# The geometry of positively curved Kähler metrics on tube domains

Gabriel Khan  Gabriel Khan. M3 Laboratory, University of Michigan, Ann Arbor,
MI 92093, USA<EMAIL_ADDRESS>,  Jun Zhang  Jun Zhang. Department of
Mathematics, University of Michigan, Ann Arbor, MI 92093, USA<EMAIL_ADDRESS>and  Fangyang Zheng  Fangyang Zheng. School of Mathematical Sciences,
Chongqing Normal University, Chongqing 401131, China<EMAIL_ADDRESS>
###### Abstract.

In this article, we study a class of Kähler manifolds defined on tube domains
in $\mathbb{C}^{n}$, and in particular those which have
$O(n)\times\mathbb{R}^{n}$ symmetry. For these, we prove a uniqueness result
showing that any such manifold which is complete and has non-negative
orthogonal bisectional curvature ($n\geq 3$) or non-negative bisectional
curvature ($n\geq 2$) is biholomorphically isometric to $\mathbb{C}^{n}$. We
also consider another curvature tensor called the “orthogonal anti-
bisectional” curvature and find necessary and sufficient conditions for a
complete $O(n)$-symmetric tube domain to have non-negative orthogonal anti-
bisectional curvature. We provide several examples of complete metrics which
satisfy this condition. These examples are also of interest to optimal
transport, as they can be used to generate new examples of cost functions
which only depend on the Euclidean distance between points and satisfy the
weak MTW condition. Finally, we discuss how the interplay between optimal
transport and complex geometry can be used to define a “synthetic” version of
curvature bounds for Kähler manifolds whose associated potential is merely
$C^{3}$.

The research of JZ is partially supported by NSF grant DMS-1401500.

The research of FZ is partially supported by a Simons Collaboration Grant
355557.

##  1\. Introduction

In this paper, we study Kähler metrics defined on tube domains, which are
domains of the form

$T\Omega=\\{z\in\mathbb{C}^{n}~{}|~{}z=x+iy,x\in\Omega\\}$

where $\Omega$ is a convex domain in $\mathbb{R}^{n}$. In particular, we focus
on metrics $\omega_{\Psi}$ whose potential $\Psi$ (in the $z$-coordinates) is
independent of $y$, and so admit a natural translation symmetry. Our primary
focus in this paper are Kähler metrics which are complete and whose curvature
is non-negative (in several different senses). In general, positive curvature
on Kähler manifolds is a very strong assumption which greatly restricts the
geometry and topology. On the other hand, positively curved metrics have very
interesting geometric properties and in many geometric or analytic
applications, an assumption of positive curvature is necessary to establish
results. Our results fall in line with both these expectations, showing that
positive curvature is quite restrictive but also has interesting applications.

Before discussing our results, it is worth noting that the tube domain
$T\Omega$ has a natural interpretation as the tangent bundle of the domain
$\Omega$. From this perspective, the $\mathbb{R}^{n}$-symmetric Kähler metric
on $T\Omega$ coincides with the so-called Sasaki metric111More generally, the
Sasaki metric is an almost-Hermitian metric defined on the tangent bundle of a
Riemannian manifold $(M,g)$ with an affine connection $\nabla$. This metric
need not be Kähler, but will be for all of the spaces of interest in this
paper. [6]. To distinguish this class of spaces from more general Kähler
manifolds, we will refer to translation-symmetric Kähler metrics on tube
domains as Kähler Sasaki metrics.

###  1.1. Positively curved $O(n)$-symmetric metrics

In the first several sections of this paper, we focus on Kähler Sasaki metrics
which are complete and $O(n)$-symmetric. In other words, we take $\Omega$ to
be the ball $B_{a}\subseteq{\mathbb{R}}^{n}$ of radius $a$ (with $a$ possibly
infinite) and suppose that rotations of $\Omega$ preserve the Kähler metric on
$T\Omega$. In Section 3, we study the bisectional and orthogonal bisectional
curvatures of these spaces. We find the following uniqueness theorem.

######  Proposition.

Let $\Psi$ be a strictly convex function on the ball
$B_{a}\subseteq{\mathbb{R}}^{n}$ of radius $0<a\leq\infty$. Assume that
$\Psi(x)$ depends solely on the norm of $x$ (i.e. the metric is rotationally
symmetric) and that the associated Kähler metric $\mathfrak{h}_{\Psi}$ on
$TB_{a}$ is complete. If either $n=2$ and the bisectional curvature is
everywhere nonnegative, or if $n\geq 3$ and the orthogonal bisectional
curvature is everywhere nonnegative, then $a=\infty$ and the metric
$\mathfrak{h}_{\Psi}$ is the flat Euclidean metric on ${\mathbb{C}}^{n}$.

Put more simply, this shows that the only Kähler Sasaki metrics which is
complete and rotationally symmetric, and whose bisectional curvature is non-
negative222Or whose orthogonal bisectional curvature is non-negative when
$n\geq 3$. is the standard flat metric on $\mathbb{C}^{n}$.

Since this result rules out any (non-trivial) examples with positive
bisectional curvature, we turn our attention to other notions of curvature. In
Section 4, we consider the anti-bisectional curvature, defined as follows:

$\mathfrak{A}(U,V)=4R^{h}(U,\overline{V},U,\overline{V})$

for two holomorphic vector fields $U$ and $V$. For Kähler Sasaki metrics,
there is a notion of “non-negative anti-bisectional curvature” (abbreviated
(NAB)) and “non-negative orthogonal anti-bisectional curvature” (abbreviated
(NOAB)), which is defined precisely in Section 4.333As a note of caution,
(NAB) (respectively (NOAB)) are weaker assumptions than assuming
$\mathfrak{A}\geq 0$ for all (respectively all orthogonal) type (1,0)-vectors.
For $O(n)$-symmetric tube domains, we find several integral-differential
inequalities which are equivalent to (NOAB) (see Proposition 6 for a precise
statement). Using this, we find several new examples of such metrics. As a
result, (NOAB) does not imply the same type of uniqueness properties as with
the bisectional curvature.

Following the theme that positively curved metrics have desirable analytic
properties, recent work by the first two authors [14] found a connection
between orthogonal anti-bisectional curvature and optimal transport. More
precisely, given a (NOAB) metric on a tube domain, it is possible to construct
a cost function $c:X\times Y\to\mathbb{R}$ (with $X,Y\subset\mathbb{R}^{n}$)
satisfying the (MTW) condition, which plays a crucial role in the regularity
theory of optimal transport (see Section 5 for more details).

As such, we can use our examples of (NOAB) Kähler Sasaki metrics to find costs
whose optimal transport has good regularity properties. For instance, we
provide the following example.

######  Example.

The cost function $c:\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}$ given by
$c(x,y)=\|x-y\|-C\log(\|x-y\|+C)$ for $C>0$ satisfies (MTW).

For Kähler Sasaki metrics which are $O(n)$-symmetric, the induced cost
function depends only on the Euclidean distance between $x$ and $y$. As such,
by finding $O(n)$-symmetric metrics with (NOAB), we can find cost functions
which have a natural geometric interpretation.

###  1.2. Synthetic curvature of Kähler Sasaki metrics

Having seen how the study of tube domains has applications to optimal
transport, we then use ideas from optimal transport to better understand the
geometry of Kähler Sasaki metrics. In particular, we study synthetic curvature
bounds when the Kähler potential is not $C^{4}$, (and thus the usual curvature
expressions are not well defined). By analogy, these results are similar to
the CAT($\kappa$)-inequality, which provides a synthetic version of lower
bounds on the sectional curvature.

As a preliminary example in Kähler geometry, we provide a synthetic
formulation for Ricci bounds on tube domains. Although simple, this is an
instructive example and serves as a proof of concept. We then use ideas from
optimal transport to provide a synthetic characterization of non-negative
orthogonal anti-bisectional curvature and non-negative anti-bisectional
curvature. Our work is based off a paper by Guillen and Kitagawa, which
introduced a condition known as quantitative quasi-convexity (abbreviated
(QQConv)) [13] and showed that it provides a synthetic version of (MTW). Using
this, we prove the following.

######  Theorem.

Suppose $\Psi:\Omega\to\mathbb{R}$ is a $C^{4}$ strongly convex function. The
associated Kähler Sasaki metric on $T\Omega$ has (NOAB) if and only if the
cost function $c(x,y)=\Psi(x-y)$ satisfies (QQConv) for all pairs of
sufficiently small Euclidean balls $B(p,\epsilon)$,
$B(q,\epsilon)\subset\Omega$.

Furthermore, we use a result of Figalli, Kim and McCann [7] to provide a
synthetic version of (NAB).

###  1.3. Organization of the paper

Section 2 provides background information on Hessian manifolds and K¨ahler
metrics on tube domains. In Section 3, we discuss the bisectional and
orthogonal bisectional curvatures of $O(n)$-symmetric tube domains. In Section
4, we study the anti-bisectional curvature of rotational symmetric tube
domains. Section 5, discuss the relationship between orthogonal anti-
bisectional curvature and optimal transport. Finally, Section 6 studies
curvature bounds on tube domains when the K¨ahler potential is not $C^{4}$.

There are also two appendices at the end of the paper. The first verifies that
a particular family of metrics satisfies (NOAB). The second proves a lemma
needed in Section 6, which also addresses a question first asked on
MathOverflow [20].

##  2\. Background on Hessian metrics and Kähler metrics on tube domains

In this section, we provide some background on Hessian manifolds and the
Kähler metrics on their tube domains. A Riemannian manifold $(\Omega,g)$ is
said to be a Hessian manifold if

  1. (1)

$\Omega$ admits a flat connection $D$,

  2. (2)

such that around every point $x\in\Omega$, there is an open neighborbood
$U_{x}\subset\Omega$,

  3. (3)

and a function $\Psi:U_{x}\to\mathbb{R}$, such that

$g=D^{2}\,\Psi.$

For convex domains in Euclidean space, we can construct Hessian metrics at
will, simply by choosing a convex function and using the connection induced by
differentiation in coordinates.444Since Hessian manifolds must be affine (i.e.
admit a flat connection), it is non-trivial to find compact examples which are
not tori. More precisely, we consider a convex domain
$\Omega\subseteq{\mathbb{R}}^{n}$ with standard coordinates
$x=(x_{1},\ldots,x_{n})$ and a strongly convex function
$\Psi:\Omega\to\mathbb{R}$. For now, we will assume that $\Psi$ is $C^{4}$,
but in Section 6 we consider less smooth potentials. Using this potential, we
define a Riemannian metric $g$ on $\Omega$, which is given (in
$x$-coordinates) by

$g=g_{\Psi}:=\sum_{i,j=1}^{n}\Psi_{ij}dx_{i}\otimes dx_{j}.$

We call this the Hessian metric with potential $\Psi$. Here and below we
denote by $\Psi_{ij}$ the partial derivative $\frac{\partial^{2}\Psi}{\partial
x_{i}\partial x_{j}}$. To define an associated Kähler manifold, we first
define the tube domain $T\Omega$. To do so, we write
$z_{i}=x_{i}+\sqrt{-1}y_{i}$ as the standard holomorphic coordinates of the
complex Euclidean space ${\mathbb{C}}^{n}$, and consider the domain
$T\Omega\subseteq{\mathbb{C}}^{n}$ given by

$T\Omega=\\{(x,y)\mid
x\in\Omega,y\in{\mathbb{R}}^{n}\\}=\Omega\times{\mathbb{R}}^{n}\subseteq{\mathbb{C}}^{n}.$

The $(1,1)$-form

$\omega_{\Psi}=\sqrt{-1}\sum_{i,j=1}^{n}\Psi_{ij}(x)dz_{i}\wedge
d\overline{z_{j}}$

defined on $T\Omega$ is positive and closed, hence gives a Kähler metric on
$T\Omega$. We will denote this metric by $\mathfrak{h}=\mathfrak{h}_{\Psi}$
and write $\omega$ for $\omega_{\mathfrak{h}}=\omega_{\Psi}$ for simplicity.
We will call $h_{\Psi}$ the Kähler Sasaki metric with potential $\Psi$. As
mentioned previously, the reason for this terminology is that it coincides
with the almost-Hermitian structure on the tangent bundle of a Riemannian
manifold with an affine connection, better known as the Sasaki metric [6].
Here, the connection $\nabla$ is suppressed, but corresponds to
differentiation with respect to the $x$-coordinates (i.e. in the
x-coordinates, the Christoffel symbols of $\nabla$ vanish).

###  2.1. Completeness and Curvature

We are primarily interested in complete metrics, so it is necessary to find a
condition which ensures that the Kähler Sasaki metric is complete. To this
end, we make the following observation, which was originally proved by Molitor
[21].

######  Lemma 1.

The Kähler manifold $(T\Omega,\mathfrak{h}_{\Psi})$ is complete if and only if
the Riemannian manifold $(\Omega,g_{\Psi})$ is complete.

