# Kähler-Ricci Flow preserves negative anti-bisectional curvature

Gabriel Khan<EMAIL_ADDRESS>and  Fangyang Zheng<EMAIL_ADDRESS>
###### Abstract.

In recent work (Pure Appl. Anal. 2 (2020), 397-426 [26]), the first named
author and J. Zhang found a connection between the regularity theory of
optimal transport and the curvature of Kähler manifolds. In particular, we
showed that the Ma-Trudinger-Wang (MTW) tensor for a cost function
$c(x,y)=\Psi(x-y)$ can be understood as the anti-bisectional curvature of an
associated Kähler metric defined on a tube domain. Here, the anti-bisectional
curvature is defined as
$R(\mathcal{X},\overline{\mathcal{Y}},\mathcal{X},\overline{\mathcal{Y}})$
where $\mathcal{X}$ and $\mathcal{Y}$ are polarized $(1,0)$ vectors and $R$ is
the curvature tensor. The correspondence between the anti-bisectional
curvature and the MTW tensor provides a meaningful sense in which the anti-
bisectional curvature can have a sign (i.e., be positive or negative).

In this paper, we study the behavior of the anti-bisectional curvature under
Kähler-Ricci flow. Somewhat unexpectedly, we find that non-positive anti-
bisectional curvature is preserved under the flow. In complex dimension two,
we also show that non-negative orthogonal anti-bisectional curvature (i.e.,
the MTW(0) condition) is preserved under the flow. We provide several
applications of these results — in complex geometry, optimal transport, and
affine geometry.

##  1\. Introduction

For the past four decades, Hamilton’s Ricci flow [24] has played a central
role in geometric analysis. Starting with an initial Riemannian manifold
$(M,g)$, this flow evolves the metric according to its Ricci curvature:

(1) $\frac{\partial g}{\partial t}=-2\,\textrm{Ric}(g).$

This acts as a reaction-diffusion equation for the curvature and can be used
to deform a Riemannian metric towards a canonical geometry.

When the underlying manifold is Kähler, the Ricci flow preserves the complex
structure and the Kählerity of the metric [8]. Furthermore, it is possible to
write the flow in terms of the Kähler potentials $\Psi(z,t)$ in a holomorphic
coordinate chart $\\{z^{i}\\}_{i=1}^{n}$. Doing so, the potential evolves
according to the parabolic complex Monge-Ampére equation

(2) $\frac{\partial\Psi}{\partial
t}=2\log\left(\operatorname{det}\left[\frac{\partial^{2}\Psi}{\partial
z^{i}\partial\bar{z}^{j}}\right]\right).$

For Kähler manifolds, this allows us to study the Ricci flow, which is
originally a weakly parabolic system of equations, in terms of a single fully
nonlinear parabolic equation.

In this paper, we study the Kähler-Ricci flow on tube domains (i.e. domains of
the form $(\Omega+\sqrt{-1}\mathbb{R}^{n})\subset\mathbb{C}^{n}$ where
$\Omega$ is a convex domain in $\mathbb{R}^{n})$, or more generally on the
tangent bundles of Hessian manifolds. Our focus is on the behavior of the
“anti-bisectional curvature,” which is defined as

(3)
$\mathfrak{A}(\mathcal{V},\mathcal{W})=R(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})$

for $(1,0)$-vectors $\mathcal{V}$ and $\mathcal{W}$.

By considering metrics which are translation symmetric in the imaginary
directions and restricting our attention to polarized holomorphic vectors,
there is a meaningful sense in which the anti-bisectional curvature can be
positive or negative. Here, we say that a $(1,0)$-vector $\mathcal{V}$ is
polarized if $\mathcal{V}=v^{i}\frac{\partial}{\partial z^{i}},$ where the
coefficients $v^{i}$ are real numbers that depend only on the point
$x\in\Omega$ (in the holomorphic coordinates
$\\{z^{i}=x^{i}+\sqrt{-1}y^{i}\\}_{i=1}^{n}$ with $x\in\Omega$ and
$y\in\mathbb{R}^{n}$). For a more precise definition of polarized vectors and
signed anti-bisectional curvature, see Subsection 2.2. We find that negativity
of the anti-bisectional curvature is preserved under Kähler-Ricci flow. With
the exception of negative Gaussian curvature for Riemann surfaces, every other
curvature condition which was previously known to be preserved under Ricci
flow is a positivity condition, so this result is somewhat unexpected.

######  Theorem 1.

Let $(T\Omega,\omega_{0})$ be a complete Kähler manifold with bounded
curvature and non-positive anti-bisectional curvature. Consider the Kähler-
Ricci flow $(T\Omega,\omega_{t})$ with initial metric $\omega_{0}$. For all
time that the flow exists, the anti-bisectional curvature remains non-
positive.

We also study the orthogonal anti-bisectional curvature (i.e., the
antibisectional curvature restricted to orthogonal polarized vectors) and show
that in complex dimension two, positivity of this quantity is also preserved
under the flow.

######  Theorem 2.

Suppose that $\Omega\subset\mathbb{R}^{2}$ is a convex domain and
$\Psi:\Omega\to\mathbb{R}$ is a strongly convex function so that the
associated Kähler manifold $(T\Omega,\omega_{0})$

  1. (1)

is complete,

  2. (2)

has bounded curvature, and

  3. (3)

has non-negative orthogonal anti-bisectional curvature.

Then the orthogonal anti-bisectional curvature remains non-negative along
Kähler-Ricci flow.

###  1.1. Applications in complex geometry

For tube domains, Theorem 1 provides a new way to study the relationship
between the holomorphic sectional curvature and Ricci curvature, which is an
important question in complex geometry. Recently, Wu and Yau [62] showed that
on any Kähler manifold $(\mathbb{M},\omega_{0})$ with negatively pinched
holomorphic sectional curvature, there exists a Kähler-Einstein metric of
negative scalar curvature which is uniformly equivalent to the original metric
$\omega_{0}$. However, the following question remains open.

######  Question 1.

When is it the case that a Kähler manifold admits a Kähler-Einstein metric
with negative holomorphic sectional curvature?

For general Kähler manifolds, this is a very difficult problem. Establishing
the existence of a Kähler-Einstein metric is already a non-trivial question
(see, e.g., [63, 64]), and the existence theorems do not provide control over
the sign of the holomorphic sectional curvature.

Although negative anti-bisectional curvature does not imply that the
holomorphic sectional curvature is negative for all $(1,0)$-vectors, it does
imply that the holomorphic sectional curvature of polarized vectors remains
non-positive under the flow. Combining Theorem 1 with work of Tong studying
the convergence of negatively pinched metrics [53], we show the following
result.

######  Corollary 1.

Suppose that $T\Omega$ admits a complete Kähler metric $\omega_{0}$ whose
holomorphic sectional curvature is negatively pinched and whose anti-
bisectional curvature is non-positive. Then the complete Kähler-Einstein
metric on $T\Omega$ with scalar curvature $-2$ has non-positive anti-
bisectional curvature.

When $\Omega$ is a bounded and strongly convex domain, it is possible to
sharpen this estimate and show that the polarized holomorphic sectional
curvature of the Kähler-Einstein metric is negatively pinched.

######  Corollary 2.

Suppose that $\Omega$ is a smooth, bounded, and strongly convex domain and
that $(T\Omega,\omega_{0})$ is a complete Kähler metric with negative anti-
bisectional curvature and negatively pinched holomorphic sectional curvature.
Then the complete Kähler-Einstein metric on $T\Omega$ has negative cost-
curvature.

Negative cost-curvature is a stronger version of non-positive anti-bisectional
curvature. The etymology of this definition comes from optimal transport, and
is described in Subsection 2.3.

######  Definition 1 (Negative cost-curvature).

A metric on a tube domain is negatively cost-curved if its anti-bisectional
curvature is non-positive and its holomorphic sectional curvature satisfies
$H(\mathcal{X})\leq K\|\mathcal{X}\|^{4}$ for some $K<0$ whenever
$\mathcal{X}$ is a polarized holomorphic vector.

For compact Hessian manifolds, it is possible to prove a version of Corollary
2 under the assumption that the first affine Chern class is negative, without
assuming that the holomorphic sectional curvature of the tangent bundle is
negatively pinched (see Definition 6 [44] for a definition of affine Chern
classes).

######  Corollary 3.

Let $(M,g_{0},D)$ be a compact Hessian manifold. Suppose that the associated
Kähler metric $\omega_{0}$ on $TM$ has non-positive anti-bisectional curvature
and that the first affine Chern class of $M$ is negative. Then, under the
normalized Kähler-Ricci flow, $TM$ converges to a Kähler-Einstein metric which
is negatively cost-curved.

####  1.1.1. Instantaneous control of the metric

Even without the assumption that the holomorphic sectional curvature of
$T\Omega$ is negatively pinched, negative anti-bisectional curvature provides
strong control over the geometry along the flow. In particular, to bound the
curvature of the positive time metrics when the initial metric has negative
anti-bisectional curvature, it is sufficient to bound a single quantity
$\mathfrak{O}$, which we call the orthogonal anti-bisectional trace curvature
(see Definition 6 for a formal definition).

