n+1$.

##  1\. Introduction

Let $k$ be a field with $\text{char}(k)\neq 2$. The $u$-invariant of $k$,
denoted by $u(k)$, is the maximal dimension of anisotropic quadratic forms
over $k$. We say that $u(k)=\infty$ if there exist anisotropic quadratic forms
over $k$ of arbitrarily large dimension. For example, $u$-invariant of an
algebraic closed field $K$, $u(K)=1$; $u(\mathbb{R})=\infty$; for $k$ a finite
field, $u(k)=2$, etc. The $u$-invariant is a positive integer if it is finite.
A key area of research in the theory of quadratic forms is to find all the
possible values this invariant can take for a fixed or varying field. For
example, it has been established in the literature (see [5, Chapter XI,
Proposition 6.8]) that the $u$-invariant cannot take values $3$, $5$, and $7$.
See [5, Chapter XIII, Section 6], for more open problems about this invariant.
Considerable progress has also been made, particularly in the computation of
the $u$-invariant of function fields of $p$-adic curves due to Parimala and
Suresh in [2], [9], and by Harbater, Hartmann, and Krashen regarding the
$u$-invariants in the case of function fields of curves over complete
discretely valued fields in [1].

Harbater, Hartmann, and Krashen defined the strong $u$-invariant of $K$,
denoted by $u_{s}(K)$, as the smallest real number $m$ such that, $u(E)\leq m$
for all finite field extensions $E/K$, and $u(E)\leq 2m$ for all finitely
generated field extensions $E/K$ of transcendence degree $1$. We say that
$u_{s}(K)=\infty$ if there exist such field extensions $E$ of arbitrarily
large $u$-invariant. In [1, Theorem 4.10], the same authors prove that for $K$
a complete discretely valued field, whose residue field $\tilde{K}$ has
characteristic unequal to 2, $u_{s}(K)=2u_{s}(\tilde{K})$.

Let $k$ be a complete non-Archimedean valued field with residue field
$\tilde{k}$ such that $\text{char}(\tilde{k})\neq 2$. Let $\sqrt{|k^{*}|}$
denote the divisible closure of the value group $|k^{*}|$. In [3], Mehmeti
shows that if $\text{dim}_{\mathbb{Q}}\sqrt{|k^{*}|}=n$, then $u_{s}(k)\leq
2^{n+1}u_{s}(\tilde{k})$, and if $|k^{*}|$ is a free $\mathbb{Z}$-module with
$\text{rank}_{\mathbb{Z}}|k^{*}|=n$, then $u_{s}(k)\leq
2^{n}u_{s}(\tilde{k})$. Mehmeti used field patching in the setting of
Berkovich analytic geometry, to prove a local-global principle, and provides
applications to quadratic forms and the $u$-invariant. The results she
obtained generalises those of [1].

Our main result concerning $u$-invariant of a complete non-Archimedean
(ultrametric) field is the following theorem.

######  Theorem 1.1 (Theorem 3.4).

Let $k$ be a complete ultrametric field with $\text{char}(\tilde{k})\neq 2$.
Suppose that $\text{dim}_{\mathbb{Q}}(\sqrt{\mid k^{*}\mid})=n$ is finite.
Then $u(k)\leq 2^{n}u(\tilde{k})$.

######  Corollary 1.2 (Theorem 4.1).

Let $k$ be a complete ultrametric field with $\text{char}(\tilde{k})\neq 2$.
Suppose the $\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $C$ be
a curve over $k$ and $F=k(C)$ the function field of $C$. Then $u(F)\leq
2^{n+1}u_{s}(\tilde{k})$.

Given a field $k$, recall the definitions of period and index of a central
simple algebra. The period (or exponent) of a central simple $k$-algebra $A$
is the order of class of $A$ in the Brauer group of $k$. The index of $A$ is
the degree of the division algebra $D_{A}$ that lies in the class of $A$
(i.e., $A$ is a matrix ring over $D_{A}$). The period and index always have
the same prime factors, and the period always divides the index [16,
Proposition 4.5.13].

The period-index problem asks whether all central simple algebras $A$ over a given field $k$ satisfy ind($A$) | per($A$)d for some fixed exponent $d$ depending only on $k$. In the spirit of [9], we make the following definition. Let $k$ be any field. For a prime $l$, define the Brauer $l$-dimension of $k$, denoted by Brldim($k$), to be the smallest integer $d\geq 0$ such that for every finite extension $L$ of $k$ and for every central simple algebra $A$ over $L$ of period a power of $l$, ind($A$) divides per($A$)d. The Brauer dimension of $k$, denoted by $\text{Brdim}(k)$, is defined as the maximum of the Brauer $l$-dimension of $k$ as $l$ varies over all primes. It is expected that this invariant should increase by one upon a finitely generated field extension of transcendence degree one.

Saltman proved some results in this direction, including the fact that the
index divides the period squared for function fields of $p$-adic curves [6].
He also developed a general mechanism to relate the Brauer dimension of curves
over discretely valued fields to that of curves over the residue field. Then
Harbater, Hartmann, and Krashen in [1, Theorem 5.5] show that for $k$ a
complete discretely valued field, its residue field $\tilde{k}$, $F$ the
function field of a curve over $k$, and $l\neq\text{char}(\tilde{k})$, they
prove that if $\text{Br}_{l}\text{dim}(\tilde{k})\leq d$ and
$\text{Br}_{l}\text{dim}(\tilde{k}(T))\leq d+1$, then
$\text{Br}_{l}\text{dim}(F)\leq d+2$.

Using [3] we get the following as a direct consequence.

######  Theorem 1.3 (Theorem 2.4).

Let $k$ be a complete ultrametric field with
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $C$ be a curve over
$k$ and $F=k(C)$ the function field of the curve. Let $A$ be a central simple
algebra over $F$ and let $V(F)$ be the set of all non-trivial rank 1
valuations of $F$. Then $\text{ind}(A)$ is the maximum of the set
$\\{\text{ind}(A\otimes\widetilde{F_{v}})\\}$ for $v\in V(F)$.

Our main result concerning the Brauer dimension are the following theorems.

######  Theorem 1.4 (Theorem 3.6).

Let $k$ be a complete ultrametric field. Let $l$ be a prime such that
$l\neq\text{char}(\tilde{k})$. Suppose
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. If the
$\text{Br}_{l}{dim}(\tilde{k})=d$, then $\text{Br}_{l}{dim}(k)\leq d+n$.

######  Theorem 1.5 (Theorem 4.3).

Let $k$ be a complete ultrametric field. Let $l$ be a prime such that
$l\neq\text{char}(\tilde{k})$. Suppose
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $C$ be a curve over
$k$ and $F=k(C)$ the function field of $C$. Suppose there exist an integer $d$
such that $\text{Br}_{l}\text{dim}(\tilde{k})\leq d$ and
$\text{Br}_{l}\text{dim}(\tilde{k}(T))\leq d+1$. Then the
$\text{Br}_{l}\text{dim}(F)\leq d+1+n$.

Remark: Note that if $k$ is a complete discretely valued field, then $n=1$.
Theorem 1.4 now gives that $\text{Br}_{l}\text{dim}(k)\leq d+1$, which is a
classical result [16, Corollary 7.1.10]. In the same case, Theorem 1.5 now
implies that if $l\neq\text{char}(\tilde{k})$, then
$\text{Br}_{l}\text{dim}(F)\leq d+2$, which is found in [1, Corollary 5.10].

###  1.1. Organisation of the manuscript

We start with some preliminaries and necessary background in Section 2, where
we talk about the known results on local-global principles from [1] and [3].

In Section 3 we prove the main decomposition lemma, which we then use to prove
that for a complete ultrametric field $k$ with residue field characteristic
$\text{char}(\tilde{k})\neq 2$ and
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$, $u(k)\leq 2^{n}u(\tilde{k})$. We
further prove the result about
$\text{Br}_{l}\text{dim}(k)\leq\text{Br}_{l}\text{dim}(\tilde{k})+n$.

In Section 4, using the results from Section 3 and the results of Mehmeti we
prove our results on $u$-invariant and Brauer $l$-dimension of function fields
of curves over complete ultrametric fields.

##  2\. Preliminaries

Let $F$ be a field. For a central simple $F$-algebra $A$, the period of $A$ is
the order of its class in the Brauer group of $F$, and the index of $A$ is the
degree of the division algebra Brauer equivalent to $A$. The index of $A$ is
denoted by ind($A$) and period of $A$ by per($A$). Also, per($A$) divides
ind($A$), and the two numbers are composed of the same prime factors.

If a linear algebraic group $G$ acts on a variety $H$ over a field $F$, we
will say that $G$ acts transitively on the points of $H$ if for every field
extension $E$ of $F$ the induced action of the group $G(E)$ on the set $H(E)$
is transitive.

The following definition is due to F. Ch$\hat{\text{a}}$telet. For a more
detailed discussion of results on Severi-Brauer varieties see Section 7.1 in
[15].

######  Definition 2.1 (Severi-Brauer variety).

Let $n$ be a positive integer. A Severi-Brauer variety of dimension $n-1$ over
a field $k$ is a twisted form of the projective space $\mathbb{P}^{n-1}_{k}$.
Equivalently, this is a $k$-variety $X$ such that there exists a field
extension $k\subset K$ and an isomorphism of $K$-varieties
$X\times_{k}K\cong\mathbb{P}^{n-1}_{k}$.

There is a natural bijection between the isomorphism classes of Severi-Brauer
varieties over a field $k$ and the isomorphism classes of central simple
$k$-algebras.

For instance, a twisted form of $\mathbb{P}^{1}_{k}$ is a smooth, projective,
geometrically integral curve $C$ of genus 0, and vice versa, any smooth plane
conic is a twisted form of $\mathbb{P}^{1}_{k}$. The automorphism group of
projective space $\mathbb{P}^{n-1}_{k}$ is the algebraic group PGLn,k [19, Ch.
II, Example 7.1.1]. The automorphism group of the matrix algebra $M_{n}(k)$ is
the projective general linear group PGL${}_{n}(k)$ [16, Corollary 2.4.2].
Galois descent then gives a bijection between the isomorphism classes of
twisted forms of $\mathbb{P}^{n-1}_{k}$ and the isomorphism classes of twisted
forms of $M_{n}(k)$ (see the discussion in [15, Section 1.3.2]), plus we know
that the matrix algebra $M_{n}(k)$ is always a central simple algebra over
$k$. Thus we get a canonical bijection of pointed sets [16, Theorem 2.4.3]

$H^{1}(k,\text{PGL}_{n,k})\cong\\{\text{central simple algebras over $k$ of
degree $n$}\\}/\text{iso}\\\ $ $\cong\\{\text{Severi-Brauer varieties over $k$
of dimension $n-1$}\\}/\text{iso}$

and a map of pointed sets $H^{1}(k,\text{PGL}_{n,k})\to\text{Br}(k)$, where a
central simple algebra $A$ of degree $n$ is sent to its class
$[A]\in\text{Br}(k)$. In the case of a Severi-Brauer variety $X$ of dimension
$n-1$, denote by $[X]\in\text{Br}(k)$ the image of the isomorphism class of
$X$ under this map.