###### Proof.

It suffices to prove the ‘if’ direction. Let $\gamma(s)=(x(s),y(s))$ be a
smooth curve in $T\Omega=\Omega\times{\mathbb{R}}^{n}$ going to infinity. We
have

$|\dot{\gamma}(s)|^{2}_{\mathfrak{h}}=\sum_{i,j=1}^{n}\left(\dot{x}_{i}\dot{x}_{j}\Psi_{ij}(x)+\dot{y}_{i}\dot{y}_{j}\Psi_{ij}(x)\right).$

Clearly, $|\dot{\gamma}(s)|_{\mathfrak{h}}\geq|x^{\prime}(s)|_{g}$, so if the
curve $x(s)$ goes to the boundary of $\Omega$, then $x(s)$, hence $\gamma(s)$,
is of infinite length as $g$ is complete. If, on the other hand, the curve
$x(s)$ is contained in a compact subset $K$ of $\Omega$, then $y(s)$ must tend
to infinity. Take ${\varepsilon}>0$ such that $g\geq{\varepsilon}\,g_{0}$ on
$K$, where $g_{0}$ is the Euclidean metric of ${\mathbb{R}}^{n}$, we have
$|\dot{\gamma}(s)|_{\mathfrak{h}}\geq|\dot{y}(s)|_{\mathfrak{h}}\geq{\varepsilon}\,|\dot{y}(s)|_{g_{0}}$.
Hence the length of $\gamma(s)$ is again infinite. This completes the proof of
the lemma. $\square$

We now turn our attention to the curvature of Kähler Sasaki metrics. For
brevity, we will not derive the full expressions for the curvature, which can
be found in full detail in Satoh [23]. Following the convention of [14], we
denote $\Psi_{ijk}=\frac{\partial^{3}\Psi}{\partial x_{i}\partial
x_{j}\partial x_{k}}$, etc., and use $\Psi^{ij}$ to denote the elements of the
matrix inverse to $\Psi_{ij}$. For the Kähler manifold
$(T\Omega,\mathfrak{h})$, under the natural frame of the holomorphic
coordinate $z$, we have $\frac{\partial}{\partial
z_{i}}=\frac{1}{2}(\frac{\partial}{\partial x_{i}}-\frac{\partial}{\partial
y_{i}})$, so we have

$\mathfrak{h}_{i\overline{j}}=\Psi_{ij},\ \ \ \
\mathfrak{h}_{i\overline{j},k}=\frac{1}{2}\Psi_{ijk},\ \ \ \
\mathfrak{h}_{i\overline{j},k\overline{\ell}}=\frac{1}{4}\Psi_{ijk\ell},\ \ \
\ \mathfrak{h}^{\overline{i}j}=\Psi^{ij}.$

The components of the curvature tensor $R^{(\mathfrak{h})}$ of $\mathfrak{h}$
are given by

$R^{(\mathfrak{h})}_{i\overline{j}k\overline{\ell}}=-\frac{1}{4}\Psi_{ijk\ell}+\sum_{p,q}\frac{1}{4}\Psi_{ipk}\,\Psi_{jq\ell}\,\Psi^{pq}$
(2.1)

In particular, for tangent vectors $u=\sum u_{i}\frac{\partial}{\partial
z_{i}}$ and $v=\sum v_{i}\frac{\partial}{\partial z_{i}}$, the bisectional
curvature of $h$ is given by

$R^{(\mathfrak{h})}_{u\overline{u}v\overline{v}}=\sum_{p,q}-\frac{1}{4}\Psi_{u\overline{u}v\overline{v}}+\frac{1}{4}\Psi_{uvp}\,\overline{\Psi_{uvq}}\,\Psi^{pq}.$
(2.2)

Here we denoted by $\Psi_{uvp}=\sum_{i,j}u_{i}v_{j}\Psi_{ijp}$, etc. for the
sake of simplicity.

###  2.2. $O(n)$-symmetric Hessian metrics

We now specialize our attention to the case when $\Psi$ is rotationally
symmetric. That is to say, we take $\Omega$ to be the ball
$B_{a}\subseteq{\mathbb{R}}^{n}$ of radius $a$ ($0<a\leq\infty$) centered at
the origin. Furthermore, we suppose that $\Psi(x)=\phi(r)$, where
$r=|x|=\sqrt{x_{1}^{2}+\cdots+x_{n}^{2}}$ and $\phi$ is a convex function.

We will write $f(r)=\frac{1}{r}\phi^{\prime}(r)$, so $f\in C^{\infty}[0,a)$
and

$\Psi_{ij}=f\delta_{ij}+\frac{f^{\prime}}{r}x_{i}x_{j}.$

For convenience, we write $h=f+rf^{\prime}$. It is straightforward to see that
$g_{\Psi}$ is a metric if and only if both $f$ and $h$ are positive on
$[0,a)$, as the eigenvalues of the matrix $(\Psi_{ij})$ are $h$ and $(n-1)$
copies of $f$.

For such metrics, we provide an alternate characterization of completeness.

######  Lemma 2.

A rotationally symmetric Hessian metric $g_{\Psi}$ is complete if and only if

$\int_{0}^{a}\sqrt{h}\,dr=\infty.$

###### Proof.

Assume that $g_{\Psi}$ is complete. Consider $\gamma(r)=(r,0,\ldots,0)$,
$0\leq r<a$. We have $\gamma^{\prime}(r)=\frac{\partial}{\partial x_{1}}$, so
$|\gamma^{\prime}(r)|^{2}_{g}=\Psi_{11}=f+rf^{\prime}=h$. Therefore the length
of $\gamma$ is equal to $\int_{0}^{a}\sqrt{h}\,dr$, which must be infinite for
the metric to be complete.

Conversely, assume that this integral is infinite. Let
$\gamma(s)=(x_{1}(s),\ldots,x_{n}(s))$, $0\leq s<b$ (where $0<b\leq\infty$),
be a smooth curve in $B_{a}$ approaching the boundary. We may assume that $s$
is the arc-length parameter in the Euclidean metric $g_{0}$, namely,
$x_{1}^{\prime 2}+\cdots+x_{n}^{\prime 2}=1$. We have

$|\gamma^{\prime}(s)|^{2}_{g}=\sum_{i,j=1}^{n}x_{i}^{\prime}x_{j}^{\prime}\Psi_{ij}=f\sum_{i=1}^{2}x_{i}^{\prime
2}+\frac{f^{\prime}}{r}\left(\sum_{i=1}^{n}x_{i}x_{i}^{\prime}\right)^{2}=f+\frac{f^{\prime}}{r}(rr^{\prime})^{2}.$

By the Cauchy-Schwartz inequality, we have

$r^{2}r^{\prime 2}=\left(\sum x_{i}x_{i}^{\prime}\right)^{2}\leq\sum
x_{i}^{2}\cdot\sum x_{i}^{\prime 2}=r^{2}\cdot 1,$

so $r^{\prime 2}\leq 1$, and

$|\gamma^{\prime}|^{2}_{g}=f+f^{\prime}rr^{\prime 2}\geq fr^{\prime
2}+f^{\prime}rr^{\prime 2}=hr^{\prime 2},$

thus $|\gamma^{\prime}|_{g}\geq\sqrt{h}\,r^{\prime}$, and we have

$\int_{0}^{b}|\gamma^{\prime}(s)|_{g}ds\geq\int_{0}^{b}\sqrt{h}r^{\prime}ds=\int_{0}^{a}\sqrt{h}\,dr=\infty.$

This proves that $g_{\Psi}$ is complete. $\square$

##  3\. $O(n)$-symmetric Kähler Sasaki metrics with non-negative bisectional
curvature

In this section, we consider $O(n)$-symmetric Kähler metrics with non-negative
bisectional curvature. For these metrics, we can simplify the curvature
formulas by computing several further derivatives of the rotationally
symmetric potential function $\Psi(x)=\phi(r)$. In particular, we find that

$\Psi_{i}=\frac{P^{\prime}}{r}x_{i}=fx_{i},\ \ \ \
\Psi_{ij}=f\delta_{ij}+\frac{f^{\prime}}{r}x_{i}x_{j},\ \ \ \
\Psi^{ij}=\frac{1}{f}\left(\delta_{ij}-\frac{f^{\prime}}{rh}x_{i}x_{j}\right).$
(3.1)

For the third and fourth derivatives, we have

$\displaystyle\Psi_{ijk}$ $\displaystyle=$
$\displaystyle\dot{f}\left(\delta_{ij}x_{k}+\delta_{jk}x_{i}+\delta_{ik}x_{j}\right)+\ddot{f}\,x_{i}x_{j}x_{k}$
(3.2) $\displaystyle\Psi_{ijk\ell}$ $\displaystyle=$
$\displaystyle\dot{f}\left(\delta_{ij}\delta_{k\ell}+\delta_{jk}\delta_{i\ell}+\delta_{ik}\delta_{j\ell}\right)+\dddot{f}\,x_{i}x_{j}x_{k}x_{\ell}\,+$
(3.3)
$\displaystyle+\,\ddot{f}\left(\delta_{ij}x_{k}x_{\ell}+\delta_{ik}x_{j}x_{\ell}+\delta_{i\ell}x_{j}x_{k}+\delta_{jk}x_{i}x_{\ell}+\delta_{j\ell}x_{i}x_{k}+\delta_{k\ell}x_{i}x_{j}\right)$

where we have used the notation

$\dot{f}=\frac{1}{r}f^{\prime},\ \ \
\ddot{f}=\frac{1}{r}\left(\frac{1}{r}f^{\prime}\right)^{\prime},\ \ \
\dddot{f}=\frac{1}{r}\left(\frac{1}{r}\left(\frac{1}{r}f^{\prime}\right)^{\prime}\right)^{\prime}.$

Now suppose that $u=(u_{1},\ldots,u_{n})$ and $v=(v_{1},\ldots,v_{n})$ are
unit vectors in ${\mathbb{C}}^{n}$ (in Euclidean norm). We denote

$\alpha_{u}=\left\langle u,\frac{x}{r}\right\rangle=\sum\frac{x_{i}u_{i}}{r},\
\ \ \alpha_{v}=\left\langle v,\frac{x}{r}\right\rangle,\ \ \
\beta=\left\langle u,v\right\rangle,\ \ \ \lambda=\left\langle
u,\overline{v}\right\rangle.$

Using this notation, we find that

$\displaystyle\Psi_{uvp}$ $\displaystyle=$ $\displaystyle\dot{f}\left(\beta
x_{p}+r\alpha_{u}v_{p}+r\alpha_{v}u_{p}\right)+r^{2}\ddot{f}\,\alpha_{u}\alpha_{v}x_{p}$
(3.4) $\displaystyle\Psi_{uvx}$ $\displaystyle=$
$\displaystyle\sum_{p=1}^{n}\Psi_{uvp}\,x_{p}\ =\
r^{2}\dot{f}\left(\beta+2\alpha_{u}\alpha_{v}\right)+r^{4}\ddot{f}\,\alpha_{u}\alpha_{v}$
(3.5) $\displaystyle\Psi_{u\overline{u}v\overline{v}}$ $\displaystyle=$
$\displaystyle\dot{f}\left(1+|\beta|^{2}+|\lambda|^{2})\right)+r^{2}\ddot{f}\left(|\alpha_{u}|^{2}+|\alpha_{v}|^{2}+2Re\\{\overline{\beta}\alpha_{u}\alpha_{v}+\lambda\overline{\alpha_{u}}\alpha_{v}\\}\right)$
(3.6) $\displaystyle+\,r^{4}\dddot{f}|\alpha_{u}|^{2}|\alpha_{v}|^{2}.$

Furthermore, it follows that

$\displaystyle\sum_{p}|\Psi_{uvp}|^{2}$ $\displaystyle=$ $\displaystyle
r^{2}\dot{f}^{2}\left(|\beta|^{2}+|\alpha_{u}|^{2}+|\alpha_{v}|^{2}+4\mathcal{B}+2\mathcal{C}\right)$
$\displaystyle+r^{6}\ddot{f}^{2}|\alpha_{u}\alpha_{v}|^{2}+r^{4}\dot{f}\ddot{f}\left(4|\alpha_{u}\alpha_{v}|^{2}+2\mathcal{B}\right)\textrm{
and }$

$\displaystyle|\Psi_{uvx}|^{2}$ $\displaystyle=$ $\displaystyle
r^{4}\dot{f}^{2}\left(|\beta|^{2}+4|\alpha_{u}\alpha_{v}|^{2}+4\mathcal{B}\right)$
$\displaystyle+\,r^{8}\ddot{f}^{2}|\alpha_{u}\alpha_{v}|^{2}+r^{6}\dot{f}\ddot{f}\left(4|\alpha_{u}\alpha_{v}|^{2}+2\mathcal{B}\right)$

where $\mathcal{B}=Re(\overline{\beta}\alpha_{u}\alpha_{v})$, and
$\mathcal{C}=Re(\lambda\overline{\alpha_{u}}\alpha_{v})$. Using the fact that
$h-r^{2}\dot{f}=f$, we get

$\displaystyle\sum_{p,q}\Psi_{uvp}\,\overline{\Psi_{uvq}}\,\Psi^{pq}$
$\displaystyle=$
$\displaystyle\frac{1}{f}\sum_{p}|\Psi_{uvp}|^{2}-\frac{\dot{f}}{fh}|\Psi_{uvx}|^{2}$
$\displaystyle=$
$\displaystyle\frac{r^{2}\dot{f}^{2}}{f}\left(|\alpha_{u}|^{2}+|\alpha_{v}|^{2}+2\mathcal{C}\right)+\frac{r^{2}\dot{f}^{2}}{h}|\beta|^{2}+$
$\displaystyle+\,|\alpha_{u}\alpha_{v}|^{2}\left(\frac{4r^{4}\dot{f}\ddot{f}+r^{6}\ddot{f}^{2}}{h}-\frac{4r^{4}\dot{f}^{3}}{fh}\right)+\mathcal{B}\,\frac{2r^{4}\dot{f}\ddot{f}+4r^{2}\dot{f}^{2}}{h}.$

Combining (3.6) with (3), we obtain

$\displaystyle 4R^{(h)}_{u\overline{u}v\overline{v}}$ $\displaystyle=$
$\displaystyle-\Psi_{u\overline{u}v\overline{v}}+\Psi_{uvp}\,\overline{\Psi_{uvq}}\,\Psi^{pq}$
(3.8) $\displaystyle=$
$\displaystyle-\dot{f}(1+|\lambda|^{2})-\frac{f\dot{f}}{h}|\beta|^{2}+\left(\frac{r^{2}\dot{f}^{2}}{f}-r^{2}\ddot{f}\right)\,\left(|\alpha_{u}|^{2}+|\alpha_{v}|^{2}+2\mathcal{C}\right)+$
$\displaystyle+\,|\alpha_{u}\alpha_{v}|^{2}\,\left(\frac{4r^{4}\dot{f}\ddot{f}+r^{6}\ddot{f}^{2}}{h}-\frac{4r^{4}\dot{f}^{3}}{fh}-r^{4}\dddot{f}\right)+\mathcal{B}\,\frac{4r^{2}\dot{f}^{2}-2r^{2}f\ddot{f}}{h}.$

In the special case when $u=\frac{x}{r}$ and $v\perp u$, we have
$\beta=\lambda=\alpha_{v}=0$, $\alpha_{u}=1$, and $\mathcal{C}=\mathcal{B}=0$,
so in this case the curvature becomes

$-\dot{f}+\frac{r^{2}\dot{f}^{2}}{f}-r^{2}\ddot{f}=-f(\log f)^{\prime\prime}.$
(3.9)

Note that in this case we have $\Psi_{u\overline{v}}=0$, so $-f(\log
f)^{\prime\prime}$ is a value of the orthogonal bisectional curvature of the
Kähler Sasaki metric $h_{\Psi}$.

