Author to whom correspondence should be addressed<EMAIL_ADDRESS>
# Generalized Boozer coordinates: a natural coordinate system for
quasisymmetry

E. Rodríguez  [  Department of Astrophysical Sciences, Princeton University,
Princeton, NJ, 08543  Princeton Plasma Physics Laboratory, Princeton, NJ,
08540  W. Sengupta  Department of Astrophysical Sciences, Princeton
University, Princeton, NJ, 08543  Princeton Plasma Physics Laboratory,
Princeton, NJ, 08540  A. Bhattacharjee  Department of Astrophysical Sciences,
Princeton University, Princeton, NJ, 08543  Princeton Plasma Physics
Laboratory, Princeton, NJ, 08540

###### Abstract

We prove the existence of a straight-field-line coordinate system we call
generalized Boozer coordinates. This coordinate system exists for magnetic
fields with nested toroidal flux surfaces provided
$\oint\mathrm{d}l/B\>(\mathbf{j}\cdot\nabla\psi)=0$, where symbols have their
usual meaning, and the integral is taken along closed magnetic field lines.
All quasisymmetric fields, regardless of their associated form of equilibria,
must satisfy this condition. This coordinate system presents itself as a
convenient form in which to describe general quasisymmetric configurations and
their properties. Insight can be gained analytically into the difference
between strong and weak forms of quasisymmetry, as well as axisymmetry, and
the interaction of quasisymmetry with different forms of equilibria.

††preprint: AIP/123-QED

##  I Introduction:

Quasisymmetric stellaratorsNührenberg and Zille (1988); Boozer (1981);
Rodríguez, Helander, and Bhattacharjee (2020) (sometimes referred to as
helically-symmetricTessarotto, Johnson, and Zheng (1995); Tessarotto  _et al._
(1996)) constitute an attractive choice for magnetic confinement fusion.
Theoretically, such designs exhibit transport properties analogous to those of
axisymmetric devicesBoozer (1983) while possessing larger three-dimensional
freedom. The latter allows avoiding some of the inherent limitations of
tokamaks. The quasisymmetric stellarator achieves this by possessing a hidden
symmetry, namely, the magnitude of the magnetic field $|\mathbf{B}|$ is
symmetric, while the full vector $\mathbf{B}$ is not.

The concept of quasisymmetry (QS) is elegant, but it appears to have a
significant theoretical limitation. It was soon realizedGarren and Boozer
(1991a) that constructing a configuration with exact QS is not possible.
Although there is no definitive proof that shows this impossibility, work on
near-axis expansionsGarren and Boozer (1991a); Landreman and Sengupta (2018)
supports this point of view. The governing system of equations is
overdetermined: there are more constraints than degrees of freedom. This
limitation does not, however, prevent designs that exhibit behavior close to
QS in a volume.Bader  _et al._ (2019); Henneberg  _et al._ (2019)

Recently, studies that explore the concept of QS more deeply have
appeared.Rodríguez, Helander, and Bhattacharjee (2020); Burby, Kallinikos, and
MacKay (2021) The main idea has been to separate the concept of QS from that
of macroscopic force balance. In these studies, QS is defined as a property of
the configuration that confers the single-particle dynamics an approximately
conserved quantity without making any statement about the form of macroscopic
equilibrium. This perspective differs significantly from the standard
approach. Prior to [Rodríguez, Helander, and Bhattacharjee, 2020] and [Burby,
Kallinikos, and MacKay, 2020] (and with very few exceptionsTessarotto  _et
al._ (1996); Burby and Qin (2013)), the concept of QS was framed in the
context of magnetohydrostatic (MHS) equilibria satisfying
$\mathbf{j}\times\mathbf{B}=\nabla p$, where
$\mathbf{j}{{}=\nabla\times\mathbf{B}}$ is the current density and $p$ is the
scalar pressure. As MHS is not intrinsic to QS, it is important to define QS
without reference to a particular form of equilibrium. Separating QS from
equilibrium allows us to investigate more deeply its meaning and limitations.

Abandoning the convenient form of MHS equilibrium, although conceptually
appropriate, comes at a cost. The magnetic field no longer needs to satisfy
the condition $\mathbf{j}\cdot\nabla p=0$ , implicitly assumed in most of the
literatureNührenberg and Zille (1988); Boozer (1983); Tessarotto, Johnson, and
Zheng (1995). Hence, one cannot guarantee the existence of Boozer
coordinatesBoozer (1981) as presently understood, even if magnetic flux
surfaces exist. Boozer coordinates are particularly convenient for studying
QS, as it presents the symmetry in an explicit, simple form. This leads to a
search for an analogous, convenient, but more general coordinate system for
quasisymmetric configurations.

In this paper, we construct such a coordinate system, which we call
generalized Boozer coordinates (GBC). This coordinate system was used to
formulate the near-axis expansion in [Rodríguez and Bhattacharjee, 2021a], in
which a QS equilibrium with anisotropic pressure was shown to avoid the
conventional problem of overdetermination. The present paper is organized as
follows. First, we introduce, develop and discuss this coordinate system
systematically and rigorously. We start by presenting a constructive proof for
GBC and the class of fields for which it exists. We then present the set of
equations describing a quasisymmetric magnetic field in this coordinate
system. This gives us the opportunity to gain an alternativeBurby, Kallinikos,
and MacKay (2020); Rodríguez, Helander, and Bhattacharjee (2020) perspective
on the distinction between the so-called weak and strong forms of
quasisymmetry, as well as a comparison to axisymmetry. We close by summarizing
the equations that link the equilibrium and the quasisymmetric field and
concluding remarks.

##  II Generalized Boozer coordinates

###  II.1 Explicitly symmetric formulation of quasisymmetry

Let us start this section by introducing the notion of QS. We consider QS from
the recent general perspective of single particle motion.Rodríguez, Helander,
and Bhattacharjee (2020); Burby, Kallinikos, and MacKay (2021) QS (and in
particular weak QS) is the property of the fields that confers the motion of
guiding-centre single particles with an approximately conserved
quantity.Rodríguez, Helander, and Bhattacharjee (2020) For the dynamics of
particles to exhibit this conservation it is necessary for the magnetic field
to have nested flux surfaces $\mathbf{B}\cdot\nabla\psi=0$ and satisfy,

$\nabla\psi\times\nabla B\cdot\nabla\left(\mathbf{B}\cdot\nabla B\right)=0.$
(1)

This form of quasisymmetry is most commonly referred to as the triple vector
product formulation. Although this is not the form that comes directly from
the single-particle analysis (that form is the one used in Sec. IIIA), it is a
succinct way to impose QS on a magnetic field.

Given that this generalized concept of QS has a single particle origin, no
notion of macroscopic equilibrium is involved. Of course, from a practical
point of view, any steady field of interest will be in some form of force
balance. With the definition of QS in (1), different forms of equilibria may
be investigated to understand how they interact with QS. One may generally
refer to the macroscopic forces by $\mathbf{F}$, and we do not attempt to
assess their origin in this paper. Instead, we focus on the requirements at a
fluid level. A complete view of the problem would require an investigation of
the kinetic basis of the forces to link the fluid forces to microscopic
behavior. This is particularly important as microscopic forces could break the
symmetry (and with it, the QS-related conserved momenta). An important example
is the electrostatic potential, whose symmetry is needed to the appropriate
order and imposes constraints even on the forces that arise from the plasma,
such as centrifugal force or anisotropic pressure. With this in mind, we shall
focus on the macroscopic aspects.

Looking at the statement of QS in the form presented in Eq. (1), the existence
of a symmetry in the problem is not apparent. In the usual context of
$\mathbf{j}\times\mathbf{B}=\nabla p$, this absence of obvious symmetry can be
amended by using Boozer coordinates. In these coordinates, a field in
magnetostatic equilibrium with well-defined flux surfaces is quasisymmetric
(aside from quasipoloidal symmetry) iff,

$B=B(\psi,\theta-\tilde{\alpha}\phi),$ (2)

where $\\{\psi,\theta,\phi\\}$ are Boozer coordinates, $\psi$ the flux surface
label, $\theta$ and $\phi$ poloidal and toroidal angles respectively, and
$\tilde{\alpha}=N/M|M,N\in\mathbb{Z}$. In order to employ Boozer coordinates,
the $\mathbf{j}\cdot\nabla\psi=0$ property of MHS equilibria is central.
Boozer coordinates have the standard straight field-line coordinate Jacobian

$\mathcal{J}=\frac{B_{\phi}+\iota B_{\theta}}{B^{2}},$ (3)

where the covariant $B_{\theta}$ and $B_{\phi}$ are flux functions,

$\mathbf{B}=B_{\theta}(\psi)\nabla\theta+B_{\phi}(\psi)\nabla\phi+B_{\psi}\nabla\psi.$

Boozer coordinates are widely used in stellarator theory and applications. The
coordinates are a natural set that simplify many of the governing equations,
including QS. In particular, construction of solutions by near-axis expansion
of three-dimensional fields is most convenient in these coordinatesGarren and
Boozer (1991b, a); Landreman and Sengupta (2018); Rodríguez and Bhattacharjee
(2021a, b).

In the context of our general quasisymmetric $\mathbf{B}$, it is not necessary
to assume that $\mathbf{j}\cdot\nabla\psi=0$. However, we demonstrate that any
quasisymmetric field must satisfy
$\oint(\mathrm{d}l/B)(\mathbf{j}\cdot\nabla\psi)=0$ (see Appendix A). The
question that naturally arises is, can we construct an appropriate straight
field line coordinate system that explicitly expresses the symmetry in QS in
the Boozer fashion?

The answer is yes. To see this, we write Eq. (1) in straight-field line
coordinates $\\{\psi,\theta,\phi\\}$ using the contravariant representation
$\mathbf{B}=\nabla\psi\times(\nabla\theta-\iota\nabla\phi)$,

$(\nabla\psi\times\nabla\theta\>\partial_{\theta}B+\nabla\psi\times\nabla\phi\>\partial_{\phi}B)\cdot\\\
\cdot\nabla(\mathcal{J}^{-1}(\partial_{\phi}B+\iota\partial_{\theta}B))=0,$

where $\mathcal{J}$ is the Jacobian associated to the coordinate system
chosen. Now assume we can construct a straight-field line coordinate system
with a Jacobian of the form $\mathcal{J}=\mathcal{J}(\psi,B)$, then the
Jacobian factor may be dropped from the equation above to obtain,

$\displaystyle(\nabla\psi\times\nabla\theta\>\partial_{\theta}B+\nabla\psi\times\nabla\phi\>\partial_{\phi}B)\cdot\nabla(\partial_{\phi}B+\iota\partial_{\theta}B)=0$
$\displaystyle\text{i.e.}\quad\left[\partial_{\phi}-\left(\frac{\partial_{\phi}B}{\partial_{\theta}B}\right)\partial_{\theta}\right](\partial_{\phi}+\iota\partial_{\theta})B=0$

assuming that $\partial_{\theta}B\neq 0$. (This makes quasi-poloidally
symmetric solutions a special case.) From near-axis expansion we know that
these solutions have a very restricted QS regionRodríguez, Helander, and
Bhattacharjee (2020).

Commuting operators, we obtain

$(\partial_{\phi}+\iota\partial_{\theta})\left(\frac{\partial_{\phi}B}{\partial_{\theta}B}\right)\partial_{\theta}B=(\mathbf{b}\cdot\nabla)\left(\frac{\partial_{\phi}B}{\partial_{\theta}B}\right)\partial_{\theta}B=0$

which implies that $B=B(\psi,\theta-\tilde{\alpha}\phi)$ or $B=B(\psi,\phi)$,
where $\tilde{\alpha}=-\partial_{\phi}B/\partial_{\theta}B$ is a flux
function. The additional requirement that $\tilde{\alpha}$ is rational to
avoid $B=B(\psi)$ at non-degenerate surfaces requires $\tilde{\alpha}$ to be
constant.Rodríguez, Helander, and Bhattacharjee (2020)

In summary, if we are able to construct a straight field line coordinate
system that has a Jacobian that can be written as a function of $\psi$ and $B$
only, then a field is QS in the weak sense if and only if $B$ depends on a
single linear combination of those coordinate angles. Note that under this
assumption, the reverse direction of the proof is straightforward.

###  II.2 Constructing generalized Boozer coordinates

We define generalized Boozer coordinates (GBC) as a set of straight field line
coordinates whose Jacobian can be written in the form
$\mathcal{J}=B_{\alpha}(\psi)/B^{2}$, where $B_{\alpha}$ is some flux function
without requiring $\mathbf{j}\cdot\nabla\psi$ to vanish identically. This
choice is more general than what is permitted by Boozer coordinates, which
separately requires $B_{\theta}$ and $B_{\phi}$ to be flux functions. We shall
now constructively explore the conditions under which such a coordinate system
exists.

Let us start with a given magnetic field, assuming that it lies on well-
defined flux surfaces. It then followsKruskal and Kulsrud (1958); Boozer
(1981); Helander (2014) that there exists some initial straight field line
coordinate system $\\{\psi,\theta,\phi\\}$, so that
$\mathbf{B}=\nabla\psi\times\nabla\theta+\iota\nabla\phi\times\nabla\psi$. The
definition of the angular straight field line coordinates is arbitrary up to a
transformation of the form

$\displaystyle\theta=\theta^{\prime}+\iota\omega,$
$\displaystyle\phi=\phi^{\prime}+\omega.$

The function $\omega=\omega(\psi,\theta,\phi)$ is a well behaved periodic
function that defines a family of transformationsHelander (2014). The
periodicity of $\omega$ preserves the toroidal and poloidal character of the
two angular coordinates.