For a central simple algebra $A$ over $k$ of degree $n$, define $X(A)$ to be
the $k$-scheme of right ideals of $A$ of rank $n$. More precisely, for any
commutative $k$-algebra $R$, the set $X(A)(R)$ is the set of right ideals of
the $k$-algebra $A\otimes_{k}R$, which are projective $R$-modules of rank $n$
and are direct summands of the $R$-module $A\otimes_{k}R$. This is a closed
subscheme of the Grassmannian of $n$-dimensional subspaces of the $k$-vector
space $A$. For more details, see the discussion in [14, Chapter 1, §1.C].

Let $A$ be a central simple $F$\- algebra of degree $n$ and let $I\subseteq A$
be a right ideal of $A$. Then the $F$-dimension of $I$, denoted by
$\text{dim}_{F}(I)$ is divisible by the degree of $A$, deg($A$). The quotient
$\text{dim}_{F}(I)/\text{deg}(A)$ is called the reduced dimension of the ideal
(see Definition 1.9 in [14]). For any integer $i$, we write SB${}_{i}(A)$ for
the $i$-th generalized Severi-Brauer variety of the right ideals in $A$ of
reduced dimension $i$. In particular, SB${}_{0}(A)$ $=$ Spec($F$) $=$
SB${}_{\text{deg}(A)}(A)$ and SB${}_{i}(A)=\phi$ for $i$ outside of the
interval $[0,\text{deg}(A)]$. The variety SB${}_{1}(A)$ is the usual Severi-
Brauer variety of A.

In the same setup let $E/F$ be a field extension and if $A_{E}$ denotes the
algebra $A\otimes_{F}E$, then the $E$-points of $\text{SB}_{i}(A)$ are in
bijection with the right ideals in $A_{E}$ which are of reduced dimension $i$
and are direct summands of $A_{E}$. Also, $\text{SB}_{i}(A)(E)\neq\phi$ if and
only if ind($A_{E}$) divides $i$ (see Chapter 1, §1.C in [14]). Since $A_{E}$
is a central simple algebra over $E$, we have $A_{E}\cong\text{M}_{m}(D)$ for
some $E$-division algebra $D$ and some $m\geq 1$. Now, the right ideals of
reduced dimension $i$ in $A_{E}$ are in natural bijection with the subspaces
of $D^{m}$ of $D$-dimension $i/\text{ind}(A_{E})$ [14, Proposition 1.12].
Thus, writing $D_{A}$ for the $F$-division algebra in the class of $A$, the
$F$-linear algebraic group GL1($A$) $=$ GLm($D_{A}$) acts transitively on the
points of the $F$-scheme $\text{SB}_{i}(A)$.

######  Definition 2.2.

[1] Let $K$ be a field. Let $X$ be a $K$-variety and $G$ a linear algebraic
group over $K$. We say that $G$ acts on $X$ and, for any field extension
$L/K$, either $X(L)=\phi$ or $G(L)$ acts transitively on $X(L)$.

A field $k$ equipped with a non-Archimedean valuation
$|\cdot|:k\to\mathbb{R}_{\geq 0}$ is called a non-Archimedean field or an
ultrametric field. We say $k$ is complete if every Cauchy sequence converges
to a limit.

In this paper, we use field patching in the setting of Berkovich analytic
geometry. By patching over analytic curves, Mehmeti proved a local-global
principle and provided applications to quadratic forms and the u-invariant.

We use the following main result of Mehmeti [3] on the local-global principle
in Berkovich theory.

######  Theorem 2.3 (Mehmeti [3]).

Let $k$ be a complete ultrametric field. Let $C$ be a normal irreducible
projective $k$-algebraic curve. Denote by $F$ the function field of $C$. Let
$X$ be an $F$-variety and $G$ a connected rational linear algebraic group over
$F$ acting strongly transitively on $X$.

Let $V(F)$ be the set of all non-trivial rank 1 valuations of $F$ which either
extend the valuation of $k$ or are trivial when restricted to $k$.

If $F$ is a perfect field or $X$ is a smooth variety, then

$X(F)\neq\phi\ \text{if and only if}\ X(F_{v})\neq\phi\ \text{for all v}\in
V(F),$

where $F_{v}$ denotes the completion of $F$ with respect to $v$.

The following theorem follows from the proof of Harbater, Hartmann, and
Krashen’s Theorem 5.1 in [1] and Mehmeti’s Corollary 3.19 in [3].

######  Theorem 2.4.

Let $k$ be a complete, non-trivially valued ultrametric field and suppose
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $C$ be a normal
irreducible projective $k$-algebraic curve. Denote by $F$ the function field
of $C$. Let $A$ be a central simple algebra over $F$ and let $V(F)$ be the set
of all non-trivial rank 1 valuations of $F$. Then $\text{ind}(A)$ is the
maximum of the set $\\{\text{ind}(A\otimes F_{v})\\}$ for $v\in V(F)$.

###### Proof.

Let $n$ be the degree of $A$. Then GL${}_{1}(A)$ is a Zariski open subset of
$\mathbb{A}_{F}^{n^{2}}$. So, it is a rational connected linear algebraic
group.

If $1\leq i<n$, then GL${}_{1}(A)$ acts transitively on the points of SB${}_{i}(A)$; and if $E$ is a field extension of $F$, then SB${}_{i}(A)(E)\neq\phi$ if and only if ind($A_{E}$) divides $i$ [14, Proposition 1.17]. So, the local-global principle proved by Mehmeti in [3] implies that ind($A$) | $i$ if and only if ind($A\otimes F_{v})|i$ for each $v\in V(F)$. Thus ind($A$) $=$ maxv∈V(F) {ind($A\otimes F_{v}$)} as claimed. ∎

We use the following notation throughout the paper.

Notation: For any valued field $E$, let $E^{\circ}$ denote the ring of
integers, $E^{\circ\circ}$ the corresponding maximal ideal, and
$\widetilde{E}$ the residue field. Let $l$ be a prime such that
$gcd(l,char(\tilde{k}))=1$.

We will need the following theorem (stated here without proof) to develop the
rest of the paper. As in Milnor’s Algebraic $K$-Theory [4], let $L$ be an
arbitrary field, $n$ a positive integer with a non-zero image in $L$. The
$K$-theory of $L$ forms a graded anti-commutative ring
$\displaystyle\oplus_{i\geq 0}K_{i}(L)$ with $K_{0}(L)\cong\mathbb{Z}$,
$K_{1}(L)\cong\text{GL}_{1}(L)$. Denote by $(a)$ for an element of $K_{1}(L)$
that corresponds with $a\in L^{*}$, such that $(a)+(b)=(ab)$. Multiplying
$(a),(b)\in K_{1}(L)$ yields an element of $K_{2}(L)$ that is written as
$(a,b)$. One calls $(a,b)$ a Steinberg symbol, or simply, a symbol.

In literature, there are two descriptions of the main theorem of Merkurjev and
Suslin, one with Galois cohomology and one with Brauer groups. Let $\mu_{n}$
be the $n$-th roots of unity in an algebraic closure $\bar{L}$ of $L$ and
denote by $L^{s}$ the separable closure of $L$ in $\bar{L}$. Now consider
$\mu_{n}$ as a module for the absolute Galois group Gal$(L^{s}/L)$. By Hilbert
Theorem 90, the Galois cohomology group $H^{1}(L,\mu_{n})$ is isomorphic with
$k_{2}(L):=K_{1}(L)/nK_{1}(L)$ (see [16, Section 8.4]). Using a theorem of
Matsumoto, and the product structures in $K$-theory and in Galois cohomology,
it gives the following homomorphism called the norm-residue homomorphism or
the Galois symbol

$\alpha_{L,n}:k_{2}(L)\to H^{2}(L,\mu_{n}^{\otimes 2}).$

The Merkurjev-Suslin theorem now simply reads

######  Theorem 2.5 (Merkurjev-Suslin [8]).

For all $L,n$ as above, $\alpha_{L,n}$ is an isomorphism.

Recall the definition of a cyclic algebra over a field. For a positive integer
$n$, let $L$ be a field in which $n$ is invertible such that $L$ contains a
primitive $n$-th root of unity $\omega$. For nonzero elements $a$ and $b$ of
$L$, the associated cyclic algebra is the central simple algebra of degree $n$
over $L$ defined by

$(a,b)_{\omega}=L\langle x,y\rangle/(x^{n}=a,y^{n}=b,xy=\omega yx).$

We know from literature that the $n$-torsion of the Brauer group,
${}_{n}\text{Br}(L)$ is isomorphic with $H^{2}(L,\mu_{n}^{\otimes 2})$ [15,
Section 1.3.4]. So, we find that $\alpha_{L,n}:k_{2}(L)\to\ _{n}\text{Br}(L)$
sends the coset of the symbol $(a,b)$ to the class of the the cyclic algebra
$(a,b)_{\omega}$, where $\omega$ is the primitive $n$-th root of unity in $L$.
We thus get the following corollary, which conveys the surjectivity of
$\alpha_{L,n}$.

######  Corollary 2.6.

Let $\mu_{n}\subset L$ and let $A$ be a central simple algebra over $L$, with
$[A]\in\ _{n}\text{Br}(L)$. Then $A$ is similar to a tensor product of cyclic
algebras $(a_{i},b_{i})_{\omega}$, $a_{i},b_{i}\in L^{*}$.

The case $n=2$ is actually Merkurjev’s theorem [11].