######  Proposition 3.

Let $\Psi$ be a strongly convex rotationally symmetry function on the ball
$B_{a}\subseteq{\mathbb{R}}^{n}$ of radius $0<a\leq\infty$. Assume that the
Kähler Sasaki metric $\mathfrak{h}_{\Psi}$ is complete. If either $n=2$ and
the bisectional curvature is everywhere nonnegative, or if $n\geq 3$ and the
orthogonal bisectional curvature is everywhere nonnegative, then $a=\infty$
and the metric $\mathfrak{h}_{\Psi}$ is the flat Euclidean metric on
${\mathbb{C}}^{n}$.

###### Proof.

Suppose that $\mathfrak{h}_{\Psi}$ were to have everywhere nonnegative
orthogonal bisectional curvature ($n\geq 3$) or nonnegative bisectional
curvature ($n=2$). If $\alpha_{u}=\alpha_{v}=0$, then
$\mathcal{C}=\mathcal{B}=0$, and

$4R^{(\mathfrak{h})}_{u\overline{u}v\overline{v}}=-\dot{f}\left(1+|\lambda|^{2}+\frac{f}{h}|\beta|^{2}\right).$

Note that when $n\geq 3$, we can choose such $u$ and $v$ so that
$\Psi_{u\overline{v}}=0$. So under the assumptions on the curvature, we always
have $\dot{f}\leq 0$ and $(\log f)^{\prime\prime}\leq 0$. For convenience,
denote $F=\log f$, which is a smooth function on $[0,a)$. We have
$F^{\prime}\leq 0$, $F^{\prime\prime}\leq 0$, and

$h=(rf)^{\prime}=(1+rF^{\prime})e^{F}>0.$

If $a<\infty$, then since $h\leq e^{F}\leq e^{F}(0)$ as $F$ is non-increasing,
we see that $\int_{0}^{a}\sqrt{h}\,dr<\infty$, contradicting with the
completeness of the metric. So we may assume that $a=\infty$. Since $h>0$, we
find that

$0\geq F^{\prime}>-\frac{1}{r},\ \ \ 0<r<\infty.$

This forces $\lim_{r\rightarrow\infty}F(r)=0$. But $F^{\prime}$ is nonpositive
and non-increasing, so must be constantly zero. This implies that $f$ is a
positive constant, hence $\mathfrak{h}_{\Psi}$ the ($f$-multiple of the
standard) flat complex Euclidean metric on ${\mathbb{C}}^{n}$. $\square$

It is worth noting that Yau’s Uniformization Conjecture (partially proven by
Liu [17]) states that any complete Kähler metric with non-negative bisectional
curvature is biholomorphic to $\mathbb{C}^{n}$. We have made strong
assumptions on the structure of our manifold (in particular
$O(n)\times\mathbb{R}^{n}$-symmetry), which is why it is possible to conclude
that the metric is actually flat, and not simply that it is biholomorphic to
$\mathbb{C}^{n}.$

##  4\. The Anti-Bisectional Curvature

For Kähler Sasaki metrics, there is also a notion of “anti-bisectional
curvature,” which is defined as follows:

$\mathfrak{A}(u,v)=\sum_{i,j,k,\ell}\left(-\Psi_{ijk\ell}+\sum_{p,q}\Psi_{ijp}\Psi_{k\ell
q}\Psi^{pq}\right)u_{i}u_{j}v_{k}v_{\ell},$ (4.1)

where $u=(u_{1},\ldots,u_{n})$, $v=(v_{1},\ldots,v_{n})$ are vectors in
${\mathbb{R}}^{n}$.

######  Definition 4 (Non-negative anti-bisectional curvature).

The Kähler metric $\mathfrak{h}_{\Psi}$ has non-negative orthogonal anti-
bisectional curvature (abbreviated (NAB)), if $\mathfrak{A}(u,v)\geq 0$ for
any $u$, $v\in{\mathbb{R}}^{n}$.

######  Definition 5 (Non-negative orthogonal anti-bisectional curvature).

The Kähler metric $\mathfrak{h}_{\Psi}$ has non-negative orthogonal anti-
bisectional curvature (abbreviated (NOAB)), if $\mathfrak{A}(u,v)\geq 0$ for
any $u$, $v\in{\mathbb{R}}^{n}$ such that
$\Psi_{uv}=\sum_{i,j}u_{i}v_{j}\Psi_{ij}=0$.

Note that if we write $U=\sum_{i}u_{i}\frac{\partial}{\partial z_{i}}$ and
$V=\sum_{i}v_{i}\frac{\partial}{\partial z_{i}}$. Then we have

$\mathfrak{A}(u,v)=4R^{(\mathfrak{h})}(U,\overline{V},U,\overline{V}).$ (4.2)

So (NOAB) and (NAB) are positivity conditions on the curvature of
$\mathfrak{h}$, but are different from requiring $\mathfrak{h}$ to have
nonnegative orthogonal bisectional curvature or non-negative bisectional
curvature. If one requires a Kähler metric $\omega$ to satisfy the curvature
condition $R(X,\overline{Y},X,\overline{Y})\geq 0$ for any type $(1,0)$
vectors $X$ and $Y$, then it is necessary for $\omega$ to have constant
holomorphic sectional curvature and so be a complex space form. As such, the
above (NOAB) and (NAB) conditions are specialized to Kähler Sasaki metrics and
do not generalize naively to arbitrary Kähler metrics. We leave it as an open
question to find meaningful generalization of these conditions for more
general Kähler metrics.

###  4.1. The orthogonal anti-bisectional curvature of $O(n)$-symmetric metric

We now return to the case where the tube domain is rotationally symmetric. We
want understand when $TB_{a}$ will have (NOAB), which means

$\mathfrak{A}(u,v)=-\Psi_{uuvv}+\sum_{p,q}\Psi_{uup}\Psi_{vvq}\Psi^{pq}\geq 0$
(4.3)

for any $u$, $v\in{\mathbb{R}}^{n}$ such that $\Psi_{uv}=0$. To find
conditions which ensure (NOAB), we will rewrite the above expression in terms
of some auxilliary functions.

Without loss of generality, we may assume that $|u|=|v|=1$ under the Euclidean
norm. Let us write $\beta=\langle u,v\rangle=\sum_{i}u_{i}v_{i}$,
$\alpha_{u}=\langle u,\frac{x}{r}\rangle$, and $\alpha_{v}$ similarly. Then
since

$0=\Psi_{uv}=\Psi_{ij}u_{i}v_{j}=f\beta+r^{2}\dot{f}\alpha_{u}\alpha_{v},$

we find

$\beta=-\frac{r^{2}\dot{f}}{f}\alpha_{u}\alpha_{v}=\left(1-\frac{h}{f}\right)\alpha_{u}\alpha_{v}.$
(4.4)

From the formulas for $\Psi^{ij}$, $\Psi_{ijk}$, and $\Psi_{ijk\ell}$, we get

$\displaystyle\Psi_{uuk}$ $\displaystyle=$
$\displaystyle\dot{f}(x_{k}+2r\alpha_{u}u_{k})+r^{2}\ddot{f}\alpha_{u}^{2}x_{k}$
(4.5) $\displaystyle\Psi_{uux}$ $\displaystyle=$ $\displaystyle
r^{2}\dot{f}(1+2\alpha_{u}^{2})+r^{4}\ddot{f}\alpha_{u}^{2}$ (4.6)
$\displaystyle\Psi_{uuvv}$ $\displaystyle=$
$\displaystyle\dot{f}(1+2\beta^{2})+r^{2}\ddot{f}(\alpha_{u}^{2}+\alpha_{v}^{2}+4\beta\alpha_{u}\alpha_{v})+r^{4}\dddot{f}\alpha_{u}^{2}\alpha_{v}^{2}.$
(4.7)

Combining Equations (4.5) and (4.6), we find

$\displaystyle\Psi_{uuk}\Psi_{vvk}$ $\displaystyle=$ $\displaystyle
r^{2}\dot{f}^{2}(1+2\alpha_{u}^{2}+2\alpha_{v}^{2}+4\beta\alpha_{u}\alpha_{v})+r^{4}\dot{f}\ddot{f}(\alpha_{u}^{2}+\alpha_{v}^{2}+4\alpha_{u}^{2}\alpha_{v}^{2})+r^{6}\ddot{f}^{2}\alpha_{u}^{2}\alpha_{v}^{2}$
(4.8) $\displaystyle\Psi_{uux}\Psi_{vvx}$ $\displaystyle=$ $\displaystyle
r^{4}\dot{f}^{2}(1+2\alpha_{u}^{2}+2\alpha_{v}^{2}+4\alpha_{u}^{2}\alpha_{v}^{2})+r^{6}\dot{f}\ddot{f}(\alpha_{u}^{2}+\alpha_{v}^{2}+4\alpha_{u}^{2}\alpha_{v}^{2})+r^{8}\ddot{f}^{2}\alpha_{u}^{2}\alpha_{v}^{2},$
(4.9)

where $k$ is summed in the first line. Since
$\beta=(1-\frac{h}{f})\alpha_{u}\alpha_{v}$ and

$\frac{1}{f}-\frac{\dot{f}}{fh}r^{2}=\frac{1}{f}-\frac{h-f}{fh}=\frac{1}{h},$

we obtain from Equations (4.8) and (4.9) that

$\displaystyle\sum_{p,q}\Psi_{uup}\Psi_{vvq}\Psi^{pq}$ $\displaystyle=$
$\displaystyle\frac{1}{f}\sum_{k}\Psi_{uuk}\Psi_{vvk}-\frac{\dot{f}}{fh}\Psi_{uux}\Psi_{vvx}$
$\displaystyle=$
$\displaystyle\frac{1}{f}\dot{f}^{2}\left(-4\frac{h}{f}\right)\alpha_{u}^{2}\alpha_{v}^{2}+\frac{1}{h}r^{2}\dot{f}^{2}(1+2\alpha_{u}^{2}+2\alpha_{v}^{2}+4\alpha_{u}^{2}\alpha_{v}^{2})+$
$\displaystyle+\
\frac{1}{h}r^{4}\dot{f}\ddot{f}(\alpha_{u}^{2}+\alpha_{v}^{2}+4\alpha_{u}^{2}\alpha_{v}^{2})+\frac{1}{h}r^{6}\ddot{f}^{2}\alpha_{u}^{2}\alpha_{v}^{2}.$

Using the above identity and Equation (4.7), we find that

$\mathfrak{A}(u,v)=A+B(\alpha_{u}^{2}+\alpha_{v}^{2})+C\alpha_{u}^{2}\alpha_{v}^{2},$
(4.10)

where

$\displaystyle A$ $\displaystyle=$
$\displaystyle\frac{1}{h}r^{2}\dot{f}^{2}-\dot{f},$ $\displaystyle B$
$\displaystyle=$
$\displaystyle\frac{2}{h}r^{2}\dot{f}^{2}+\frac{1}{h}r^{4}\dot{f}\ddot{f}-r^{2}\ddot{f},$
$\displaystyle C$ $\displaystyle=$
$\displaystyle-\frac{4h}{f^{2}}r^{2}\dot{f}^{2}+\frac{4}{h}r^{2}\dot{f}^{2}+\frac{4}{h}r^{4}\dot{f}\ddot{f}+\frac{1}{h}r^{6}\ddot{f}^{2}-\
2(1-\frac{h}{f})^{2}\dot{f}-4(1-\frac{h}{f})r^{2}\ddot{f}-r^{4}\dddot{f}.$

Using the relation $\ r^{2}\dot{f}=h-f$ to simplify $A,B$ and $C$, we find

$\displaystyle A$ $\displaystyle=$ $\displaystyle-\frac{f}{h}\dot{f},$ (4.11)
$\displaystyle B$ $\displaystyle=$
$\displaystyle\frac{r^{2}}{h}(2\dot{f}^{2}-f\ddot{f}),$ (4.12) $\displaystyle
C$ $\displaystyle=$
$\displaystyle\left(-4\frac{f}{h}+2+8\frac{h}{f}-6\frac{h^{2}}{f^{2}}\right)\dot{f}+4\left(\frac{h}{f}-\frac{f}{h}\right)r^{2}\ddot{f}+\frac{r^{6}}{h}\ddot{f}^{2}-r^{4}\dddot{f}.$
(4.13)

Now let us try to understand the (NOAB) condition for these metrics. Recall
that this says that $\mathfrak{A}(u,v)\geq 0$ for any unit vectors
$u,v\in{\mathbb{R}}^{n}$ such that $\Psi_{uv}=0$ (or equivalently, vectors for
which $\beta=(1-\frac{h}{f})\alpha_{u}\alpha_{v}$). Fixing the unit vector
$\frac{x}{r}$ and denoting its orthogonal complement in ${\mathbb{R}}^{n}$ by
$W$, we write

$u=\alpha_{u}\frac{x}{r}+u^{\prime},\ \ \ v=\alpha_{v}\frac{x}{r}+v^{\prime},$

where $u^{\prime}$, $v^{\prime}\in W$. Since $u$, $v$ are unit vectors, we
have $|u^{\prime}|^{2}=1-\alpha_{u}^{2}$ and
$|v^{\prime}|^{2}=1-\alpha_{v}^{2}$. Also,