######  Corollary 4.

Suppose that $(T\Omega,\omega_{0})$ is a Kähler metric whose scalar curvature
is bounded below by a constant $-K$ and whose anti-bisectional curvature is
non-positive. Consider the positive time metric $(T\Omega,\omega_{t_{0}})$ and
$\mathcal{E}=\\{\mathcal{E}_{i}\\}_{i=1}^{n}$ be a polarized unitary frame at
time $t_{0}$ and $\mathcal{X}$ be a polarized holomorphic unit vector with
respect to $\omega_{t_{0}}$. Then there exists a constant $C(n)$ depending
only on the dimension $n$ so that along Kähler-Ricci flow, the polarized
holomorphic sectional curvature satisfies the estimate

(4) $0\geq
H(\mathcal{X})\geq-C(n)\left(\frac{n}{t+\frac{n}{K}}+2\mathfrak{O}(\mathcal{E})\right).$

Note that for any $K\geq 0$, we have the uniform bound

(5) $-\frac{C(n)}{t+\frac{n}{K}}<-\frac{C(n)}{t}.$

As such, the previous corollary provides a bound on the polarized holomorphic
sectional curvature which is independent of the initial scalar curvature and
only depends on $\mathfrak{O}(\mathcal{E})$ at time $t_{0}$. It is possible to
show that for metrics with negative anti-bisectional curvature, the curvature
tensor can be bounded from above and below in terms of the polarized
holomorphic sectional curvature and the orthogonal anti-bisectional trace
curvature (see Appendix B for details). Thus, this corollary shows that in
order to control the geometry of the metric at a time $t_{0}>0$, we need only
control $\mathfrak{O}$.

To prove Corollary 4, we establish a quantitative version of Berger’s formula
for the scalar curvature [1]. This estimate has other applications in complex
geometry, so we mention it now.

######  Proposition 3.

Suppose $\mathbb{M}^{n}$ is a Kähler manifold with non-positive holomorphic
sectional curvature. Then there exists a constant $C_{n}$ depending only on
the dimension so that for all unit $(1,0)$ tangent vectors $\mathcal{X}$, the
scalar curvature $S$ satisfies the inequality

(6) $S\leq C_{n}H(\mathcal{X}).$

For metrics with non-negative holomorphic sectional curvature, we obtain the
same result with the final inequality reversed.

###  1.2. Surfaces with non-negative orthogonal anti-bisectional curvature and
optimal transport

Apart from its applications in complex geometry, anti-bisectional curvature
plays an important role in optimal transport. In particular, given a Kähler
metric on a tube domain whose orthogonal anti-bisectional curvature is non-
negative, we can define an associated cost function which satisfies the MTW(0)
condition (see Subsection 2.3 for more details). This condition plays an
important role in the regularity theory of optimal transport, in that it
precludes any local obstructions to the smoothness of transport maps.

For this reason, it is worthwhile to determine whether the non-negativity of
orthogonal anti-bisectional curvature is preserved under the flow. Theorem 2
shows that in complex dimension two (but in no other dimension), this is
indeed the case. Due to the correspondence between the MTW tensor and the
orthogonal anti-bisectional curvature, this result allows us to generalize the
MTW(0) condition for cost functions of the form $c(x,y)=\Psi(x-y)$ where
$\Psi$ is a convex function on an open domain in $\mathbb{R}^{2}$.

######  Definition 2 (KR weak regularity).

A cost function $c(x,y)=\Psi_{0}(x-y)$ is KR weakly regular if there exists a
solution to the parabolic flow

(7) $\begin{cases}\frac{\partial}{\partial
t}\Psi(x,t)=2\log\left(\det[\operatorname{Hess}\Psi(x,t)]\right)\\\
\Psi(x,0)=\Psi_{0}(x)\end{cases}$

which has non-negative orthogonal anti-bisectional curvature for positive
time.

Using this flow, we prove an interior Hölder estimate for optimal transport
which only relies on a $W^{2,p}$ estimate for $\Psi_{0}$ (see Corollary 6 for
a precise statement). At present, this result is limited in that to define the
above flow we must consider complete Kähler metrics. One natural question is
whether these results can be extended to incomplete Kähler metrics, which
often appear in optimal transport.

######  Question 2.

Can we find suitable boundary conditions so that the above flow is well
defined and Theorem 2 holds when $\Psi_{0}$ does not induce a complete metric?

###  1.3. Examples

At first, one might worry that negative anti-bisectional curvature is an
overly strong assumption and that the results of the previous sections may be
vacuous. To show that this is not the case, in Section 6, we provide several
example of metrics which satisfy this assumption. Of particular interest is
the following example.

######  Example.

Let $\mathbb{B}$ be the unit ball in $\mathbb{R}^{n}$ and consider the tube
domain $T\mathbb{B}$. Let $\omega_{0}$ be the Kähler metric induced by the
potential

(8) $\Psi(z)=-\log\left(1-\sum_{i=1}^{n}x_{i}^{2}\right),$

where $z^{i}=x^{i}+\sqrt{-1}y^{i}$ and $\Psi$ is independent of the fiber
directions $y$.

This is a negative cost curved metric on $T\mathbb{B}$ with negative
holomorphic sectional curvature. Appealing to Corollary 2, this shows that the
Kähler-Einstein metric on $T\mathbb{B}$ (which was originally studied by
Calabi [6]) also has negative cost-curvature.

##  2\. Background

We now provide background material on Hessian manifolds, anti-bisectional
curvature, optimal transport, and Kähler-Ricci flow. This section is not
intended to be a complete reference; we focus on results that are needed in
this paper. However, we have included references which provide a more thorough
background for the interested reader.

###  2.1. Background on Kähler-Sasaki metrics

In this paper, we will primarily study tube domains. However, our results can
be generalized to the tangent bundles of compact or parallelizable Hessian
manifolds (we reserve $\Omega$ for convex domains, so denote Hessian manifolds
by $M$). We refer to such metrics as Kähler-Sasaki metrics. For a Riemannian
manifold $(M,g)$ with an affine connection $D$, the Sasaki metrics is an
associated almost-Hermitian structures on the tangent bundle $TM$ (see [16,
47] for a more complete reference). In this paper, we restrict our attention
to the case where $(M,g,D)$ is a Hessian manifold, in which case the Sasaki
metric will be Kähler, hence the term Kähler-Sasaki.

######  Definition 3 (Hessian manifolds).

A Riemannian manifold $(M,g)$ with an affine connection $D$ is said to be a
Hessian manifold if

  1. (1)

$D$ is a flat (curvature- and torsion-free) connection,

  2. (2)

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

  3. (3)

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

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

At first, it might seem unnecessary to consider convex domains as Hessian
manifolds. However, there is one conception advantage to using this approach;
by considering a convex domain $\Omega$ as a Hessian manifold, we can change
coordinates in an affine way. Doing so will play an important role in the
proof of Theorem 1, so we take some time to discuss this more general setting.

To define a Kähler metric on the tangent bundle of a Hessian manifold, we
observe that the connection $D$ induces an atlas $\mathcal{A}$ of coordinates
charts whose transition maps are affine functions. We consider such a chart
(denoted $x$) and construct coordinates on the tangent bundle $TM$ by $(x,y)$,
where the fiber coordinates $y$ are defined as
$y^{j}(\mathcal{X})=\mathcal{X}^{j}$ for a vector
$\mathcal{X}=\mathcal{X}^{i}\partial_{x^{i}}$. Finally, we consider the
complex coordinates $z=x+\sqrt{-1}y$. This coordinate chart is defined on
$TU_{x}$. Furthermore, it is possible to construct such coordinates in a
neighborhood of every point in $M$. Since the transition maps of the charts in
$\mathcal{A}$ are affine, the transition maps of their associated charts on
$TM$ are holomorphic. As such, the tangent bundle of an affine manifold is a
complex manifold.

To define a Kähler metric on the tangent bundle, we lift the Hessian potential
$\Psi$ to the tangent bundle $TM$ using the horizontal lift, defined as

(10) $\Psi^{h}(z)=\Psi(x)$

for $z=x+\sqrt{-1}y$. It is worth noting that a Kähler-Sasaki metric is
complete if and only if the underlying Hessian manifold $(M,g,D)$ is [39].

Given a convex domain $\Omega$ in Euclidean space, it is possible to construct
a Hessian metric simply by choosing a convex potential function and using the
connection induced by differentiation in coordinates. In this case, the
tangent bundle of $\Omega$ is simply the tube domain $T\Omega$. Furthermore,
the Kähler metric has a natural $\mathbb{R}^{n}$ symmetry induced by
translations in the $y$-coordinate.

####  2.1.1. Lifting vectors

It is possible to lift tangent vectors of $\Omega$ to tangent vectors of the
tube domain $T\Omega$. Conceptually, this is quite simple (at least for tube
domains), but it is essential to the notion of signed anti-bisectional
curvature. As such, we take some time to define the concept carefully. Before
we do so, we remark on some notational conventions.

To discuss lifting vectors, we must consider the tangent bundle of $\Omega$
separately from the tube domain $T\Omega$. To denote the tangent bundle of
$\Omega$, we use the notation $T^{real}\Omega$. We will generally denote
elements of $T^{real}\Omega$ with lower case letters (e.g. “$v$”) and tangent
vectors on $T\Omega$ (i.e., elements of $TT\Omega$) with uppercase
calligraphic letters (e.g. “$\mathcal{V}$”). When it is clear from context, we
drop the notation for which base point we are using. In other words, $v\in
T^{real}\Omega$ is used as shorthand for $v\in T^{real}_{x}\Omega$.
Furthermore, in order to avoid over-using the letter $T$, we will often simply
use the letter $\mathbb{M}$ to refer to Kähler manifolds, Kähler-Sasaki or
otherwise.