To define a cyclic algebra over a ring $R$, let $n$ be an invertible positive
integer in $R$. Suppose $R$ contains a primitive $n$-th root of unity
$\omega$. For $a,b\in R^{*}$, the associated cyclic algebra is an Azumaya
algebra of degree $n$ over $R$ defined by

$(a,b)_{\omega}=R\langle x,y\rangle/(x^{n}=a,y^{n}=b,xy=\omega yx).$

The notion of central simple algebra over a field generalises to that of an
Azumaya algebra over a commutative ring. The following theorem is due to
Azumaya (over a local ring), Auslander and Goldman (over an arbitrary
commutative ring), and Grothendieck (over a scheme).

######  Definition 2.7 (Azumaya algebra).

Let $R$ be a commutative ring and $A$ a ring over $R$ that is finitely
generated and projective (equivalently, locally free) as an $R$-module. Let
$A_{S}:=S\otimes_{R}A$ for a homomorphism of rings $R\to S$. A ring $A$
satisfying any of the following equivalent conditions is called an Azumaya
algebra over $R$.

  * (i)

The map $A\otimes_{R}A^{\text{op}}\to\text{End}_{R}(A)$ is an isomorphism.

  * (ii)

For every algebraically closed field $k$ and a homomorphism $R\to k$,
$A_{k}\simeq\text{Mat}_{n}(k)$.

  * (iii)

For every maximal ideal $\mathfrak{m}\subset R$, let $k=R/\mathfrak{m}$; the
ring $A_{k}$ is a central simple algebra over $k$.

An Azumaya algebra over a commutative ring $R$ is an algebra over $R$ that has
an inverse up to Morita equivalence. That is, $A$ is an Azumaya algebra if
there is an $R$-algebra $B$ such that $B\otimes_{R}A$ is Morita equivalent to
$R$, which is the unit for the tensor product of $R$-algebras. Thus, Morita
equivalence classes of Azumaya algebras over $R$ form a group, which is called
the Brauer group of $R$, denoted by $\text{Br}(R)$.

We also record here the following well-known result.

######  Proposition 2.8.

Let $R$ be a complete local domain with field of fractions $F$ and residue
field $k$. Let $A$ be an Azumaya algebra over $R$. Then

$ind(A\otimes_{R}F)\leq ind(A\otimes_{R}k).$

###### Proof.

Note that $A\otimes_{R}k$ is a central simple algebra over $k$ [18, Corollary
2.6]. Let $n=$ ind$(A\otimes_{R}k)$. Then there exists a separable field
extension $L/k$ of degree $n$ such that $(A\otimes_{R}k)\otimes L\simeq
M_{n}(L)$ [16, Proposition 2.25]. Since $L/k$ is a finite separable extension,
$L=k(\alpha)$ for some $\alpha\in L$. Let $f(x)\in k[X]$ be the monic minimal
polynomial of $\alpha$ over $k$. Let $g(X)\in R[X]$ be a monic polynomial of
degree $n$ which maps to $f(X)$ in $k[X]$. Since $f(X)$ is irreducible in
$k[X]$, $g(X)$ is irreducible in $R[X]$. Let $S=R[X]/(g(X))$ and
$E=F[X]/(g(X))$. Since $R$ is a complete local ring, $S$ is also a complete
local ring with residue field $L$ (see the proof of [7, Theorem 6.3]). Since
the map $Br(S)\to Br(L)$ is injective [7, Corollary 6.2] and
$(A\otimes_{R}S)\otimes_{S}L\simeq(A\otimes_{R}k)\otimes_{k}L$ is a matrix
algebra, $A\otimes_{R}S$ represents the trivial element in $Br(S)$. Hence
$(A\otimes_{R}F)\otimes E$ is a matrix algebra. In particular
ind$(A\otimes_{R}F)\leq[E:F]=n$. ∎

The following discussion can be found in Commutative Algebra by Bourbaki [12,
Chap. 6, $\S$10, n◦2]. For an abelian group $\Gamma$, define the rational rank
of $\Gamma$, denoted by rat.rank$(\Gamma)$, to be the
$\text{dim}_{\mathbb{Q}}(\Gamma\otimes_{\mathbb{Z}}\mathbb{Q})$. The rational
rank is an element of $\mathbb{N}\cup\\{+\infty\\}$. If $\Gamma^{\prime}$ is a
subgroup of $\Gamma$, then since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module,
we have that rat.rank$(\Gamma)$ $=$ rat.rank$(\Gamma^{\prime})$ $+$
rat.rank$(\Gamma/\Gamma^{\prime})$. The rational rank of a group $\Gamma$ is
zero if and only if $\Gamma$ is a torsion group. If $\Gamma$ is the value
group of a valuation, then its rational rank is zero if and only if the
valuation is the trivial valuation. Also, if the field $L$ is an algebraic
extension of the valued field $k$, then the quotient group
$\displaystyle|L^{*}|/|k^{*}|$ is a torsion group, and the residue field
$\tilde{L}$ is an algebraic extension of $\tilde{k}$ [17, Proposition 1.16].

##  3\. Complete ultrametric fields

Let $M\subset\mathbb{R}^{*}$ be a subgroup and let
$\sqrt{M}:=\\{a\in\mathbb{R}^{*}|\ a^{z}\in M\ \text{for some}\
z\in\mathbb{Z}_{\neq 0}\\}$ be the divisible closure of $M$. It is a
$\mathbb{Q}$-vector space, and suppose $dim_{\mathbb{Q}}(\sqrt{M})$ is finite.
For a prime $l$, set $M^{l}=\\{m^{l}|m\in M\\}$.

We want to understand what do elements of $M^{*}/{M^{*}}^{l}$ look like.

There exists a $\mathbb{Q}$-basis $z_{1},z_{2},\dots,z_{n}$ of $\sqrt{M}$. By
definition of $\sqrt{M}$, there exist non-zero integers
$\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ such that
$z_{1}^{\alpha_{1}},z_{2}^{\alpha_{2}},\dots,z_{n}^{\alpha_{n}}$ are in $M$.
Let $t_{i}:=z_{i}^{\alpha_{i}}\in M$.

Then $t_{1},\dots,t_{n}$ is also a basis. Thus, for any $t\in M$, there exist
unique $p_{1},p_{2},\dots,p_{n}\in\mathbb{Q}$ such that

$t=\prod_{i=1}^{n}{t_{i}}^{p_{i}}.$

######  Definition 3.1.

Let $t\in M$. Let $t=\displaystyle\prod_{i=1}^{n}{t_{i}}^{s_{i}/r_{i}}$ for
$\displaystyle\frac{s_{i}}{r_{i}}\in\mathbb{Q}$ with $gcd(s_{i},r_{i})=1$ for
$i=1,2,\dots,n$. Let $r$ be the lcm of $r_{i},i=1,2,\dots,n$. We will say that
$r$ is the order of $t$.

Let $k$ be an ultrametric field and $|\cdot|$ be the non-Archimedean valuation
on $k$. Suppose that dim${}_{\mathbb{Q}}\sqrt{|k^{*}|}$ (i.e., the rational
rank of $|k^{*}|$) is finite, say $n=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})$.

In particular, from the above discussion we get that when $M=|k^{*}|$, there
exist $\pi_{1},\pi_{2},\dots,\pi_{n}\in k^{*}$ with $|\pi_{i}|=t_{i}$ such
that for any $t\in\sqrt{|k^{*}|}$, there exist unique
$p_{1},p_{2},\dots,p_{n}\in\mathbb{Q}$ such that

$t=\prod_{i=1}^{n}{|\pi_{i}|}^{p_{i}}=\prod_{i=1}^{n}t_{i}^{p_{i}}.$

Let us fix such elements $\pi_{1},\pi_{2},\dots,\pi_{n}$ of $k^{*}$.

######  Lemma 3.2.

Let $k$ be a field and $m,n$ be positive integers. Suppose $m$ is coprime to
$n$. Let $a\in k^{*}$. Then there exists an integer $m^{\prime}$ such that
$a=a^{mm^{\prime}}\in k^{*}/(k^{*})^{n}$.

###### Proof.

Since $m$ and $n$ are coprime, there exists
$m^{\prime},n^{\prime}\in\mathbb{Z}$ such that $mm^{\prime}+nn^{\prime}=1$.
Hence $a=a^{mm^{\prime}+nn^{\prime}}=a^{mm^{\prime}}\in k^{*}/(k^{*})^{n}$. ∎

######  Theorem 3.3 (Main Decomposition Lemma).

Let $k$ be an ultramteric valued field, $rank_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$,
and $l$ a prime such that $gcd(l,char(\tilde{k}))=1$. Let
$a_{1},a_{2},\dots,a_{s}\in k^{*}$, there exist elements
$c_{1},c_{2},\dots,c_{n}$ in $k^{*}$ such that any $a_{i}=\displaystyle
u_{i}\prod_{j=1}^{n}c_{j}^{\mu_{ij}}(b_{i})^{l}$, $\mu_{ij}\in\mathbb{Z}$,
$0\leq\mu_{ij}\leq l-1$ , and $b_{i}\in k$.

###### Proof.

Let us fix elements $\pi_{1},\pi_{2},\dots,\pi_{n}$ in $k^{*}$ such that
$\\{t_{i}=|\pi_{i}|\\}$ is a basis of $\sqrt{|k^{*}|}$.

Since $a_{i}\in k^{*}$,$|a_{i}|\in\sqrt{|k^{*}|}$. So there exist unique
$p_{1},p_{2},\dots,p_{n}\in\mathbb{Q}$ such that,
$|a_{i}|=\displaystyle\prod_{j=1}^{n}|\pi_{j}|^{p_{ij}}=\prod_{j=1}^{n}t_{j}^{p_{ij}}=t_{1}^{r_{i1}}\prod_{j=2}^{n}t_{j}^{r_{ij}}$,
where $r_{ij}\in\mathbb{Q}$. Let $y_{i1}$ be the highest power of $l$ that
divides the denominator of $r_{i1}$. Then we can write
$|a_{i}|=t_{1}^{x_{i1}/l^{y_{i1}}z_{i1}}\prod_{j=2}^{n}t_{j}^{r_{ij}}$, where
$x_{ij},y_{i1},z_{i1}\in\mathbb{Z}$, and $r_{ij}\in\mathbb{Q}$.

Since $l\nmid z_{i1}$, from the previous lemma, $a_{i}\equiv
a_{i}^{z_{i1}}(\mod{(k^{*})^{l}})$.