$\beta=\langle u,v\rangle=\alpha_{u}\alpha_{v}+\langle
u^{\prime},v^{\prime}\rangle,$

so we get

$-\frac{h}{f}\alpha_{u}\alpha_{v}=\langle
u^{\prime},v^{\prime}\rangle=|u^{\prime}|\,|v^{\prime}|\cos\theta$

where $\theta$ is the angle between $u^{\prime}$ and $v^{\prime}$. When $n=2$,
$W$ is one dimensional, hence we get

$\frac{h^{2}}{f^{2}}\alpha_{u}^{2}\alpha_{v}^{2}=(1-\alpha_{u}^{2})(1-\alpha_{v}^{2})$
(4.14)

When $n\geq 3$, we find instead that

$\frac{h^{2}}{f^{2}}\alpha_{u}^{2}\alpha_{v}^{2}\leq(1-\alpha_{u}^{2})(1-\alpha_{v}^{2})$
(4.15)

Writing $s=\alpha_{u}^{2}$ and $t=\alpha_{v}^{2}$, we know that (NOAB) simply
means that the function $F(s,t)=A+B(s+t)+Cst$ is nonnegative for all
$s,t\in[0,1]$ such that $\frac{h^{2}}{f^{2}}st=(1-s)(1-t)$ when $n=2$ or
$\frac{h^{2}}{f^{2}}st\leq(1-s)(1-t)$ when $n\geq 3$. Let $\gamma$ be the
curve segment:

$\frac{h^{2}}{f^{2}}st=(1-s)(1-t),\ \ \ 0\leq s,t\leq 1$

in the $st$-plane. If $h=f$, then $\gamma$ is the line segment $s+t=1$. If
$h\neq f$, then $\gamma$ is the segment of the hyperbola contained in the unit
square, and it goes from $(1,0)$ to $(0,1)$. The intersection of $\gamma$ with
the diagonal line is $(s_{0},s_{0})$, where
$s_{0}=\left(1+\frac{h}{f}\right)^{-1}$. Let us define

$D=\left(1+\frac{h}{f}\right)^{2}A+2\left(1+\frac{h}{f}\right)B+C,$ (4.16)

where $A$, $B$, and $C$ are given by Equations 4.11-4.13. We have the
following.

######  Proposition 6.

For a rotationally symmetric convex function $\Psi$ on the ball
$B_{a}\subseteq{\mathbb{R}}^{n}$, $(TB_{a},\mathfrak{h}_{\Psi})$ has (NOAB) if
and only if

  1. (1)

when $n=2$: $A+B$, $D$ are everywhere non-negative

  2. (2)

when $n\geq 3$: $A$, $A+B$, $D$ are everywhere non-negative.

Here, A, B and D are given by 4.11, 4.12 and 4.16, respectively.

###### Proof.

As noted above, (NOAB) means that $F(s,t)=A+B(s+t)+Cst\geq 0$ for any $0\leq
s,t\leq 1$ satisfying the above constraint condition, which means that $(s,t)$
lies on the curve segment $\gamma$ if $n=2$ and $(s,t)$ lies in the sub-region
$\Omega$ in the unit square bounded by the coordinate axes and $\gamma$.

First let us assume that $n=2$. Since $F(1,0)=A+B$ and
$F(s_{0},s_{0})=s_{0}^{2}D$, we know that both $A+B$ and $D$ are non-negative
when (NOAB) is satisfied. Conversely, if both $A+B$ and $D$ are non-negative,
then it is easy to see that $F$ is non-negative along the entire curve segment
$\gamma$, hence (NOAB) is satisfied.

Now assume that $n\geq 3$. If (NOAB) is satisfied, then $A=F(0,0)$,
$A+B=F(1,0)$, $D=F(s_{0},s_{0})$ are all non-negative. Conversely, suppose
that $A$, $A+B$, $D$ are all non-negative. Then it is not hard to see that $F$
is non-negative along the boundary of the domain $\Omega$. We claim that $F$
is also non-negative in the interior of $\Omega$. To see this, suppose that
$(s_{1},t_{1})$ is a critical point of $F$ in the interior of $\Omega$. Then
we have $B+Cs_{1}=B+Ct_{1}=0$ and $s_{1},t_{1}\in(0,1)$. So $B$ and $C$ are
non-zero and with opposite sign.

We consider two separate cases.

  1. (1)

If $C>0$, then $s_{1}=t_{1}=-\frac{B}{C}$ so $0<-B<C$. On the other hand,
since $A+B\geq 0$, we have $-B\leq A$. So $B^{2}\leq AC$, hence

$F(s_{1},t_{1})=A-\frac{B^{2}}{C}=\frac{1}{C}(AC-B^{2})\geq 0.$

  2. (2)

On the other hand, if $C<0$, then $AC-B^{2}\leq AC\leq 0$, hence
$F(s_{1},t_{1})=\frac{1}{C}(AC-B^{2})\geq 0$.

In either case, $F(s_{1},t_{1})\geq 0$, so the orthogonal anti-bisectional
curvature is non-negative. This completes the proof of the proposition.
$\square$

The above proposition says that to find $O(n)$-symmetric metrics with (NOAB),
we need only to focus on the derived functions $A$, $A+B$, and $D$. As such,
we will analyze these terms more carefully. First let us try to understand the
second order derivative term $A+B$. To make the equation more tractable, it is
helpful to define further auxilliary functions. To this end, we define

$\ell=\frac{1}{f},\ \ \ \lambda=\ell-r\ell^{\prime}$ (4.17)

Recall that in terms of the original convex function $\phi:\mathbb{R}_{\geq
0}\to\mathbb{R}$ used to define a radially symmetric potential,
$\ell=\frac{r}{\phi^{\prime}(r)}$.

It follows immediately that $\ell$ is a positive function. From the formula
$h=(rf)^{\prime}=\frac{\ell-r\ell^{\prime}}{\ell^{2}}=\frac{\lambda}{\ell^{2}}$,
we know that $\lambda$ is also a positive function. By a straight forward
computation, we find

$\displaystyle A$ $\displaystyle=$
$\displaystyle\frac{\ell^{\prime}}{r\lambda\ell}$ (4.18) $\displaystyle A+B$
$\displaystyle=$ $\displaystyle\frac{\ell^{\prime\prime}}{\lambda\ell}$ (4.19)
$\displaystyle D$ $\displaystyle=$
$\displaystyle\left(\\!-\frac{1}{r\lambda\ell}+\frac{1}{r\ell^{2}}-\frac{3\lambda}{r\ell^{3}}+\frac{6\lambda^{2}}{r\ell^{4}}\right)\ell^{\prime}+\left(\\!-\frac{7}{\lambda\ell^{2}}+\frac{6}{\ell^{3}}+\frac{8\lambda}{\ell^{4}}\right)\ell^{\prime
2}$ (4.20) $\displaystyle+\
\left(\frac{4r}{\lambda\ell^{3}}+\frac{6r}{\ell^{4}}\right)\ell^{\prime
3}+\frac{4r^{2}}{\lambda\ell^{4}}\ell^{\prime
4}+\left(\frac{6}{\lambda\ell}-\frac{1}{\ell^{2}}-\frac{4\lambda}{\ell^{3}}\right)\ell^{\prime\prime}$
$\displaystyle-\,\left(\frac{2r}{\lambda\ell^{2}}+\frac{6r}{\ell^{3}}\right)\ell^{\prime}\ell^{\prime\prime}-\frac{4r^{2}}{\lambda\ell^{3}}\ell^{\prime
2}\ell^{\prime\prime}+\frac{r^{2}}{\lambda\ell^{2}}\ell^{\prime\prime
2}+\frac{r}{\ell^{2}}\ell^{\prime\prime\prime}.$

From this, we get the following:

######  Proposition 7.

Let $\Psi$ be a strictly convex rotationally symmetric function on the ball
$B_{a}\subseteq{\mathbb{R}}^{n}$ (where $0<a\leq\infty$) so that the Kähler
Sasaki metric $g_{\Psi}$ is complete and satisfies (NOAB). If $n\geq 3$, then
$a=\infty$.

###### Proof.

By the previous proposition, we have $A$, $A+B$, $D$ all non-negative. Thus
$\ell^{\prime}\geq 0$, $\ell^{\prime\prime}\geq 0$. So
$h=\frac{\lambda}{\ell^{2}}$ is non-increasing since
$\lambda^{\prime}=-r\ell^{\prime\prime}\leq 0$. The completeness of $g_{\Psi}$
means that the integral $\int_{0}^{a}\sqrt{h}dr=\infty$. So $a$ must be
$\infty$. $\square$

In particular, any such metric must be biholomorphic to $\mathbb{C}^{n}$ with
its standard complex structure. This analysis also allows us to find several
examples of such metrics, two of which we provide here.

######  Example 8.

Consider the function $\ell=c+r$ on $[0,\infty)$, where $c>0$ is a constant.
We have $\ell^{\prime}=1$, $\ell^{\prime\prime}=0$, and $\lambda=c$. Hence

$\displaystyle r\ell^{4}D$ $\displaystyle=$
$\displaystyle\frac{1}{c}\left(4r^{3}+4r^{2}\ell-7r\ell^{2}-\ell^{3}\right)+\left(6r^{2}+6r\ell+\ell^{2}\right)+\
c\left(8r-3\ell\right)+6c^{2}$ $\displaystyle=$
$\displaystyle-\frac{1}{c}\left(13cr^{2}+10c^{2}r+c^{3}\right)+\left(13r^{2}+8cr+c^{2}\right)+c(5r-3c)+6c^{2}$
$\displaystyle=$ $\displaystyle 3cr+3c^{2}\ =\ 3c\,\ell.$

So the metric has (NOAB). Also, since $h=\frac{c}{\ell^{2}}$, we have
$\int_{0}^{\infty}\sqrt{h}dr=\infty$, hence $g_{\Psi}$ is complete.

We provide a second example, although verifying it is more involved and we
postpone a proof that $D\geq 0$ to Appendix A.

######  Example 9.

Consider the function

$\ell=r+\frac{1}{L^{2}},\ \ \ L=\log(c+r)$

for $c\log(c)^{3}\geq 2$. In this case, if we write $R=c+r$, then we have that

$\lambda=\frac{1}{L^{2}}+\frac{2r}{RL^{3}}>0,\ \ \
\ell^{\prime}=1-\frac{2}{RL^{3}}>0,\ \ \
\ell^{\prime\prime}=\frac{2}{R^{2}L^{3}}+\frac{6}{R^{2}L^{4}}>0,$

and $\sqrt{h}=\frac{\sqrt{\lambda}}{\ell}\sim\frac{1}{RL}$, so
$\int_{0}^{\infty}\sqrt{h}dr=\infty$ and the metric is complete. Furthermore,
$D\geq 0$ so this is another examples of an $O(n)$-symmetric complete metric
with (NOAB).

##  5\. The MTW tensor and the Regularity of Optimal Transport

Apart from complex geometry, the primary motivation for considering anti-
bisectional curvature arises from optimal transport. To explain this, we first
discuss some preliminary background on optimal transport. For a more complete
reference on this topic, we refer the reader to the survey paper of
DePhilippis and Figalli [5] or the book by Villani [25].

The original transport problem was considered by Monge in 1781 [22]. In his
work, he sought to find the most cost-efficient way to transport rubble
(déblais) into a desired configuration to build a fortification (remblais). In
the modern setting, this problem is formalized in terms of the Kantorovich
formulation.

Given probability spaces $(X,\mu)$ and $(Y,\nu)$ and a lower semi-continuous
cost function555In Monge’s work, the cost of transporting a unit of mass from
$x$ to $y$ was taken to be $|x-y|$. $c(x,y):X\times Y\to\mathbb{R}$, the
Kantorovich problem seeks to find a coupling $\pi$ of $\mu$ and $\nu$ which
achieves

$\inf_{\pi\in\Pi(\mu,\nu)}\int_{X\times Y}c(x,y)d\pi.$ (5.1)

Here, $\Pi(\mu,\nu)$ is the set of all couplings of $(X,\mu)$ and $(Y,\nu)$
(i.e. probability measures on $X\times Y$ whose marginals are $\mu$ and $\nu$,
respectively). Under mild regularity assumptions on $\mu,\nu$ and $c$, such a
coupling exists, which intuitively describes how mass from $X$ is transported
to $Y$. In general, the optimal coupling may split mass at a single point and
distribution it throughout $Y$. However, when $X$ and $Y$ are domains in
Euclidean space (or domains in a smooth manifold) and certain technical
conditions hold, Gangbo and McCann [9] showed that the optimal coupling is
induced by a map $T:X\to Y$. More precisely, they showed the following result,
which is based off an earlier work of Brenier [2] for the cost function
$c(x,y)=|x-y|^{2}$.

######  Theorem.

Let $X$ and $Y$ be two open domains of $\mathbb{R}^{n}$ and consider a cost
function $c:X\times Y\to\mathbb{R}$. Suppose that $d\mu$ is a smooth
probability density supported on $X$ and that $d\nu$ is a smooth probability
density supported on $Y$. Suppose that the following conditions hold:

  1. (1)

The cost function $c$ is of class $C^{4}$ with $\|c\|_{C^{4}(X\times
Y)}<\infty$

  2. (2)

The following two conditions hold, which are collectively called (Twist):

    1. (a)

For any $x\in X$, the map $Y\ni y\to D_{x}c(x,y)\in\mathbb{R}^{n}$ is
injective.

    2. (b)

For any $y\in Y$, the map $X\ni x\to D_{y}c(x,y)\in\mathbb{R}^{n}$ is
injective.

  3. (3)

The mixed Hessian matrix $c_{i,j}=\frac{\partial^{2}}{\partial x^{i}\partial
y^{j}}c(x,y)$ is invertible for all $(x,y)\in X\times Y$. In other words,
$\det(c_{i,j})(x,y)\neq 0$. Through the rest of the paper, we will denote this
condition as (NonDeg).