It is possible to lift vectors from $T^{real}\Omega$ (or $T^{real}M$ for a
general Hessian metric) to $T\mathbb{M}$ in several different ways. In this
paper, we will only consider a single lift, which takes a tangent vector

(11) $v=\sum_{i}v^{i}\frac{\partial}{\partial{x^{i}}}\in
T^{real}_{x_{0}}\Omega$

to the $(1,0)$ vector

(12) $\mathcal{V}=v^{pol.}:=\sum_{i}v^{i}\frac{\partial}{\partial{z^{i}}}\in
T_{(x_{0},y)}^{(1,0)}\mathbb{M}.$

As before, $\\{z^{i}\\}_{i=1}^{n}$ are the holomorphic coordinates
$x^{i}+\sqrt{-1}y^{i}$. In this formula, we are free to pick the fiber
coordinate $y$ at will. Since the metric is translation symmetric in the
fibers, this choice will not affect the geometry. We say that the lifted
vectors are polarized, since their components are restricted to an
$n$-dimensional subspace of $T\mathbb{M}$. Furthermore, we can use the space
of polarized (1,0) vectors to make sense of the notion of positive or negative
anti-bisectional curvature.

###  2.2. Anti-bisectional curvature

Given two $(1,0)$ vectors $\mathcal{V}$ and $\mathcal{W}$, the anti-
bisectional curvature is defined to be

(13)
$\mathfrak{A}(\mathcal{V},\mathcal{W})=R(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}}).$

At first, it does not seem meaningful to say that the anti-bisectional
curvature is either positive or negative. While the quantity
$R(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})$ is
well defined for any two $(1,0)$-vectors $\mathcal{V}$ and $\mathcal{W}$, if
we simply consider the vectors $\sqrt{-1}\mathcal{V}$ and $\mathcal{W}$, we
have
$R(\mathcal{V},\overline{\mathcal{V}},\mathcal{V},\overline{W})=-R(\sqrt{-1}\mathcal{V},\overline{\mathcal{W}},\sqrt{-1}\mathcal{V},\overline{\mathcal{W}})$.

However, for Kähler-Sasaki metrics, by restricting our attention to polarized
vectors, it is meaningful to talk about this quantity being positive or
negative.

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

The Kähler metric $\mathfrak{h}_{\Psi}$ has non-positive orthogonal anti-
bisectional curvature, if $\mathfrak{A}(\mathcal{V},\mathcal{W})\leq 0$ for
any $\mathcal{V}=v^{pol.}$, $\mathcal{W}=w^{pol.}$ with $v,w\in
T^{real}\Omega$.

By taking $v=w$, we can see that negative anti-bisectional curvature implies
negative holomorphic sectional curvature for vectors which are polarized.
However, negative anti-bisectional curvature does not imply negative
holomorphic sectional curvature for arbitrary holomorphic vectors. It also
does not imply negative Ricci curvature (see Subsection 4.3 for some
consequences of this).

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

The Kähler metric $\mathfrak{h}_{\Psi}$ has non-negative orthogonal anti-
bisectional curvature (abbreviated (NOAB)), if
$\mathfrak{A}(\mathcal{V},\mathcal{W})\geq 0$ for any $\mathcal{V}=v^{pol.}$,
$\mathcal{W}=w^{pol.}$ with $v,w\in T^{real}\Omega$ such that $g(v,w)=0$.

We also define the orthogonal anti-bisectional trace curvature to be the trace
of the orthogonal anti-bisectional curvature.

######  Definition 6 (Orthogonal anti-bisectional trace curvature).

Let $\\{\mathcal{E}_{i}\\}\subset T_{z}T\Omega$ be a unitary frame of
polarized holomorphic vectors. The orthogonal anti-bisectional trace curvature
is defined to be the quantity

(14) $\mathfrak{O}=\sum_{i\neq
j}R(\mathcal{E}_{i},\overline{\mathcal{E}_{j}},\mathcal{E}_{i},\overline{\mathcal{E}_{j}}).$

The orthogonal anti-bisectional trace curvature generally depends on the
choice of polarized unitary frame (which corresponds to an orthonormal frame
of $T^{real}\Omega$). As such, we can interpret the orthogonal anti-
bisectional trace curvature as a map from the orthonormal frame bundle of
$\Omega$ to $\mathbb{R}$. In this paper, we will often fix such a frame and
consider this as a scalar function. Non-negative orthogonal anti-bisectional
curvature does not imply any sign for the holomorphic sectional curvature or
the scalar curvature.

####  2.2.1. The curvature of tube domains

Heuristically, we expect invariant metrics on tube domains to be negatively
curved. This heuristic can be made precise in various ways. For instance, when
$\Omega$ is a smooth strongly convex domain that does not contain any line,
$T\Omega$ is biholomorphic to a smooth bounded strongly pseudoconvex domain.
As such, by a result of Klembeck [31], it admits a complete metrics of
strongly negative holomorphic sectional curvature. Furthermore, we can use
results of Cheng and Yau [12] (or alternatively the recent work of Wu and Yau
[62]) to show that such tube domains admit Kähler-Einstein metrics with
negative scalar curvature.

With these results in mind, it is reasonable to try to find complete metrics
of negative anti-bisectional curvature (in that
$R(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{W})<0$ for all
$\mathcal{V}=v^{pol.},\mathcal{W}=w^{pol.}$) or negative cost-curvature.
However, when the complex dimension is greater than one, we are not aware of
any examples with strongly negative anti-bisectional curvature (in that

(15)
$R(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{W})<-\kappa|V|^{2}|W|^{2}$

for some $\kappa>0$). In particular, any complex space form has vanishing
orthogonal anti-bisectional curvature. Furthermore, for a smooth and strictly
pseudo-convex tube domain, the Kähler-Einstein metric converges to a complex
hyperbolic metric near its boundary, which implies that its orthogonal anti-
bisectional curvature attenuates. As such, negative cost-curvature is the
correct way to strengthen the hypothesis of negative anti-bisectional
curvature so that the assumption does not become vacuous.

###  2.3. Optimal transport and the role of anti-bisectional curvature

Optimal transport studies the most economical way to transport resources. This
problem was originally studied in 1781 [40] and combines aspects from
analysis, probability and geometry. For two comprehensive references on the
subject, we refer the reader [46] and [58]. An understanding of this subject
is not necessary to understand the main results of this paper, so this
subsection may be skipped by readers whose primary interest is complex
geometry. However, the regularity theory of optimal transport provides the
original motivation for studying anti-bisectional curvature and we will use
several results about optimal transport in Subsection 5.1.

The Monge problem of optimal transport is the following. We are given an
initial probability space $(X,\mu)$, a target probability space $(Y,\nu)$ and
a cost function $c:X\times Y\to\mathbb{R}$, which provides the cost of
transporting a unit of mass from a point $x\in X$ to a point $y\in Y$. We then
seek to find a $\mu$-measurable map (i.e., a transport plan) $T:X\to Y$ such
that such that the total cost is minimized.

(16) $\int_{X}c(x,T(x))d\mu=\inf_{S_{\sharp}\mu=\nu}\int_{X}c(x,S(x))d\mu$

Here, the notation $T_{\sharp}\mu=\nu$ means that $\mu[T^{-1}(Y)]=\nu[Y]$ for
each Borel set $Y\subset{\mathbb{R}^{n}}$.

This is a special case of the Monge-Kantorovich problem of optimal transport,
which tries to minimize the integral

(17) $\min_{\pi\in\Pi\left(\mu,\nu\right)}\int_{X\times Y}c(x,y)d\pi(x,y),$

where $\Pi\left(\mu,\nu\right)$ denotes the space of all couplings between
$\mu$ and $\nu$. The Monge-Kantorovich problem can be solved in great
generality and has a dual formulation as a linear program. We will make use of
this generalization in Corollary 6, so we will mention it now mention it now.

Under fairly general conditions, the solution to the Monge-Kantorovich problem
will be unique and also be a solution to the Monge problem. For instance, this
is the case whenever

  1. (1)

$X$ and $Y$ are domains of $\mathbb{R}^{n}$,

  2. (2)

$\mu$ and $\nu$ are absolutely continuous with respect to the Lebesgue
measure, and

  3. (3)

$c$ satisfies some smoothness and non-degeneracy assumptions.

When $c$ and the measures are sufficiently smooth, the transport map $T$ is
given by the $c$-subdifferential of a potential function $u:X\to\mathbb{R}$,
which weakly solves the Monge-Ampère type equation:

(18) $\det(\nabla^{2}u+A(x,\nabla u))=B(x,\nabla u).$

Here, $A$ is a matrix-valued function which depends on the cost function and
$B$ is a scalar valued function which depends on the cost function and the two
probability measures. Furthermore, the boundary conditions force the image of
the $c$-subdifferential of $u$ on $X$ to be $Y$. For a more complete overview
on the existence theory of the Monge problem, we refer the reader to Chapter
10 of Villani’s text [58] (see also Brenier [4] and Gangbo-McCann [18] for the
original results).

Once the existence of an optimal transport map has been established, it is
natural to ask is whether the transport map is continuous, in that nearby
points in initial configuration $(X,\mu)$ get sent to nearby points in the
target $(Y,\nu)$. This problem was first studied in the special case
$c(x,y)=|x-y|^{2}/2$ (see, e.g., [5] [13] [56]).

For general cost functions, the interior regularity of $u$ was studied in a
landmark 2005 paper of Ma, Trudinger and Wang [35], and the global regularity
was addressed by Trudinger and Wang several years later [55]. In particular,
Ma, Trudinger and Wang showed that for smooth cost functions, in order to
obtain a $C^{2}$-estimate on $u$ (and thus a Lipschitz estimate on $T$), it is
necessary to make the following assumption.

######  Definition 7 (MTW(0) Condition).

Let $\xi\in\mathbb{R}^{n}$ be a vector and $\eta\in(\mathbb{R}^{n})^{\ast}$ be
a covector with $\eta(\xi)=0$. We say a cost function satisfies the MTW(0)
condition if the following inequality holds

(19)
$\mathfrak{S}(\xi,\eta):=\sum_{i,j,k,l,p,q,r,s}(c_{ij,p}c^{p,q}c_{q,rs}-c_{ij,rs})c^{r,k}c^{s,l}\xi^{i}\xi^{j}\eta^{k}\eta^{l}\geq
0.$

Here, we use the notation $c_{I,J}=\frac{\partial^{|I|+|J|}}{\partial
x^{I}\partial y^{J}}$ for any pair of multi-indices $I,J$ and use the notation
$c^{i,j}$ to denote the matrix inverse of the mixed Hessian $c_{i,j}$.