Without loss of generality, we can say that any $a_{i}\in k^{*}$ has
$\displaystyle|a_{i}^{z_{i1}}|=t_{1}^{x_{i1}/l^{y_{i1}}}\prod_{j=2}^{n}t_{j}^{r_{ij}}$,
where $r_{ij}\in\mathbb{Q}$. Note that gcd($x_{i1},l^{y_{i1}}$) $=1$.

Case(i): Suppose $y_{i1}=0$ for all $i$. So,
$\displaystyle|a_{i}|=t_{1}^{x_{i1}}\prod_{j=2}^{n}t_{j}^{r_{ij}}$.
Considering the element $\pi_{1}^{-x_{i1}}a_{i}\in k^{*}$,
$\displaystyle|\pi_{1}^{-x_{i1}}a_{i}|=\prod_{j=2}^{n}t_{j}^{r_{ij}}$, where
$r_{ij}\in\mathbb{Q}$.

Case(ii): Suppose $y_{i1}\geq 1$ for some $i$. Without loss of generality, say
$y_{11}=\max_{i}\\{y_{i1}\\}\geq 1$. So,
$\displaystyle|a_{1}|=t_{1}^{x_{11}/l^{y_{11}}}\prod_{j=2}^{n}t_{j}^{r_{1j}}$.
We also know that $l\nmid x_{11}$, so we can choose
$x_{11}^{\prime},x_{11}^{\prime\prime}\in\mathbb{Z}$ such that $l\nmid
x_{11}^{\prime}$ and
$x_{11}x_{11}^{\prime}+l^{y_{11}+1}x_{11}^{\prime\prime}=1$.

So, $\displaystyle a_{1}\equiv a_{1}^{x_{11}^{\prime}}(\mod{(k^{*})^{l}})$,
implying
$\displaystyle|a_{1}^{x_{11}^{\prime}}|=t_{1}^{x_{11}x_{11}^{\prime}/l^{y_{11}}}\prod_{j=2}^{n}t_{j}^{r_{1j}x_{11}^{\prime}}$.
Now consider the exponent of $t_{1}$:

$\frac{x_{11}x_{11}^{\prime}}{l^{y_{11}}}=\frac{1-l^{y_{11}+1}x_{11}^{\prime\prime}}{l^{y_{11}}}=\frac{1}{l^{y_{11}}}-lx_{11}^{\prime\prime}.$

So $\displaystyle|a_{1}|\equiv|a_{1}^{x_{11}^{\prime}}|\equiv
t^{1/l^{y_{11}}}\prod_{j=2}^{n}t_{j}^{\lambda_{ij}}(\mod{(k^{*})^{l}})$.
Without loss of generality, say
$\displaystyle|a_{1}|=t_{1}^{1/l^{y_{11}}}\prod_{j=2}^{n}t_{j}^{r_{1j}}$ and
$\displaystyle|a_{2}|=t_{1}^{x_{21}/l^{y_{21}}}\prod_{j=2}^{n}t_{j}^{r_{2j}}$.
Now consider the element
$\displaystyle|a_{1}^{-x_{21}l^{y_{11}-y_{21}}}a_{2}|=\displaystyle
t_{1}^{\frac{-l^{y_{11}-y_{21}}}{l^{y_{11}}}+\frac{x_{21}}{l^{y_{21}}}}\prod_{j=2}^{n}t_{j}^{\lambda_{j}}$,
where $\lambda_{j}\in\mathbb{Q}$.

Consider again the exponent of $t_{1}$:

$\frac{-x_{21}l^{y_{11}-y_{21}}}{l^{y_{11}}}+\frac{x_{21}}{l^{y_{21}}}=0.$

So for $x_{21}l^{y_{11}-y_{21}}\in\mathbb{Z}$, we have
$\displaystyle|a_{1}^{-x_{21}l^{y_{11}-y_{21}}}a_{2}|=\prod_{j=2}^{n}t_{j}^{\lambda_{j}}$,
where $\lambda_{j}\in\mathbb{Q}$. Choose $c_{1}:=a_{1}$. So for all $i$,
$\displaystyle|a_{1}^{-x_{i1}l^{y_{11}-y_{i1}}}a_{i}|=\prod_{j=2}^{n}t_{j}^{\lambda_{ij}}$,
where $\lambda_{ij}\in\mathbb{Q}$.

Continuing this process, we get $c_{2},\dots,c_{n}$, $b_{i}\in k^{*}$ and
$\mu_{ij}\in\mathbb{Z}$ such that

$\mid\\!a_{i}b_{i}^{\ell}\\!\mid=\mid\\!\prod_{j=1}^{n}c_{j}^{\mu_{ij}}\\!\mid.$

Since for any $u\in k^{*}$, $u$ is a unit in the valuation ring of $k$ if and
only if $\mid\\!u\\!\mid=1$, we have

$a_{i}=u_{i}b_{i}^{\ell}\prod_{j}c_{j}^{\mu_{ij}}$

for some units $u_{i}\in k^{\circ}$, the valuation ring of $k$, $b_{i}\in
k^{*}$ and $\mu_{ij}\in\mathbb{Z}$.

∎

######  Theorem 3.4.

Let $k$ be a complete ultrametric field with $\text{char}(\tilde{k})\neq 2$.
Suppose that $\text{dim}_{\mathbb{Q}}(\sqrt{\mid k^{*}\mid})=n$ is finite.
Then $u(k)\leq 2^{n}u(\tilde{k})$.

###### Proof.

Given a quadratic form $\displaystyle q=\langle a_{1},\dots,a_{s}\rangle$ over
$k$ of dimension $s>2^{n}u(\tilde{k})$, such that $\mid a_{i}\mid\in\sqrt{\mid
k^{*}\mid}$ for all $i$. From the main decomposition theorem, we have elements
$c_{1},\dots,c_{n}$ in $k^{*}$ such that each $a_{i}=\displaystyle
u_{i}\prod_{j=1}^{n}{c_{j}}^{\lambda_{ij}}b_{i}^{2}$, for $u_{i}$ unit in
$(k^{\circ})^{*}$, $b_{i}\in k^{*}$, and $\lambda_{ij}\in\\{0,1\\}$.

So there are $2^{n}$ possibilities for
$\displaystyle\prod_{j=1}^{n}c_{j}^{\lambda_{ij}}$. Let’s call them
$\theta_{1},\theta_{2},\dots,\theta_{2^{n}}$. Thus each $a_{i}\equiv
u_{i}\theta_{j}(\mod{(L^{*})^{2}})$, for some $j$ with $1\leq j\leq 2^{n}$.

So after re-indexing, we have, $\displaystyle\langle
a_{1},\dots,a_{s}\rangle\cong\langle
u_{1},\dots,u_{s_{1}}\rangle\theta_{1}\perp\langle
u_{s_{1}+1},\dots,u_{s_{2}}\rangle\theta_{2}\perp\dots\perp\langle
u_{s_{2^{n-1}+1}},\dots,u_{s_{2^{n}}}\rangle\theta_{2^{n}}$, where all the
$s_{i}$’s add up to $s$. Consequently, for any $\theta_{i}$ as above, there
exists a diagonal quadratic form $\displaystyle\sigma_{\theta_{i}}$ with
coefficients in $(k^{\circ})^{*}$ such that $q$ is $k$-isometric to
$\displaystyle\perp_{\sigma\in Q}\theta_{i}\cdot\sigma_{\theta_{i}}$.

Since $s>2^{n}u(\tilde{k})$, there exists $i$ such that
$\displaystyle\text{dim}(\sigma_{\theta_{i}})=\text{dim}(\langle
u_{s_{i}+1},\dots,u_{s_{i+1}}\rangle)=s_{i+1}-s_{i}>u(\tilde{k})$. Thus
$\sigma_{\theta_{i}}$ is isotropic over $\tilde{k}$. So $\sigma_{\theta_{i}}$
is isotropic over $k$. Thus $q$ is isotropic over $k$, making $u(k)\leq
2^{n}u(\tilde{k})$. ∎

This gives us a refinement of the results on $u$-invariant from Mehmeti’s
paper [3].

######  Theorem 3.5.

Let $k$ be an ultrametric field such that $\text{dim}_{\mathbb{Q}}(\sqrt{\mid
k^{*}\mid})=n$ is finite. Let $\tilde{k}$ be the residue field of $k$ and $l$
such that $l\neq\text{char}(k)$. Let $\mu_{l}\subset k$. Let $A\in\
_{l}\text{Br}(k)$. Then there exists $a_{1},a_{2},\dots,a_{n}$ and
$b_{1},b_{2},\dots,b_{n}$ in $k$; $A_{0}\in\ _{l}\text{Br}(k^{\circ})$ such
that

$\displaystyle A=A_{0}\otimes\prod_{i=1}^{n}(a_{i},b_{i})_{l}.$

###### Proof.

Let $A\in\ _{l}\text{Br}(k)$. Merkurjev-Suslin’s theorem implies that
$A=\displaystyle\prod_{i=1}^{m}(\alpha_{i},\beta_{i})_{l}$ for some
$m\in\mathbb{N}$. By Theorem 3.3, there exists $c_{1},c_{2},\dots,c_{n}$ in
$k^{*}$ such that $\alpha_{i}=u_{i}\prod c_{j}^{r_{ij}}$ and
$\beta_{i}=v_{i}\prod c_{j}^{s_{ij}}$, where $r_{ij}$ and $s_{ij}$ are
integers and $u_{i},v_{i}$ are units.

So,

$\displaystyle(\alpha_{i},\beta_{i})=(u_{i}\prod c_{j}^{r_{ij}},v_{i}\prod
c_{j}^{s_{ij}})=(u_{i},v_{i})\otimes\sum(u_{i},c_{j}^{s_{ij}})\otimes\sum(c_{j}^{r_{ij}},v_{i})\otimes\sum(\prod
c_{j}^{r_{ij}},c_{j}^{s_{ij}}).$

Let $\prod c_{j}^{r_{ij}}=d_{i}$. So we get that

$\displaystyle(\alpha_{i},\beta_{i})=(u_{i},v_{i})\otimes\sum(u_{i},c_{j}^{s_{ij}})\otimes\sum(c_{j}^{r_{ij}},v_{i})\otimes\sum(d_{i},c_{j}^{s_{ij}})$
$\displaystyle=(u_{i},v_{i})\otimes\sum(u_{i}^{s_{ij}},c_{j})\otimes\sum(v_{i}^{-r_{ij}},c_{j})\otimes\sum(d_{i}^{s_{ij}},c_{j}),$

the last equality coming from properties of symbols, i.e.,
$(\alpha,\beta^{j})=(\alpha^{j},\beta)$ and
$(\alpha,\beta)=-(\beta,\alpha)=(\beta^{-1},\alpha)$. Now combining the terms,
we get the following equality

$\displaystyle(\alpha_{i},\beta_{i})=(u_{i},v_{i})\otimes\displaystyle\sum_{j=1}^{n}(u_{i}^{s_{ij}}v_{i}^{-r_{ij}}d_{i}^{s_{ij}},c_{j}).$

Let $x_{j}=u_{i}^{s_{ij}}v_{i}^{-r_{ij}}d_{i}^{s_{ij}}$ in $k^{*}$. Thus, we
have
$A=\displaystyle\prod_{i=1}^{m}\big{(}(u_{i},v_{i})\otimes\sum_{j=1}^{n}(x_{j},c_{j})\big{)}$.