Then:

  1. (1)

There exists a unique solution to the Kantorovich problem (5.1).

  2. (2)

This solution is induced by a measurable map $T:X\to Y$ satisfying
$T_{\sharp}\mu=\nu$, which is injective $d\mu$-a.e.

  3. (3)

There exists a function $u:X\to\mathbb{R}$ such that
$T_{u}(x):=c\textrm{-}\exp_{x}(\nabla u(x))$, where $c\textrm{-}\exp$ is the
so called $c$-exponential map.

  4. (4)

The potential $u$ satisfies the following Monge-Ampere type equation

$|\det(\nabla T_{u}(x))|=\frac{d\mu(x)}{d\nu(T_{u}(x))}\hskip
14.45377ptd\mu-a.e.$ (5.2)

In other words, the optimal transport is solved by a transport map, which
sends each point in $X$ to a unique point in $Y$. Furthermore, we can solve
for this map by solving a fully non-linear equation of Monge-Ampère type. In
this case, it is of interest to determine the continuity properties of $T$,
which is to ask whether nearby points in $X$ are sent to nearby points in $Y$.
Even for smooth costs and measures, a priori this potential is merely
Lipschitz666A deep theorem of De Phillipis and Figalli [4] shows that for
smooth costs and measures, the transport is smooth away from a singular set of
measure zero., and so the transport map may be discontinuous. Determining the
regularity of $T$ is known as the regularity problem of optimal transport, and
is an active area of research.

In the years after Brenier’s initial work, much of the focus for the
regularity problem was for the cost function $c(x,y)=|x-y|^{2}$, in which case
the Monge-Ampere equation 5.2 takes a simple form. For this cost function,
Caffarelli and others established a priori interior $C^{2}$-estimates for weak
solutions to (5.2) when

  1. (1)

$X$ and $Y$ are strictly convex domains, and

  2. (2)

the associated densities $d\mu$ and $d\nu$ are bounded away from $0$ and
$\infty$.

These estimates imply a $C^{1}$-estimate for $T$ (and also higher-order
estimates using elliptic bootstrapping for the linearized operator). However,
this work did not address the regularity for more general cost functions.

In 2005, breakthrough work of Ma, Trudinger and Wang found two structural
conditions, one on the cost function and another on the domains $X$ and $Y$
that are sufficient777Loeper [18] showed that these conditions are essentially
necessary. to prove regularity for the optimal transport. We will discuss the
condition on $X$ and $Y$ in Section 6, but for now, we will focus on the
condition on the cost function, which is that the so-called MTW tensor is non-
negative. In order to define the MTW tensor, we first introduce some notation.
In the following, we use $c_{I,J}$ to denote $\frac{\partial^{|I|}}{\partial
x^{I}}\frac{\partial^{|J|}c}{\partial y^{J}}$ for multi-indices $I$ and $J$
and $c^{i,j}$ to denote the matrix inverse of the mixed Hessian $c_{i,j}$.
Using this notation, for a $C^{4}$ cost function which satisfies (NonDeg), the
MTW tensor (denoted $\mathfrak{S}$) is defined as follows:

$\mathfrak{S}(\xi,\eta)=\sum_{i,j,k,l,p,q,r,s}(c_{ij,p}c^{p,q}c_{q,rs}-c_{ij,rs})c^{r,k}c^{s,l}\xi^{i}\xi^{j}\eta^{k}\eta^{l}$
(5.3)

In this formula, $\eta$ is a vector and $\xi$ is a covector. The MTW tensor is
a fourth-order quantity which scales quadratically in $\xi$ and $\eta$.
Although it is not immediately obvious, this expression transforms tensorially
under change of coordinates.

######  Definition.

A $C^{4}$ cost function with invertible mixed Hessian satisfies:

  1. (1)

(MTW) If $\mathfrak{S}(\xi,\eta)\geq 0$ for all vector-covector pairs
satisfying $\eta(\xi)=0$. Such cost functions are also said to be weakly-
regular.

  2. (2)

(MTW($\kappa)$) If $\mathfrak{S}(\xi,\eta)\geq\kappa|\xi|^{2}|\eta|^{2}$ for
all vector-covector pairs satisfying $\eta(\xi)=0$. This is also known as
strong MTW non-negativity.

  3. (3)

(NNCC) If $\mathfrak{S}(\xi,\eta)\geq 0$ for all vector-covector pairs, not
necessarily orthogonal. This condition is also known as non-negative cost
curvature.

As mentioned previously, to prove regularity for optimal transport, it is
necessary to prove a priori estimates for equations of the form (5.2). A full
overview of this line of research would take us too far from the main focus of
this paper. However, to motivate our considerations, we present one example of
such a regularity result, proven by Figalli, Kim, and McCann [8].

######  Theorem ([8], Theorem 2.1).

Let $X$ and $Y$ be two domains in $\mathbb{R}^{n}$ and let $c$ be a cost
function $c:X\times Y\to\mathbb{R}$. Consider two probability densities $f(x)$
and $g(y)$ supported on $X$ and $Y$ and suppose that the following conditions
hold:

  1. (1)

The cost function $c$ is of class $C^{4}$ with $\|c\|_{C^{4}(X\times
Y)}<\infty$

  2. (2)

The cost function satisfies (Twist) and (NonDeg)

  3. (3)

The density $f$ is bounded from above on $X$ and the density $g$ bounded away
from both zero and infinity on $Y$.

  4. (4)

The domains $X$ and $Y$ are uniformly relatively $c$-convex. (See Definition
13)

  5. (5)

The cost function satisfies (MTW).

Then the optimal transport from $f$ to $g$ is induced by a map $T\in
C^{\alpha}(\overline{X^{\prime}})$ $X^{\prime}\subset X$ is an open set with
$f$ bounded uniformly away from zero.

###  5.1. The Anti-Bisectional Curvature and the MTW tensor.

To relate the MTW tensor to the curvature of tube domains, we specialize our
attention to cost functions of the form $c(x,y)=\Psi(x-y)$ for some strongly
convex function $\Psi:\Omega\to\mathbb{R}$ (henceforth  $\Psi$-costs). Such
cost functions were first studied by Gangbo and McCann [9], although they did
not use this terminology.

For a $C^{4}$ $\Psi$-cost, the MTW tensor is proportional to the orthogonal
anti-bisectional curvature of the associated Kähler Sasaki metric (Theorem 6
of [14]). As a result of this, a $\Psi$-cost satisfies (MTW) iff the
associated Kähler Sasaki metric satisfies (NOAB). Furthermore, the cost-
curvature is proportional to the anti-bisectional curvature, so (NNCC) for a
$\Psi$-cost corresponds to (NAB) for the associated Kähler Sasaki metric.

From this observation, we can use our examples to generate $\Psi$-costs with
(MTW). Furthermore, these costs will satisfy a growth condition at infinity,
which corresponds to completeness for the Kähler Sasaki metric (or
equivalently completeness of the underlying Hessian manifold). As a shorthand
for this, we say that a $\Psi$-cost is complete if the associated Hessian
manifold is complete as a Riemannian manifold.

######  Example 10.

The cost function $c(x,y)=\|x-y\|-c\log(\|x-y\|+c)$ is a complete cost
function which satisfies (MTW).

This is the cost function corresponding to Example 8 (after integrating out to
solve for $\phi$). Unfortunately, it is not possible to write out a closed
form cost associated with Example 9, as $r\cdot f$ does not have elementary
anti-derivative.

##  6\. Synthetic Notions of Curvature in Complex Geometry

In the previous section, we used analysis from complex geometry to find cost
functions of interest to optimal transport. Here, we do the converse, and use
optimal transport to better understand the complex geometry. More precisely,
we show how optimal transport can be used to define synthetic curvature bounds
in Kähler geometry. We use the term “synthetic curvature” in the sense of
defining curvature for low-regularity metric spaces, which may not be smooth
enough for the Riemann curvature tensor to be defined. This notion has also
been called “coarse curvature” (see, e.g., [1]), but we will not use this
terminology. Although the concept is perhaps best understood by analogy, we
will use the following definition for synthetic curvature bounds.

######  Definition (Synthetic curvature bounds).

A condition $Q_{\kappa}$ is a synthetic lower bound for a curvature tensor $S$
if the following two conditions hold:

  1. (1)

On a smooth manifold $M$ where $S$ is defined,

$S\geq\kappa\iff Q_{\kappa}.$

  2. (2)

The condition $Q_{\kappa}$ is well-defined for spaces with low regularity
(where $S$ is not well-defined).

One can define synthetic curvature upper bounds analogously. Note that we have
purposely left the condition $Q_{\kappa}$ and the curvature tensor $S$
ambiguous, so as to make this definition as general as possible. Depending on
the context, $S$ might be the sectional curvature, Ricci curvature, scalar
curvature, or any other sort of curvature. The main goal of this section is to
define synthetic versions of (NOAB) and (NAB) on tube domains. However, it is
instructive to first consider several examples of synthetic curvature.

To motivate the definition of a synthetic curvature bound, it is worth
considering the CAT($\kappa$)-inequality, which is the prototypical example.

######  Theorem (CAT($\kappa$)-inequality).

Suppose $M$ is a Riemannian manifold with sectional curvature $S$ satisfying
$S\geq\kappa$. Denote the distance function on $M$ by $d$. Let $\triangle pqr$
be a geodesic triangle in $M$ (i.e. a triangle whose sides are geodesics) such
that

  1. (1)

the sides $\overline{pq},\overline{pr}$ and $\overline{qr}$ are minimal, and

  2. (2)

if $\kappa>0$, all of the sides have length at most
$\frac{\pi}{\sqrt{\kappa}}$.

For comparison, let $M_{\kappa}$ be a simply connected space of constant
curvature $\kappa$ and consider $\triangle p^{\prime}q^{\prime}r^{\prime}$ a
geodesic triangle in $M_{\kappa}$ with

  1. (1)

$\textrm{length}(\overline{pq})=\textrm{length}(\overline{p^{\prime}q^{\prime}})$,

  2. (2)

$\textrm{length}(\overline{pr})=\textrm{length}(\overline{p^{\prime}r^{\prime}})$,
and

  3. (3)

$\textrm{length}(\overline{qr})=\textrm{length}(\overline{q^{\prime}r^{\prime}})$.

For any pair of points $(x,y)\in\overline{pq}\times\overline{pr}\subset
M\times N$, consider the pair
$(x^{\prime},y^{\prime})\in\overline{p^{\prime}q^{\prime}}\times\overline{p^{\prime}r^{\prime}}\subset
M_{\kappa}\times M_{\kappa}$ satisfying
$d(p,x)=d^{\prime}(p^{\prime},x^{\prime})$ and
$d(p,y)=d^{\prime}(p^{\prime},y^{\prime})$.

Then the following inequality holds.

$d(x,y)\leq d^{\prime}(x^{\prime},y^{\prime}).$ (6.1)

In fact, this result characterizes sectional curvature bounds, in that
whenever a Riemannian manifold has some sectional curvature smaller than
$\kappa$, it is possible to find a small geodesic triangle where inequality
(6.1) fails. Furthermore, if we use Inequality 6.1 as the definition for
sectional curvature bounds, this has the additional advantage in that it is
well-defined on spaces which are not smooth manifolds. In this vein, a
complete geodesic space which satisfies the inequality 6.1 is said to be a
$CAT(\kappa)$-spaces [11] and play an important role in metric geometry and
geometric group theory.

###  6.1. Synthetic Ricci bounds on Kähler manifolds

For a simple though instructive example of this idea in complex geometry, we
now discuss a synthetic formulation for Ricci bounds. On a smooth Kähler
manifold, the Ricci form is given by the formula

$\rho=-\sqrt{-1}\partial\bar{\partial}\log\det\partial\bar{\partial}\Psi,$
(6.2)

where $\Psi$ is the Kähler potential (i.e. The Kähler form $\omega$ satisfies
$\omega=\partial\bar{\partial}\Psi$). The Ricci curvature is bounded below
(respectively above) by a constant $\kappa$ if

$\rho\geq\kappa\omega\textrm{ (respectively }\leq\kappa\omega\textrm{).}$
(6.3)

The above inequality should be interpreted in the sense of $(1,1)$-forms. That
is to say, given a holomorphic vector $X$, the above inequality implies that
$rho(X,\overline{X})\geq\kappa\omega(X,\overline{X})$. To rephrase this in
synthetic terms, we consider the function
$Q_{\kappa}=\log\det\partial\bar{\partial}\Psi+\kappa\Psi$ and say that the
Ricci curvature is bounded above (or below) by $\kappa$ whenever $Q_{\kappa}$
is plurisubharmonic (plurisuperharmonic)888This is a slight abuse of notation
from our definition of synthetic curvature bounds, where $Q_{\kappa}$ was a
condition, instead of a function, but this is not important.. When $\Psi$ is
$C^{4}$, this is equivalent to Ricci bounds in the normal sense. However, this
definition does not require $Q_{\kappa}$ to be $C^{2}$, so we are able to
define Ricci curvature bounds when the potential is only $C^{3}$ (which is the
natural regularity so that the Kähler condition $d\omega=0$ is well-defined).

For Kähler Sasaki metrics on tube domains, we can simplify this further. For
these metrics, the Ricci form simplies to

$\rho_{i\bar{j}}=\frac{\partial^{2}}{\partial x^{i}\partial
x^{j}}\log\det\left[\frac{\partial^{2}\Psi}{\partial x^{k}\partial
x^{l}}\right].$

Therefore, if we define

$Q_{\kappa}=\log\det\left[\frac{\partial^{2}\Psi}{\partial x^{k}\partial
x^{l}}\right]+\kappa\Psi$ (6.4)

we can see that the Ricci curvature of a Sasaki metric is bounded below by
$\kappa$ if and only if $Q_{\kappa}$ is convex (on $\Omega$). This immediately
implies the following proposition.