The quantity $\mathfrak{S}$ is known as the MTW tensor and plays a central
role in the regularity theory of optimal transport. More specifically, this
assumption prevents local obstructions for smoothness of the transport map.
Loeper found a geometric perspective for this condition and showed that for
costs which do not satisfy MTW(0), it is possible to find smooth
measures111There is also a global obstruction to regularity, which is avoided
whenever the spaces $X$ and $Y$ are relatively $c$-convex ([35] Definition
2.3). We will use this assumption in Corollary 6, but will not need it
anywhere else in this paper. for which the solution to the Monge optimal
transport problem is discontinuous [34]. As such, cost functions which satisfy
the MTW(0) condition are said to be weakly regular. Furthermore, cost
functions which satisfy the inequality while dropping the orthogonality of
$\eta$ and $\xi$ are said to have non-negative cost-sectional curvature. For
costs of the form $c(x,y)=\Psi(x-y)$ (henceforth $\Psi$-costs), the
relationship between the $\mathfrak{S}$ and the anti-bisectional curvature was
observed by the first author and Zhang.

######  Theorem ([26] Theorem 6).

Let $c$ be a cost function of the form $c(x,y)=\Psi(x-y)$ for some convex
function $\Psi:\Omega\to\mathbb{R}$. Then the MTW tensor of $c$ is
proportional to the anti-bisectional curvature of the tube domain $T\Omega$
whose Kähler potential is $\Psi^{h}$.

In view of this result, we see that Theorem 2 is relevant to optimal
transport. In particular, for KR weakly regular two-variable $\Psi$-costs
whose convex potential induces a complete Hessian metric, we can flow in time
to find new costs which are weakly regular.

####  2.3.1. Parabolic Flows and Optimal Transport

The solution to the Monge problem involves solving an elliptic equation, so
most of the research has focused on elliptic equations. However, there have
been several works which have studied parabolic flows in the context of
optimal transport. In particular, Kitagawa introduced a parabolic flow of
potentials which converges to the solution of a Monge problem [30]. For the
cost function $c(x,y)=\frac{1}{2}\|x-y\|^{2}$ (better known as the squared-
distance cost), the highest order term of this flow corresponds to Kähler-
Ricci flow if we lift the potentials to obtain a flow on the tube domain
$T\Omega$. There are extra drift and potential terms to ensure that the flow
converges to the Monge solution and non-linear boundary conditions that serve
to preserve the mass. However, the formal correspondence to Kähler-Ricci flow
is very suggestive.

There is one important conceptual difference between the flow in Kitagawa’s
work and the one we consider here. In the former, the potential evolves to
converge to the solution of an optimal transport problem. In this paper, the
flow instead deforms the _cost function_.

Using Ricci flow to evolve the cost function has also been studied [36, 54].
However, these works take a somewhat different perspective in that the cost
function is either the square-distance cost $c(x,y)=d(x,y)^{2}/2$ or
Perelman’s reduced length functional $\mathcal{L}$ [43]. Further more, the
Ricci flow is used to evolve the Riemannian metric, so the evolution of the
cost function is implicit. In this paper, we instead consider costs which are
induced by a convex potential and the flow directly evolves the cost function.

From an optimal transport perspective, this might seem unusual at first.
However, there are reasons to think that this deformation is useful. First,
Kähler-Ricci flow tends to make the geometry of a space more homogeneous. As
such, if we consider a two dimensional $\Psi$-cost which is KR weakly regular
but has points where the MTW tensor is strongly positive, after a short period
of time we can expect the MTW tensor to be strictly positive. Furthermore, the
Ricci potential (i.e., $\log\left(\det[\operatorname{Hess}\Psi]\right))$ has
an optimal transport interpretation, in that it gives a quantitative measure
for how strongly the cost $c(x,y)=\Psi(x-y)$ is twisted (see [58] Chapter 12
for a in-depth discussion of twisting for cost functions).

####  2.3.2. Completeness of $\Psi$-costs

At present, there is a limitation in this theory, in that Kähler-Ricci flow is
only uniquely defined for $\Psi$-costs where $(\mathbb{M},\omega_{\Psi})$ is
complete as a Kähler manifold. We refer to such $\Psi$-costs as _complete_.
Completeness is a natural assumption in Kähler geometry. However, it is not as
important in optimal transport. As such, there are many $\Psi$-costs which are
weakly regular but not complete. Notable among these is the cost

(20) $c(x,y)=\log\left(1+\sum_{i=1}^{n}\exp(x_{i}-y_{i})\right).$

This cost function appears in mathematical finance [42]. From a geometric
perspective, it induces a metric of constant positive holomorphic sectional
curvature which is not complete. This implies that the cost (20) is weakly
regular and has non-negative cost curvature. However, since the associated
Kähler metric is not complete, there are many possible Ricci flows with this
initial condition, many of which will not preserve the weak regularity. It is
of interest to find appropriate boundary conditions for incomplete metrics so
that Theorem 2 continues to hold.

###  2.4. Background on Kähler-Ricci flow

We now provide some background on Kähler-Ricci flow. In this section, we will
focus only on the results that we will need for this paper. For a much more
complete overview, we refer to the book by Boucksom et. al. [3]. For complete
Kähler manifolds, the flow is given by the parabolic flow

(21) $\begin{cases}\frac{\partial}{\partial
t}\omega(t)=-2\operatorname{Ric}(\omega(t))-\lambda\omega(t)\\\
\omega(0)=\omega_{0}.\end{cases}$

Here, $\lambda$ is a parameter which serves to normalize the flow. When
$\lambda$ is zero, this is the unnormalized Kähler-Ricci flow. We will also
consider the case $\lambda=2$, which we refer to as the normalized Kähler-
Ricci flow. Our main focus in this paper are curvature conditions which are
preserved under the flow.

######  Definition 8 (Preserved curvature conditions).

A curvature condition $\kappa$ is preserved under Kähler-Ricci flow if
whenever $\omega_{0}$ satisfies $\kappa$, the metrics $\omega_{t}$ also
satisfy $\kappa$ for $t>0$.

For compact manifolds, Hamilton proved a tensor maximum principle which states
that a curvature condition is preserved under the flow whenever it satisfies a
null-vector condition [23]. As such, to find invariant curvature conditions we
must understand how the curvature evolves under the Kähler-Ricci flow. To do
so, we consider holomorphic coordinates $\\{z^{i}\\}_{i=1}^{n}$ and write out
the curvature tensor as

(22) $R_{i\overline{j}k\overline{\ell}}=R\left(\frac{\partial}{\partial
z^{i}},\frac{\partial}{\partial\overline{z}^{j}},\frac{\partial}{\partial
z^{k}},\frac{\partial}{\partial\overline{z}^{\ell}}\right).$

Computing the time-evolution of this tensor, we find that

(24) $\displaystyle\frac{\partial}{\partial
t}R_{i\overline{j}k\overline{\ell}}$ $\displaystyle=$ $\displaystyle\Delta
R_{i\overline{j}k\overline{\ell}}-\lambda R_{i\overline{j}k\overline{\ell}}$
$\displaystyle+2R_{i\overline{j}p\overline{r}}R_{s\overline{q}k\overline{\ell}}h^{\overline{p}q}h^{r\overline{s}}+2R_{i\overline{\ell}p\overline{r}}R_{s\overline{q}k\overline{j}}h^{\overline{p}q}h^{r\overline{s}}-2R_{i\overline{q}k\overline{r}}R_{p\overline{j}s\overline{\ell}}h^{\overline{p}q}h^{r\overline{s}}$
$\displaystyle-\left(\textrm{Ric}_{i\overline{q}}R_{p\overline{j}k\overline{\ell}}h^{\overline{p}q}+\textrm{Ric}_{p\overline{j}}R_{i\overline{q}k\overline{\ell}}h^{\overline{p}q}+\textrm{Ric}_{k\overline{q}}R_{i\overline{j}p\overline{\ell}}h^{\overline{p}q}+R_{p\overline{\ell}}\textrm{Ric}_{i\overline{j}k\overline{q}}h^{\overline{p}q}\right).$

From this, we can see that Kähler-Ricci flow behaves like a non-linear
reaction-diffusion equation for the curvature. The Laplacian term acts to
diffuse the curvature throughout the space and the rest of the terms are
reaction terms.

The term involving $\lambda$ comes from the renormalization of the metric and
vanishes when $\lambda=0$. If we consider curvature conditions which are
scale-invariant, this term will play no role, so we ignore it in our analysis.
From a conceptual perspective, the terms in line (24) are an artifact of the
coordinates, and do not come from any intrinsic geometric evolution. In other
words, it is possible to find a gauge transformation which makes these terms
vanish (see Equation (29) for the precise gauge transformation).

Thus, the crucial reaction terms are those in line (24). Hamilton’s maximum
principle states than to find an invariant curvature tensors, we can discard
the diffusion terms and study the system of the ordinary differential
equations

(25) $\frac{\partial}{\partial
t}R_{i\overline{j}k\overline{\ell}}=2R_{i\overline{j}p\overline{r}}R_{s\overline{q}k\overline{\ell}}h^{\overline{p}q}h^{r\overline{s}}+2R_{i\overline{\ell}p\overline{r}}R_{s\overline{q}k\overline{j}}h^{\overline{p}q}h^{r\overline{s}}-2R_{i\overline{q}k\overline{r}}R_{p\overline{j}s\overline{\ell}}h^{\overline{p}q}h^{r\overline{s}},$

which defines a flow of the space of algebraic curvature tensors. To show that
a curvature condition is preserved, we must only verify that these reaction
terms are “inward-pointing” for any algebraic curvature tensor on the boundary
of the curvature condition.