Starting with some given straight-field line coordinate system, we want to
understand how to construct a set with a Jacobian of the GBC form. Thus, we
need to know how to transform the coordinate Jacobian induced by $\omega$.
This reads

$\mathcal{J}^{\prime-1}=\nabla\psi\times\nabla(\theta-\iota\omega)\cdot\nabla(\phi-\omega).$

Here $\\{\psi,\theta^{\prime},\phi^{\prime}\\}$ represents the newly defined
straight field line coordinates whose associated Jacobian is
$\mathcal{J}^{\prime}$. This equation may be recast into the form of a
magnetic differential equation,

$\mathbf{B}\cdot\nabla\omega=\frac{1}{\mathcal{J}}-\frac{1}{\mathcal{J}^{\prime}}.$
(4)

Now, let us require111A more general form
$\mathcal{J}^{\prime}=\mathcal{J}^{\prime}(\psi,B)$ could have been demanded.
However, one may show in that case that the system may always be cast into the
form here employed.

$\mathcal{J}^{\prime}=\frac{B_{\alpha}(\psi)}{B^{2}}.$ (5)

In order for the magnetic differential equation Eq. (4) to have a single-
valued solution for the transformation function $\omega$, Eq. (4) must satisfy
Newcomb’s criterionNewcomb (1959). According to Newcomb, for a magnetic
differential equation $\mathbf{B}\cdot\nabla f=s$ to have a single-valued
solution for $f$, the source term $s$ must satisfy the line-integral condition
along a field line,

$\oint s\frac{\mathrm{d}l}{B}=0.$ (6)

For our problem $s=1/\mathcal{J}-1/\mathcal{J}^{\prime}$. Start with,

$\oint\frac{1}{\mathcal{J}}\frac{\mathrm{d}l}{B}=\oint\mathbf{B}\cdot\nabla\phi\frac{\mathrm{d}l}{B}=\oint\mathrm{d}\phi=2\pi
n.$

Here we have used the definition of the Jacobian in terms of the magnetic
field
$\mathbf{B}\cdot\nabla\phi=\nabla\psi\times\nabla\theta\cdot\nabla\phi=1/\mathcal{J}$,
where $\phi$ and $\theta$ are, respectively, toroidal and poloidal angular
coordinates that increase by $2\pi$ in going around the long or short toroidal
circular paths. We also considered the rotational transform $\iota=n/m$, where
$n,m\in\mathbb{Z}$ and are coprime.

Now, consider Newcomb’s condition on the last term on the right-hand-side of
Eq. (4), and define the closed line integral $\mathcal{I}$ so that,

$\oint\frac{1}{\mathcal{J}^{\prime}}\frac{\mathrm{d}l}{B}=\frac{\mathcal{I}}{B_{\alpha}(\psi)}.$
(7)

For Newcomb’s condition on Eq. (4) to hold, the following must be true:

$B_{\alpha}(\psi)=\frac{\mathcal{I}}{2\pi n}.$ (8)

With this choice

$\oint\mathbf{B}\cdot\nabla\omega\frac{\mathrm{d}l}{B}=0.$ (9)

Then there must exist a single-valued solution $\omega$ that enacts the
coordinate transformation into the set associated with the Jacobian (5).

For Eq. (8) to hold, it is necessary for $\mathcal{I}$ to be a flux function.
This condition may be written in the form,

$\mathcal{I}(\psi)=\oint\mathbf{B}\cdot\mathrm{d}\mathbf{r},$ (10)

Figure 1: Ribbon surface. Ribbon defined by two closely lying rational
magnetic field lines labelled by $C_{1}$ and $C_{2}$.

along a magnetic field line. Focus on the case of rational field lines, for
which the condition is most stringent. To proceed further, consider a non-
self-intersecting ribbon over a flux surface bounded by two adjacent magnetic
field lines (see Fig. 1). Denote the line integrals along each of these lines
by $C_{1}$ and $C_{2}$. We may then write, using Stoke’s theorem,

$\oint_{C_{1}}\mathbf{B}\cdot\mathrm{d}\mathbf{r}-\oint_{C_{2}}\mathbf{B}\cdot\mathrm{d}\mathbf{r}=\int_{\mathrm{rib}}\mathbf{j}\cdot\mathrm{d}\mathbf{S},$
(11)

where the surface element is taken to be perpendicular to the flux surface.
The integral over the surface may be written as,Helander (2014)

$\int_{\mathrm{rib}}\mathbf{j}\cdot\mathrm{d}\mathbf{S}=\int_{\alpha_{0}}^{\alpha_{0}+\delta\alpha}\mathrm{d}\alpha\oint\frac{\mathrm{d}l}{B}\mathbf{j}\cdot\nabla\psi.$

Here $\alpha$ labels the field lines on the surface. Now, if $\mathcal{I}$ is
to be truly a flux function, then following (11) the last surface integral
must vanish. And it must do so for all field line labels $\alpha_{0}$ and
adjacent field lines $\delta\alpha$. This gives the necessary and sufficient
condition,

$\oint\frac{\mathrm{d}l}{B}\mathbf{j}\cdot\nabla\psi=0.$ (12)

This subclass of magnetic fields grant the required form of $\mathcal{I}$, and
thus the single-valued solution to Eq. (4). This subclass includes the
stronger constraint $\mathbf{j}\cdot\nabla\psi=0$ (for which the coordinates
reduce to Boozer coordinates), or more to our concern, the QS scenario (see
Appendix A). Note that magnetic fields that possess nested surfaces have the
property $\langle\mathbf{j}\cdot\nabla\psi\rangle=0$, following an application
of $\nabla\cdot(\mathbf{B}\times\nabla\psi)$. However, the Newcomb condition
(12) is more stringent.Newcomb (1959)

Restricting ourselves to this subclass of fields, we can always choose
$B_{\alpha}$ as in (8). This choice might appear artificial, especially given
the presence of the discrete $n$. However, we may relate this to the surface
average over irrational flux surfaces. To see this, we write,

$B_{\alpha}=\frac{1}{2\pi n}\oint
B^{2}\frac{\mathrm{d}l}{B}=\frac{1}{4\pi^{2}n}\int_{0}^{2\pi}\mathrm{d}\alpha\underbrace{\oint
B^{2}\frac{\mathrm{d}l}{B}}_{n~{}\mathrm{turns}}=\\\
=\frac{1}{4\pi^{2}}\int_{0}^{2\pi}\mathrm{d}\alpha\underbrace{\oint
B^{2}\frac{\mathrm{d}l}{B}}_{1~{}\mathrm{turn}}=\frac{V^{\prime}}{4\pi^{2}}\langle
B^{2}\rangle,$ (13)

where
$V^{\prime}=\int_{0}^{2\pi}\mathrm{d}\alpha\int\mathrm{d}l/B=V^{\prime}\langle
1\rangle$ is the usual volume $\psi$ derivative. $B_{\alpha}$ is, therefore, a
well behaved quantity for both rational and irrational surfaces.

As it stands, with an appropriate choice of $B_{\alpha}$, and restricting
ourselves to the subclass of fields satisfying Eq. (12), one may perform a
coordinate transformation that provides a Jacobian of the desired form (5). It
remains to be shown that this new coordinate system is well-behaved. By this
we mean that the Jacobian does not vanish nor diverge. To see this, it is most
useful to rewrite $\mathcal{J}^{\prime}$ in terms of (13),

$\mathcal{J}^{\prime}=\frac{B_{\alpha}}{B^{2}}=\frac{V^{\prime}}{4\pi^{2}}\frac{\langle
B^{2}\rangle}{B^{2}}.$ (14)

It is clear that $\mathcal{J}^{\prime}$ has a definite sign, as given by the
sign of $V^{\prime}$. Thus, the Jacobian will never vanish nor diverge, given
that $B^{2}>0$.

###  II.3 Magnetic field in generalized Boozer coordinates

We have shown that under the assumption that the magnetic field is
quasisymmetric, we have a straight-field-line coordinate system, GBC, with a
Jacobian $J=B_{\alpha}(\psi)/B^{2}$. In this coordinate system and following
Sec. II.1, a quasisymmetric field is one whose magnitude can be expressed in
GBC as $B=B(\psi,\theta-\tilde{\alpha}\phi)$. This form is analogous to the
Boozer formulation of QS but is only subordinated to the configuration being
QS and not $\mathbf{j}\cdot\nabla\psi=0$.

Before proceeding to analyze some of the properties of QS and other governing
equations in GBC, we explicitly write the covariant and contravariant forms
for $\mathbf{B}$. The covariant form is

$\displaystyle\mathbf{B}=B_{\theta}\nabla\theta+(B_{\alpha}-\iota
B_{\theta})\nabla\phi+B_{\psi}\nabla\psi.$ (15)

The usual covariant function $B_{\phi}$ has been deprecated for $B_{\alpha}$.
This is the flux function that appears in the Jacobian (5). The simplicity of
(15) shows why it was convenient to choose the Jacobian of GBC to have the
particular form in (5). To obtain (15), it is sufficient to take its scalar
product with the contravariant representation,

$\mathbf{B}=\nabla\psi\times\nabla\theta+\iota\nabla\phi\times\nabla\psi,$
(16)

and capitalize on the definition of $\mathcal{J}$. Compared to Boozer
coordinates, the covariant function $B_{\theta}$ in GBC is not necessarily a
flux function. Thus, GBC is an extension of Boozer coordinates. When
$\mathbf{j}\cdot\nabla\psi=0$, however, and as previously pointed, GBCs reduce
to Boozer coordinates. So far, the forms in (15) and (16) have only required
of the existence of GBC, and not of QS per se –other than to guarantee (12).
To enforce the latter, one needs to specify $|\mathbf{B}|$ as a symmetric
function.Rodríguez and Bhattacharjee (2021a, b); Sengupta  _et al._ (2021)

##  III Describing weak quasisymmetry in GBC

Having developed GBC, let us see what this coordinate system can teach us
about weak QS. We first write down the complete set of equations that describe
a weakly quasisymmetric magnetic field. The first relevant equation equates
the covariant (15) and contravariant (16) forms of the magnetic field. The
equation reads,

$B_{\theta}\nabla\theta+(B_{\alpha}-\iota
B_{\theta})\nabla\phi+B_{\psi}\nabla\psi=\\\
=\nabla\psi\times\nabla\theta+\iota\nabla\phi\times\nabla\psi.$ (17)

To specify that we are using GBC, and to introduce the quasisymmetric
condition, we require

$\nabla\psi\times\nabla\theta\cdot\nabla\phi=\frac{B_{\alpha}(\psi)}{B^{2}(\psi,\theta-\tilde{\alpha}\phi)}.$
(18)

This set of equations (17) and (18) is entirely self-contained. It describes a
general magnetic field (a vector field satisfying
$\nabla\cdot\mathbf{B}=0=\mathbf{B}\cdot\nabla\psi$, without a particular form
of equilibrium) that is quasisymmetric —no more, no less. Such equations have
been recently used in near-axis expansions with anisotropic pressure
equilibria.Rodríguez and Bhattacharjee (2021a, b) Equations (17) and (18),
referred to as the co(ntra)variant and Jacobian equations respectively, were
there expanded systematically. A more thorough and complete exploration of the
expansion of the magnetic equations and its implications will be presented in
a separate publication.

Equations (17) and (18) apply beyond near-axis expansions. Alternatively, one
could study the behavior of this system in a perturbative sense but around a
surface rather than the axis.Sengupta  _et al._ (2021) This could shed some
light to standard optimization approaches to QSHenneberg  _et al._ (2019);
Bader  _et al._ (2019).

Beyond the set of equations (17) and (18), the formulation of weak QS in terms
of GBC can be used to compare weak QS to other (quasi)symmetric forms. This
comparison to strong and axisymmetry will help frame the notion of weak QS in
the larger space of configurations.

###  III.1 Comparison to strong quasisymmetry

Strong QS is a more resrictive form of QS compared to its weak formBurby,
Kallinikos, and MacKay (2020); Rodríguez, Helander, and Bhattacharjee (2020).
Weak QS is the necessary and sufficient condition for the leading guiding
centre dynamics to have an approximately conserved momentum to leading gyro-
centre order.Rodríguez, Helander, and Bhattacharjee (2020) This condition can
be written as in (1), but is most naturally given as,

$\displaystyle\mathbf{u}\cdot\nabla B=0,$ (19)
$\displaystyle\mathbf{B}\times\mathbf{u}=\nabla\Phi(\psi),$ (20)
$\displaystyle\nabla\cdot\mathbf{u}=0.$ (21)

The vector field $\mathbf{u}$ is defined by these equations and points in the
direction of symmetry. In strong QS the conserved momentum for the particle
dynamics is exact for the first-order guiding centre Lagrangian.Burby,
Kallinikos, and MacKay (2020); Rodríguez, Helander, and Bhattacharjee (2020);
Tessarotto  _et al._ (1996) In the notation that explicitly introduces the
symmetry vector $\mathbf{u}$, strong QS is equivalent to weak QS – i.e., Eqs.
(19)-(21)– plus the constraint

$\mathbf{j}\times\mathbf{u}+\nabla C=0,$ (22)

where $C=\mathbf{B}\cdot\mathbf{u}$. Note that only the $\mathbf{b}$ component
of this additional constraint is contained in the weak formulation of the
problem. We now explore the significance of (22) in the context of GBC.

To begin, we need to construct $\mathbf{u}$. From Eqs. (19)-(21),Rodríguez,
Helander, and Bhattacharjee (2020)

$\displaystyle\mathbf{u}=\bar{\iota}\frac{\mathbf{\nabla}\psi\times\mathbf{\nabla}\chi}{\mathbf{B}\cdot\mathbf{\nabla}\chi},$
(23)

where $\chi=\theta-\tilde{\alpha}\phi$, and it is convenient to use $\chi$ as
part of the coordinate triplet $\\{\psi,\chi,\phi\\}$. We have made the choice
$\Phi^{\prime}=\bar{\iota}$ so that $\mathbf{u}\cdot\nabla\phi=1$, for
simplicity. Scaling of the flux label for $\mathbf{u}$ leaves the weak QS
conditions unchanged (see Appendix B for further discussion on the gauge and
the particular choice). This form of $\mathbf{u}$ is equivalent to Eqs.
(19)-(21) only if we enforce $B=B(\psi,\chi)$ and the coorrdinate system
$\mathbf{B}\cdot\nabla\chi=\bar{\iota}\mathcal{J}^{-1}$ with the Jacobian in
(5). The parameter $\bar{\iota}=\iota-\tilde{\alpha}$ has been defined.