######  Proposition 11.

A $C^{4}$ Kähler Sasaki metric has Ricci curvature bounded below by $\kappa$
iff the function $Q_{\kappa}$ satisfies

$\lambda Q_{\kappa}(p_{1})+(1-\lambda)Q_{\kappa}(p_{2})\geq Q_{\kappa}(\lambda
p_{1}+(1-\lambda)p_{2})$

for all $p_{1},p_{2}\in\Omega$ and $0<\lambda<1$

As such, convexity (or concavity) of $Q_{\kappa}$ gives a way to define bounds
on the Ricci curvature. For $C^{4}$ potentials, this is equivalent to Ricci
curvature bounds in the usual sense, but has the advantage of being defined
for less smooth potentials. It is worth noting that there are many other ways
of defining synthetic bounds for Ricci curvature, several of which can be
defined for less regular metric-measure spaces. We refer to the work of
Villani [26] and Ache and Warren [1] for some references on this topic.

###  6.2. A synthetic formulation of (NOAB)

We now come to the main focus of this section, which is to provide a synthetic
version of non-negative orthogonal bisectional curvature. In order to do so,
it is first necessary to provide some additional background on Hessian
manifolds and optimal transport.

####  6.2.1. Dual coordinates and $c$-segments

Given a Hessian manifold on a domain $\Omega$ with metric

$g_{ij}=\frac{\partial^{2}\Psi}{\partial x^{i}\partial x^{j}},$

there are a set of dual coordinates $\theta^{i}\in\Omega^{\ast}$ which are
given by

$\theta^{i}:=\frac{\partial\Psi}{\partial{u^{i}}}.$ (6.5)

As implied by the name, these functions form coordinates for the Hessian
manifold. Furthermore, in these coordinates, the metric is also given by the
second derivative of a convex function. The potential function in the dual
coordinates is the Legendre dual $\Psi^{\ast}$, which satisfies

$\Psi^{\ast}(\theta)=\sup_{x\in\Omega}\left\langle\theta,x\right\rangle-\Psi(x).$

These dual coordinates play an essential role in the study of Hessian
manifolds, and also in the optimal transport of $\Psi$-costs. To explain this,
we first provide a definition of $c$-segments and relative $c$-convexity.

######  Definition 12 ($c$-segment).

For a $C^{1}$ cost function $c:X\times Y\to\mathbb{R}$, a $c$-segment in $X$
with respect to a point $y$ is a solution set $\\{x\\}$ to
$D_{y}c(x,y)\in\ell$ for $\ell$ a line segment in $\mathbb{R}^{n}$. A
$c^{*}$-segment in $Y$ with respect to a point $x$ is a solution set $\\{y\\}$
to $D_{x}c(x,y)\in\ell$ where $\ell$ is a line segment in $\mathbb{R}^{n}$.

######  Definition 13 ($c$-convexity).

A set $X$ is $c$-convex relative to a set $Y$ if for any two points
$x_{0},x_{1}\in X$ and any $y\in Y$, the $c$-segment relative to $y$
connecting $x_{0}$ and $x_{1}$ lies in $X$. Similarly we say $Y$ is
$c^{*}$-convex relative to $Y$ if for any two points $y_{0},y_{1}\in Y$ and
any $x\in X$, the $c^{*}$-segment relative to $x$ connecting $y_{0}$ and
$y_{1}$ lies in $Y$.

For a cost function of the form $\Psi(x-y)$, $c$-segments correspond to line
segments in the $\theta$-coordinates. As such, for costs of this form, we have
the following result.

######  Proposition ([14], Proposition 8).

For a $\Psi$-cost, a set $Y$ is $c$-convex relative to $X$ if and only if, for
all $x\in X$, the set $x-Y\subset\Omega$ is convex in terms of the dual
coordinates $\theta$.

####  6.2.2. A Synthetic Version of (MTW)

With the notions of dual coordinates and $c$-segments in hand, we now provide
a synthetic version of MTW non-negativity. To this end, we introduce the
condition known as quantitative quasiconvexity, which was introduced by
Guillen and Kitagawa [13].

######  Definition (Quantitative Quasiconvexity).

A cost function is said to be quantitatively quasiconvex (denoted (QQConv)) if
there is a universal constant $M\geq 1$ such that for any points
$x,x_{0},x_{1}\in X$ and $y,y_{0},y_{1}\in Y$

$\displaystyle-c(x,y(t))+c(x,y_{0})-$
$\displaystyle(-c(x_{0},y(t))+c(x_{0},y_{0}))$ $\displaystyle\leq Mt$
$\displaystyle(-c(x,y_{1})+c(x,y_{0})-(-c(x_{0},y_{1})+c(x_{0},y_{0}))),\hskip
43.36243pt\forall t\in[0,1]$ $\displaystyle-c(x(s),y)+c(x_{0},y)-$
$\displaystyle(-c(x(s),y_{0})+c(x_{0},y_{0}))$ $\displaystyle\leq Ms$
$\displaystyle(-c(x_{1},y)+c(x_{0},y)-(-c(x_{1},y_{0})+c(x_{0},y_{0}))),\hskip
43.36243pt\forall s\in[0,1].$

Here, $y(t)$ is the c-segment with respect to $x_{0}$ from $y_{0}$ to $y_{1}$,
and $x(s)$ is the c-segment with respect to $y_{0}$ from $x_{0}$ to $x_{1}$.
If this inequality holds with $M=1$, then we say that the function is
quantitatively convex, which we denote (QuantConv).

Quantitative quasiconvexity only requires $C^{1}$ smoothness for the cost
function (so that $c$-segments are well defined), and so plays an important
role in low-regularity optimal transport.

######  Lemma ([13], Lemma 2.23).

Suppose $c$ is a $C^{4}$ cost function999The actual regularity required of $c$
is slightly less, but $C^{4}$-regularity is sufficient for the MTW tensor to
be defined. and that $c$, $X$, and $Y$ satisfy the following conditions.

  1. (1)

$c$ satisfies (Twist).

  2. (2)

$c$ satisfies (Nondeg) (i.e. $\det\left(D_{x}D_{y}c(x,y)\right)\neq 0$ )

  3. (3)

The domains $X$ and $Y$ are uniformly relatively $c$-convex. (i.e. $X$ and $Y$
satisfy (DomConv) )

Then the cost $c$ satisfies (MTW) iff it satisfies (QQConv).

We refer to [13] for a proof of the lemma. For our purposes, this immediately
implies the following corollary.

######  Corollary 14.

Suppose $\Psi:\Omega\to\mathbb{R}$ is a $C^{4}$ convex function and that $X$
and $Y$ are subsets of $\mathbb{R}^{n}$ with $S=X-Y\subset\Omega$. Suppose
that for all $x_{0}\in X$, $y_{0}\in Y$, the sets $x_{0}-Y$ and $X-y_{0}$ are
convex when viewed in the dual coordinates $\theta$. Then $\Psi$ induces a
Kähler Sasaki metric with (NOAB) on $TS$ if and only if the associated
$\Psi$-cost satisfies (QQConv).

###### Proof.

To prove this, it suffices to show that the (Twist) and (Nondeg) assumptions
in the previous lemma are satisfied. Note for a $\Psi$-cost, the map
$y\mapsto-D_{x_{0}}c(x_{0},y)$ simplifies to $y\mapsto-
D_{x_{0}}\Psi(x_{0}-y)$, which is simply the point $x_{0}-y$ expressed in
terms of the dual $\theta$ coordinates. For a strongly convex function, the
transition maps from primal to dual coordinates are invertible, which is
equivalent to (Twist).

Furthermore, when $\Psi$ is strongly convex, $Hess(\Psi)>0$. This immediately
implies that the cost is non-degenerate as well. Finally, for a $\Psi$-cost,
(DomConv) is equivalent to convexity of $x-Y$ and $X-y$ within the dual
coordinates, which is exactly the second assumption in this corollary. From
this, we can apply the result of Guillen and Kitagawa to show that (QQConv) is
equivalent to (MTW) for the cost $\Psi(x-y)$, which furthermore implies that
$\Psi$ induces a Kähler Sasaki metric with (NOAB). $\square$

####  6.2.3. The role of (DomConv)

In Corollary 14, the assumption of (DomConv) is somewhat awkward. It plays a
crucial role in the regularity of optimal transport, but is a bit out of place
from the perspective of curvature bounds. When the potential is strongly
convex and bounded in $C^{4}$, it is possible to choose $X$ and $Y$ so that
(DomConv) is always satisfied. To do this, we use the following lemma, proven
in the appendix (Section B).

######  Lemma 15.

Suppose $\Omega,g$ is a Hessian manifold with local potential $\Psi$ that is
strongly convex and bounded in $C^{4}$. Then there exists an $\epsilon_{0}>0$
so that all balls $B(p,\epsilon)$ ($p\in\Omega$, $\epsilon<\epsilon_{0}$) in
the primal coordinates $x$ are convex when expressed in terms of the dual
coordinates $\theta$.

Using this result, for a point $p\in\Omega$, we choose $x_{0}$ and $y_{0}$ so
that $p=x_{0}-y_{0}$. We then set $X$ and $Y$ so that $X-y_{0}$ and $x_{0}-Y$
are small balls of radius $\epsilon$ in the primal coordinates. For
sufficiently small balls, the previous lemma shows that $x-Y$ and $X-y$ are
convex sets in dual coordinates for every $x\in X$ and $y\in Y$, which implies
$X$ and $Y$ are relatively $c$-convex. Combining these results, we obtain the
following theorem.

######  Theorem 16.

Suppose $\Psi:\Omega\to\mathbb{R}$ is a $C^{4}$ strongly convex function. The
associated Kähler Sasaki metric on $T\Omega$ satisfies (NOAB) if and only if
there exists an $\epsilon>0$ so that the $\Psi$-cost $c(x,y)=\Psi(x-y)$
satisfies (QQConv) for all pairs of Euclidean balls $B(x_{1},\epsilon)$,
$B(x_{2},\epsilon))$ for $x_{1},x_{2}\in\Omega$.

Strictly speaking, this is not a synthetic version of (NOAB), as Lemma 15
requires $C^{4}$-regularity of $\Psi$ to guarantee pairs of small Euclidean
balls are relatively $c$-convex. However, it may be possible to refine this
result to give a version of (NOAB) when $\Psi$ is less regular than $C^{4}$.
At least $C^{3}$-regularity is necessary for the Kähler form to satisfy
$d\omega=0$. Furthermore, Guillen and Kitagawa were able to establish
continuity for the solution to the Monge problem when the cost function is
$C^{3}$ and satisfies (QQConv). This suggests that $C^{3}$ smoothness of $c$
is a critical threshold for optimal transport.

We can also provide a synthetic version of non-negative anti-bisectional
curvature. This is based on the equivalence between (NNCC) and (QuantConv) for
$C^{4}$ cost functions, which was proven by Figalli, Kim and McCann (see Lemma
6.1 of [7]). We thank Professor Kitagawa for informing us of this result.
Using this observation, we have the following.

######  Corollary 17.

Suppose $\Psi:\Omega\to\mathbb{R}$ is a $C^{4}$ strongly convex function. Then
$\Psi$ induces a Kähler Sasaki metric with non-negative bisectional curvature
on $T\Omega$ if and only if there exists an $\epsilon>0$ so that the
$\Psi$-cost $c(x,y)=\Psi(x-y)$ satisfies (QuantConv) for all pairs of
Euclidean balls $B(x_{1},\epsilon)$, $B(x_{2},\epsilon))$ for
$x_{1},x_{2}\in\Omega$.

We leave as a future question to determine a synthetic version of strong MTW
non-negativity (i.e. (MTW($\kappa$))), and in turn strong lower bounds for the
orthogonal anti-bisectional curvature. It would also be of interest to modify
$(QuantConv)$ to find a synthetic version of non-negative holomorphic
sectional curvature (possibly by restricting the choices of $x$ and $x_{0}$).

As a final remark, we note that there is also a natural pseudo-Riemannian
framework of optimal transport [15] [16]. In this framework, the MTW tensor is
the sectional curvature of light-like planes. As such, we expect that there
are versions of Theorem 16 and Corollary 17 which provide synthetic curvature
bounds for pseudo-Riemannian metrics. In this paper, we have focused on the
complex framework, so will not address the pseudo-Riemannian framework
further.

##  7\. Acknowledgements

The first author would like to thank Jun Kitagawa for his helpful comments and
discussions.

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##  Appendix A Derivation of Example 9

In this section, we prove that Example 9 on page 12 has non-negative anti-
bisectional curvature.

Recall that in this example, we set

$\ell=r+\frac{1}{L^{2}},\ \ \ L=\log(c+r)\textrm{ for $c\log(c)^{3}\geq 2$.}$

Here, $\ell=\frac{r}{\phi^{\prime}(r)}$ where $\phi$ is the convex function
satisfying $\Psi(x)=\phi(|x$). A tedious but straightforward computation shows
the following.