The terms in Equation (25) are quadratic in the curvature. Heuristically, they
also tend to be positive (see [61] for a Lie-algebraic perspective on this
phenomena). For this reason, the curvature conditions which are preserved are
generally positivity conditions (see [2, 9, 21, 23, 38, 41] for examples). In
fact, the only previously known example of an invariant negativity condition
was negative curvature for Riemann surfaces. For this reason, our main result
is somewhat unexpected.

Before moving on, it is worth noting that we are primarily concerned with
complete Kähler manifolds. As such, we will not be able to use Hamilton’s
maximum principle directly. Instead, we will use a version which applies to
complete manifolds and was originally proven by Shi [48].

####  2.4.1. Analytic Preliminaries

In order to study the behavior of Kähler-Ricci flow on tube domains, we must
first know that solutions exist and are unique. Fortunately, this problem has
been studied extensively in the literature and we will be able to use existing
results to establish existence, uniqueness and convergence.

When Hamilton introduced the Ricci flow for compact manifolds, his proof of
the existence and uniqueness used the Nash-Moser inverse function theorem.
Deturck later found a much simpler proof by conjugating Ricci flow by a time
dependent diffeomorphism [15]. However, these results concern compact
manifolds, and so do not directly apply to our setting.

For complete manifolds, it is no longer possible to directly use the standard
maximum principle, which makes establishing existence and uniqueness much more
difficult. The pioneering work for complete non-compact spaces was done by
Shi, who established the following result.

######  Theorem ([48] Theorems 2.1 and 5.1).

Let $(\mathbb{M},\omega_{0})$ be a complete Kähler manifolds with bounded
curvature. The Kähler-Ricci flow exists for some positive time $T$.
Furthermore, it preserves the complex structure and the metrics remain Kähler
under the flow.

For a tube domain with a complete Kähler metric of bounded curvature, this
result shows that the Kähler-Ricci flow exists. However, a priori one might
worry that if one starts with a Kähler-Sasaki metric, the solution might fail
to be Kähler-Sasaki after some time. To address this concern, we can use a
result of Chen and Zhu which shows that for metrics of bounded curvature, the
flow is unique and that it preserves the isometry group.

######  Theorem ([11] Corollary 1.2).

Suppose $(M^{n},g_{ij}(x))$ is a complete Riemannian manifold and suppose
$g_{ij}(x,t)$ is a solution to the Ricci flow with bounded curvature on
$M^{n}\times[0,T]$ and with $g_{ij}(x)$ as initial data. If $G$ is the
isometry group of $(M^{n},g_{ij}(x))$, then $G$ remains and isometric subgroup
of $(M^{n},g_{ij}(x,t))$ for $t\in[0,T]$.

Taken together, the previous two results show the following.

######  Corollary 5.

Let $(M,g,D)$ be a complete Hessian manifold whose tangent bundle is trivial.
Suppose also that the curvature of $\mathbb{M}=TM$ with the Kähler metric
$\omega_{0}$ induced by $g$ and $D$ is bounded. Then there exists a unique
Kähler-Ricci flow on $\mathbb{M}=TM$ with initial metric $\omega_{0}$.
Furthermore, this flow preserves the $\mathbb{R}^{n}$ symmetry and so is
induced by a flow of the underlying convex potential
$\Psi:\Omega\to\mathbb{R}$.

For Kähler-Sasaki metrics where the underlying Hessian manifold $M$ is compact
(but whose tangent bundle need not be trivial), we can also establish short
time existence and uniqueness of this flow. To do so, we pass to the affine
universal cover $\Omega$ of $M$. Doing so, we make use of the following result
of Shima.

######  Theorem ([49] Theorem B).

Let $M$ be a connected Hessian manifold. If there exists a subgroup $G$ of
automorphisms of $M$ such that $G\backslash M$ is quasicompact, then the
universal covering manifold of $M$ is a convex domain in a real affine space.

Since the affine universal cover is a convex domain, its tangent bundle is a
tube domain, so we can make use of the previous existence and uniqueness
result. Furthermore, the Chen-Zhu theorem implies that the deck
transformations remain isometries, so we quotient them out to solve the
Kähler-Ricci flow on the original Hessian manifold $M$. Alternatively, we can
establish existence and uniqueness for compact Hessian manifolds by using the
Hesse-Koszul flow, which we discuss in the next subsection.

These results establish short-time existence and uniqueness for Kähler-Ricci
flow on the spaces we will study in this paper, so we turn our attention to
understanding the behavior in the large-time limit.

Using the intuition that Kähler-Ricci flow behaves similarly to a reaction-
diffusion equation for Riemannian metrics, we might hope to show that the
normalized flow converges to a steady state (i.e., a Kähler-Einstein metric).
For an arbitrary Kähler metric, this need not be the case (see work of Song
and Tian [52] studying singularity formation). However, in a recent paper
[53], Tong showed that this is the case for complete Kähler metrics which have
negatively pinched holomorphic sectional curvature.

######  Theorem ([53] Theorem 1.1).

Suppose $(\mathbb{M}^{2n},\omega_{0})$ is a complete Kähler manifold whose
holomorphic sectional curvature is negatively pinched. Then the Kähler-Ricci
flow with initial condition $\omega_{0}$ exists for all time and the
normalized flow converges to a Kähler-Einstein metric of scalar curvature $-2$
which is uniformly equivalent to the original metric.

Using this result, Tong found an alternate proof for Wu and Yau’s result on
the existence of Kähler-Einstein metrics for negatively pinched spaces [62].
In this paper, we will use this theorem as a sufficient condition for the
Kähler-Ricci flow to converge. This result does not imply that negative
holomorphic sectional curvature is preserved under the flow. However, Wu and
Yau showed that the holomorphic sectional curvature remains negatively pinched
(with a weaker constant) for some definite amount of time.

It is worth emphasizing that Tong’s result uses the assumption that the
holomorphic sectional curvature is negatively pinched for all $(1,0)$ vectors.
It seems likely that a version of this result holds for tube domains with
negatively pinched polarized holomorphic sectional curvature given a
sufficiently sharp lower bound on the orthogonal antibisectional curvature.
Without the latter assumption, it does not seem possible to establish Royden’s
lemma [45], which plays an essential role in Tong’s proof.

###  2.5. The Hesse-Koszul Flow and affine geometry

We can understand the Kähler-Ricci flow on tube domains in terms of a flow on
the underlying Hessian manifold. Mirghafouri and Malek [37] studied this flow,
which evolves a Hessian manifold $(M,g_{t},D)$ by deforming the potential by
the parabolic Monge-Ampère equation222In their paper, the flow is expressed in
terms of Koszul forms in affine geometry, but we will not use this
terminology:

(26) $\frac{\partial\Psi}{\partial
t}=2\log\left(\det\left[\frac{\partial\Psi}{\partial x^{i}\partial
x^{j}}\right]\right).$

Following the convention in [44], we call this the Hesse-Koszul flow. Prior to
this work, existence and uniqueness for the Hesse-Koszul flow had only been
established for compact Hessian manifolds.333Corollary 5 provides short-time
existence and uniqueness for the Hesse-Koszul flow for complete parallelizable
Hessian manifolds whose curvature is bounded. The long-time behavior of the
Hesse-Koszul flow was recently studied by Puechmorel and Tô [44], who showed
the following result.

######  Theorem ([44] Theorem 2).

Let $(M,g_{0},D)$ be a compact Hessian manifold. Assume that the first affine
Chern class of $M$ is negative. Starting from any Hessian metric g0, the
normalized Hesse-Koszul flow exists for all time and converges in $C^{\infty}$
to a Hesse-Einstein metric444Hesse-Einstein metrics are the affine geometric
equivalent of a Kähler-Einstein metrics.

By combining this result and Corollary 2, we obtain the following result,
which is simply a restatement of Corollary 3 in the language of affine
geometry.

######  Corollary.

Let $(M,g_{0},D)$ be a compact Hessian manifold. Suppose that $g_{0}$ has non-
positive anti-bisectional curvature and that the first affine Chern class of
$M$ is negative. Then under the normalized Hesse-Koszul flow, $(M,g_{t},D)$
converges to a Hesse-Einstein metric $g_{\infty}$ which is negatively cost-
curved.

For complete Hessian manifolds with bounded curvature555Bounded curvature here
means that the tangent bundle has bounded curvature, not simply the underlying
Hessian manifold. but non-trivial tangent bundle, the existence and uniqueness
of the Hesse-Koszul flow has not been established. Presumably, this can be
done by adapting Shi’s work to this setting, after which one can then apply
the Chen-Zhu uniqueness theorem immediately. However, we will not consider
these spaces in this paper.

####  2.5.1. Advantages and disadvantages of the Hesse-Koszul flow

For readers whose primary interest is affine geometry, it might be more
natural to rephrase our results in terms of Hessian manifolds and the Hesse-
Koszul flow. Indeed, the distinction between Kähler-Ricci flow for tube
domains and the Hesse-Koszul flow is mainly a matter of terminology. Using
affine geometry, it is simpler to define the notion of signed anti-bisectional
curvature. Furthermore, some of the calculations simplify when stated in terms
of Hesse-Koszul flow, as there is no need to pass back and forth from the
tangent bundle to the underlying Hessian manifold. However, there are two
disadvantages of working directly in terms of the Hesse-Koszul flow.