From (15) and (23), we find,

$C=B_{\alpha}-\bar{\iota}B_{\theta}.$ (24)

This relation (24), together with (23) and the definition of $C$ can be used
to write the gauge-independent form,

$\mathbf{B}\cdot\nabla\psi\times\nabla
B=\frac{B_{\alpha}-\bar{\iota}B_{\theta}}{\bar{\iota}}\mathbf{B}\cdot\nabla
B.$ (25)

The magnetohydrostatic form of this equation has been used previouslyPaul  _et
al._ (2020); Helander (2014).

The contravariant form of $\mathbf{u}$ together with (16) give simple
expressions for directional derivatives in GBC. The differential operators can
be simply written as partial derivatives,

$\displaystyle\mathbf{B}\cdot\mathbf{\nabla}=\mathcal{J}^{-1}\left(\bar{\iota}\partial_{\chi}+\partial_{\phi}\right),\quad\mathbf{u}\cdot\mathbf{\nabla}=\partial_{\phi}.$
(26)

Since the Jacobian is quasisymmetric from (5), the two operators
$\mathbf{B}\cdot\mathbf{\nabla}$ and $\mathbf{u}\cdot\mathbf{\nabla}$ commute
with each otherBurby, Kallinikos, and MacKay (2021). This commutation property
is made manifest in the GBC.

The symmetry field $\mathbf{u}$ can also be written in the following covariant
form:

$\displaystyle\mathbf{u}=u_{\psi}\mathbf{\nabla}\psi+u_{\chi}\mathbf{\nabla}\chi+u_{\phi}\mathbf{\nabla}\phi.$
(27)

Taking the scalar product with $\mathbf{u}$ and $\mathbf{B}$ we obtain

$\displaystyle
u_{\phi}=u^{2},\quad\bar{\iota}u_{\chi}+u_{\phi}=\frac{C}{\mathcal{J}^{-1}}.$
(28)

To complete the family of vectors required for the strong quasisymmetric
condition (22), we need a closed form for $\mathbf{j}$ in GBC. From the curl
of the covariant form of $\mathbf{B}$ in Eq. (15), we obtain

$\mathbf{j}=\mathbf{\nabla}B_{\psi}\times\mathbf{\nabla}\psi+\mathbf{\nabla}B_{\theta}\times\mathbf{\nabla}\chi+\mathbf{\nabla}(B_{\alpha}-\bar{\iota}B_{\theta})\times\mathbf{\nabla}\phi.$
(29)

Using (29) and (24), we can show that

$\mathbf{j}\times\mathbf{u}+\mathbf{\nabla}C=(\mathbf{u}\cdot\mathbf{\nabla}B_{\psi})\mathbf{\nabla}\psi+(\mathbf{u}\cdot\mathbf{\nabla}B_{\theta})\left(\mathbf{\nabla}\chi-\bar{\iota}\mathbf{\nabla}\phi\right).$
(30)

The simplicity of (30) is due to the choice of GBC.

Recall that strong QS requires the expression
$\mathbf{j}\times\mathbf{u}+\mathbf{\nabla}C$ to be identically zero. This
means that all the terms on the right side of (30) need to vanish. That is to
say, the covariant functions $(B_{\psi},B_{\theta})$ are required to be
quasisymmetric. If $B_{\theta}$ is quasisymmetric, then $C$ is automatically
so from (24). In an explicit coordinate representation, using (26), we may
write $B_{\theta}(\psi,\chi)$ and $B_{\psi}(\psi,\chi)$.

Thus, the GBC representation provides an elegant way to formulate strong QS,
which can now be understood as weak quasisymmetry plus the conditions that
$B_{\psi}$ and $B_{\theta}$ are QS. In other words, not only is $B$ QS but so
are $B_{\theta}$ and $B_{\psi}$.

#### Implications for near-axis expansion

. We refer to [Rodríguez and Bhattacharjee, 2021a] and [Rodríguez and
Bhattacharjee, 2021b] for a detailed treatment using near-axis expansions of
the weakly quasisymmetric problem. The procedure is based on expanding all
governing equations describing the weak quasisymmetric field in some form of
equilibrium in powers of the distance from the magnetic axis. To do so
efficiently, both the field and equations –including Eqs. (17) and (18)– are
expressed in GBC. It was shown there that when equilibria with anisotropic
pressure are considered, the common overdetermination problemGarren and Boozer
(1991a); Landreman and Sengupta (2018) that limits the expansion is overcome.
The number of governing equations becomes the same as that of functions to be
solved.

Extending the expansion to strong QS is straightforward. From the discussion
above, the only difference is that the covariant functions $B_{\theta}$ and
$B_{\psi}$ are both quasisymmetric rather than general functions of space. In
practical terms, this simply leads to more restricted Taylor-Fourier
expansions of those functions; the coefficients that were functions of $\phi$
become constants. This restriction in freedom once again leads us back to the
Garren-Boozer overdetermination problem. In fact, it does so the same way as
it did in the case of MHS equilibrium. The restriction on the covariant
functions imposes very severe constraints on the allowed geometry. The only
way to escape this impasse is to assume axisymmetry ($\phi$ independent). Once
again, consistent with what we observed in [Rodríguez and Bhattacharjee,
2021a], the asymmetry of the covariant representation of $\mathbf{B}$ appears
to be vital to the construction of QS solutions.

###  III.2 Comparison to axisymmetry

We have seen that the strong formulation of QS is more constraining than its
weak form. We would also like to compare QS to the limiting case of
axisymmetry. We shall think of this case as a symmetry generated by rotation
in space: the system is invariant under rotations about an axis. In Euclidean
space, a rotation is an isometry, and it is generated by a vector field known
as a Killing vector. Using the notion of a Killing vector, we want to explore
how ‘far’ the weak concept of QS is from this ‘true’ symmetry.

A measure of the departure of a symmetry generator from a Killing vector is
the so-called deformation metric.Burby, Kallinikos, and MacKay (2020) Taking
$\mathbf{u}$ to represent the symmetry vector for QS, the idea is to see how
far it is from being a Killing vector. A vector field $\mathbf{v}$ is Killing
if and only if the deformation tensor $\mathcal{L}_{\mathbf{v}}g=0$. Here
$\mathcal{L}_{u}$ denotes the Lie derivative along $\mathbf{u}$ and $g$ is the
Euclidean metric. In 3D, this may be written as,

$\mathcal{L}_{\mathbf{u}}g=\nabla\mathbf{u}+(\nabla\mathbf{u})^{T}.$ (31)

Evaluating this tensor for a quasisymmetric configuration should then provide
information regarding the closeness to an isometry. It is convenientBurby,
Kallinikos, and MacKay (2020) to evaluate this rank-2 tensor in a basis
defined by $\\{\mathbf{B},\mathbf{u},\nabla\psi\\}$, a triplet that we shall
take to be independent. Then,

$\displaystyle\left[\nabla\mathbf{u}+(\nabla\mathbf{u})^{T}\right]\cdot\mathbf{B}=$
$\displaystyle\mathbf{j}\times\mathbf{u}+\nabla C,$ (32)
$\displaystyle\left[\nabla\mathbf{u}+(\nabla\mathbf{u})^{T}\right]\cdot\mathbf{u}=$
$\displaystyle\nabla u^{2}-\mathbf{u}\times\nabla\times\mathbf{u},$ (33)
$\displaystyle\left[\nabla\mathbf{u}+(\nabla\mathbf{u})^{T}\right]\cdot\nabla\psi=$
$\displaystyle\nabla\psi\times\nabla\times\mathbf{u}+2\nabla\psi\cdot\nabla\mathbf{u},$
(34)

where the weak quasisymmetric properties have been used where necessary.
Equation (33) is what Burby et al. call $\mathbf{w}$.Burby, Kallinikos, and
MacKay (2021) We shall explore this vector $\mathbf{w}$ in more detail after
obtaining an explicit form for $\mathcal{L}_{\mathbf{u}}g$. Using (32)-(34),
and projecting once again onto the non-orthogonal basis triplet, we obtain

$\mathcal{L}_{\mathbf{u}}g=\begin{pmatrix}0&\mathbf{u}\cdot\nabla
C&\nabla\psi\cdot(\mathbf{j}\times\mathbf{u}+\nabla C)\\\
\dots&\mathbf{u}\cdot\mathbf{w}&\mathbf{w}\cdot\nabla\psi\\\
\dots&\dots&-\mathbf{u}\cdot\nabla|\nabla\psi|^{2}\end{pmatrix}.$ (35)

The matrix is symmetric by construction. The content of its elements can be
made clearer using GBC explicitly.

The top row, corresponding to (32), has already been dealt with, as it is
precisely the piece corresponding to strong QS. We made the observation that
for this condition to be satisfied, (30) required $\mathbf{u}\cdot\nabla
B_{\theta}=0=\mathbf{u}\cdot\nabla B_{\psi}$.

For the other components, the vector field
$\mathbf{w}=\mathbf{\omega_{u}}\times\mathbf{u}+\mathbb{\nabla}u^{2},\quad\mathbf{\omega_{u}}=\mathbf{\nabla}\times\mathbf{u}$
is key. Using the covariant form of $\mathbf{u}$ we obtain the curl of the
vector $\mathbf{u}$ in the form

$\displaystyle\mathbf{\omega_{u}}=\mathbf{\nabla}u_{\psi}\times\mathbf{\nabla}\psi+\mathbf{\nabla}u_{\chi}\times\mathbf{\nabla}\chi+\mathbf{\nabla}u_{\phi}\times\mathbf{\nabla}\phi.$
(36)

Taking the cross product with $\mathbf{u}$, using the orthogonality of
$\mathbf{u}$ with $\mathbf{\nabla}\psi$ and $\mathbf{\nabla}\chi$, and (24)
and (28) we get

$\mathbf{w}=(\mathbf{u}\cdot\mathbf{\nabla}u_{\psi})\mathbf{\nabla}\psi+(\mathbf{u}\cdot\mathbf{\nabla}u^{2})\left(\mathbf{\nabla}\phi-\frac{1}{\bar{\iota}}\mathbf{\nabla}\chi\right)+\\\
-\mathcal{J}(\mathbf{u}\cdot\mathbf{\nabla}B_{\theta})\mathbf{\nabla}\chi,$
(37)

which implies that

$\displaystyle\mathbf{B}\cdot\mathbf{w}=$
$\displaystyle-\bar{\iota}\mathbf{u}\cdot\mathbf{\nabla}B_{\theta},$ (38)
$\displaystyle\mathbf{u}\cdot\mathbf{w}=$ $\displaystyle\mathbf{u}\cdot\nabla
u^{2}.$ (39)

Most importantly, a vanishing $\mathbf{w}$ implies that the covariant
components of the symmetry vector as well as $B_{\theta}$ are quasisymmetric.

To complete the simplification of the metric tensor, we invoke
$B^{2}=(C^{2}+|\nabla\psi|^{2})/u^{2}$, which follows from the definition of
$\mathbf{u}$. This means,

$\mathbf{u}\cdot\nabla|\nabla\psi|^{2}=B^{2}\mathbf{u}\cdot\nabla
u^{2}-\mathbf{u}\cdot\nabla C^{2}.$ (40)

With this coordinate representation, the dependence of the various metric
pieces is made explicit. We may then schematically present the dependence of
$\mathcal{L}_{\mathbf{u}}g$ as follows,

$\mathcal{L}_{\mathbf{u}}g\sim\begin{pmatrix}0&\boxed{\partial_{\phi}B_{\theta}}&\boxed{\begin{subarray}{c}\partial_{\phi}B_{\theta},~{}\partial_{\phi}B_{\psi}\end{subarray}}\\\
\dots&\boxed{\partial_{\phi}u^{2}}&\boxed{\begin{subarray}{c}\partial_{\phi}u^{2},~{}\partial_{\phi}u_{\psi}\\\
\partial_{\phi}B_{\theta}\end{subarray}}\\\
\dots&\dots&\boxed{\begin{subarray}{c}\partial_{\phi}B_{\theta},~{}\partial_{\phi}u^{2}\end{subarray}}\end{pmatrix}.$
(41)

The boxed expressions are meant to indicate that the corresponding tensor
component vanishes if the expressions there do. If the tensor (41) is to
vanish, then the symmetry vector would correspond to a rotation. This is not
surprising if one looks at what it means for the components in (41) to vanish.
Axisymmetry is reached when the covariant components of the magnetic field and
the symmetry vector are themselves symmetric. The latter is intimately related
to the geometry, as we may see when writing
$\mathbf{u}\propto\partial_{\phi}\mathbf{x}|_{\chi,\psi}$.

From (41), it follows that in some sense, weak QS is far from being an
isometry. This is so because only one of the components of the tensor exactly
vanishes. The $\phi$ dependence of the functions $B_{\psi}$, $B_{\theta}$,
$u^{2}$, and $u_{\psi}$ takes the configuration away from axisymmetry. These
apparent four degrees of freedom (especially those involving $\mathbf{u}$) may
not be independent and involve highly non-linear combinations—they should
ultimately be related through the quasisymmetric magnetic equations. Anyhow,
the field-line dependence is key in distinguishing the weakly quasisymmetric
form from, say, an axisymmetric tokamak.

To make a comparative measurement of the departure from axisymmetry, consider
now the case of strong QS. In this case, following (22), the first whole row
(and thus also column) of (41) drop. The remaining dependence also simplifies,
and the system is precluded from being an isometry, a priori, through the
$\phi$ dependence of $u^{2}$ and $u_{\psi}$ only, which is consistent with the
work by Burby et al.Burby, Kallinikos, and MacKay (2020)

Imposing additional properties to the field may also affect the form of the
deformation tensor. An example would be a particular form of force balance. We
now explore how the magnetics and equilibria are linked.