$A=\frac{\log[c+r]^{2}(-2+(c+r)\log[c+r]^{3})}{r(2r+(c+r)\log[c+r])(1+r\log[c+r]^{2})}$

$B=\frac{\log[c+r](6r+2(c+2r)\log[c+r]-(c+r)^{2}\log[c+r]^{4})}{r(c+r)(2r+(c+r)\log[c+r])(1+r\log[c+r]^{2})}$

$C=\frac{\left(\begin{aligned}
-60r^{3}-4r^{2}(19c+26r)\log[c+r]-2r(17c^{2}+49cr+2r^{2}(17+3r))\log[c+r]^{2}\\\
-2(3c^{3}+12c^{2}r+4r^{3}(2+3r)+cr^{2}(17+6r))\log[c+r]^{3}-2r^{2}(5c^{2}+17cr+14r^{2})\log[c+r]^{4}\\\
+2r(2c^{3}+3c^{2}r-2cr^{2}-3r^{3})\log[c+r]^{5}+3(c+r)^{2}(c^{2}+2cr+r^{2}-4r^{3})\log[c+r]^{6}\\\
-2r^{2}(c+r)^{2}(c+3r)\log[c+r]^{7}+2r(c+r)^{4}\log[c+r]^{8}+r^{2}(c+r)^{4}\log[c+r]^{10}\end{aligned}\right)}{r(c+r)^{3}\log[c+r](2r+(c+r)\log[c+r])(1+r\log[c+r]^{2})^{3}}$

$D=\frac{\left(\begin{aligned}
-60r^{3}-4r^{2}(15c+22r)\log[c+r]-2r(9c^{2}+29cr+2r^{2}(11+3r))\log[c+r]^{2}\\\
-2(3c^{3}+8c^{2}r+3c(3-2r)r^{2}+4r^{3})\log[c+r]^{3}+2r^{2}(15c^{2}+27cr+10r^{2})\log[c+r]^{4}\\\
+2r(6c^{3}+21c^{2}r+22cr^{2}+7r^{3})\log[c+r]^{5}+3(c+r)^{4}\log[c+r]^{6}\end{aligned}\right)}{r(c+r)^{3}\log[c+r](2r+(c+r)\log[c+r])(1+r\log[c+r]^{2})^{3}}$

Whenever $c\log[c]^{3}\geq 2$, we have that $A$ and $A+B$ are greater than 0,
so what remains to show is that $D\geq 0$.

We start by consider the denominator of $D$

$Denom[D]=r(c+r)^{3}\log[c+r](2r+(c+r)\log[c+r])(1+r\log[c+r]^{2})^{3}.$

This is positive for $r\geq 0$ whenever $c\geq 1$, so the denominator is
positive. As such, what remains to show is that the numerator is also
positive.

$Numer[D]=\left(\begin{aligned}
-60r^{3}-4r^{2}(15c+22r)\log[c+r]-2r(9c^{2}+29cr+2r^{2}(11+3r))\log[c+r]^{2}\\\
-2(3c^{3}+8c^{2}r+3c(3-2r)r^{2}+4r^{3})\log[c+r]^{3}+2r^{2}(15c^{2}+27cr+10r^{2})\log[c+r]^{4}\\\
+2r(6c^{3}+21c^{2}r+22cr^{2}+7r^{3})\log[c+r]^{5}+3(c+r)^{4}\log[c+r]^{6}\end{aligned}\right)$

Momentarily treating powers of $\log[c+r]$ as if they were constants, then
this appears to be a quartic polynomial in $r$. Arranging the terms in this
fashion, we find

$Numer[D]=\mathcal{A}_{0}+\mathcal{A}_{1}r+\mathcal{A}_{2}r^{2}+\mathcal{A}_{3}r^{3}+\mathcal{A}_{4}r^{4},\textrm{
where,}$ $\displaystyle\mathcal{A}_{0}$ $\displaystyle=$ $\displaystyle
3c^{3}\log[c+r]^{3}(-2+c\log[c+r]^{3}),$ $\displaystyle\mathcal{A}_{1}$
$\displaystyle=$
$\displaystyle-18c^{2}\log[c+r]^{2}-16c^{2}\log[c+r]^{3}+12c^{3}\log[c+r]^{5}+12c^{3}\log[c+r]^{6},$
$\displaystyle\mathcal{A}_{2}$ $\displaystyle=$
$\displaystyle-60c\log[c+r]-58c\log[c+r]^{2}-18c\log[c+r]^{3}$
$\displaystyle+30c^{2}\log[c+r]^{4}+42c^{2}\log[c+r]^{5}+18c^{2}\log[c+r]^{6},$
$\displaystyle\mathcal{A}_{3}$ $\displaystyle=$
$\displaystyle-60-88\log[c+r]-44\log[c+r]^{2}-8\log[c+r]^{3}$
$\displaystyle+12c\log[c+r]^{3}+54c\log[c+r]^{4}+44c\log[c+r]^{5}+12c\log[c+r]^{6},$
$\displaystyle\mathcal{A}_{4}$ $\displaystyle=$
$\displaystyle-12\log[c+r]^{2}+20\log[c+r]^{4}+14\log[c+r]^{5}+3\log[c+r]^{6}.$

We will now show that each of these terms are non-negative, which completes
the proof that $D\geq 0$.

  1. $\mathcal{A}_{0}\geq 0$.

Note that $c^{3}\log[c+r]^{3}>0$ and we have assumed that $c\log[c]^{3}\geq
2$, so both factors are non-negative. As such, $\mathcal{A}_{0}\geq 0$.

  2. $\mathcal{A}_{1}>0$.

Divide $\mathcal{A}_{1}$ by $c^{2}\log[c+r]^{2}>0$ to obtain

$\mathcal{A}^{\prime}_{1}=2(-9-8\log[c+r]+6c\log[c+r]^{3}+6c\log[c+r]^{4}).$

By the assumption that $c\log[c]^{3}\geq 2$, we have that

$\mathcal{A}^{\prime}_{1}\geq 2(-9-8\log[c+r]+12+12\log[c+r])=3+4\log[c+r]>0.$

As such, we have that $\mathcal{A}_{1}>0$.

  3. $\mathcal{A}_{2}>0$.

Simplifying $\mathcal{A}_{2}$ by dividing out $c\log[c+r]$, we obtain

$\frac{\mathcal{A}_{2}}{c\log[c+r]}=2\left(\begin{aligned}
-30-29\log[c+r]-9\log[c+r]^{2}+15c\log[c+r]^{3}\\\
+21c\log[c+r]^{4}+9c\log[c+r]^{5}\end{aligned}\right).$

Once again using our assumption on $c$, we have that

$\displaystyle\frac{\mathcal{A}_{2}}{c\log[c+r]}$ $\displaystyle\geq$
$\displaystyle
2\left(-29\log[c+r]-9\log[c+r]^{2}+42\log[c+r]+18\log[c+r]^{2}\right)$
$\displaystyle=$ $\displaystyle 2(13\log[c+r]+9\log[c+r]^{2})>0$

As such, $\mathcal{A}_{2}>0$.

  4. $\mathcal{A}_{3}>0$.

To show that this term is positive, we cannot simply bound each term by below,
as we did for the previous terms. Instead, consider $\mathcal{A}_{3}(r,c)$ as
a function of $r$ and $c$. We first show that when $c$ satisfies
$c\log[c]^{3}\geq 2$, $\mathcal{A}_{3}(0,c)>0$. To see this, observe that

$\displaystyle\mathcal{A}_{3}(0,c)$ $\displaystyle=$
$\displaystyle-60-88\log[c]-44\log[c]^{2}-8\log[c]^{3}+12c\log[c]^{3}$
$\displaystyle+54c\log[c]^{4}+44c\log[c]^{5}+12c\log[c]^{6}$

We estimate this term from below as follows.

$\displaystyle\mathcal{A}_{3}(0,c)$ $\displaystyle\geq$
$\displaystyle-60-88\log[c]-44\log[c]^{2}-8\log[c]^{3}$
$\displaystyle+24+108\log[c]+88\log[c]^{2}+24\log[c]^{3}$ $\displaystyle=$
$\displaystyle-36+20\log[c]+44\log[c]^{2}+16\log[c]^{3}$

From the fact that $c\log[c]^{3}\geq 2,$ we have that $\log[c]>.87$.101010Note
that $.87^{3}\exp[.87]<.87^{3}e<.87^{3}\cdot 3=1.975509<2,$ which is how we
obtain this lower bound.  Plugging in $.87$ as a lower bound for $\log[c]$ in
the previous inequality, we have that

$\mathcal{A}_{3}(0,c)>-36+20\cdot.87+44\cdot(.87)^{2}+16\cdot(.87)^{3}=25.239648>0.$

Finally, we show that $(r+c)\frac{\partial}{\partial
r}\mathcal{A}_{3}(r,c)>0$.

$\displaystyle(r+c)\frac{\partial}{\partial r}\mathcal{A}_{3}(r,c)$
$\displaystyle=$ $\displaystyle 4\left(\begin{aligned}
-22-22\log[c+r]+(-6+9c)\log[c+r]^{2}\\\
+54c\log[c+r]^{3}+55c\log[c+r]^{4}+18c\log[c+r]^{5}\end{aligned}\right)$
$\displaystyle\geq$ $\displaystyle 4\left(\begin{aligned}
-22-22\log[c+r]+(-6+9c)\log[c+r]^{2}\\\
+108+110\log[c+r]+36\log[c+r]^{2}\end{aligned}\right)$ $\displaystyle=$
$\displaystyle 4(86+88\log[c+r]+(27+9c)\log[c+r]^{2})>0.$

This implies that $\frac{\partial}{\partial r}\mathcal{A}_{3}(r,c)>0$, which
further implies that $\mathcal{A}_{3}(r,c)>\mathcal{A}_{3}(0,c)$ for all
$r>0$. Since $\mathcal{A}_{3}(0,c)>0$, this implies that $\mathcal{A}_{3}>0$.

  5. $\mathcal{A}_{4}>0$.

Simplifying $\mathcal{A}_{4}$ by dividing out $\log[c+r]^{2}$, we obtain

$\mathcal{A}^{\prime}_{4}=-12+20\log[c+r]^{2}+14\log[c+r]^{3}+3\log[c+r]^{4}.$

Now, using the fact that $\log[c+r]>.87$ (as shown above), we have that

$\displaystyle\mathcal{A}^{\prime}_{4}$ $\displaystyle>$
$\displaystyle-12+20(.87)^{2}+14(.87)^{3}+3(.87)^{4}$ $\displaystyle=$
$\displaystyle 14.07573483>0$

Since all these terms are non-negative, this implies that

$Numer[D]=\mathcal{A}_{0}+\mathcal{A}_{1}r+\mathcal{A}_{2}r^{2}+\mathcal{A}_{3}r^{3}+\mathcal{A}_{4}r^{4}\geq
0.$

Since the denominator of $D$ is also positive, this implies that $D>0$, and so
this metric has non-negative anti-bisectional curvature.

##  Appendix B Convexity of geodesic balls in scale-controlled coordinates

In this section, we prove two lemmas which address relatively $c$-convex
subsets. In short, we show that for a $\Psi$-cost with strongly convex and
$C^{3}$ potential, sufficiently small balls in the primal coordinates are
relatively $c$-convex. This shows that the (DomConv) assumption can be avoided
by simply considering sufficiently small local neighborhoods.

To begin, we first prove the following more general lemma about convexity of
geodesic balls in scale controlled coordinates.

######  Lemma 18.

Suppose that $g_{ij}$ is a Riemannian metric tensor on an open subset
$\Omega\subset\mathbb{R}^{n}$. Suppose further that $g_{ij}$ is
scale-$C^{2}$-controlled in the following way:

There exists constants $r>0$, $Q>1$ so that for all $p\in\Omega$, on the
Euclidean ball $B_{\Omega}(p,r)$,

  1. (1)

$Q^{-1}\delta_{ij}\leq g_{ij}\leq Q\delta_{ij}$ (in the sense of positive
definite matrices),

  2. (2)

$r|\frac{\partial g_{ij}}{\partial x^{k}}|\leq Q-1$,

  3. (3)

and $r^{2}|\frac{\partial^{2}g_{ij}}{\partial x^{k}\partial x^{l}}|\leq Q-1$

Then there exists an $\epsilon_{0}(r,Q)$ so that whenever
$\epsilon<\epsilon_{0}$, the geodesic ball (with respect to the metric $g$)
$B_{g}(p,\epsilon)$ is convex in $\Omega$ as a subset of Euclidean space.

###### Proof.

We start by fixing a point $p$, which will be the center of the ball
throughout. For this proof, it is necessary to consider two different
Riemannian metrics on $\Omega$. The first, (denoted $g$) is the Riemannian
metric of interest. The second, which we denote $g_{0}$, is the flat metric
satisfying $(g_{0}){}_{ij}(x)=g_{ij}(p)$ for all $x\in\Omega$ (i.e. we
consider the metric at $p$ and do not change the components throughout
$\Omega$). We then consider two separate distance functions. The first, which
we denote $d$, is the distance from $p$ in the $g$ metric. The second, which
we denote $\delta_{0}$, is the distance from $p$ in the $g_{0}$ metric.

As a broad overview, the goal is to show that $d$ and $\delta_{0}$ are
$C^{2}$-close. Then, using the implicit function theorem, we show that their
level sets are also $C^{2}$-close. However, since the level sets of
$\delta_{0}$ are ellipsoids with bounded eccentricity, its level sets are
uniformly strongly convex, which implies that the level sets of $d$ must also
be convex.

We first show that in a small ball, the functions $d$ and $\delta_{0}$ are
close in the $C^{0}$-sense. To see this, consider a second point $q$. We
consider two separate paths from $p$ to $q$. One, which we denote $\gamma$, is
the geodesic with respect to $g$. The other, denoted $\gamma_{0}$, is the
geodesic with respect to $g_{0}$ (and is simply a straight line in the
coordinates). Assume that both are parametrized so that
$\gamma(0)=\gamma_{0}(0)=p$ and $\gamma(1)=\gamma_{0}(1)=q$.

Using the bounds on the metric, we have that

$\displaystyle d(q)^{2}$ $\displaystyle=$
$\displaystyle\int_{0}^{1}g\left(\dot{\gamma},\dot{\gamma}\right)ds$
$\displaystyle\geq$ $\displaystyle
Q^{-1}\int_{0^{1}}g_{0}\left(\dot{\gamma},\dot{\gamma}\right)ds$
$\displaystyle\geq$ $\displaystyle
Q^{-1}\inf_{\\{\tilde{\gamma}~{}|~{}\tilde{\gamma}(0)=p,\tilde{\gamma}(1)=q\\}}\int_{0}^{1}g_{0}\left(\frac{d\tilde{\gamma}}{ds},\frac{d\tilde{\gamma}}{ds}\right)ds$
$\displaystyle=$ $\displaystyle Q^{-1}\delta_{0}(q)^{2}$

Similarly, by consider the curve $\gamma_{0}$ and repeating the same argument,
we can show that $d(q)^{2}\leq Q\delta_{0}(q)^{2}$. This shows that for $q$
close to $p$, the two distance functions are close in $C^{0}$-sense.