First, the analytic results that we rely on were proven for Ricci flow. In
order to work with Hesse-Koszul flow, it would necessary to translate these
results into the language of affine differential geometry. In particular,
rephrasing Tong’s result in terms of affine geometry is quite awkward.

Second, one of our primary motivations for considering these flows are to
understand the interaction between Ricci curvature and holomorphic sectional
curvature. There is an extensive body of literature studying this question in
Kähler geometry but no corresponding body of literature studying the
relationship between the second Koszul form and the Hessian sectional
curvature for Hessian manifolds666Shima defines the Hessian sectional
curvature to be the negative of what we call the polarized holomorphic
sectional curvature [50].. For these two reasons, we will state our results in
terms of Kähler-Ricci flow and complex geometry.

##  3\. Negative anti-bisectional curvature is preserved by Kähler-Ricci Flow

In this section, we prove Theorem 1, which shows that Kähler-Ricci flow
preserves negative anti-bisectional curvature. We want a version that applies
to compact and non-compact spaces alike. Because of this, instead of simply
checking the null-vector condition, we adapt an argument by Shi, which
originally showed that non-negative bisectional curvature is preserved along
Kähler-Ricci flow (Theorem 5.3 of [48]). For readers who are interested in the
case of compact Hessian manifolds, it is possible to skip Subsection 3.2 and
simply use $\widetilde{R}$ in place of $A$ in Subsection 3.3

###  3.1. Uhlenbeck’s trick

Let $(\mathbb{M},\omega_{t})$ be a family of complete Kähler-Sasaki metrics
which evolve according to the unnormalized Kähler-Ricci flow. As discussed in
Subsection 2.4, the curvature of $\omega_{t}$ evolves via the equation

(28) $\displaystyle\frac{\partial}{\partial
t}R_{i\overline{j}k\overline{\ell}}$ $\displaystyle=$ $\displaystyle\Delta
R_{i\overline{j}k\overline{\ell}}$
$\displaystyle+2\left(R_{i\overline{j}p\overline{r}}R_{s\overline{q}k\overline{\ell}}h^{\overline{p}q}h^{r\overline{s}}+R_{i\overline{\ell}p\overline{r}}R_{s\overline{q}k\overline{j}}h^{\overline{p}q}h^{r\overline{s}}-R_{i\overline{q}k\overline{r}}R_{p\overline{j}s\overline{\ell}}h^{\overline{p}q}h^{r\overline{s}}\right)$
$\displaystyle-\left(R_{i\overline{q}}R_{p\overline{j}k\overline{\ell}}h^{\overline{p}q}+R_{p\overline{j}}R_{i\overline{q}k\overline{\ell}}h^{\overline{p}q}+R_{k\overline{q}}R_{i\overline{j}p\overline{\ell}}h^{\overline{p}q}+R_{p\overline{\ell}}R_{i\overline{j}k\overline{q}}h^{\overline{p}q}\right)$

The reaction terms in this expression (lines (28) and (28)) are particularly
unwieldy, so we make use of a technique known as Uhlenbeck’s trick to simplify
the expression. We consider an abstract vector bundle $V$ which is isomorphic
to the complex tangent bundle $T_{\mathbb{C}}\mathbb{M}$. Furthermore, we fix
a metric $\tilde{h}_{AB}$ on the fibers of $V$ which is
$\mathbb{R}^{n}$-symmetric777With Uhlenbeck’s trick, normally one can choose
$\tilde{h}$ arbitrarily, Here, it is important to respect the translation
symmetry.. Using this symmetry, we also have a time invariant metric on
$\Omega$ as well, which we denote $\tilde{g}$.

We choose an isometry $\mathcal{U}=\left\\{\mathcal{U}_{i}^{j}\right\\}$
between $V$ and $T\mathbb{M}$ and let this isometry evolve by the equation

(29) $\frac{\partial}{\partial
t}\mathcal{U}_{j}^{i}=h^{ik}\textrm{Ric}_{k\ell}\mathcal{U}_{j}^{\ell}.$

Furthermore, we can pull over the Levi-Civita connection of the metric $h$ to
the abstract vector bundle $V$ and compute the curvature tensor, which we
denote $\widetilde{R}$. Although this is conceptually more complicated, the
advantage of doing this is that it greatly simplifies the curvature evolution
equations.

$\frac{\partial}{\partial
t}\widetilde{R}_{i\overline{j}k\overline{\ell}}=\Delta\widetilde{R}_{i\overline{j}k\overline{\ell}}+2\left(\widetilde{R}_{i\overline{j}p\overline{r}}\widetilde{R}_{s\overline{q}k\overline{\ell}}\widetilde{h}^{\overline{p}q}\widetilde{h}^{r\overline{s}}+\widetilde{R}_{i\overline{\ell}p\overline{r}}\widetilde{R}_{s\overline{q}k\overline{j}}\widetilde{h}^{\overline{p}q}\widetilde{h}^{r\overline{s}}-\widetilde{R}_{i\overline{q}k\overline{r}}\widetilde{R}_{p\overline{j}s\overline{\ell}}\widetilde{h}^{\overline{p}q}\widetilde{h}^{r\overline{s}}\right)$

We now pick holomorphic coordinates $\\{z\\}$ so that $\tilde{h}$ is the
identity at a point $p\in\mathbb{M}$. We also insist that the real parts of
$z$ give affine coordinates of $\Omega$ (i.e., the coordinates respect the
polarization of $\mathbb{M}$). In other words, we pick coordinates which
respect the affine structure and translation symmetry of $\mathbb{M}$. Using
the fact that Ricci flow preserves the Kählerness of the metric (Theorem 5.1
of Shi [48]), the curvature evolution further simplifies (at $p$) to the
following:

(30) $\displaystyle\frac{\partial}{\partial
t}\widetilde{R}_{i\overline{j}k\overline{\ell}}$ $\displaystyle=$
$\displaystyle\Delta\widetilde{R}_{i\overline{j}k\overline{\ell}}+2\widetilde{R}_{i\overline{j}p\overline{q}}\widetilde{R}_{q\overline{p}k\overline{\ell}}+2\widetilde{R}_{i\overline{\ell}p\overline{q}}\widetilde{R}_{q\overline{p}k\overline{j}}-2\widetilde{R}_{i\overline{p}k\overline{q}}\widetilde{R}_{p\overline{j}q\overline{\ell}}$

We now define the quadratic $Q$ to be

(31)
$Q(\widetilde{R})_{i\overline{j}k\overline{\ell}}=2\left(\widetilde{R}_{i\overline{j}p\overline{q}}\widetilde{R}_{q\overline{p}k\overline{\ell}}+\widetilde{R}_{i\overline{\ell}p\overline{q}}\widetilde{R}_{q\overline{p}k\overline{j}}-\widetilde{R}_{i\overline{p}k\overline{q}}\widetilde{R}_{p\overline{j}q\overline{\ell}}\right),$

so that the curvature evolution can be written as

(32) $\frac{\partial}{\partial
t}\widetilde{R}_{i\overline{j}k\overline{\ell}}=\Delta\widetilde{R}_{i\overline{j}k\overline{\ell}}+Q(\widetilde{R})_{i\overline{j}k\overline{\ell}}.$

By definition, the anti-bisectional curvature is non-positive if and only if
for any $v,w\in T^{real}\Omega$,

(33) $R(v^{pol.},\overline{w}^{pol.},v^{pol.},\overline{w}^{pol.})\leq 0.$

From the fact that our choice of coordinates on $V$ respect the tube domain
structure and polarization, we have that the following:

(34) $R(v^{pol.},\overline{w}^{pol.},v^{pol.},\overline{w}^{pol.})\leq
0\iff\widetilde{R}(v^{pol.},\overline{w}^{pol.},v^{pol.},\overline{w}^{pol.})\leq
0.$

As such, it remains to show that the anti-bisectional curvature of
$\widetilde{R}$ is non-positive. To do this, we make use of a maximum
principle for complete spaces that was proven by Shi.

###  3.2. Shi’s Maximum Principle

For complete manifolds with bounded curvature that evolve under Ricci flow,
Shi established the following maximum principle.

######  Theorem ([48] Theorem 4.8).

Suppose that $(\mathbb{M}^{n},g(t))$ is a complete Ricci flow with uniformly
bounded curvature at each time-slice. Suppose that $\varphi(z,t)$ is a
$C^{\infty}$ function on $\mathbb{M}\times[0,T]$ such that

$\left\\{\begin{aligned} \frac{\partial\varphi}{\partial
t}=\Delta\varphi+C_{1}\left|\nabla_{k}\varphi\right|^{2}+Q(\varphi,z,t)&\text{
on }\mathbb{M}\times[0,T]\\\ \varphi(z,t)\leq C_{2}<+\infty,\quad&\text{ on
}\mathbb{M}\times[0,T]\\\ \varphi(z,0)\leq 0,\quad&\text{ on }\mathbb{M}\\\
Q(\varphi,z,t)\leq C_{3}\varphi,\quad&\text{ if }\varphi\geq
0,\end{aligned}\right.$

where $C_{1},\,C_{2}$ and $C_{3}$ are non-negative constants. Then we have

(35) $\varphi(z,0)\leq 0,\quad\text{ on }\mathbb{M}\times[0,T]$

In order to apply this result, we must find an appropriate function $\varphi$
to apply this result. At each point in space-time
$(z,t)\in\mathbb{M}\times[0,T)$, we consider the space of polarized unit
vectors

(36) $S(z,t):=\\{\mathcal{V}=v^{pol.}~{}|~{}v\in T^{real}\Omega\textrm{ with
}\widetilde{g}_{z}(v,v)=1\\},$

Note that the norm here is given with respect to the fixed metric
$\widetilde{g}$, which is invariant in time. We then define a function

(37)
$\varphi(z,t)=-\sup\\{\theta\in\mathbb{R}~{}|~{}A(v^{pol.},\overline{w^{pol.}},v^{pol.},\overline{w^{pol.}})(\theta,z,t)\leq
0\textrm{ for all }v,w\in S(z,t)\\}$

where the tensor $A(\theta,z,t)$ is defined as

(38)
$A(\theta)_{i\overline{j}k\overline{\ell}}=\widetilde{R}_{i\overline{j}k\overline{\ell}}+\theta\tilde{g}_{ik}\tilde{g}_{\overline{j}\overline{\ell}}.$

Note that the definition of $A(\theta,z,t)$ is slightly different from the
definition of $A$ used by Shi. This is due to the difference between
bisectional and anti-bisectional curvature. Also, note that we are passing
freely between vectors on $\Omega$ and their lifts on $\mathbb{M}$.