Let $\displaystyle A_{0}=\otimes(u_{i},v_{i})_{k^{\circ}}$. Since each cyclic
algebra $\displaystyle(u_{i},v_{i})_{k^{\circ}}$ is an Azumaya algebra over
$k^{\circ}$, $A_{0}\in\text{Br}(k^{\circ})$ and we thus have $\displaystyle
A=A_{0}\otimes\prod_{i=1}^{n}(a_{i},b_{i})$.

∎

Recall the following definition. Let $K$ be any field. For a prime $l$, define
the Brauer $l$-dimension of $K$, denoted by Brldim($K$), to be the smallest
integer $d\geq 0$ such that for every finite extension $L$ of $K$ and for
every central simple algebra $A$ over $L$ of period a power of $l$, ind($A$)
divides per($A$)d.

######  Theorem 3.6.

Let $k$ be a complete ultrametric field. Suppode
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $l$ be a prime such
that $l\neq\text{char}(\tilde{k})$ and $\mu_{l}\subset k$. Then the
$\text{Br}_{l}\text{dim}(k)\leq\text{Br}_{l}\text{dim}(\tilde{k})+n$.

###### Proof.

Consider the $l$-torsion of Br$(k)$, ${}_{l}\text{Br}(k)$. Then for a given
$A\in\ _{l}\text{Br}(k)$, by Theorem 3.5, there exists $A_{0}\in\
_{l}\text{Br}(k^{\circ})$ and $c_{1},c_{2},\dots,c_{n}\in k^{*}$ such that
$(A-A_{0})\otimes\
k(\displaystyle\sqrt[l]{c_{1}},\sqrt[l]{c_{2}},\dots,\sqrt[l]{c_{n}})$ is
split. Let
$L=k(\displaystyle\sqrt[l]{c_{1}},\sqrt[l]{c_{2}},\dots,\sqrt[l]{c_{n}})$, so
$(A-A_{0})\otimes L$ is split.

This implies that $\text{ind}(A)$ divides $\text{ind}(A_{0})l^{n}$. Since
$k^{\circ}$ is complete, it follows from Proposition 2.8 that
$\displaystyle\text{ind}(A_{0})\leq\text{ind}(A_{0}\otimes_{k^{\circ}}\tilde{k})$,
hence divides $\text{Br}_{l}\text{dim}(\tilde{k})$. So,
$\text{Br}_{l}\text{dim}(k)\leq\text{Br}_{l}\text{dim}(\tilde{k})+n$.

∎

######  Corollary 3.7.

Let $k$ be a complete ultrametric field. Suppose
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $l$ be a prime such
that $l\neq\text{char}(\tilde{k})$. Then the
$\text{Br}_{l}\text{dim}(k)\leq\text{Br}_{l}\text{dim}(\tilde{k})+n$.

###### Proof.

Let $A\in\ _{l}\text{Br}(k)$. Let $\rho_{l}$ be the $l$-th primitive root of
unity. So, the degree of the extension $k(\rho_{l})$ over $k$ is at most
$l-1$, i.e., $l\nmid[k(\rho_{l}):k]$. Thus, ind($A$) $=$ ind($A\otimes
k(\rho_{l})$) and per($A$) $=$ per($A\otimes k(\rho_{l})$), by [10],
Propositions 13.4(vi) and 14.4b(v). Hence the corollary follows from Theorem
3.6.

∎

##  4\. Function field of curves on complete ultrametric fields

Recall the discussion at the end of Section 2 regarding the rational rank of
abelian groups. In our situation, consider a complete ultrametric valued field
$k$. Let $L$ be a valued field extension of $k$. Let $|L^{*}|$, $|k^{*}|$ be
the value groups of $L$ and $k$, respectively. We thus have
$\text{dim}_{\mathbb{Q}}(\sqrt{|L^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+\text{dim}_{\mathbb{Q}}(\frac{|L^{*}|}{|k^{*}|}\otimes_{\mathbb{Z}}\mathbb{Q})$.
If $L$ is an algebraic extension of $k$, then the quotient group
$|L^{*}|/|k^{*}|$ is a torsion group. Thus
$\text{dim}_{\mathbb{Q}}(|L^{*}|/|k^{*}|\otimes_{\mathbb{Z}}\mathbb{Q})=0$,
which implies that the two divisorial closures $\sqrt{|L^{*}|}$ and
$\sqrt{|k^{*}|}$ have the same dimension over $\mathbb{Q}$.

######  Theorem 4.1.

Let $k$ be a complete ultrametric field with $\text{char}(\tilde{k})\neq 2$.
Suppose $\text{dim}_{\mathbb{Q}}(\sqrt{\mid k^{*}\mid})=n$ is finite. Let $C$
be a curve over $k$ and $F=k(C)$ the function field of the curve. Let $q$ be a
quadratic form over $F$ with dimension $d$. If $d>2^{n+1}u_{s}(\tilde{k})$,
then $q$ is isotropic. In particular, $u(F)\leq 2^{n+1}u_{s}(\tilde{k})$.

###### Proof.

Let $V(F)$ be the set of all non-trivial rank 1 valuations of $F$ which either
extend the valuation of $k$ or are trivial when restricted to $k$. Let $F_{v}$
denote the completion of $F$ with respect to $v$. We will first show that for
a quadratic form $q$ over $F$ with $\text{dim}(q)>2^{n+1}u_{s}(\tilde{k})$,
$q$ is isotropic over $F_{v}$ for all $v\in V(F)$. Then we will apply
Corollary 3.19 from [3] to say that $q$ is isotropic over $F$.

Since we want to say that $q$ is isotropic over $F_{v}$ for all $v\in V(F)$,
we have two cases. If $v\in V(F)$ is such that $v$ restricted to $k$ is the
trivial valuation, then $F_{v}$ is a discrete valued field. Let
$\widetilde{F_{v}}$ be the residue field of $F_{v}$. By Springer’s theorem on
nondyadic complete discrete valuation fields (see [5], VI.1.10 and XI.6.2(7)),
we have $u(F_{v})=2u(\widetilde{F_{v}})$. Since $\widetilde{F_{v}}$ is a
finite extension of $k$, $\widetilde{F_{v}}$ is a complete ultrametric field
with residue field $\widetilde{\widetilde{F_{v}}}$ a finite field extension of
$\tilde{k}$. Using the fact that $\widetilde{F_{v}}$ is an algebraic extension
of $k$, then the quotient group $|\widetilde{F_{v}}^{*}|/|k^{*}|$ is a torsion
group. Thus
$\text{dim}_{\mathbb{Q}}(|\widetilde{F_{v}}^{*}|/|k^{*}|\otimes_{\mathbb{Z}}\mathbb{Q})=0$,
which in turn implies that the two divisorial closures
$\sqrt{|\widetilde{F_{v}}^{*}|}$ and $\sqrt{|k^{*}|}$ have the same dimension
over $\mathbb{Q}$. It then follows from Theorem 3.4 that
$u(\widetilde{F_{v}})\leq 2^{n}u(\widetilde{\widetilde{F_{v}}})\leq
2^{n}u_{s}(\tilde{k})$. Thus, $u(F_{v})\leq 2^{n+1}u_{s}(\tilde{k})$.

In the case that $v$ restricted to $k$ is the valuation on $k$, then
$\widetilde{F_{v}}$ is an extension of $\tilde{k}$ of transcendence degree
$\leq 1$. Let $\displaystyle s=\text{dim}_{\mathbb{Q}}\Big{(}\frac{\mid
F_{v}^{*}\mid}{\mid k^{*}\mid}\otimes_{\mathbb{Z}}\mathbb{Q}\Big{)}$ and
$t=\text{tr deg}_{\tilde{k}}(\widetilde{F_{v}})$. By Abhyankar’s inequality
[13], we have $0\leq s+t\leq 1$.

Suppose $t=0$. Then $\widetilde{F_{v}}$ is a finite extension of $\tilde{k}$,
and hence $u(\widetilde{F_{v}})\leq u_{s}(\tilde{k})$. Since $t=0$ and $0\leq
s+t\leq 1$, we have $s\leq 1$. When $s=0$, the discussion before Theorem 4.1
tells us that,
$\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+\text{dim}_{\mathbb{Q}}(\frac{|F_{v}^{*}|}{|k^{*}|}\otimes_{\mathbb{Z}}\mathbb{Q})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+s$.
Thus
$\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$.
From Theorem 3.4, we thus have $u(F_{v})\leq 2^{n}u(\widetilde{F_{v}})\leq
2^{n}u_{s}(\tilde{k})$. When $s=1$, again the discussion before Theorem 4.1
implies that,
$\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+s$.
Thus $\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=n+1$. Applying Theorem 3.4
again, we have $u(F_{v})\leq 2^{n+1}u(\widetilde{F_{v}})\leq
2^{n+1}u_{s}(\tilde{k})$.

For the case when $t=1$ and $s=0$, the extension $\widetilde{F_{v}}$ over
$\tilde{k}$ is finitely generated of transcendence degree $1$. Thus,
$u(\widetilde{F_{v}})\leq 2u_{s}(\tilde{k})$. Once again applying Theorem 3.4,
we get that $u(F_{v})\leq 2^{n}u(\widetilde{F_{v}})\leq
2^{n+1}u_{s}(\tilde{k})$.

Now applying 2.3, we get that $q$ is isotropic over $F$. This proves the claim
in the theorem that $u(F)\leq 2^{n+1}u_{s}(\tilde{k})$. ∎

######  Corollary 4.2.

Let $k$ be a complete ultrametric field with $\text{char}(\tilde{k})\neq 2$.
Suppose $\text{dim}_{\mathbb{Q}}(\sqrt{\mid k^{*}\mid})=n$ is finite. Then
$u_{s}(k)\leq 2^{n}u_{s}(\tilde{k})$.