We now show that the functions are close in $C^{1}$-sense as well. To do so,
we bound the acceleration of the $g$-geodesics in the $x$-coordinates. This
then implies that for $p$ and $q$ close, the gradients of $d(q)$ and
$\delta_{0}(q)$ are very similar. Consider a point $q$ which is close111111We
can assume that $p$ and $q$ are close in terms of $d$ or $\delta_{0}$, since
we have shown that these distances control each other. to $p$. The geodesic
from $p$ to $q$ in the $g_{0}$ metric is a straight line whose length (in the
sense of $g_{0}$) is $\delta_{0}(q)$. Furthermore, the gradient of
$\delta_{0}$ at $q$ is of unit norm (again in the sense of $g_{0}$) and points
in the same direction as the line segment from $p$ to $q$.

On the other hand, the unit speed geodesic for the metric $g$ from $p$ to $q$
satisfy the equations

$\frac{d^{2}\gamma^{i}}{ds^{2}}+\Gamma^{i}_{jk}\frac{d\gamma^{j}}{ds}\frac{d\gamma^{k}}{ds}=0$
(B.1)

where

$\Gamma^{i}_{jk}=\frac{1}{2}g^{i\ell}\left(\frac{\partial g_{j\ell}}{\partial
x^{k}}+\frac{\partial g_{k\ell}}{\partial x^{j}}-\frac{\partial
g_{jk}}{\partial x^{\ell}}\right).$ (B.2)

However, from the scale-$C^{1}$-control, we can estimate that

$\displaystyle\left|\frac{d^{2}\gamma^{i}}{ds^{2}}\right|$ $\displaystyle\leq$
$\displaystyle\sum_{j,k}\frac{Q}{2}\frac{3(Q-1)}{r}\left|\frac{d\gamma^{j}}{ds}\right|\left|\frac{d\gamma^{k}}{ds}\right|$

along the entire geodesic. In order to make the estimates readable, we will
absorb any constants involving $n,r$ and $Q$ using the notation $f_{1}\lesssim
f_{2}$ whenever $f_{1}\leq Cf_{2}$ for some constant $C$ depending only on
$n,Q,$ and $r$. We will also use the notation

$\left.\left\|\frac{d^{k}\gamma}{ds^{k}}\right\|_{L^{1}}\right|_{s=\tau}=\sum_{i=1}^{n}\left|\frac{d^{k}\gamma^{i}(\tau)}{ds^{k}}\right|$

and the same corresponding for the $L^{2}$ norm as well.

In this notation, we find that

$\left|\frac{d^{2}\gamma^{i}}{ds^{2}}\right|\lesssim\left|\frac{d\gamma^{j}}{ds}\right|\left|\frac{d\gamma^{k}}{ds}\right|.$

Summing over the $i$ index, this implies that

$\displaystyle\left.\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}\right|_{s=\tau}$
$\displaystyle\lesssim$
$\displaystyle\left.\left\|\frac{d\gamma}{ds}\right\|_{L^{2}}^{2}\right|_{s=\tau}$
$\displaystyle\lesssim$
$\displaystyle\left\|1+\int_{0}^{\tau}\left.\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}\right|_{s=t}\,dt\right\|_{L^{2}}^{2}$
$\displaystyle\leq$ $\displaystyle
2+2\left(\int_{0}^{\tau}\left.\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}\right|_{s=t}\,dt\right)^{2}$

We now define
$F(\tau)=\int_{0}^{\tau}\left.\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}\right|_{s=t}\,dt$.
When written in terms of $F$, the above estimate shows that

$\displaystyle\frac{dF}{d\tau}$ $\displaystyle\lesssim$ $\displaystyle
1+F^{2}$

Dividing both sides by $1+F^{2}$ and integrating, we find that

$\arctan{F(\tau)}\lesssim\tau.$

For $\tau$ small, this provides an upper bound for $F$. However, since
$\gamma$ is a unit speed geodesic, this shows that for small $d$ (equivalently
$\delta_{0}$), the acceleration (in coordinates) of $\gamma$ is very small. As
a result, $\gamma$ is $C^{1}$-close to a line segment from $p$ to $q$ (in
coordinates). Therefore, the gradient of $d$ at $q$ is close to the gradient
of $\delta_{0}$ at $q$, which implies that the functions are $C^{1}$-close as
well. Since the choice of $q$ was arbitrary, this shows that
$\|d-\delta_{0}\|_{C^{1}\left(B_{g_{0}}(p,\epsilon)\backslash\\{p\\}\right)}$
can be made arbitrarily small by taking $\epsilon$ small.

We can use this same idea to show that $d$ and $\delta_{0}$ are $C^{2}$-close
as well. Since the argument is essentially identical to the $C^{1}$ estimate,
we will provide a sketch rather than a fully detailed argument. When it is
unambiguous, we will also drop the subscript indicating which particular value
of $s$ to consider.

  1. (1)

First differentiate Equations B.1 with respect to $s$. Differentiating the
Christoffel symbols (Equations B.2) and using the scale-$C^{2}$-control, we
obtain an estimate of the form

$\left\|\frac{d^{3}\gamma}{ds^{3}}\right\|_{L^{1}}\lesssim\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}\left\|\frac{d\gamma}{ds}\right\|_{L^{1}}^{2}+\left\|\frac{d\gamma}{ds}\right\|_{L^{1}}^{3}.$

  2. (2)

Using Young’s inequality, this shows that

$\displaystyle\left\|\frac{d^{3}\gamma}{ds^{3}}\right\|_{L^{1}}$
$\displaystyle\lesssim$
$\displaystyle\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}^{2}+\left\|\frac{d\gamma}{ds}\right\|_{L^{1}}^{4}+\left\|\frac{d\gamma}{ds}\right\|_{L^{1}}^{3}.$

  3. (3)

Using the bound on $F$ from the $C^{1}$ estimate, for small time we can
control $\left\|\frac{d\gamma}{ds}\right\|$ by a constant. This then implies
that

$\displaystyle\left\|\dfrac{d^{3}\gamma}{ds^{3}}\right\|_{L^{1}}$
$\displaystyle\lesssim\left\|\dfrac{d^{2}\gamma}{ds^{2}}\right\|_{L^{1}}^{2}+1.$

  4. (4)

Using the fact that

$\left.\left\|\frac{d^{2}\gamma}{ds^{2}}\right\|\right|_{s=\tau}\leq\int_{0}^{\tau}\left.\left\|\frac{d^{3}\gamma}{ds^{3}}\right\|\right|_{s=t}dt,$

we find that

$\displaystyle\left.\left\|\frac{d^{3}\gamma}{ds^{3}}\right\|\right|_{s=\tau}$
$\displaystyle\lesssim\left(\int_{0}^{\tau}\left.\left\|\dfrac{d^{3}\gamma}{ds^{3}}\right\|\right|_{s=t}dt\right)^{2}+1.$

  5. (5)

From this estimate, we can integrate out the differential inequality and bound

$\int_{0}^{\tau}\left.\left\|\dfrac{d^{3}\gamma}{ds^{3}}\right\|\right|_{s=t}dt\lesssim\tan(\tau).$

For sufficiently small times $\tau$, this provides a small bound on the total
jerk of $\gamma$.

  6. (6)

As a result, $\gamma$ is $C^{2}$-close to a line segment from $p$ to $q$.
Therefore, the gradient of $d$ is $C^{1}$-close to the gradient of
$\delta_{0}$ at q, which implies that the two distance functions are
$C^{2}$-close.

With these estimate, we can show that $B_{g_{0}}(p,\epsilon)$ is convex in the
$x$-coordinates. Consider a point $q$ with $d(p,q)$ small and the hyperplane
(in $x$-coordinates) $V$ through $q$ which is perpendicular to the line
segment from $p$ to $q$. Near $q$, both the functions $d$ and $\delta_{0}$
have gradients which are transverse to this hyperplane. As such, by the
implicit function theorem we can locally find two functions
$\ell_{1},\ell_{2}:V\to\mathbb{R}$ so that $d(v,\ell_{1}(v))=d(q)$ and
$d(v,\ell_{2}(v))=\delta_{0}(q)$. In other words, we use the implicit function
theorem to express the level sets of $d$ and $\delta_{0}$ as graphs in a small
neighborhood of $q$. Furthermore, we can write the derivatives of $\ell_{1}$
and $\ell_{2}$ in terms of the derivatives of $d(q)$ and $\delta_{0}(q)$,
respectively.

Since $d(q)$ and $\delta_{0}(q)$ are $C^{2}$-close, it follows that $\ell_{1}$
and $\ell_{2}$ are also $C^{2}$-close. However, the level sets of $\delta_{0}$
are ellipsoids with bounded eccentricity, and thus $\ell_{2}$ is a uniformly
strongly convex function. For small $d$, this implies that the level sets of
$d$ are also convex, since $\ell_{1}$ is $C^{2}$-close to a strongly convex
function (and so must have non-negative definite Hessian). This then implies
that the ball $B_{g}(p,d)$ is convex as a subset of Euclidean space. $\square$

This lemma addresses a question posed by Macbeth [20] on Mathoverflow, so may
be of independent interest. Originally, she asked for explicit estimates on
the radius of geodesic balls whose image in coordinates is convex. In theory,
it is possible to find closed form estimates by carefully keeping track of all
of the constants involving $Q,n$ and $r$ and making the estimates from the
implicit function theorem explicit. However, we have not done this here.

For our purposes, the previous lemma shows that for a Hessian manifold whose
potential is strongly convex and bounded in $C^{4}$, small balls in the primal
coordinates are convex in the dual coordinates.

######  Lemma 19.

Suppose $\Omega$ is a Hessian manifold with local potential $\Psi$ that is
strongly convex and bounded in $C^{4}$. Then there exists an $\epsilon>0$ so
that all balls $B(p,r)$ ($p\in\Omega$, $r<\epsilon$) in the primal coordinates
$x$ are convex in the dual coordinates $\theta$.

###### Proof.

Consider the primal $x$-coordinates with domain $\Omega$. We induce $\Omega$
with its natural inner product as an open domain of Euclidean space, which we
denote by $\tilde{g}$. More precisely,

$\tilde{g}\left(\frac{\partial}{\partial x^{i}},\frac{\partial}{\partial
x^{j}}\right)=\delta_{ij}.$

Note that this is not the metric on $\Omega$ as a Hessian manifold, which is
why we have included the tilde. We now show that the dual coordinates $\theta$
are $C^{2}$-scale-controlled with respect to $\tilde{g}$. To see this, note
that we can express the primal coordinates in terms of the dual coordinates as

$x^{i}=\frac{\partial\Psi^{\ast}}{\partial\theta^{i}},$

where $\Psi^{\ast}$ is the Legendre dual of $\Psi$. As such, we have the
following expression for $\tilde{g}$ in the $\theta$ coordinates.

$\displaystyle\tilde{g}\left(\frac{\partial}{\partial\theta^{i}},\frac{\partial}{\partial\theta^{j}}\right)$
$\displaystyle=$
$\displaystyle\frac{\partial^{2}\Psi^{\ast}}{\partial\theta^{i}\partial^{k}}\delta_{kl}\frac{\partial^{2}\Psi^{\ast}}{\partial\theta^{j}\partial^{l}}$
$\displaystyle=$
$\displaystyle\frac{\partial^{2}\Psi^{\ast}}{\partial\theta^{i}\partial^{k}}\frac{\partial^{2}\Psi^{\ast}}{\partial\theta^{j}\partial^{k}}$

Using the symmetry of mixed partials, this shows that as a matrix, $\tilde{g}$
satisfies

$\tilde{g}=[\operatorname{Hess}_{\theta}\Psi^{\ast}]^{\top}[\operatorname{Hess}_{\theta}\Psi^{\ast}]$

where
$[\operatorname{Hess}_{\theta}\Psi^{\ast}]_{ij}=\frac{\partial^{2}\Psi^{\ast}}{\partial\theta^{i}\partial^{j}}$.
As such, to show that $\tilde{g}$ is $C^{2}$-scale-controlled in the
$\theta$-coordinates, we need a lower bound on the singular values of $h$ and
a $C^{4}$-estimate on $\Psi^{\ast}$.

To bound the singular values, we return our attention to the primal
coordinates $x^{i}$ and note the following. Since $\Psi$ is strongly convex
and $C^{4}$, there exists a $Q>1$ so that

$Q^{-1}\delta_{ij}\leq\left[\operatorname{Hess}_{x}\Psi\right]\leq
Q\delta_{ij},$

where $\operatorname{Hess}_{x}\Psi=\frac{\partial^{2}\Psi}{\partial
x^{i}\partial x^{j}}$ Here, the first inequality follows from strong
convexity, and the second inequality follows from the $C^{4}$-estimate (which
implies a $C^{2}$-estimate).

Now, in the $\theta$ coordinates, it is a well known fact about Legendre
duality that $[\operatorname{Hess}_{\theta}\Psi^{\ast}]$ is given by the
inverse of $[\operatorname{Hess}_{x}\Psi]$. From this, we immediately obtain
that

$Q^{-1}\delta_{ij}\leq[\operatorname{Hess}_{\theta}\Psi^{\ast}]\leq
Q\delta_{ij},$

which implies that

$Q^{-2}\delta_{ij}\leq\tilde{g}\leq Q^{2}\delta_{ij}.$

Furthermore, we can express the third and fourth derivatives of $\Psi^{\ast}$
in terms of the first four derivatives of $\Psi$ and the inverse of the
Hessian of $\Psi$. Using the $C^{4}$-estimate on $\Psi$ and the bound from
strong convexity, this provides a $C^{4}$-estimate on $\Psi^{\ast}$. As such,
$\tilde{g}$ is scale controlled in the $\theta$-coordinates. which implies
from the previous lemma that small balls (in the sense of $\tilde{g}$) are
convex in the $\theta$-coordinates.

$\square$