We then define the tensor $A(z,t)$ to be

(39) $A:=A(-\varphi(z,t),z,t).$

In other words, at each point of space-time which choose the supremum of
$\theta$ so that $A(\theta,z,t)$ is non-positive definite at that point. The
unusual sign convention for $\varphi$ is done so that we can use Shi’s maximum
principle verbatim without reversing inequalities. To use the maximum
principle, much of the proof follows Theorem 5.3 from [48] with slight
modifications. However, the proof of the null-vector condition (i.e., that $Q$
is “inward pointing”) is somewhat different, so we provide the details for
that step.

###  3.3. Verification of the null-vector condition

######  Lemma 4.

Suppose $\\{A_{i\overline{j}k\overline{\ell}}\\}$ is a tensor which has the
same symmetries as $\\{R_{i\overline{j}k\overline{\ell}}\\}$. As before, we
set

(40)
$Q(A)_{i\overline{j}k\overline{\ell}}=2\left(A_{i\overline{j}p\overline{q}}A_{q\overline{p}k\overline{\ell}}+A_{i\overline{\ell}p\overline{q}}A_{q\overline{p}k\overline{j}}-A_{i\overline{p}k\overline{q}}A_{p\overline{j}q\overline{\ell}}\right),$

where we assume $\tilde{g}$ is the identity at a point $z$ (and that the
metric respects the translation symmetry). Suppose for a fixed point
$z\in\mathbb{M}$, we have that

  1. (1)

$A(\mathcal{X},\overline{\mathcal{Y}},\mathcal{X},\overline{\mathcal{Y}})\leq
0$ for all $\mathcal{X}=x^{pol.},\mathcal{Y}=y^{pol.}$ with $x,y\in
T^{real}\Omega$, and

  2. (2)

$A(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})=0$
for some $\mathcal{V}=v^{pol.},\mathcal{W}=w^{pol.}$.

Then

(41)
$Q(A)(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})\leq
0.$

###### Proof.

To prove this result, we adapt an argument of Cao [9] (but which was
originally shown Mok [38]). We start with two small lemmas.

######  Lemma 5.

For any vector $\mathcal{X}=x^{pol.}$,

(42)
$A_{\mathcal{V}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{W}}}=A_{\mathcal{X}\overline{\mathcal{W}}\mathcal{V}\overline{\mathcal{W}}}=0.$

To see this, consider
$G(s)=A(\mathcal{V}+s\mathcal{X},\overline{\mathcal{W}},\mathcal{V}+s\mathcal{X},\overline{\mathcal{W}})$.
$G(s)$ is maximized at $0$, so $G^{\prime}(0)=0$. Computing $G^{\prime}(0)$,
we find the following:

(43)
$G^{\prime}(0)=R_{\mathcal{X}\overline{\mathcal{W}}\mathcal{V}\overline{\mathcal{W}}}+R_{\mathcal{V}\overline{\mathcal{W}}\mathcal{X}\overline{\mathcal{W}}}=2R_{\mathcal{X}\overline{\mathcal{W}}\mathcal{V}\overline{\mathcal{W}}}.$

Similarly, we can show the same for
$\overline{G}(s)=R(\mathcal{V},\overline{\mathcal{W}}+s\overline{\mathcal{X}},\mathcal{V},\overline{\mathcal{W}}+s\overline{\mathcal{X}})$.

######  Lemma 6.

For any two vectors $\mathcal{X}=x^{pol.}$ and $\mathcal{Y}=y^{pol.}$,

(44)
$A_{\mathcal{Y}\overline{\mathcal{W}}\mathcal{Y}\overline{\mathcal{W}}}+A_{\mathcal{V}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{X}}}+4A_{\mathcal{Y}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{W}}}\leq
0.$

We consider the function

(45)
$H(s):=A(\mathcal{V}+s\mathcal{Y},\overline{\mathcal{W}}+s\overline{\mathcal{X}},\mathcal{V}+s\mathcal{Y},\overline{\mathcal{W}}+s\overline{\mathcal{X}}).$

Since $H$ is maximized at $s=0$, we have that $H(0)=H^{\prime}(0)=0$ and
$H^{\prime\prime}(0)\leq 0$. Calculating $H^{\prime\prime}$, we find that

(46) $\displaystyle H^{\prime\prime}(0)$ $\displaystyle=$ $\displaystyle
A_{\mathcal{Y}\overline{\mathcal{W}}\mathcal{Y}\overline{\mathcal{W}}}+A_{\mathcal{V}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{X}}}$
$\displaystyle+A_{\mathcal{Y}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{W}}}+A_{\mathcal{V}\overline{\mathcal{W}}\mathcal{Y}\overline{\mathcal{X}}}+A_{\mathcal{Y}\overline{\mathcal{W}}\mathcal{V}\overline{\mathcal{X}}}+A_{\mathcal{V}\overline{\mathcal{X}}\mathcal{Y}\overline{\mathcal{W}}}$
$\displaystyle=$ $\displaystyle
A_{\mathcal{Y}\overline{\mathcal{W}}\mathcal{Y}\overline{\mathcal{W}}}+A_{\mathcal{V}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{X}}}+4A_{\mathcal{Y}\overline{\mathcal{X}}\mathcal{V}\overline{\mathcal{W}}}.$

We can use this $H$ to define a semi-positive definite bilinear form
$\mathcal{H}$. To do so, we consider an an orthonormal basis $\\{e_{i}\\}$ of
$T^{real}\Omega$ and its lifts $\\{\mathcal{E}_{i}=e_{i}^{pol.}\\}$, which
form a polarized unitary basis of $T\mathbb{M}$. For vectors
$\Xi=\xi^{i}\mathcal{E}_{i}$ and $\mathcal{Z}=\zeta^{i}\mathcal{E}_{i}$ (where
all the coefficients are real), we define $\mathcal{H}(\Xi,\mathcal{Z})$ as
follows:

(47)
$\mathcal{H}(\Xi,\mathcal{Z})=A(\mathcal{E}_{i},\overline{\mathcal{W}},\mathcal{E}_{j},\overline{\mathcal{W}})\xi^{i}\xi^{j}+A(\mathcal{V},\overline{\mathcal{E}}_{i},\mathcal{V},\overline{\mathcal{E}}_{j})\overline{\zeta}^{i}\overline{\zeta}^{j}+4A(\mathcal{E}_{i},\overline{\mathcal{E}}_{j},\mathcal{V},\overline{\mathcal{W}})\xi^{i}\overline{\zeta}^{j}$

The estimate on $H^{\prime\prime}$ shows that

(48) $\mathcal{H}(\Xi,\mathcal{Z})\leq 0.$

We now make use of the following lemma to show that $Q$ is non-positive. This
lemma was originally stated by Cao ([7] Lemma 4.1) and is attributed to
Hamilton. However, there are some typos in the original proof888 In
particular, the matrices $G_{2}$ and $G_{3}$ defined by Cao need not be semi-
positive definite. For instance, if we set
$G_{1}=\left[\begin{array}[]{cccc}0&0&0&0\\\ 0&1&1&0\\\ 0&1&1&0\\\ 0&0&0&0\\\
\end{array}\right],$ both $G_{2}$ and $G_{3}$ will have negative eigenvalues.,
so we provide a detailed argument.

######  Lemma 7.

Let $M_{1}$ and $M_{2}$ be two $n\times n$ real symmetric semi-positive
definite matrices, and let $N$ be a real $n\times n$ matrix such that the
$2n\times 2n$ real symmetric matrix

(49) $G_{1}=\left[\begin{array}[]{cc}M_{1}&N\\\ N^{T}&M_{2}\\\
\end{array}\right]$

is semi-positive definite. Then, we have that

(50) $\operatorname{Tr}(M_{1}M_{2}-N^{2})\geq 0.$

###### Proof.

Consider the matrix

(51) $G_{2}=\left[\begin{array}[]{cc}M_{2}&-N^{T}\\\ -N&M_{1}\\\
\end{array}\right].$

If we conjugate $G_{1}$ by the skew-symmetric orthogonal matrix

(52) $J=\left[\begin{array}[]{cc}0&Id_{n}\\\ -Id_{n}&0\\\ \end{array}\right],$

we find that the matrix $G_{2}$ is similar to $G_{1}$;

(53) $G_{2}=JG_{1}J^{T}.$

As a result, $G_{1}$ and $G_{2}$ have the same characteristic polynomial.
Furthermore, since both matrices are symmetric, $G_{2}$ is semi-positive
definite whenever $G_{1}$ is.