###### Proof.

Let $K/k$ be a finite field extension. Then $K$ is an ultrametric field and
$\tilde{K}/\tilde{k}$ is a finite extension. Let $q$ be a $K$-quadratic form
of dimension $d>2^{n}u_{s}(\tilde{k})$. Since $\text{char}(\tilde{k})\neq 2$,
we may assume $q$ to be diagonal. Further,
$\text{dim}_{\mathbb{Q}}(\sqrt{|K^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$
is finite. This equality follows from the discussion before Theorem 4.1, and
noting that since $K$ is a finite extension of $k$, the quotient group
$|K^{*}|/|k^{*}|$ is torsion. Then applying Theorem 3.4, we have $u(K)\leq
2^{n}u(\tilde{K})\leq 2^{n}u_{s}(\tilde{k})$.

Let $C$ be a curve over $k$ and $F=k(C)$ the function field of $C$. Then it
follows from Theorem 4.1 that $u(F)\leq 2^{n+1}u_{s}(\tilde{k})$. Then by the
definition of strict $u$-invariant, we have that $u_{s}(k)\leq
2^{n}u_{s}(\tilde{k})$. ∎

######  Theorem 4.3.

Let $k$ be a complete ultramteric field. Suppose
$\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$ is finite. Let $l$ be a prime such
that $l\neq\text{char}(\tilde{k})$. Let $C$ be a curve over $k$ and $F=k(C)$
the function field of the curve. Suppose there exist an integer $d$ such that
$\text{Br}_{l}\text{dim}(\tilde{k})\leq d$ and
$\text{Br}_{l}\text{dim}(\tilde{k}(T))\leq d+1$. Then the
$\text{Br}_{l}\text{dim}(F)\leq d+1+n$.

###### Proof.

Let $V(F)$ be the set of all non-trivial rank 1 valuations of $F$ which either
extend the valuation of $k$ or are trivial when restricted to $k$. Let $F_{v}$
denote the completion of $F$ with respect to $v$.

If $v\in V(F)$ is such that $v$ restricted to $k$ is the trivial valuation,
then $F_{v}$ is a discrete valued field. Let $\widetilde{F_{v}}$ be the
residue field of $F_{v}$. Since $\widetilde{F_{v}}$ is a finite extension of
$k$, it is a complete ultrametric field with residue field
$\widetilde{\widetilde{F_{v}}}$ a finite extension of $\tilde{k}$. By the
definition of $\text{Br}_{l}\text{dim}$, we have that
$\text{Br}_{l}\text{dim}(\widetilde{\widetilde{F_{v}}})\leq\text{Br}_{l}\text{dim}(\tilde{k})$.
Note that $F_{v}$ is a complete discretely valued field, thus
$\text{Br}_{l}\text{dim}(F_{v})\leq\text{Br}_{l}\text{dim}(\widetilde{F_{v}})+1$,
which follows from [1, Theorem 5.5]. Now utilising the fact that
$\widetilde{F_{v}}$ is an algebraic extension of $k$, the quotient group
$|\widetilde{F_{v}}^{*}|/|k^{*}|$ is a torsion group. Thus,
$\text{dim}_{\mathbb{Q}}(|\widetilde{F_{v}}^{*}|/|k^{*}|\otimes_{\mathbb{Z}}\mathbb{Q})=0$,
which in turn implies that the two divisorial closures
$\sqrt{|\widetilde{F_{v}}^{*}|}$ and $\sqrt{|k^{*}|}$ have the same dimension
over $\mathbb{Q}$. Applying Theorem 3.6 to $\widetilde{F_{v}}$, we have that
$\text{Br}_{l}\text{dim}(\widetilde{F_{v}})\leq\text{Br}_{l}\text{dim}(\widetilde{\widetilde{F_{v}}})+n$.
Putting it all together gives
$\text{Br}_{l}\text{dim}(F_{v})\leq\text{Br}_{l}\text{dim}(\widetilde{F_{v}})+1\leq\text{Br}_{l}\text{dim}(\widetilde{\widetilde{F_{v}}})+n+1\leq\text{Br}_{l}\text{dim}(\tilde{k})+n+1\leq
d+n+1$.

In the case that $v$ restricted to $k$ is the valuation on $k$, then
$\widetilde{F_{v}}$ is an extension of $\tilde{k}$ of transcendence degree
$\leq 1$. Let $\displaystyle s=\text{dim}_{\mathbb{Q}}\Big{(}\frac{\mid
F_{v}^{*}\mid}{\mid k^{*}\mid}\otimes_{\mathbb{Z}}\mathbb{Q}\Big{)}$ and
$t=\text{tr deg}_{\tilde{k}}(\widetilde{F_{v}})$. By Abhyankar’s inequality
[13], we have $0\leq s+t\leq 1$.

Suppose $t=0$. Then $\widetilde{F_{v}}$ is a finite extension if $\tilde{k}$.
Thus,
$\text{Br}_{l}\text{dim}(\widetilde{F_{v}})\leq\text{Br}_{l}\text{dim}(\tilde{k})$.
Since $t=0$ and $0\leq s+t\leq 1$, we have $s\leq 1$. When $s=0$, the
discussion at the beginning of Section 4 implies that
$\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+\text{dim}_{\mathbb{Q}}(\frac{|F_{v}^{*}|}{|k^{*}|}\otimes_{\mathbb{Z}}\mathbb{Q})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+s$.
Thus
$\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})=n$.
Applying Theorem 3.6 now gives that,
$\text{Br}_{l}\text{dim}(F_{v})\leq\text{Br}_{l}\text{dim}(\widetilde{F_{v}})+n\leq\text{Br}_{l}\text{dim}(\tilde{k})+n\leq
d+n$. When $s=1$, once again the discussion at the beginning of Section 4
tells us that,
$\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=\text{dim}_{\mathbb{Q}}(\sqrt{|k^{*}|})+s$.
Thus $\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})=n+1$. By Theorem 3.6,
$\text{Br}_{l}\text{dim}(F_{v})\leq\text{Br}_{l}\text{dim}(\widetilde{F_{v}})+\text{dim}_{\mathbb{Q}}(\sqrt{|F_{v}^{*}|})\leq\text{Br}_{l}\text{dim}(\tilde{k})+n+1\leq
d+n+1$.

For the case when $t=1$ and $s=0$, the extension $\widetilde{F_{v}}$ over
$\tilde{k}$ is finitely generated of transcendence degree $1$, i.e.,
$\widetilde{F_{v}}$ is a finite extension of $\tilde{k}(T)$. So,
$\text{Br}_{l}\text{dim}(\widetilde{F_{v}})\leq\text{Br}_{l}\text{dim}(\tilde{k}(T))$.
By Theorem 3.6,
$\text{Br}_{l}\text{dim}(F_{v})\leq\text{Br}_{l}\text{dim}(\widetilde{F_{v}})+n\leq\text{Br}_{l}\text{dim}(\tilde{k}(T))+n\leq
d+1+n$.

Thus for all $v\in V(F)$, we have $\text{Br}_{l}\text{dim}(F_{v})\leq d+n+1$.
From Theorem 2.4, we see that
$\text{Br}_{l}\text{dim}(F)\leq\text{Br}_{l}\text{dim}(F_{v})$, proving the
claim of the theorem.

∎

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# Probabilistic Ray-Tracing Aided Positioning at mmWave frequencies

Viet-Hoa Nguyen, Vincent Corlay, Nicolas Gresset, and Cristina Ciochina
Mitsubishi Electric R&D Centre Europe, Rennes, France
Email: {v.nguyen, v.corlay, n.gresset<EMAIL_ADDRESS>
###### Abstract

We consider the following positioning problem where several base stations (BS)
try to locate a user equipment (UE): The UE sends a positioning signal to
several BS. Each BS performs Angle of Arrival (AoA) measurements on the
received signal. These AoA measurements as well as a 3D model of the
environment are then used to locate the UE. We propose a method to exploit not
only the geometrical characteristics of the environment by a ray-tracing
simulation, but also the statistical characteristics of the measurements to
enhance the positioning accuracy.

###### Index Terms:

Positioning, ray tracing, Bayesian, Angle of Arrival, non line-of-sight,
mmWave.

##  I Introduction

Standard positioning systems rely on line-of-sight (LoS) measurements. As an
example, if several BS estimate the AoA of a reference signal transmitted by a
UE, the regular triangulation method can be implemented as follows. A server,
knowing the position of the BS and the measured AoA, looks for the
intersection of the rays leaving the BS in the measured AoA directions. The
estimated position of the UE is this intersection.

Nevertheless, the received signals at the BS may be non line-of-sight (NLoS).
It means that the paths followed by the signals have at least one reflection.
In this case, the measured AoA does not indicate the direction of the UE but
the direction of the last reflecting object. As a result, the above standard
triangulation method is not sufficient for the positioning operation. One
solution to address this issue is to use the knowledge of the environment to
improve the positioning operation.

One can for instance, in an offline phase, create a database of measurements
(at the BS) corresponding to each possible position of the UE. Then, in the
online phase, a measurement is performed and the most similar measurement in
the database is searched. Once found, the position corresponding to the
measurement is the estimated position. Similarly, neural network can be
trained with the database to infer the UE position based on measurements. This
class of methods is called fingerprinting. The database can be established via
real measurements or by using a digital twin of the environment combined with
ray tracing. This latter approach is for instance considered in [13][5]. It is
also currently studied at the 3rd Generation Partnership Project (3GPP) Radio
Access Network (RAN) 1, in the scope of release 18, within the study item
“AI/ML for positioning accuracy enhancement” [15].

Alternatively, instead of using a database, online ray tracing can be
considered: Given measured AoA, ray tracing in these AoA directions is
performed. The intersection of the rays is then the estimated position. This
approach, which can be called “reverse ray tracing”, is for instance presented
in [6]. A similar method with the notion of “virtual BS” is described in [11].
In this latter work, geometric equations rather than conventional ray tracing
are considered.

Of course, the measurements may not be perfectly accurate, and the measured
AoA may be corrupted by some noise. In [6], the authors propose to launch
several rays in an interval around the measured AoA. The considered width of
the interval is twice the estimated standard deviation of the estimated AoA
value. A least square solution, assuming i.i.d. Gaussian noise on the
estimated AoA, is used in [11].