##  IV Quasisymmetry and equilibria

Let us consider the force balance part of the problem. Generally, a magnetic
equilibrium with some arbitrary force $\mathbf{F}$ reads,

$\mathbf{j}\times\mathbf{B}=\mathbf{F}.$ (42)

As we argued in Sec. II, we are concerned in this work with a general fluid
force $\mathbf{F}$. Its connection to the microphysics is not considered. Let
us express the left-hand side of (42) in GBC. Using the contravariant form for
the current (29) together with (24), we obtain

$\mathbf{j}\times\mathbf{B}=\left[\mathbf{B}\cdot\mathbf{\nabla}B_{\psi}-\mathcal{J}^{-1}\left(B_{\alpha}^{\prime}-\bar{\iota}^{\prime}B_{\theta}\right)\right]\mathbf{\nabla}\psi+\\\
+(\mathbf{B}\cdot\mathbf{\nabla}B_{\theta})\left(\mathbf{\nabla}\chi-\bar{\iota}\mathbf{\nabla}\phi\right),$
(43)

which is an explicit coordinate representation of
$\mathbf{j}\times\mathbf{B}$. The form of (43) mirrors the form of (30). In
this case, the magnetic differential operators substitute the directional
derivatives along $\mathbf{u}$. We note that (43) does not have any component
along $\mathbf{B}$, as can be checked by taking the dot product with
$\mathbf{B}$.

The form of (43) puts constraints on the allowable forms for $\mathbf{F}$. As
already noted, $\mathbf{F}\cdot\mathbf{B}=0$ must hold true. Otherwise the
system would be imbalanced, as the magnetic field is unable to exerts forces
along $\mathbf{B}$. Because of this reduction in the dimensionality of
$\mathbf{F}$ and in view of (43), it is convenient to write

$\displaystyle\mathbf{F}=F_{\psi}\nabla\psi+F_{\alpha}\left(\nabla\chi-\bar{\iota}\nabla\phi\right).$
(44)

An alternative form would be to use the contravariant form of $\mathbf{B}$
(16) in (42) to get

$\displaystyle\mathbf{F}=\left(\mathbf{J\cdot\nabla}\chi-\bar{\iota}\mathbf{J\cdot\nabla}\phi\right)\nabla\psi-(\mathbf{J\cdot\nabla}\psi)\left(\nabla\chi-\bar{\iota}\nabla\phi\right).$
(45)

Substituting (43) and (44) into (42) we get two magnetic differential
equations

$\displaystyle\mathbf{B}\cdot\mathbf{\nabla}B_{\psi}=F_{\psi}+\mathcal{J}^{-1}\left(B_{\alpha}^{\prime}-\bar{\iota}^{\prime}B_{\theta}\right),$
(46a)
$\displaystyle\mathbf{B}\cdot\mathbf{\nabla}B_{\theta}=F_{\alpha}=-\mathbf{J\cdot\nabla}\psi.$
(46b)

Therefore, the generalized force-balance condition is equivalent to two
magnetic differential equations (MDEs) and $\mathbf{B\cdot F}=0$. If solutions
to these equations can be found together with the magnetic equations (17) and
(18), we will have obtained a quasisymmetric configuration in equilibrium.

Let us describe in more detail the implications of these equations. First look
at the simpler (46b). This equation has two pieces to it. First, and
regardless of the assumed form for $F_{\alpha}$, it follows from weak QS that
$\mathbf{B}\cdot\nabla B_{\theta}=-\mathbf{j}\cdot\nabla\psi$ (see Appendix A
as well). This imposes the condition (II.2) on the field. Secondly, the
component $F_{\alpha}$ of the forcing $\mathbf{F}$ directly sets the off-
surface current. This means that,Newcomb (1959)

$\oint\frac{d\ell}{B}F_{\alpha}=0.$ (47)

We may not choose $F_{\alpha}$ arbitrarily. It must satisfy (47) if the force
is to be consistent with QS –condition like (12). Then the magnetic
differential equation can be satisfied, and one can directly relate
$B_{\theta}$ and $F_{\alpha}$ up to a flux function. In Fourier
representation, it is clear that the $\phi$ content of $B_{\theta}$ will be
non-zero only if that of $F_{\alpha}$ is (and vice versa). In the light of
(41), choosing $\partial_{\phi}(\mathbf{j}\cdot\nabla\psi)=0$ brings the
quasisymmetric configuration closer to an isometry. This freedom in the form
of $B_{\theta}$ does not exist in the strong formulation of QS, for which
$\mathbf{j}\cdot\nabla\psi$ is independent of the field line label.

A similar analysis is suitable for (46a). The appropriate Newcomb condition in
this case is,

$\displaystyle\oint\frac{d\ell}{B}\left[F_{\psi}+\mathcal{J}^{-1}\left(B_{\alpha}^{\prime}-\bar{\iota}^{\prime}B_{\theta}\right)\right]=0.$
(48)

This condition may be understood as an averaged radial equilibrium equation. A
similar solvability condition can be found, for the special case of MHS, in
[Tessarotto, Johnson, and Zheng, 1995], where the notion of a QS field is
presented as one that satisfies the Newcomb conditions. Given Eq. (48), Eq.
(46a) relates $B_{\psi},~{}B_{\theta}$ (or $F_{\alpha}$) and $F_{\psi}$. Once
again, we see the close relationship between the forcing, the magnetic
covariant representation, and the deviation from axisymmetry. A $\phi$
dependence on $B_{\psi}$ will bring a finite deviation of $\mathbf{u}$ from
being a Killing vector. However, it will also force $F_{\alpha}$ and
$F_{\psi}$ to have a $\phi$ dependence which may require very particular
shaping of the forces. This observation is consistent with the Constantin-
Drivas-Ginsberg theoremConstantin, Drivas, and Ginsberg (2021), in which the
forcing is seen to be intimately related to the deviation from an isometry.
Here the asymmetric geometry, quasisymmetry, and the forcing are all
intimately connected.

When the magnetic differential equations imposing force balance are brought
together with the magnetic equations from the previous section, it is not
obvious how the system of equations is to be interpreted: what is to be taken
as an input and what should be solved for. Just as an analogy, in the Grad-
Shafranov equation, it is clear that $p$ and $F$ are inputs, and $\psi$ is the
output. In the present problem, we have the construction of GBC in addition to
the various magnetic covariant forms and the components of $\mathbf{F}$.
Motivated by the treatment in [Rodríguez, Helander, and Bhattacharjee, 2020]
(which deals with a special case of the above), we propose as a possibility
for a well-posed problem (46b) to be solved for $B_{\theta}$ given
$F_{\alpha}$, while $F_{\psi}$ is the output of (46a), with the function
$B_{\psi}$ specified from the magnetic equations. It is not a trivial matter
to determine a convenient way in which to formulate the problem. A more
elaborate discussion on this procedure and its implications on constructing
solutions is left to future work.

###  IV.1 Ideal MHD: $\mathbf{j}\times\mathbf{B}=\mathbf{\nabla}p(\psi)$

Let us now revisit the limit of ideal MHD without flows,
$\mathbf{j}\times\mathbf{B}=\mathbf{\nabla}p$. More general forms will be
discussed elsewhere, together with a more systematic treatment of the
quasisymmetric system of equations.

In ideal MHD, from $\mathbf{j}\cdot\nabla p(\psi)=0$ it follows that

$\displaystyle\mathbf{B}\cdot\mathbf{\nabla}B_{\theta}=0.$ (49)

Thus taking $F_{\alpha}=0$ forces $B_{\theta}$ into a flux function. Of
course, this also means that $\mathbf{u}\cdot\mathbf{\nabla}B_{\theta}=0$.
Furthermore, as $F_{\psi}=p^{\prime}$, in static ideal MHD, (43) leads to the
magnetic differential equation for $B_{\psi}$,

$\displaystyle\mathbf{B}\cdot\mathbf{\nabla}B_{\psi}=p^{\prime}(\psi)+\mathcal{J}^{-1}\left(B_{\alpha}^{\prime}-\bar{\iota}^{\prime}B_{\theta}\right)$
(50)

Since $B_{\psi}$ must be a single-valued function the flux-surface average of
(50) gives

$\displaystyle p^{\prime}(\psi)+\frac{\langle
B^{2}\rangle}{B_{\alpha}}\left(B_{\alpha}^{\prime}-\bar{\iota}^{\prime}B_{\theta}\right)=0.$
(51)

If we choose the forms of $B$, $p$, and $\iota$, this pins the form of
$B_{\alpha}$ down. Now, looking back to (50), every term on the right-hand
side is quasisymmetric. Therefore, $B_{\psi}$ must also be quasisymmetric if
it is to satisfy the force-balance equation. Note that we have already
recognized this constraining requirement on the form of $B_{\psi}$ as the
origin of the Garren-Boozer overdetermination problem Rodríguez and
Bhattacharjee (2021a). The Newcomb condition on this equation can be
recognized as the condition to avoid Pfirsch-Schlüter current singularities.

The simplifications due to ideal static MHD leads to the vanishing of (30).
Therefore, in this limit weak QS is identical to strong QS.

One can further show using (29), (15), (23) and (50),

$\displaystyle\mathbf{j}=-\frac{1}{\bar{\iota}}\partial_{\psi}(B_{\alpha}-\bar{\iota}B_{\theta})\mathbf{B}-\frac{1}{\bar{\iota}}p^{\prime}(\psi)\mathbf{u},$
(52)

where the gauge choice for $\mathbf{u}$ has been made
$\Phi^{\prime}=\bar{\iota}$. The expression for $\mathbf{j}$ ought to be
$\mathbf{u}$-gauge independent, as it is a physical quantity. The $\mathbf{B}$
piece as written is gauge independent, but the $\mathbf{u}$ term is not. The
$\bar{\iota}$ factor in the latter is to be interpreted as $\Phi^{\prime}$.
This equation had been obtained previouslyBurby, Kallinikos, and MacKay (2020,
2021) using coordinate free, differential forms (see Appendix C). Two special
cases of ideal MHD, a) vacuum ($\mathbf{j}=0$) and b) force-free
($\mathbf{j}=\lambda(\psi,\alpha)\mathbf{B}$), are worth pointing to for their
importance in plasma physics. For both these cases $p^{\prime}(\psi)=0$. From
(52) we see that for the magnetic field to be curl-free (vacuum) and
quasisymmetric $C^{\prime}(\psi)=0$; i.e., $C(\psi)$ must be a constant. For
quasisymmetric force-free fields, we must have $\lambda=-C^{\prime}(\psi)$.
Note that in strong QS these conclusions follow directly from the equation
$\mathbf{j\times u}+\mathbf{\nabla}C=0$ with $\mathbf{j}=0$ and
$\mathbf{j}=\lambda\mathbf{B}$.

##  V Conclusions

In this paper, we have presented, defined, and discussed a straight field line
coordinate system which is natural for the representation of general-
equilibria quasisymmetric magnetic fields: generalized Boozer coordinates. We
proved the existence of the said coordinate system for the subset of fields
for which $\oint\mathbf{j}\cdot\nabla\psi\mathrm{d}l/B=0$, to which
quasisymmetric fields belong. These coordinates reduce to Boozer coordinates
when $\mathbf{j}\cdot\nabla\psi=0$.

The explicit form of the symmetry in this coordinate representation enables a
simple formulation of the quasisymmetric problem. We explicitly construct the
governing equations setting clearly the foundation for future investigations,
including expansionRodríguez and Bhattacharjee (2021a, b) and global
approaches. Exploiting GBC, we explicitly show the essential differences
between the weak and strong formulations of QS and between quasisymmetry and
axisymmetry. Weak QS generally lies far from axisymmetry, for which many
functions describing the field and symmetry need to be symmetric.

We also included a set of simple magnetic differential equations that fully
describe equilibrium with an arbitrary macroscopic force to complete the
treatment. The property of QS, together with the force-balance structure,
imposes requirements on the forcing terms in the form of Newcomb conditions.
In addition, the equations establish clear connections between QS, forcing,
and departures from axisymmetry.

## Acknowledgements

We are grateful to J. Burby, N. Duigan, J. Meiss, and D. Ginsburg for
stimulating discussions. This research is supported by grants from the Simons
Foundation/SFARI (560651, AB) and DoE Contract No DE-AC02-09CH11466.

## Data availability

Data sharing is not applicable to this article as no new data were created or
analyzed in this study.

##  Appendix A Off-surface current and QS

In this appendix we directly show how the triple vector formulation of QS in
Eq. (1) imposes a constraint on the off-normal component of the current at
every magnetic surface.

Let us start by noting that $\mathbf{B}\cdot\nabla B=f(\psi,B)$ is a
consequence of (1). In that sense, we can reshape the triple vector condition
in the convenient form,

$\nabla\psi\times\nabla B\cdot\nabla\left(\frac{B^{2}}{\mathbf{B}\cdot\nabla
B}\right)=0.$ (53)

(For now we shall not worry about $\mathbf{B}\cdot\nabla B=0$.) We may express
this as a divergence as $\nabla\cdot(\nabla\psi\times\nabla B)=0$. Separating
the argument of the divergence into a component parallel to the magnetic field
and perpendicular to it, we may rewrite (53),

$\nabla\cdot\left[\frac{\nabla\psi\times\nabla
B\cdot\mathbf{B}}{\mathbf{B}\cdot\nabla
B}\mathbf{B}+\nabla\psi\times\mathbf{B}\right]=0.$ (54)

It is customary to define $C=\nabla\psi\times\nabla
B\cdot\mathbf{B}/(\mathbf{B}\cdot\nabla B)$. This can therefore be rewritten,
using $\nabla\cdot\mathbf{B}=0$,

$\mathbf{B}\cdot\nabla C=\mathbf{j}\cdot\nabla\psi.$ (55)

It is then clear that the Newcomb conditionNewcomb (1959),

$\oint\mathbf{j}\cdot\nabla\psi\frac{\mathrm{d}l}{B}=0,$ (56)

with the line integral taken along closed magnetic field lines. For irrational
field lines, this amounts to $\langle\mathbf{j}\cdot\nabla\psi\rangle=0$,
which is true of all magnetic fields with flux surfaces
$\mathbf{B}\cdot\nabla\psi=0$. The condition (56) is a more constraining
condition than the flux surface average oneNewcomb (1959).

The division by $\mathbf{B}\cdot\nabla B$ seems to be ill-defined at the
extrema of the magnetic field along the field lines. However, the above holds,
first, arbitrarily close to these extrema, and thus would expect to hold by
continuity. The fact that the condition holds can be seen by looking at the
fundamental formulation of QS.Rodríguez, Helander, and Bhattacharjee (2020) As
for the behavior of $C$, this is a physical quantity related to the width of
banana orbits of bouncing particles. If deeply trapped and barely trapped
particles are to be kept, then physically $C$ must be finite at the extrema.
Once we express everything in terms of GBC, the apparent divergence of $C$
disappears as both the numerator and denominator are proportional to
$\partial_{\chi}B=0$ when $\mathbf{B}\cdot\nabla B=0$.

##  Appendix B Gauge freedom in weak $\mathbf{u}$

As noted, the definition of $\mathbf{u}$ by the weak quasisymmetry equations
(19)-(21) is invariant to the relabelling of $\Phi$. A rescaling of the flux
surface label together with $\mathbf{u}$ leaves the equations invariant, and
should therefore have no physical implication in the description of
quasisymmetry. Keeping the general gauge label $\Phi(\psi)$ of (20), Eq. (23)
reads,

$\mathbf{u}=\frac{\nabla\Phi\times\nabla\chi}{\mathbf{B}\cdot\nabla\chi}.$
(57)

Then it folllows,

$C=\frac{\Phi^{\prime}}{\bar{\iota}}(B_{\alpha}-\bar{\iota}B_{\theta}).$ (58)

and

$\mathbf{u}\cdot\nabla=\frac{\Phi^{\prime}}{\bar{\iota}}\partial_{\phi}.$ (59)

In Sec. III.1 we then went on to evaluate how the symmetry vector $\mathbf{u}$
as defined by weak QS compares to strong QS. To do so, we needed to evaluate
(30), the third of the strong QS conditions. The first thing to note is that
this equation is not gauge-invariant. This is, of course, an added
complication to the problem. In a sense, this gauge-symmetry breaking leaves
no unique form for $\mathbf{j}\times\mathbf{u}+\nabla C$, but rather a whole
family.

This can be seen explicitly if instead of restricting ourselves to the special
case $\Phi^{\prime}=\bar{\iota}$, we keep $\Phi$ general. Then, (30) can be
written in the form,

$\mathbf{j}\times\mathbf{u}+\mathbf{\nabla}C=\left[\mathbf{u}\cdot\mathbf{\nabla}B_{\psi}-C\frac{\Phi^{\prime}}{\bar{\iota}}\partial_{\psi}\left(\frac{\bar{\iota}}{\Phi^{\prime}}\right)\right]\mathbf{\nabla}\psi+\\\
+(\mathbf{u}\cdot\mathbf{\nabla}B_{\theta})\left(\mathbf{\nabla}\chi-\bar{\iota}\mathbf{\nabla}\phi\right).$
(60)

So the question may be reformulated: what are the conditions for this
expression to vanish? As before one piece is $\mathbf{u}\cdot\nabla
B_{\theta}=0$. Now, the other component has the form of a differential
equation along the direction of symmetry. We may write it explicitly as,

$\partial_{\phi}B_{\psi}-(B_{\alpha}-\bar{\iota}B_{\theta})\partial_{\psi}\ln(\bar{\iota}/\Phi^{\prime})=0.$
(61)

For the equation to be consistent it must then hold that either
$\bar{\iota}=a\Phi^{\prime}$ where $a$ is a constant, or
$\int(B_{\alpha}-\bar{\iota}B_{\theta})\mathrm{d}\phi=0$. The latter is a
condition on the physically relevant part of the problem. Thus, it is minimal
to choose $\Phi^{\prime}=\bar{\iota}$ to satisfy this condition. Then we are
simply left with $\mathbf{u}\cdot\nabla B_{\psi}=0$, which is the choice made
in the main text.

##  Appendix C Coordinate independent form of quasisymmetric current

In this appendix we present a brief derivation of (52) without using
coordinates. Start from,

$\mathbf{B}\times\mathbf{u}=\nabla\Phi$ (62)

with $\Phi$ some single-valued function. With magnetic flux surfaces labelled
by $\Phi$, to which $\mathbf{B}$ as well as the vector field $\mathbf{u}$ are
tangent. From this condition, one may also write the symmetry vector field in
the following form,

$\mathbf{u}=\frac{1}{|\mathbf{B}|^{2}}(C\mathbf{B}-\mathbf{B}\times\nabla\Phi).$
(63)

Here $C=\mathbf{B}\cdot\mathbf{u}$ is some arbitrary function of space, as the
constraint equation only describes the perpendicular part of $\mathbf{u}$.

Because of magnetostatic equilibrium, we hav

$\mathbf{j}\times\mathbf{B}=\nabla p$ (64)

where $\mathbf{j}=\nabla\times\mathbf{B}$ is the plasma current density.
Because $\mathbf{B}\cdot\nabla p=0$, it follows that $p=p(\psi)$, again
assuming that $\mathbf{B}$ is irrational. By continuity, we argue that this
must be true for the rational cases as well (provided the rotational transform
is not constant $\iota^{\prime}\neq 0$).

Because the plasma pressure is also a flux function, the current density
$\mathbf{j}$ also lives on magnetic flux surfaces. Furthermore, requiring
$\nabla p$ to be nowhere vanishing (except the axis), the magnetic field and
the currents are guaranteed to be non-collinear. Therefore, at every point on
a flux surface, we may define a tangent plane spanned by $\mathbf{j}$ and
$\mathbf{B}$.

Because $\mathbf{u}$ also exists in this subspace, we can express it in the
chosen basis as,

$\mathbf{u}=a\mathbf{j}+\tilde{\beta}\mathbf{B}$

Taking the cross product of this form of $\mathbf{u}$ with $\mathbf{B}$, and
making use of (62) and (64), it is clear that
$a=a(\psi)=-\Phi^{\prime}/p^{\prime}$, where the prime denotes a derivative
with respect to the flux $\psi$.

Let us now apply (21), namely $\nabla\cdot\mathbf{u}=0$, on our form of
$\mathbf{u}$,

$\nabla\cdot\mathbf{u}=\mathbf{B}\cdot\nabla\tilde{\beta}=0~{}~{}\therefore~{}~{}\tilde{\beta}=\tilde{\beta}(\psi)$

so that,

$\mathbf{j}=-\frac{p^{\prime}(\psi)}{\Phi^{\prime}}\mathbf{u}+\beta(\psi)\mathbf{B}.$
(65)

Using the form (63), we can then rewrite the expression for $\mathbf{j}$ in
the form,

$\mathbf{j}=\left(\beta(\psi)-\frac{p^{\prime}(\psi)C}{B^{2}\Phi^{\prime}}\right)\mathbf{B}+\frac{p^{\prime}(\psi)}{B^{2}}\mathbf{B}\times\nabla\psi$
(66)

The gauge-independent combination
$C/\Phi^{\prime}=(\mathbf{B}\cdot\nabla\psi\times\nabla
B)/(\mathbf{B}\cdot\nabla B)$ here is the same as appears in (25), and it is a
flux function. There we found a closed form for this combination in terms of
the covariant representation of $\mathbf{B}$ in GBC. Equation (66), together
with $\nabla\cdot\mathbf{B}=0$ and $\mathbf{B}\cdot\nabla\psi=0$ constitute
the whole set of governing equations for the QS, MHS field. The problem in
this case can be formulated by providing $\beta$, $C/\Phi^{\prime}$ and $p$ as
input flux functions and solving for $\mathbf{B}$.

Assuming that we have strong QS, $\beta=-C^{\prime}/\Phi^{\prime}$, obtaining
an expression equivalent to that obtained in (52). The assumption of strong QS
was, however, not necessary when working with the coordinate representation.
The adoption of some form of covariant representation of $\mathbf{B}$ seems to
be necessary to obtain such a relation. To that end, GBC is convenient, though
not necessary. In fact, [Burby, Kallinikos, and MacKay, 2021] employs the
$B^{\flat}$ form (the generalisation of the covariant representation of the
magnetic field), without specifying coordinates, to prove (52). Note, however,
that the difference in the form of $\beta$ is merely a matter of how the flux
function is defined. The problem has $\beta$ or $\Phi^{\prime}$ as free flux
function inputs.

## References

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# Determination of the distance from a projection to nilpotents

Masaki Izumi  Graduate School of Science, Kyoto University, Sakyo-ku, Kyoto
606-8502, Japan<EMAIL_ADDRESS>and  Michiya Mori  Graduate School
of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba, Meguro-ku,
Tokyo, 153-8914, Japan; Interdisciplinary Theoretical and Mathematical
Sciences Program (iTHEMS), RIKEN, 2-1 Hirosawa, Wako, Saitama 351-0198, Japan.
<EMAIL_ADDRESS>
###### Abstract.

In this note, we study the distance from an arbitrary nonzero projection $P$
to the set of nilpotents in a factor $\mathcal{M}$ equipped with a normal
faithful tracial state $\tau$. We prove that the distance equals
$(2\cos\frac{\tau(P)\pi}{1+2\tau(P)})^{-1}$. This is new even in the case
where $\mathcal{M}$ is the matrix algebra. The special case settles a
conjecture posed by Z. Cramer.

###### Key words and phrases:

projection, nilpotent, distance

###### 2020 Mathematics Subject Classification:

Primary 15A60, 47A30.

The first author was supported in part by JSPS KAKENHI Grant Number 20H01805.
The second author was supported in part by JSPS KAKENHI Grant Number 22K13934.

##  1\. Introduction

Let ${\mathcal{M}}$ be a factor (a von Neumann algebra with trivial center)
acting on a separable Hilbert space. A _nilpotent_ is an operator
$X\in{\mathcal{M}}$ satisfying $X^{n}=0$ for some positive integer $n$. For
$X\in{\mathcal{M}}$, let $\nu_{\mathcal{M}}(X)$ denote the distance from $X$
to the set of nilpotents in ${\mathcal{M}}$. In other words,

$\nu_{\mathcal{M}}(X)=\inf\\{\lVert X-N\rVert\mid N\text{ is a nilpotent in
}{\mathcal{M}}\\},$

where $\lVert\cdot\rVert$ denotes the operator norm.

In 1970, Paul Halmos proposed a list of 10 problems in operator theory [6].
Problem 7 in the list asks the following. Let $\mathbb{B}(\mathcal{H})$ denote
the von Neumann algebra (type I∞ factor) of all bounded linear operators on a
separable infinite-dimensional complex Hilbert space $\mathcal{H}$. Let
$X\in\mathbb{B}(\mathcal{H})$ be a _quasinilpotent_ operator, that is, an
operator whose spectrum $\sigma(X)$ consists of only one point $0$. Is it true
that $\nu_{\mathbb{B}(\mathcal{H})}(X)=0$?

The answer to this problem is affirmative. In fact, the theorem below is given
by Apostol, Foias, Voiculescu in [1], which completely characterizes the
condition for an operator $X$ to satisfy $\nu_{\mathbb{B}(\mathcal{H})}(X)=0$.

######  Theorem 1.

Let $X\in\mathbb{B}(\mathcal{H})$. The equation
$\nu_{\mathbb{B}(\mathcal{H})}(X)=0$ holds if and only if the following three
conditions hold.

  * •

The spectrum $\sigma(X)$ of $X$ is connected and $0\in\sigma(X)$.

  * •

The essential spectrum $\sigma_{e}(X)$ of $X$ is connected and
$0\in\sigma_{e}(X)$.

  * •

If $\lambda\in\mathbb{C}$ and $\lambda I-X$ is semi-Fredholm, then the
Fredholm index of $\lambda I-X$ equals $0$.

It is natural to get interested in the value $\nu_{\mathcal{M}}(X)$ for a more
general factor ${\mathcal{M}}$ and an operator $X\in{\mathcal{M}}$. In this
note, we completely compute $\nu_{\mathcal{M}}(P)$ for a projection $P$ in a
general factor ${\mathcal{M}}$. Recall that a _projection_ is an operator
$P\in{\mathcal{M}}$ satisfying $P=P^{2}=P^{*}$. Let
${\mathcal{P}}({\mathcal{M}})$ denote the set of projections in
${\mathcal{M}}$. For $P\in{\mathcal{P}}({\mathcal{M}})$, we use the symbol
$P^{\perp}:=I-P$. It is clear that $\nu_{\mathcal{M}}(0)=0$.

It appears that the oldest result on $\nu_{\mathcal{M}}(P)$ for a projection
$P$ is given by Hedlund in 1972 [7]. Herrero [8, Corollary 9] completely
computed $\nu_{\mathcal{M}}(P)$ when ${\mathcal{M}}$ is a type I∞ factor. From
this result, it is not difficult to get the following.

######  Theorem 2 ([8, Corollary 9], see also [16, Section 8]).

Let ${\mathcal{M}}$ be an infinite factor and
$P\in{\mathcal{P}}({\mathcal{M}})\setminus\\{0\\}$.

  * •

If $P^{\perp}\in{\mathcal{P}}({\mathcal{M}})$ is an infinite projection, then
$\nu_{\mathcal{M}}(P)=1/2$.

  * •

If $P^{\perp}\in{\mathcal{P}}({\mathcal{M}})$ is a finite projection, then
$\nu_{\mathcal{M}}(P)=1$.

Therefore, we restrict our attention to finite factors. In the case of matrix
algebras, one may find some research in the literature. Let $n$ be a positive
integer and let $\mathbb{M}_{n}$ denote the algebra of $n\times n$ complex
matrices, which is a type In factor. MacDonald [11, Theorem 1] proved that
$\nu_{\mathbb{M}_{n}}(P)=(2\cos\frac{\pi}{n+2})^{-1}$ for every projection
$P\in{\mathcal{P}}(\mathbb{M}_{n})$ of rank $1$. Cramer [3, Theorem 3.6]
proved that $\nu_{\mathbb{M}_{n}}(P)=(2\cos\frac{\pi}{\frac{n}{n-1}+2})^{-1}$
for every projection $P\in{\mathcal{P}}(\mathbb{M}_{n})$ of rank $n-1$. He
conjectured that
$\nu_{\mathbb{M}_{n}}(P)=(2\cos\frac{\pi}{\frac{n}{m}+2})^{-1}$ for every
projection $P\in{\mathcal{P}}(\mathbb{M}_{n})$ of rank $m$, $1\leq m\leq n$
[3, Conjecture 5.1]. The second author recently proved that
$\nu_{\mathbb{M}_{n}}(P)\geq(2\cos\frac{\pi}{n+2})^{-1}$ for every nonzero
projection $P\in{\mathcal{P}}(\mathbb{M}_{n})$ [13], settling a conjecture
posed by MacDonald in [11]. For more information concerning related topics,
see [9, Chapter 2], [3], [12]. See also [16], [17], where the distance to the
set of (quasi)nilpotent operators is studied in various settings of C∗\- or
von Neumann algebras.