To the best of our knowledge, the statistics of the measurements are not taken
into account in a Bayesian manner in conventional ray-tracing aided
positioning. Consequently, we propose to modify the conventional reverse ray-
tracing method to take into account the AoA statistics.

##  II Problem statement and statistical modelling

###  II-A Problem statement

Figure 1: Left: Standard triangulation approach via LoS-based AoA at the BS.
Right: Situation where some LoS AoA are not available for several BS. In the
example, two of the three LoS paths are blocked by clutters. In this case,
reverse ray tracing using a digital twin can be considered.

As mentioned in the introduction, this paper copes with the UE positioning
problem, based on the AoA measurements. The UE broadcasts a radio signal and
several BS, with fixed and known positions, receive the signal and measure the
AoA. In this considered AoA-based positioning framework, two following
situations occur:

  * •

LoS situation: If all UE-BS links are LoS, the UE lies on the line that
crosses the BS with an angle equal to the AoA. Therefore, the intersection of
all said lines is the UE position. The standard triangulation technique can be
used as illustrated on Figure 1 (left).

  * •

NLoS situation: If a UE-BS link is NLoS, the UE does not lie on the line that
crosses this BS with an angle equal to the AoA. An example is provided on
Figure 1 (right). In this case, the standard AoA positioning algorithm cannot
solve the positioning problem.

In this paper, we focus on indoor positioning scenarios where mostly NLoS
situations are encountered. Indeed, in a typical indoor environments, there
are many objects of different sizes, shapes, materials, etc. Therefore, the UE
is rarely in a LoS situation with the BS. Nevertheless, it is sometimes
possible to have a 3D map of the considered indoor environment. This motivates
the proposed approach.

###  II-B Statistical modelling

Let $X$ be a random variable representing the position of a UE. Let $\Theta$
be a random variable representing the true AoA of the signal, and $Y$ a random
variable representing the BS measurement(s) of the AoA.

The connection between $Y$ and $\Theta$ depends on the antenna hardware and
software used by the BS. The BS can have different software and hardware
configurations for the AoA measurement. In terms of software, different angle
estimation algorithms such as Delay-and-Sum [7], MUSIC [12], MVDR [8], ESPRIT
[4], induce different error profiles. On the other hand, a distinct hardware
configuration, such as the antenna type, the number of antennas, the RF
filter, also imposes a particular accuracy of AoA measurement. Hence, each BS
may have a specific error model.

For instance, with the standard Gaussian model we have:

$\displaystyle Y=\ \Theta+W,$ (1)

where $W\sim\mathcal{N}\left(0,\sigma^{2}\right)$. The value of $\sigma^{2}$
may for example depend on the antenna quality. Moreover, if we assume that
$\Theta$ is equiprobable, then $p\left(\Theta=\theta\middle|Y=y\right)\propto
p\left(Y=y\middle|\Theta=\theta\right)\sim\mathcal{N}\left(\theta,\sigma^{2}\right)$,
where $\propto$ means “proportional to”. Note that $\Theta$ may not always be
equiprobable, especially if the BS are located in corners.

The possible values $\theta$ for $\Theta$ is denoted by $\mathcal{C}$. For
instance, $\mathcal{C}=[0,2\pi[$ in the 2D case or $[0,\ 2\pi[\times[0,\ \pi[$
in the 3D case. Note that the interval can be discretized if needed.

Regarding the notations, we use $p\left(\theta\middle|y\right)$ for
$p\left(\Theta=\theta|Y=y\right)$ and similarly $p\left(x\middle|y\right)$ for
$p\left(X=x|Y=y\right)$.

We let $n$ be the number of measurements such that
$\textbf{y}=[y_{1},y_{2},\ldots,y_{i},\ldots,y_{n}]$. The vector y comprises
the measurement performed by all BS. For the sake of simplicity, we assume
that there is one measurement per BS, and therefore $n$ BS.

In the considered problem, the goal is to compute

$\displaystyle p\left(x|\textbf{y}\right).$ (2)

The estimated position $\hat{x}$ is then computed from these estimated
probabilities. It can be either estimated by finding the arg max:

$\displaystyle\hat{x}=\text{arg\ max}_{x}\ p\left(x|\textbf{y}\right),$ (3)

or the mean

$\displaystyle\hat{x}=\int_{x}x\cdot p\left(x|\textbf{y}\right)dx.$ (4)

##  III Proposed Bayesian approach with ray tracing

###  III-A Standard Bayesian equation

In this section, we derive the Bayesian equation to take into account the AoA
statistics (i.e., the $p(\theta|y_{i})$) in the estimation of
$p(x|\textbf{y})$. We let
$\boldsymbol{\theta}=[\theta_{1},\theta_{2},\ldots,\theta_{i},\ldots,\theta_{n}]$.
This simply consists in marginalizing with respect to $\boldsymbol{\theta}$.

Coming back to (2), with the law of total probability we have

$\displaystyle\begin{split}p\left(x\middle|\textbf{y}\right)&=\int_{\boldsymbol{\theta}\in\mathcal{C}^{n}}p\left(x\middle|\boldsymbol{\theta},\textbf{y}\right)p\left(\boldsymbol{\theta}\middle|\textbf{y}\right)d\boldsymbol{\theta},\\\
&=\int_{\boldsymbol{\theta}\in\mathcal{C}^{n}}p\left(x\middle|\boldsymbol{\theta}\right)p\left(\boldsymbol{\theta}\middle|\textbf{y}\right)d\boldsymbol{\theta}.\end{split}$
(5)

As the AoA measurement errors of the BS are independent, the probability
$p(\boldsymbol{\theta}|\mathbf{y})$ is the product of the
$p(\theta_{i}|y_{i})$:

$\displaystyle\int_{\boldsymbol{\theta}\in\mathcal{C}^{n}}p\left(x\middle|\boldsymbol{\theta}\right)p\left(\boldsymbol{\theta}\middle|\textbf{y}\right)d\boldsymbol{\theta}=\int_{\boldsymbol{\theta}\in\mathcal{C}^{n}}p\left(x\middle|\boldsymbol{\theta}\right)\prod_{i=1}^{n}{p\left(\theta_{i}\middle|y_{i}\right)}d\boldsymbol{\theta}.$
(6)

###  III-B Ray tracing

The aim of this section is to explain the above equation in a NLoS situation
using ray tracing. Indeed, in the LoS situation the term
$p\left(x\middle|\boldsymbol{\theta}\right)$ is trivial to compute via the
standard triangulation method.

We first introduce the notion of ray tracing.

####  III-B1 Reverse ray tracing

Let us formally introduce the notion of ray. In this paper, a ray is defined
as a path, characterized by a 3D scene, starting from a given location and
with an angle of departure (AoD)111The AoA of the received signal becomes the
AoD of the ray in reverse ray tracing. $\theta$. We use the notation
path($\theta$) to refer to a ray.

If only NLoS AoA measurements are available, a digital twin is required to
find the paths of the rays and thus their intersections. This approach is
called reverse ray tracing and is illustrated on Figure 1 (right). The term
“reverse” is used because the rays are launched from the receiver (the BS who
received the UE signal). This approach holds as the considered channel is
reciprocal, as justified by the following paragraph.

Regarding the propagation of the rays, we make the following geometric
assumption. Due to smaller wavelength, the diffraction phenomenon is less
important at millimeter-wave (mmWave) frequencies compared to centimeter-wave
frequencies [2]. Consequently, many channel models consider only specular
reflections at the former higher frequencies [10]. As a result, unlike sub-6
GHz signals, the propagation of mmWave in specific environments can be studied
by ray-tracing simulations, as done e.g., in [3][9][10].

####  III-B2 Interpretation of Equation (6)

First, the term $p\left(\theta_{i}\middle|y_{i}\right)$ corresponds to the
statistics of the measurement error of the AoA. It also represents the
distribution of the rays as it enables to compute the probability of a given
ray path($\theta_{i}$).

Second, the term $p\left(x\middle|\boldsymbol{\theta}\right)$ is given by the
ray-tracing model. Assuming a perfect knowledge of the scene and an unlimited
number of rays, $p(x|\boldsymbol{\theta})$ becomes an indicator function:

$\displaystyle\begin{split}&p\left(x\middle|\boldsymbol{\theta}\right)=\\\
&1\\{\text{If the }n\text{ rays defined by
}\boldsymbol{\theta}=[\theta_{1},\theta_{2},...,\theta_{n}]\text{ cross at
}x\\}.\end{split}$ (7)

$\displaystyle\begin{split}p\left(x\middle|\boldsymbol{\theta}\right)=0\\{\text{If
the rays do not cross at x}\\}.\end{split}$

However, the model for $p\left(x\middle|\boldsymbol{\theta}\right)$ may not be
perfect and the number of rays may be finite. As a result, the rays may never
cross at one exact location. In this case, a simple solution is to consider
squares (or cubes in the 3D case) $\mathcal{X}$ and to modify (6) and
(LABEL:eq_proba_1) into (8) and (9), respectively, as follows. Let
$\mathcal{C}_{i}^{\prime}$ be the set of AoD chosen to launch the rays from
the $i$-th BS. We get:

$\displaystyle
p\left(\mathcal{X}\middle|\boldsymbol{y}\right)\approx\sum_{\boldsymbol{\theta}\in\mathcal{C}_{1}^{\prime}\times\mathcal{C}_{2}^{\prime}\times...}{p\left(\mathcal{X}\middle|\boldsymbol{\theta}\right)\prod_{i=1}^{n}{p\left(\theta_{i}\middle|y_{i}\right)}},$
(8)

where:

$\displaystyle\begin{split}&p\left(\mathcal{X}\middle|\boldsymbol{\theta}\right)=\\\
&1\\{\text{If all }n\text{ rays defined by
}\boldsymbol{\theta}=[\theta_{1},\theta_{2},\ ...,\theta_{n}]\text{ go through
}\mathcal{X}\\}\end{split}$ (9)
$\displaystyle\begin{split}p\left(\mathcal{X}\middle|\boldsymbol{\theta}\right)=0\\{\text{If
not all rays go through }\mathcal{X}\\}.\end{split}$

Note that the digital twin could also be used to estimate the distribution of
$\Theta$. However, for the sake of simplicity, we assume that it is
equiprobable in our simulations.

##  IV Proposed algorithms

A first natural approach to compute (8) via ray tracing is as follows. First,
we let $\mathcal{C}^{\prime}_{i}$, for all $i$, be a uniformly discretized
version of the possible AoA $\mathcal{C}$. Then, for each BS a ray is launched
in each resulting direction. Finally, the probabilities are computed via (8),
where each ray is weighted by the probability $p(\theta_{i}|y_{i})$. A pseudo
code of this approach is provided in Appendix VII.

An alternative manner to compute (8) is via Monte Carlo sampling. The main
idea is to sample the angles from the distribution
$p\left(\theta_{i}\middle|y_{i}\right)$ and launch the rays accordingly.
Hence, $\mathcal{C}_{i}^{\prime}$ is the set of sampled angles at the $i$-th
BS. Then, (8) becomes:

$\displaystyle
p\left(\mathcal{X}\middle|\boldsymbol{y}\right)\approx\sum_{\boldsymbol{\theta}\in\mathcal{C}_{1}^{\prime}\times\mathcal{C}_{2}^{\prime}\times...}{p\left(\mathcal{X}\middle|\boldsymbol{\theta}\right)}.$
(10)

In other words, once the rays are launched, we simply count the number of
times a set of rays crosses the square $\mathcal{X}$ of interest. The pseudo
code for this approach is provided in Algorithm 1.

In the case of a limited number of rays, the Monte Carlo approach is more
adapted than the uniform approach (Algorithm 3): More samples (i.e., rays) are
available at the likely AoA with the former approach than with the latter one.
Consequently, one is more likely to have rays crossing in the squares
corresponding to high position probabilities. This approach is chosen for the
simulations.

Algorithm 1 Monte Carlo method to estimate the position via reverse ray
tracing

Inputs: AoA measurements y, AoA statistics
$p(\boldsymbol{\theta}|\textbf{y})$, and 3D model of environment (digital
twin).

1: For each of the $n$ BS, sample $l$ angles according to
$p\left(\theta|y_{i}\right)$.

2: Launch the $n\ \times l$ rays in the sampled angle directions.

3: for each distinct $\mathcal{X}_{k}$ in the scene do

4: Collect all the rays crossing $\mathcal{X}_{k}$.

5: Find $m$, the number possible combinations of $n$ rays (where each of the
$n$ rays comes from a different BS). Set $\beta_{k}=m$ to obtain (8).

6: end for

7: The square with the estimated highest probability is $\mathcal{X}_{k}$
where ${k=\text{arg \ max}_{k}\beta}_{k}$.

##  V Simulation results

###  V-A Simulation environment

We use our own Matlab-implemented ray-launching software for the simulation.
The considered scene is shown in Figure 2. The dimensions are the following:
width 8m, length 18m, and height 2.5m. The scene was drawn via the software
Fusion 360. This environment is inspired from the recommendations of the 5G
Alliance for Connected Industries and Automation (5G-ACIA) for indoor
industrial scenario [14]. It includes both an open area and alleys.

Figure 2: Considered scene for the simulations.

Figure 3 shows an example of rays launched from a given position. Only the
rays crossing both the UE (cyan circle) and at least one BS (yellow squares)
are shown. The maximum number of bounces is set to 5. This value is high
enough as the radio signal loses a lot of energy after each bounce at mmWave
frequencies. The 4 BS, located in the top corners of the scene, are shown by
the yellow squares. We see that one BS receives no ray.

Figure 3: Example of ray tracing between the BS and the UE in the considered
scene. The positions of the BS are shown by the yellow square and the one of
the UE by the blue ball.

###  V-B Noiseless AoA and error model

Since we do not have a real deployment of the scene, the noiseless AoA
resulting from the transmission between the UE and the BS must also be
established in the digital twin. The AoA for a given position is established
as follows: Many rays are launched from the given position. Only a subset of
these rays crosses a given BS. We then perform binning of the AoA of the
crossing rays. Finally, the angle corresponding to the center of the bin
having the highest number of crossing rays is chosen as the AoA for the
considered position. This process has to be performed in an offline manner for
every considered position.

Regarding the error model, we consider a simple Gaussian noise. Let
$\theta_{i}^{k}$ be the established AoA for a $i$-th BS and a UE position
$x_{j}$. The measurement $y^{j}_{i}$ is obtained as in (1):
$y^{j}_{i}=\theta^{j}_{i}+w$ where $w$ is sampled from a centered Gaussian
distribution with variance $\sigma^{2}$. Hence, the distribution
$p(\theta_{i}|y_{i})$ is also Gaussian.

Algorithm 2 Benchmark reverse ray tracing algorithm

Inputs: AoA measurements y, Noise variance $\sigma^{2}$, and 3D model of
environment (digital twin).

1: For each of the $n$ BS, generate $l$ angles in the range
$[-\sigma,+\sigma]$ discretized uniformly.

2: Launch the $n\ \times l$ rays in the according to the generated angles.

3: for each distinct $\mathcal{X}_{k}$ in the scene do

4: Collect all the rays crossing $\mathcal{X}_{k}$.

5: Find $m$, the number possible combinations of $n$ rays (where each of the
$n$ rays comes from a different BS). Set $\beta_{k}=m$.

6: end for

7: The square with the estimated highest probability is $\mathcal{X}_{k}$
where ${k=\text{argmax}_{k}\beta}_{k}$.

###  V-C Benchmark algorithm

We compare our algorithm with a benchmark algorithm similar to the one
proposed in [6] (mentioned in the introduction). With this benchmark
algorithm, the rays are launched uniformly in a cone whose width depends on
the variance of the error but where the rays are not weighted by
probabilities. It is summarized in Algorithm 2.

Note that once the rays are launched, the counting is done as in Algorithm 1.
Hence, the only difference between these two algorithms lies in the way the
rays are launched. The complexity of both methods is therefore the same.

###  V-D Simulation parameters

We consider only the 2D $(x,y)$ location problem. The height of the UE to
locate is assumed fixed (and known) at 1m above the floor. Several random UE
positions are considered.

The size of the squares $\mathcal{X}$ should be neither too large nor too
small: In the former case, the positioning accuracy would be reduced as we
have no indication of the position within a square. In the latter case, if the
number of rays is not large enough, we may have no square where at least $n$
rays cross, as required by (9) to have a non-zero probability. As a result,
there is a trade-off to find. Of course, the larger the number of rays, the
smaller the size of square and the better the performance, but also the higher
the complexity. In our simulations, we choose squares of length equal to 10cm.

Finally, we recall that Algorithm 1 is considered for these simulations.

###  V-E Results

Figure 4: CDF of the positioning error with different number of rays both with
the proposed method and the benchmark method. The standard deviation of the
AoA measurement error is $\sigma=0.5$ degree.

We show the cumulative density function (CDF) of the positioning error. It is
obtained as follows: For one position, we sample several AoA errors according
to (1). For each sample, we compute the positioning error with both the
proposed method and the benchmark method. We repeat the process for several
positions. Moreover, we use the $Q\left(0.9\right)$ value (such that $90\%$ of
the errors are under $Q\left(0.9\right)$) to compare the positioning
performance of the algorithms. The positioning error is denoted by
$\mathcal{L}=||\hat{x}_{k}-x_{k}||$, where $x_{k}$ is the true position and
$\hat{x}_{k}$ (center of $\mathcal{X}_{k}$) the estimated position.

We investigate the impact of the number of rays as well as the impact of the
standard deviation of the measurement error $\sigma$.

The results with different number of rays are provided on Figure 4. The
standard deviation of the angle error is $\sigma=0.5$ degree. First, we see
with the solid lines that the number of rays has a significant impact on the
positioning error. The $Q(0.9)$ value decreases from 4m down to 50cm when the
number of rays launched per BS increases from 500 to 1700. Note that the
positioning performance with the low number of rays could probably be improved
by increasing the size of the squares, as discussed in the previous section.
We let this optimization for a future work. Second, we observe that the
proposed method offers significant improvement compared to the benchmark
method. The $Q(0.9)$ values decreases from 5m down to 50cm with 1700 rays.

The accuracy achieved by the proposed methods with $\sigma=0.5$ is similar
(even slightly better) to the fingerprint methods based on the channel impulse
response (CIR) and neural networks. In [15][1], $Q(0.9)$ values around 1m are
reported when neural networks are trained on dataset with inter-position
spacing of 0.3m. Note that this latter performance is achieved assuming
perfect estimation of the CIR while we consider measurement errors.

The results with different standard deviation of the measurement error
$\sigma$ are provided on Figure 5. Unsurprisingly, the performance of both
techniques decreases with a higher $\sigma$. Nevertheless, we observe that the
proposed probabilistic method maintains a clear advantage over the benchmark
method, especially for moderate $Q$ values. Thus, the method is relevant for
several noise levels.

Figure 5: CDF of the positioning error with different values of standard
deviation $\sigma$ of the measurement error both with the proposed method and
the benchmark method. The number of rays per BS is 1700.

##  VI Conclusions

In this paper, we addressed the positioning problem based on AoA measurements
in a NLoS environment. A reverse ray-tracing solution, using a digital twin of
the scene, is considered. We proposed to take into account the statistics of
the AoA measurement errors in a Bayesian manner to improve the reverse ray-
tracing algorithm. Simulation results show great improvements with the
proposed method compared to the benchmark reverse ray-tracing method, where
the statistics of the measurements are not taken into account in a Bayesian
manner.

##  VII Appendix

Algorithm 3 Uniform method to estimate the position via reverse ray tracing

Inputs: AoA measurements y, AoA statistics
$p(\boldsymbol{\theta}|\textbf{y})$, and 3D model of environment (digital
twin).

1: For each of the $n$ BS, discretize uniformly $\mathcal{C}$ to have $l$
angles.

2: Launch the $n\ \times l$ rays in the discretized angle directions.

3: for each distinct $\mathcal{X}_{k}$ in the scene do

4: Collect all the rays crossing $\mathcal{X}_{k}$.

5: For all $m$ possible combinations of $n$ rays (where each of the $n$ rays
comes from a different BS), compute the corresponding
$\alpha_{j}=\prod_{i=1}^{n}{p\left(\theta_{i}\middle|y_{i}\right)}$.

6: Compute $\beta_{k}=\sum_{j=1}^{m}\alpha_{j}$ to obtain (8).

7: end for

8: The square with the highest probability is $\mathcal{X}_{k}$ where
${k=\text{arg \ max}_{k}\beta}_{k}$.

## References

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