In what follows, let ${\mathcal{M}}$ be a finite factor. Then ${\mathcal{M}}$
is equipped with a unique normal faithful tracial state $\tau$. As one may see
from the preceding paragraph, the computation of the exact value of
$\nu_{\mathcal{M}}(P)$ was not complete even in the case of matrix algebras,
and little was known about the case of type II1 factors. The goal of this note
is to prove

######  Theorem 3.

If $0\neq P\in{\mathcal{P}}({\mathcal{M}})$, then
$\nu_{\mathcal{M}}(P)=(2\cos\theta)^{-1}$, where
$\theta=\displaystyle\frac{\tau(P)\pi}{1+2\tau(P)}$.

This completes the program of the determination of $\nu_{\mathcal{M}}(P)$ for
a general factor ${\mathcal{M}}$ and a general projection
$P\in{\mathcal{P}}({\mathcal{M}})$. Note that $\tau(X)=\mathrm{tr}\,X/n$ when
${\mathcal{M}}=\mathbb{M}_{n}$ and $X\in{\mathcal{M}}$. It follows that
Theorem 3 applied to a type In factor settles Cramer’s conjecture.

##  2\. Results

For $X\in{\mathcal{M}}$, the projection $E$ onto the closure of the range of
$X$ belongs to ${\mathcal{P}}(M)$. We call $E$ the range projection of $X$.

######  Lemma 4.

Let $n\geq 1$. An operator $N\in{\mathcal{M}}$ satisfies $N^{n}=0$ if and only
if there are projections $0=P_{0}\leq P_{1}\leq\cdots\leq P_{n}=I$ in
${\mathcal{M}}$ satisfying $P_{k-1}^{\perp}NP_{k}=0$ for every $1\leq k\leq
n$.

###### Proof.

If $N^{n}=0$, then define $P_{k}$ to be the range projection of $N^{n-k}$. The
converse is easier. ∎

######  Lemma 5.

Let $0=P_{0}\leq P_{1}\leq\cdots\leq P_{n}=I$ be in
${\mathcal{P}}({\mathcal{M}})$ and $X\in{\mathcal{M}}$. Set
$\mathcal{N}:=\\{N\in{\mathcal{M}}\mid P_{k-1}^{\perp}NP_{k}=0\text{ for every
}1\leq k\leq n\\}$. Then there is a nilpotent $N_{0}\in\mathcal{N}$ satisfying

$\lVert X-N_{0}\rVert=\inf_{N\in\mathcal{N}}\lVert X-N\rVert=\max_{1\leq k\leq
n}\lVert P_{k-1}^{\perp}XP_{k}\rVert.$

###### Proof.

This is a version of the so-called Arveson’s distance formula [2]. See [14,
Theorem 1] and [15]. ∎

For $X\in{\mathcal{M}}$, let $\delta(X)$ denote the trace of the range
projection of $X$.

######  Lemma 6.

If $X,Y\in{\mathcal{M}}$, then $\delta(XY),\delta(YX)\leq\delta(X)$ and
$\delta(X+Y)\leq\delta(X)+\delta(Y)$ hold.

###### Proof.

It is clear that $\delta(XY)\leq\delta(X)$. If $Z=V\lvert Z\rvert$ is the
polar decomposition of an operator $Z\in{\mathcal{M}}$, then $VV^{*}$ is the
range projection of $Z$ and $V^{*}V$ is the range projection of $Z^{*}$, hence
$\delta(Z)=\tau(VV^{*})=\tau(V^{*}V)=\delta(Z^{*})$. Thus we have
$\delta(YX)=\delta((YX)^{*})=\delta(X^{*}Y^{*})\leq\delta(X^{*})=\delta(X)$.

Let $E_{1},E_{2},E_{3}$ denote the range projections of $X+Y,X,Y$,
respectively. Then we have $E_{1}\leq E_{2}\vee E_{3}$. Since $E_{2}\vee
E_{3}-E_{2}$ is Murray–von Neumann equivalent to $E_{3}-E_{2}\wedge E_{3}$, we
have $\tau(E_{2}\vee E_{3}-E_{2})=\tau(E_{3}-E_{2}\wedge E_{3})$, which
implies $\tau(E_{2}\vee E_{3})\leq\tau(E_{2})+\tau(E_{3})$. Thus
$\delta(X+Y)\leq\delta(X)+\delta(Y)$ holds. ∎

######  Lemma 7.

Let $X\in{\mathcal{M}}$ satisfy $X\geq 0$. Let $n\geq 1$ and let $0=P_{0}\leq
P_{1}\leq\cdots\leq P_{n}=I$ be in ${\mathcal{P}}({\mathcal{M}})$. For each
$k$, put $A_{k}=X^{1/2}P_{k}X^{1/2}$. Then

$\lVert P_{k-1}^{\perp}XP_{k}\rVert=\lVert(X-A_{k-1})^{1/2}\cdot
A_{k}^{1/2}\rVert.$

Moreover,

(1) $0=A_{0}\leq A_{1}\leq\cdots\leq A_{n}=X\text{ and
}\sum_{k=1}^{n}\delta(A_{k}-A_{k-1})\leq 1$

hold.

###### Proof.

We have

$\begin{split}\lVert P_{k-1}^{\perp}XP_{k}\rVert&=\lVert
P_{k-1}^{\perp}X^{1/2}\cdot X^{1/2}P_{k}\rVert\\\
&=\lVert(X^{1/2}P_{k-1}^{\perp}X^{1/2})^{1/2}\cdot(X^{1/2}P_{k}X^{1/2})^{1/2}\rVert\\\
&=\lVert(X-X^{1/2}P_{k-1}X^{1/2})^{1/2}\cdot(X^{1/2}P_{k}X^{1/2})^{1/2}\rVert\\\
&=\lVert(X-A_{k-1})^{1/2}\cdot A_{k}^{1/2}\rVert.\end{split}$

It is clear that $0=A_{0}\leq A_{1}\leq\cdots\leq A_{n}=X$. Since
$\delta(A_{k}-A_{k-1})\leq\delta(P_{k}-P_{k-1})=\tau(P_{k}-P_{k-1})$, we get
$\sum_{k=1}^{n}\delta(A_{k}-A_{k-1})\leq 1$. ∎

Conversely, the following holds.

######  Lemma 8.

Let $X\in{\mathcal{M}}$ satisfy $X\geq 0$. Let $n\geq 1$ and
$A_{0},A_{1},\dots,A_{n}\in{\mathcal{M}}$ satisfy (1). Then there are
projections $0=P_{0}\leq P_{1}\leq\cdots\leq P_{n}=I$ in ${\mathcal{M}}$
satisfying $X^{1/2}P_{k}X^{1/2}=A_{k}$ for every $k$.

###### Proof.

Since $\displaystyle\sum_{k=1}^{n}\delta(A_{k}-A_{k-1})\leq 1$, one may choose
operators $Y_{1},Y_{2},\ldots,Y_{n}\in{\mathcal{M}}$, whose range projections
$Q_{1},Q_{2},\ldots,Q_{n}$ are pairwise orthogonal, with the property
$Y_{k}^{*}Y_{k}=A_{k}-A_{k-1}$. Then $Y:=Y_{1}+Y_{2}+\cdots+Y_{n}$ satisfies
$Y^{*}(Q_{1}+\cdots+Q_{k})Y=A_{k}$ for every $k$. Since ${\mathcal{M}}$ is of
finite type and $Y^{*}Y=X$, one may take a unitary operator
$U\in{\mathcal{M}}$ with $Y=UX^{1/2}$. Thus $P_{k}=U^{*}(Q_{1}+\cdots
Q_{k})U$, $k\leq n-1$, and $P_{n}=I$ satisfy the desired property. ∎

Therefore, we get the following.

######  Proposition 9.

Let $X\in{\mathcal{M}}$ and $X\geq 0$. Then

$\nu_{\mathcal{M}}(X)=\inf\max_{1\leq k\leq n}\lVert(X-A_{k-1})^{1/2}\cdot
A_{k}^{1/2}\rVert,$

where the infimum in the right-hand side is taken over the collection of all
$n\geq 1$ and $(A_{0},A_{1},\dots,A_{n})\in{\mathcal{M}}^{n+1}$ satisfying
(1).

In what follows, we study the case $0\neq X=P\in{\mathcal{P}}({\mathcal{M}})$.
Let $n\geq 1$ and $(A_{0},A_{1},\dots,A_{n})\in{\mathcal{M}}^{n+1}$ satisfy
(1). Set $\displaystyle\theta:=\frac{\tau(P)\pi}{1+2\tau(P)}$ and
$\varphi:=\theta/2$. To get Theorem 3, we need to compare $(2\cos\theta)^{-1}$
with

$\max_{1\leq k\leq n}\lVert(P-A_{k-1})^{1/2}A_{k}^{1/2}\rVert=\max_{1\leq
k\leq n}\lVert(I-A_{k-1})^{1/2}A_{k}^{1/2}\rVert.$

Observe that the mapping

$t\mapsto\frac{1}{2\cos\theta}\cdot\frac{\sin t}{\sin(t+\theta)}$

is a monotone increasing bijection from $[0,\pi-2\theta]$ onto $[0,1]$. Since
$0\leq A_{k}\leq P$, by functional calculus we may uniquely take an operator
$B_{k}\in{\mathcal{M}}$ satisfying $0\leq B_{k}\leq(\pi-2\theta)P$ and

(2) $A_{k}=\frac{1}{2\cos\theta}\cdot\frac{\sin B_{k}}{\sin(B_{k}+\theta I)}.$

(Note that the right-hand side makes sense because $\sin(B_{k}+\theta I)$ is
invertible and the pair $\sin B_{k}$, $\sin(B_{k}+\theta I)$ commutes. In what
follows we frequently use such fractional expressions.) Since $A_{0}\leq
A_{1}\leq\cdots\leq A_{n}$, we have

$\frac{\sin B_{0}}{\sin(B_{0}+\theta I)}\leq\frac{\sin
B_{1}}{\sin(B_{1}+\theta I)}\leq\cdots\leq\frac{\sin B_{n}}{\sin(B_{n}+\theta
I)}.$

######  Lemma 10.

Let $B,{B^{\prime}}\in{\mathcal{M}}$ satisfy $0\leq B\leq(\pi-2\theta)I$,
$0\leq{B^{\prime}}\leq(\pi-2\theta)I$. Then $\displaystyle\frac{\sin
B}{\sin(B+\theta I)}\leq\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta I)}$
holds if and only if $\cot(B+{\varphi}I)\geq\cot({B^{\prime}}+{\varphi}I)$.
Moreover, we have

$\delta\left(\frac{\sin B}{\sin(B+\theta
I)}-\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta
I)}\right)=\delta(\cot(B+{\varphi}I)-\cot({B^{\prime}}+{\varphi}I)).$

###### Proof.

By applying the formula
$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$ to mutually
commuting operators, we get

$\begin{split}\sin(B+{\varphi}I\pm{\varphi}I)&=\sin(B+{\varphi}I)\cos({\varphi}I)\pm\cos(B+{\varphi}I)\sin({\varphi}I)\\\
&=\cos{\varphi}\sin(B+{\varphi}I)\pm\sin{\varphi}\cos(B+{\varphi}I).\end{split}$

Thus

$\frac{\sin B}{\sin(B+\theta
I)}=\frac{\cos{\varphi}\sin(B+{\varphi}I)-\sin{\varphi}\cos(B+{\varphi}I)}{\cos{\varphi}\sin(B+{\varphi}I)+\sin{\varphi}\cos(B+{\varphi}I)}=\frac{2I}{I+\tan{\varphi}\cot(B+{\varphi}I)}-I.$

We also get

$\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta
I)}=\frac{2I}{I+\tan{\varphi}\cot({B^{\prime}}+{\varphi}I)}-I.$

Therefore, $\displaystyle\frac{\sin B}{\sin(B+\theta
I)}\leq\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta I)}$ holds if and only
if

$\frac{2I}{I+\tan{\varphi}\cot(B+{\varphi}I)}\leq\frac{2I}{I+\tan{\varphi}\cot({B^{\prime}}+{\varphi}I)}.$

Since $X\leq Y\iff X^{-1}\geq Y^{-1}$ for positive invertible operators, we
see that the above condition is equivalent to
$\cot(B+{\varphi}I)\geq\cot({B^{\prime}}+{\varphi}I)$.

Using the identity $X^{-1}-Y^{-1}=X^{-1}(Y-X)Y^{-1}$ for invertible operators,
we also get the equality concerning $\delta$. ∎

######  Lemma 11.

Let $B,{B^{\prime}}\in{\mathcal{M}}$ satisfy $0\leq B\leq(\pi-2\theta)I$,
$0\leq{B^{\prime}}\leq(\pi-2\theta)I$. Put

$A=\frac{1}{2\cos\theta}\cdot\frac{\sin B}{\sin(B+\theta I)},\quad
A^{\prime}=\frac{1}{2\cos\theta}\cdot\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta
I)}.$

Then
$\displaystyle\lVert(I-A)^{1/2}(A^{\prime})^{1/2}\rVert\leq(2\cos\theta)^{-1}$
if and only if $\cot({B^{\prime}}+{\varphi}I)\geq\cot(B+{3\varphi}I)$.

###### Proof.

Since

$\sin B+\sin(B+2\theta I)=2\sin(B+\theta I)\cos(\theta
I)=2\cos\theta\sin(B+\theta I),$

we get

$I-\frac{1}{2\cos\theta}\cdot\frac{\sin B}{\sin(B+\theta
I)}=\frac{1}{2\cos\theta}\cdot\frac{\sin(B+2\theta I)}{\sin(B+\theta I)}.$

It follows that

$\lVert(I-A)^{1/2}(A^{\prime})^{1/2}\rVert\leq(2\cos\theta)^{-1}\iff\left\lVert\left(\frac{\sin(B+2\theta
I)}{\sin(B+\theta
I)}\right)^{1/2}\left(\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta
I)}\right)^{1/2}\right\rVert\leq 1.$

Using the argument as in the proof of the preceding lemma, we get

$\frac{\sin(B+2\theta I)}{\sin(B+\theta
I)}=\frac{I+\tan{\varphi}\cot(B+{3\varphi}I)}{I-\tan{\varphi}\cot(B+{3\varphi}I)}=\frac{I+S}{I-S}$

and

$\frac{\sin{B^{\prime}}}{\sin({B^{\prime}}+\theta
I)}=\frac{I-\tan{\varphi}\cot({B^{\prime}}+{\varphi}I)}{I+\tan{\varphi}\cot({B^{\prime}}+{\varphi}I)}=\frac{I-T}{I+T},$

where $S=\tan{\varphi}\cot(B+{3\varphi}I)$ and
$T=\tan{\varphi}\cot({B^{\prime}}+{\varphi}I)$. (Observe that $-I\leq S\leq I$
and $-I\leq T\leq I$ hold. Note that $I-S$ and $I+T$ are invertible, while
$I+S$ and $I-T$ are not necessarily invertible.) Hence

$\begin{split}\lVert(I-A)^{1/2}(A^{\prime})^{1/2}\rVert\leq(2\cos\theta)^{-1}&\iff\left\lVert\left(\frac{I+S}{I-S}\right)^{1/2}\left(\frac{I-T}{I+T}\right)^{1/2}\right\rVert\leq
1\\\
&\iff\left(\frac{I-T}{I+T}\right)^{1/2}\left(\frac{I+S}{I-S}\right)\left(\frac{I-T}{I+T}\right)^{1/2}\leq
I\\\ &\iff(I-T)^{1/2}\left(\frac{2I}{I-S}-I\right)(I-T)^{1/2}\leq I+T\\\ &\iff
2(I-T)^{1/2}(I-S)^{-1}(I-T)^{1/2}\leq 2I\\\
&\iff(I-T)^{1/2}(I-S)^{-1}(I-T)^{1/2}\leq I\\\
&\iff\lVert(I-S)^{-1/2}(I-T)^{1/2}\rVert\leq 1\\\
&\iff(I-S)^{-1/2}(I-T)(I-S)^{-1/2}\leq I\\\ &\iff I-T\leq I-S\\\ &\iff S\leq
T\\\ &\iff\cot(B+{3\varphi}I)\leq\cot({B^{\prime}}+{\varphi}I).\end{split}$

∎

######  Lemma 12.

Let $X,Y,Z\in{\mathcal{M}}$ be self-adjoint operators satisfying $X\geq Y\geq
Z$. If $X-Z$ is invertible, then there uniquely exists a self-adjoint operator
$W\in{\mathcal{M}}$ satisfying $X\geq Y\geq W\geq Z$,
$\delta(X-Y)=\delta(X-W)$, and $\delta(X-W)+\delta(W-Z)=1$.

###### Proof.

One may assume $Z=0$ without loss of generality. Then $X\geq Y\geq 0$ and $X$
is invertible. By multiplying $X^{-1/2}$ from both sides of each operator, we
may also assume $X=I$ without loss of generality. Then we have $I\geq Y\geq
0$. Assume that a self-adjoint operator $W\in{\mathcal{M}}$ satisfies the
desired properties. From $\delta(I-W)+\delta(W)=1$, we see that $W$ is a
projection. This together with $I\geq Y\geq W\geq 0$,
$\delta(I-Y)=\delta(I-W)$ shows that $W=\chi_{\\{1\\}}(Y)$. Conversely, it is
easy to see that $W=\chi_{\\{1\\}}(Y)$ satisfies the desired properties. ∎

######  Lemma 13.

Let $X,Y\in{\mathcal{M}}$ be self-adjoint operators satisfying $X\leq Y$. If
$f\colon\sigma(X)\cup\sigma(Y)\to\mathbb{R}$ is a monotone decreasing
function, then $\tau(f(X))\geq\tau(f(Y))$.

###### Proof.

By considering the pair $X+\alpha I$, $Y+\alpha I$ instead of $X$, $Y$ for a
suitable real number $\alpha$, one may assume
$\sigma(X),\sigma(Y)\subset(0,\infty)$ without loss of generality. One may
also assume that the range of $f$ is contained in $(0,\infty)$ by adding some
constant function. For a real number $t\in(0,1]$ and a positive operator
$Z\in{\mathcal{M}}$, we define

$\mu_{t}(Z)=\inf\\{s\geq 0\mid\tau(\chi_{(s,\infty)}(Z))\leq t\\}.$

By [5, Lemma 2.5 (iii)], we have $\mu_{t}(X)\leq\mu_{t}(Y)$ for every
$t\in(0,1]$. This together with [5, Remark 3.3] and the assumption on $f$
implies

$\tau(f(X))=\int_{0}^{1}f(\mu_{t}(X))\,dt\geq\int_{0}^{1}f(\mu_{t}(Y))\,dt=\tau(f(Y)).$

∎

######  Lemma 14.

Let $U,V\in{\mathcal{M}}$ be unitaries, and $r,s$ real numbers with $0\leq
r\leq 1$, $0<s<2\pi$. If $\delta(U-V)=r$ and $\delta(U-e^{is}V)=1-r$, then
there is a projection $Q\in{\mathcal{P}}({\mathcal{M}})$ satisfying
$\tau(Q)=r$ and $U=Ve^{isQ}$.

###### Proof.

It follows from $\delta(U-V)=r$ and $\delta(U-e^{is}V)=1-r$ that
$\delta(V^{*}U-I)=r$ and $\delta(V^{*}U-e^{is}I)=1-r$. Considering the
spectral distribution of the unitary $V^{*}U$, we see that there is a
projection $Q\in{\mathcal{P}}({\mathcal{M}})$ satisfying $\tau(Q)=r$ and
$V^{*}U=e^{isQ}$. ∎

######  Lemma 15.

Let $B,{B^{\prime}}\in{\mathcal{M}}$ satisfy $0\leq B\leq(\pi-2\theta)I$,
$0\leq{B^{\prime}}\leq(\pi-2\theta)I$, and

$\cot(B+{\varphi}I)\geq\cot({B^{\prime}}+{\varphi}I)\geq\cot(B+{3\varphi}I).$

Set $r=\delta(\cot(B+{\varphi}I)-\cot({B^{\prime}}+{\varphi}I))$. Then
$\tau({B^{\prime}})-\tau(B)\leq r\theta$. Moreover,
$\tau({B^{\prime}})-\tau(B)=r\theta$ holds if and only if
$\delta(\cot({B^{\prime}}+{\varphi}I)-\cot(B+{3\varphi}I))=1-r$.

###### Proof.

Since $\cot(B+{\varphi}I)-\cot(B+{3\varphi}I)$ is invertible, by Lemma 12 we
may find an operator $C\in{\mathcal{M}}$ satisfying
$\delta(\cot(B+{\varphi}I)-C)=r$, $\delta(C-\cot(B+{3\varphi}I))=1-r$, and

$\cot(B+{\varphi}I)\geq\cot({B^{\prime}}+{\varphi}I)\geq
C\geq\cot(B+{3\varphi}I).$

One may take an operator $D\in{\mathcal{M}}$ with ${\varphi}I\leq
D\leq(\pi-{\varphi})I$ and $C=\cot D$. Then

$\cot(B+{\varphi}I)\geq\cot({B^{\prime}}+{\varphi}I)\geq\cot
D\geq\cot(B+{3\varphi}I).$

Since the inverse function of $\cot$ is monotone decreasing, Lemma 13 implies
that

$\tau(B+{\varphi}I)\leq\tau({B^{\prime}}+{\varphi}I)\leq\tau(D)\leq\tau(B+{3\varphi}I).$

Hence

$\tau({B^{\prime}})-\tau(B)=\tau({B^{\prime}}+{\varphi}I)-\tau(B+{\varphi}I)\leq\tau(D)-\tau(B+{\varphi}I).$

Moreover, we see that $\tau({B^{\prime}}+{\varphi}I)=\tau(D)$ holds if and
only if $\cot({B^{\prime}}+{\varphi}I)=\cot D$, which is in turn equivalent to
$\delta(\cot({B^{\prime}}+{\varphi}I)-\cot(B+{3\varphi}I))=1-r$. Consequently,
to get the desired conclusion, it suffices to show that
$\tau(D)-\tau(B+{\varphi}I)=r\theta$. By $\delta(\cot(B+{\varphi}I)-C)=r$ and
the equality

(3) $\begin{split}\cot(B+{\varphi}I)-C&=\cot(B+{\varphi}I)-\cot D\\\
&=i\left(\frac{2I}{e^{2i(B+\varphi
I)}-I}+I\right)-i\left(\frac{2I}{e^{2iD}-I}+I\right)\\\ &=2i((e^{2i(B+\varphi
I)}-I)^{-1}-(e^{2iD}-I)^{-1})\\\ &=2i(e^{2i(B+\varphi
I)}-I)^{-1}(e^{2iD}-e^{2i(B+\varphi I)})(e^{2iD}-I)^{-1},\end{split}$

we get

(4) $\delta(e^{2iD}-e^{2i(B+\varphi I)})=r.$

Similarly, $\delta(C-\cot(B+{3\varphi}I))=1-r$ implies

(5) $\delta(e^{2iD}-e^{2i(B+3\varphi I)})=1-r.$

The equations (4) and (5) and Lemma 14 imply that there is a projection
$Q\in{\mathcal{P}}({\mathcal{M}})$ satisfying $\tau(Q)=r$ and
$e^{2iD}=e^{2i(B+\varphi I)}\cdot e^{4i\varphi Q}$. Thus Lemma 18 below
implies that $\tau(2D)=\tau(2(B+\varphi I))+\tau(4\varphi Q)$, hence
$\tau(D)-\tau(B+\varphi I)=2\tau(Q)\varphi=r\theta$. ∎

######  Lemma 16 (Lemma 3 in Section I.6.11 of [4]).

Let $\mathcal{U}\subset\mathbb{C}$ be an open set and
$f\colon\mathcal{U}\to\mathbb{C}$ an analytic function. Let
$g\colon[0,1]\to\mathcal{M}$ be a norm-differentiable function. Assume that
the spectra $\sigma(g(t))$, $t\in[0,1]$, are contained in a common compact
subset of $\mathcal{U}$. Then

$\frac{d}{dt}\tau(f\circ g(t))=\tau(f^{\prime}(g(t))g^{\prime}(t))$

holds for every $t\in[0,1]$.

######  Lemma 17.

Let $X,Y\in{\mathcal{M}}$ be positive operators. Then

$\sigma(e^{iX}e^{iY})\subset\\{e^{i\alpha}\mid 0\leq\alpha\leq\lVert
X\rVert+\lVert Y\rVert\\}.$

###### Proof.

This is trivial when $\lVert X\rVert+\lVert Y\rVert\geq 2\pi$ (note that
$e^{iX}e^{iY}$ is unitary). Assume that $\lVert X\rVert+\lVert
Y\rVert<\alpha<2\pi$. We need to show that $e^{iX}e^{iY}-e^{i\alpha}I$ is
invertible. Observe that the convex hull of $\sigma(e^{iX})$ and that of
$\sigma(e^{i\alpha}e^{-iY})$ do not intersect. Let $\mathcal{H}$ be the
Hilbert space on which ${\mathcal{M}}$ acts. Observe that $\langle
Zh,h\rangle$ lies in the convex hull of $\sigma(Z)$ for every normal operator
$Z\in{\mathcal{M}}$ and every unit vector $h$. It follows that there is a
positive real number $\varepsilon>0$ satisfying
$\lvert\langle(e^{iX}-e^{i\alpha}e^{-iY})h,h\rangle\rvert\geq\varepsilon$ for
every unit vector $h\in\mathcal{H}$. This implies that
$\lVert(e^{iX}-e^{i\alpha}e^{-iY})h\rVert\geq\varepsilon$ and
$\lVert(e^{iX}-e^{i\alpha}e^{-iY})^{*}h\rVert\geq\varepsilon$ for every unit
vector $h\in\mathcal{H}$. Hence $e^{iX}-e^{i\alpha}e^{-iY}$ is invertible, so
$e^{iX}e^{iY}-e^{i\alpha}I$ is also invertible. ∎

######  Lemma 18 (See also Lemma 4 in Section I.6.11 of [4]).

Let $X,Y,Z\in{\mathcal{M}}$ be positive operators satisfying $\lVert
X\rVert+\lVert Y\rVert<2\pi$, $\lVert Z\rVert<2\pi$, and
$e^{iZ}=e^{iX}e^{iY}$. Then $\tau(Z)=\tau(X)+\tau(Y)$.

###### Proof.

Consider the mapping $g\colon[0,1]\to\mathcal{M}$ defined by
$g(t)=e^{iX}e^{itY}$. By Lemma 17, we get
$\sigma(g(t))\subset\\{e^{i\alpha}\mid 0\leq\alpha\leq\lVert X\rVert+\lVert
Y\rVert\\}$ for every $t\in[0,1]$. Fix $\beta\in(\lVert X\rVert+\lVert
Y\rVert,2\pi)$. Let $f\colon\\{re^{i\alpha}\mid
r>0,\,\beta-2\pi<\alpha<\beta\\}\to\mathbb{C}$ be the inverse mapping of the
exponential function on $\\{a+bi\mid a,b\in\mathbb{R},\
\beta-2\pi<b<\beta\\}$. Observe that $g^{\prime}(t)=e^{iX}e^{itY}\cdot
iY=g(t)\cdot iY$ and $f^{\prime}(z)=1/z$. Thus Lemma 16 implies that

$\frac{d}{dt}\tau(f\circ g(t))=\tau(g(t)^{-1}g^{\prime}(t))=i\tau(Y).$

By integrating over $[0,1]$, we get

$\tau(f\circ g(1))-\tau(f\circ g(0))=i\tau(Y).$

Since $g(1)=e^{iX}e^{iY}=e^{iZ}$ and $g(0)=e^{iX}$, we obtain $f\circ g(1)=iZ$
and $f\circ g(0)=iX$. Thus the desired conclusion follows. ∎

Now we are ready to prove one-half of Theorem 3.

###### Proof of $\nu_{\mathcal{M}}(P)\geq(2\cos\theta)^{-1}$.

Let $A_{k}$ and $B_{k}$ be as above. Assume that

$\max_{1\leq k\leq
n}\lVert(P-A_{k-1})^{1/2}A_{k}^{1/2}\rVert<(2\cos\theta)^{-1}.$

By Lemmas 10, 11, 15, we have
$\tau(B_{k})-\tau(B_{k-1})<\delta(A_{k}-A_{k-1})\theta$ for every $k$. Thus we
obtain $\tau(B_{n})-\tau(B_{0})<\theta$. Since $B_{0}=0$ and
$B_{n}=(\pi-2\theta)P$, we get $(\pi-2\theta)\tau(P)<\theta$, so
$\displaystyle\theta>\frac{\tau(P)\pi}{1+2\tau(P)}$, a contradiction. ∎

We continue with the other half of the proof. We first study the case where
${\mathcal{M}}=\mathbb{M}_{n}$ is the $n\times n$ matrix algebra and
$\tau(X)=\mathrm{tr}\,X/n$, $X\in\mathbb{M}_{n}$.

######  Lemma 19.

Let $X,Y,Z\in\mathbb{M}_{n}$ be self-adjoint ($=$hermitian) matrices. If $X-Z$
is positive and invertible, and if
$\mathrm{rank}\,(X-Y)+\mathrm{rank}\,(Y-Z)=n$, then $X\geq Y\geq Z$.

###### Proof.

As in the proof of Lemma 12, one may assume $Z=0$ and $X=I$ without loss of
generality. In that case, we have $\mathrm{rank}\,(I-Y)+\mathrm{rank}\,Y=n$,
which shows that $Y$ is a projection, so $I\geq Y\geq 0$, as desired. ∎

######  Proposition 20.

Let $0\neq P\in{\mathcal{P}}(\mathbb{M}_{n})$ and
$\displaystyle\theta:=\frac{\tau(P)\pi}{1+2\tau(P)}$. Let $0=A_{0}\leq
A_{1}\leq\cdots\leq A_{n}=P$ be in $\mathbb{M}_{n}$ satisfying
$\mathrm{rank}(A_{k}-A_{k-1})=1$ for every $k\in\\{1,2,\dots,n\\}$. Take
$B_{k}\in\mathbb{M}_{n}$ satisfying $0\leq B_{k}\leq(\pi-2\theta)P$ and the
equation (2). Then $\displaystyle\max_{1\leq k\leq
n}\lVert(P-A_{k-1})^{1/2}A_{k}^{1/2}\rVert\leq(2\cos\theta)^{-1}$ holds if and
only if $\mathrm{tr}\,B_{k}=k\theta$ holds for every $k\in\\{1,2,\dots,n\\}$.

###### Proof.

Assume that $\displaystyle\max_{1\leq k\leq
n}\lVert(P-A_{k-1})^{1/2}A_{k}^{1/2}\rVert\leq(2\cos\theta)^{-1}$ holds. By
Lemma 15, we have $\tau(B_{k})-\tau(B_{k-1})\leq\theta/n$, or equivalently,
$\mathrm{tr}\,(B_{k}-B_{k-1})\leq\theta$ for each $k$. This together with
$\mathrm{tr}\,B_{0}=0$ and
$\mathrm{tr}\,B_{n}=\mathrm{tr}\,((\pi-2\theta)P)=n\theta$ shows that
$\mathrm{tr}\,B_{k}=k\theta$.

Conversely, assume that $\mathrm{tr}\,B_{k}=k\theta$ holds for every
$k\in\\{1,2,\dots,n\\}$. We show $\displaystyle\max_{1\leq k\leq
n}\lVert(P-A_{k-1})^{1/2}A_{k}^{1/2}\rVert\leq(2\cos\theta)^{-1}$. By Lemma
11, it suffices to show that
$\cot(B_{k}+{\varphi}I)\geq\cot(B_{k-1}+{3\varphi}I)$ for every
$k\in\\{1,2,\dots,n\\}$. Let $k\in\\{1,2,\dots,n\\}$. By Lemma 10, we know
that

(6)
$\mathrm{rank}\,(\cot(B_{k-1}+{\varphi}I)-\cot(B_{k}+{\varphi}I))=\mathrm{rank}\,(A_{k-1}-A_{k})=1.$

By imitating the calculation in (3), we get

$\begin{split}&\quad\cot(B_{k-1}+{\varphi}I)-\cot(B_{k}+{\varphi}I)\\\
&=2i(e^{2i(B_{k-1}+\varphi I)}-I)^{-1}(e^{2i(B_{k}+\varphi
I)}-e^{2i(B_{k-1}+\varphi I)})(e^{2i(B_{k}+\varphi I)}-I)^{-1}\\\
&=2ie^{2i\varphi}(e^{2i(B_{k-1}+\varphi
I)}-I)^{-1}(e^{2iB_{k}}-e^{2iB_{k-1}})(e^{2i(B_{k}+\varphi
I)}-I)^{-1},\end{split}$

which implies $\mathrm{rank}\,(e^{2iB_{k}}-e^{2iB_{k-1}})=1$ and thus
$\mathrm{rank}\,(e^{-2iB_{k-1}}e^{2iB_{k}}-I)=1$. On the other hand, we know
$\mathrm{tr}\,(B_{k}-B_{k-1})=\theta$ from the assumption. Recall that $\det
e^{X}=e^{\mathrm{tr}\,X}$ holds for every $X\in\mathbb{M}_{n}$. Thus we get

$\det(e^{-2iB_{k-1}}e^{2iB_{k}})=e^{\mathrm{tr}\,(-2iB_{k-1})}\cdot
e^{\mathrm{tr}\,(2iB_{k})}=e^{2i\theta}.$

It follows that there is a projection $Q\in\mathbb{M}_{n}$ of rank one
satisfying $e^{-2iB_{k-1}}e^{2iB_{k}}=e^{2i\theta Q}$. Then

$\mathrm{rank}\,(e^{2iB_{k}}-e^{2i(B_{k-1}+\theta
I)})=\mathrm{rank}\,(e^{2iB_{k-1}}(e^{2i\theta Q}-e^{2i\theta}I))=n-1.$

By imitating the calculation in (3), we also get

$\begin{split}&\quad\cot(B_{k-1}+{3\varphi}I)-\cot(B_{k}+{\varphi}I)\\\
&=2i(e^{2i(B_{k-1}+3\varphi I)}-I)^{-1}(e^{2i(B_{k}+\varphi
I)}-e^{2i(B_{k-1}+3\varphi I)})(e^{2i(B_{k}+\varphi I)}-I)^{-1}\\\
&=2ie^{2i\varphi}(e^{2i(B_{k-1}+3\varphi
I)}-I)^{-1}(e^{2iB_{k}}-e^{2i(B_{k-1}+\theta I)})(e^{2i(B_{k}+\varphi
I)}-I)^{-1}.\end{split}$

Thus we get

(7) $\mathrm{rank}\,(\cot(B_{k-1}+{3\varphi}I)-\cot(B_{k}+{\varphi}I))=n-1.$

Since $\cot(B_{k-1}+{\varphi}I)-\cot(B_{k-1}+{3\varphi}I)$ is positive and
invertible, Lemma 19 with (6) and (7) shows that

$\cot(B_{k-1}+{\varphi}I)\geq\cot(B_{k}+{\varphi}I)\geq\cot(B_{k-1}+{3\varphi}I),$

which completes the proof. ∎

We will need the following well-known fact in matrix analysis.

######  Lemma 21 (See for example [10, Theorem 4.3.26]).

Let

$\alpha_{1}^{\prime}\geq\alpha_{1}\geq\alpha_{2}^{\prime}\geq\alpha_{2}\geq\cdots\geq\alpha_{n}^{\prime}\geq\alpha_{n}$

be real numbers. Let $A\in\mathbb{M}_{n}$ be a self-adjoint matrix whose
eigenvalues are $\alpha_{1}\geq\alpha_{2}\geq\cdots\geq\alpha_{n}$. Then there
is a self-adjoint matrix $A^{\prime}\in\mathbb{M}_{n}$ whose eigenvalues are
$\alpha_{1}^{\prime}\geq\alpha_{2}^{\prime}\geq\cdots\geq\alpha_{n}^{\prime}$
with $A\leq A^{\prime}$ and $\mathrm{rank}\,(A^{\prime}-A)\leq 1$.

######  Lemma 22.

Let $0\neq P\in{\mathcal{P}}(\mathbb{M}_{n})$ and
$\displaystyle\theta:=\frac{\tau(P)\pi}{1+2\tau(P)}$. Then
$\nu_{\mathbb{M}_{n}}(P)\leq(2\cos\theta)^{-1}$.

###### Proof.

Fix real numbers $\beta_{kl}\in[0,\pi-2\theta]$ ($0\leq k\leq n$, $1\leq l\leq
n$) satisfying the following conditions:

  1. (a)

$\displaystyle\sum_{l=1}^{n}\beta_{kl}=k\theta$ for every
$k\in\\{0,1,\dots,n\\}$.

  2. (b)

$\beta_{k,1}\geq\beta_{k-1,1}\geq\beta_{k,2}\geq\beta_{k-1,2}\geq\cdots\geq\beta_{k,n}\geq\beta_{k-1,n}$
for every $k\in\\{1,2,\dots,n\\}$.

  3. (c)

$\beta_{nl}=\pi-2\theta$ for every $l\leq\mathrm{rank}\,P$, and $\beta_{nl}=0$
for every $l>\mathrm{rank}\,P$.

For example, we may choose
$\beta_{kl}=\max\\{\min\\{\pi-2\theta,k\theta-(l-1)(\pi-2\theta)\\},0\\}$.

We set

$\alpha_{kl}=\frac{1}{2\cos\theta}\cdot\frac{\sin\beta_{kl}}{\sin(\beta_{kl}+\theta)}\in[0,1].$

Since the mapping $t\mapsto\sin t/\sin(t+\theta)$ is monotone on
$[0,\pi-2\theta]$, we see from (b) that

$\alpha_{k,1}\geq\alpha_{k-1,1}\geq\alpha_{k,2}\geq\alpha_{k-1,2}\geq\cdots\geq\alpha_{k,n}\geq\alpha_{k-1,n}$

for every $k\in\\{1,2,\dots,n\\}$. Therefore, Lemma 21 implies that there are
$0=A_{0}\leq A_{1}\leq\cdots\leq A_{n}$ in $\mathbb{M}_{n}$ such that
$\mathrm{rank}(A_{k}-A_{k-1})=1$ and
$\alpha_{k1}\geq\alpha_{k2}\geq\cdots\geq\alpha_{kn}$ are the eigenvalues of
$A_{k}$ for every $k\in\\{1,2,\dots,n\\}$. By (c), $A_{n}$ is a projection
that has the same rank as $P$. Thus $A_{n}$ is unitarily equivalent to $P$,
which implies $\nu_{\mathbb{M}_{n}}(A_{n})=\nu_{\mathbb{M}_{n}}(P)$. Take
$B_{k}\in\mathbb{M}_{n}$ satisfying $0\leq B_{k}\leq(\pi-2\theta)A_{n}$ and
the equation (2). Then (a) implies that $\mathrm{tr}\,B_{k}=k\theta$ for every
$k\in\\{1,2,\dots,n\\}$. Therefore, by Propositions 9 and 20, we obtain
$\nu_{\mathbb{M}_{n}}(A_{n})\leq(2\cos\theta)^{-1}$. ∎

Let us get back to the general situation and complete the proof of Theorem 3.

###### Proof of $\nu_{\mathcal{M}}(P)\leq(2\cos\theta)^{-1}$.

If ${\mathcal{M}}$ is of type I, then Lemma 22 gives the desired conclusion.
Assume that ${\mathcal{M}}$ is of type II1. Let $n\geq 1$ be an integer and
take the unique integer $a_{n}\in\\{1,2,\dots,n\\}$ with
$(a_{n}-1)/n<\tau(P)\leq a_{n}/n$. Since ${\mathcal{M}}$ is a type II1 factor,
one may take a projection $Q\in{\mathcal{P}}({\mathcal{M}})$ satisfying $P\leq
Q$ and $\tau(P)=\tau(Q)\cdot a_{n}/n$. One may also take a system of matrix
units $\\{E_{ij}\\}_{1\leq i,j\leq n}$ in the algebra $QMQ$ with
$E_{11}+\cdots+E_{a_{n}a_{n}}=P$. Let ${\mathcal{M}}_{1}$ be the algebra
generated by $\\{E_{ij}\\}_{1\leq i,j\leq n}$. Observe that
$P\in{\mathcal{M}}_{1}$ and that ${\mathcal{M}}_{1}$ can be identified with
$\mathbb{M}_{n}$. Set
$\displaystyle\theta_{n}:=\frac{(a_{n}/n)\pi}{1+2(a_{n}/n)}$. By Lemma 22, we
obtain $\nu_{{\mathcal{M}}_{1}}(P)\leq(2\cos\theta_{n})^{-1}$. It is clear
that $\nu_{{\mathcal{M}}}(P)\leq\nu_{{\mathcal{M}}_{1}}(P)$. Since
$\theta_{n}\to\theta$ as $n\to\infty$, we get
$\nu_{\mathcal{M}}(P)\leq(2\cos\theta)^{-1}$, as desired. ∎

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