As a result, whenever $G_{1}$ is semi-positive definite, we have that

(54) $\operatorname{Tr}(G_{1}G_{2})\geq 0.$

Calculating this explicitly, we find that

(55) $G_{1}G_{2}=\left(\begin{array}[]{cc}M_{1}M_{2}-NN&-M_{1}N^{T}+NM_{1}\\\
N^{T}M_{2}-M_{2}N&M_{2}M_{1}-N^{T}N^{T}\end{array}\right).$

Taking the trace (and noting that
$\operatorname{Tr}(N^{2})=\operatorname{Tr}(N^{T}N^{T})$, we obtain the
desired result. ∎

To make use of Lemma 7, we set

  1. (1)

$(M_{1})_{ij}=-A(\mathcal{E}_{i},\overline{\mathcal{W}},\mathcal{E}_{j},\overline{\mathcal{W}})$

  2. (2)

$(M_{2})_{ij}=-A(\mathcal{V},\overline{\mathcal{E}}_{i},\mathcal{V},\overline{\mathcal{E}}_{j})$,
and

  3. (3)

$(N)_{ij}=-2A(\mathcal{V},\overline{\mathcal{W}},\mathcal{E}_{i},\overline{\mathcal{E}}_{j})$.

Since $M_{1},M_{2}$ and $-\mathcal{H}$ are semi-positive definite, Equation
(50) implies that

(56)
$A(\mathcal{E}_{i},\overline{\mathcal{W}},\mathcal{E}_{j},\overline{\mathcal{W}})A(\mathcal{V},\overline{\mathcal{E}}_{i},\mathcal{V},\overline{\mathcal{E}}_{j})\geq
4A(\mathcal{V},\overline{\mathcal{W}},\mathcal{E}_{i},\overline{\mathcal{E}}_{j})A(\mathcal{V},\overline{\mathcal{W}},\mathcal{E}_{j},\overline{\mathcal{E}}_{i}).$

Furthermore, we have that

(57)
$A(\mathcal{E}_{i},\overline{\mathcal{W}},\mathcal{E}_{j},\overline{\mathcal{W}})A(\mathcal{V},\overline{\mathcal{E}}_{i},\mathcal{V},\overline{\mathcal{E}}_{j})\geq
0,$

since this is the trace of the product of semi-positive definite matrices.
Combining (40), (56) and (57), we see that

(58)
$Q(A)(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})\leq
0.$

Note that (56) implies the the sharper estimate

(59)
$Q(A)(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})+A(\mathcal{E}_{i},\overline{\mathcal{W}},\mathcal{E}_{j},\overline{\mathcal{W}})A(\mathcal{V},\overline{\mathcal{E}}_{i},\mathcal{V},\overline{\mathcal{E}}_{j})\leq
0.$

Although we did not use this stronger inequality here, it will play an
important role in Corollary 2. ∎

From here, it is possible to follow the rest of Shi’s argument for Theorem 5.3
verbatim without making any further changes. For brevity, we refer to that
paper for the details.

##  4\. Geometric Consequences

In this section, we provide several geometric consequences of Theorem 1.

###  4.1. Positive time control in terms of the orthogonal anti-bisectional
trace curvature

We first show that if we start with a metric of negative anti-bisectional
curvature, we have control of the holomorphic sectional curvature at positive
times solely in terms of the orthogonal anti-bisectional trace curvature.

######  Corollary.

If $\mathbb{M}$ has negative anti-bisectional curvature and the scalar
curvature at time $t=0$ satisfies $S\geq-K$ for some $K>0$, then the
holomorphic sectional curvature satisfies the estimate

(60) $0\geq
H(\mathcal{X})\geq-C(n)\left(\frac{n}{t+\frac{n}{K}}+2\mathfrak{O}\right).$

####  4.1.1. Chen’s Estimate

To prove this result, we start by using a result of Chen ([10] Corollary 2.3 )
to observe that along the Ricci flow, the scalar curvature $S$ satisfies the
estimate

(61) $S\geq-\frac{n}{t+\frac{n}{K}}.$

Here, $K$ is a lower bound of the scalar curvature at the initial time. For
compact manifolds, this inequality is a straightforward application of the
maximum principle, but Chen proved it for complete manifolds as well. On its
own, this estimate on its own does not give uniform control over the geometry,
as one expects the scalar curvature to become very large and positive.
However, since the polarized holomorphic sectional curvature remains negative,
we can use the following lemma (which is a quantitative version of Berger’s
estimate on the scalar curvature) to prove Corollary 4.

####  4.1.2. A Modified Berger’s Lemma

######  Lemma 8.

Suppose $\mathbb{M}$ is a Kahler manifold with non-positive polarized
holomorphic sectional curvature. Then, there exists a constant $C(n)$
depending only on the dimension so that for all polarized unit $(1,0)$ tangent
vectors $\mathcal{X}$, the scalar curvature satisfies the inequality

(62) $S+2\mathfrak{O}\leq C(n)(H(\mathcal{X})).$

Using the polarized unitary basis $\\{\mathcal{E}_{i}\\}$ as before, the
scalar curvature is given by

(63) $S=\sum R_{i\overline{i}j\overline{j}}$

and the orthogonal anti-bisectional trace curvature is given by

(64) $\mathfrak{O}=\sum_{i\neq j}R_{i\overline{j}i\overline{j}}$

We first prove a modified version of Berger’s lemma, [1] which shows that the
scalar curvature is the average of the holomorphic sectional curvatures. We
let $d\theta(\mathcal{Z})$ be the uniform probability measure on
$\mathbb{S}^{n-1}\subset\left(T^{real}_{z}\Omega\right)^{pol.}$ and consider
the integral

(65)
$\int_{|\mathcal{Z}|=1,\mathcal{Z}\in\left(T_{z}^{real}\Omega\right)^{pol.}}R_{i\overline{j}k\overline{\ell}}\mathcal{Z}_{i}\mathcal{Z}_{j}\mathcal{Z}_{k}\mathcal{Z}_{\ell}\,d\theta(\mathcal{Z}).$

When we integrate polynomials over the sphere [19], we find that

(66) $\int_{S^{n-1}}|\mathcal{Z}_{j}|^{4}\hskip
0.85358ptd\theta(\mathcal{Z})=\frac{2\Gamma[5/2]\,\Gamma[1/2]^{n-2}}{\Gamma[\frac{3+n}{2}]},\quad
j=1,\dots,n,$

and

(67) $\int_{S^{n-1}}|\mathcal{Z}_{j}|^{2}|\mathcal{Z}_{k}|^{2}\hskip
0.85358ptd\theta(\mathcal{Z})=\frac{2\Gamma[3/2]^{2}\,\Gamma[1/2]^{n-3}}{\Gamma[\frac{3+n}{2}]},\quad
1\leq j\neq k\leq n.$

As such, we have that

(68) $\int_{S^{n-1}}|\mathcal{Z}_{j}|^{4}\hskip
0.85358ptd\theta(\mathcal{Z})=2\int_{S^{n-1}}|\mathcal{Z}_{j}|^{2}|\mathcal{Z}_{k}|^{2}\hskip
0.85358ptd\theta(\mathcal{Z}).$

As a result, when we compute (65), all of the terms vanish except those where
the indices appear in pairs (or are all the same). This implies the following:

(69)
$\displaystyle\int_{|\mathcal{Z}|=1,\mathcal{Z}\in\left(T_{z}^{real}\Omega\right)^{pol.}}R_{i\overline{j}k\overline{\ell}}\mathcal{Z}_{i}\mathcal{Z}_{j}\mathcal{Z}_{k}\mathcal{Z}_{\ell}\,d\theta(\mathcal{Z})$
$\displaystyle=$ $\displaystyle
C(n)\sum_{i,j}R_{iijj}(2\delta_{ij}+(1-\delta_{ij}))$
$\displaystyle+2C(n)\sum_{i\neq j}R_{i\overline{j}i\overline{j}}$
$\displaystyle=$ $\displaystyle C(n)\left(S(z)+2\sum_{i\neq
j}R_{i\overline{j}i\overline{j}}\right).$

This is where the terms on the left hand side of Inequality (62) come from.

####  4.1.3. Bounding the holomorphic sectional curvature from below

To finish the proof of lemma 8, it suffices to show inequality (62) for the
polarized holomorphic tangent vector $\mathcal{X}$ which minimizes the
holomorphic sectional curvature. As such, we must establish that the minimum
holomorphic sectional curvature cannot be too much smaller than the average.

Suppose $\mathcal{X}$ is a polarized vector which minimizes the holomorphic
sectional curvature at $z$. Let $\mathcal{X}^{\perp}$ be a polarized unit
holomorphic tangent vector perpendicular to $\mathcal{X}$. Then, we consider
the polarized unit holomorphic tangent vector
$\mathcal{X}_{\theta}=\cos\theta\mathcal{X}+\sin\theta\mathcal{X}^{\perp}$.

Define the function

(70)
$f(\theta):=R(\mathcal{X}_{\theta},\overline{\mathcal{X}_{\theta}},\mathcal{X}_{\theta},\overline{\mathcal{X}_{\theta}}).$

We write out $f(\theta)$ as a trigonometric polynomial. Since the curvature
tensor is linear in each entry, the polynomial must be of the form

(71)
$f(\theta)=a_{0}+a_{1}\cos(2\theta)+a_{2}\sin(2\theta)+a_{3}\cos(4\theta)+a_{4}\sin(4\theta).$

Since $f$ is minimized at $\theta=0$, it follows that

(72) $a_{2}+2a_{4}=0,$ (73) $a_{1}+4a_{3}\leq 0,\textrm{ and }$ (74)
$a_{1}\leq 0.$

The first two inequalities follow from computing the first and second
derivatives of $f$. The third follows from the fact that $f(0)\leq f(\pi/2)$,
and noting that the only term that changes in the trigonometric polynomial is
$a_{1}$. By possibly switching $\mathcal{X}^{\perp}$ to
$-\mathcal{X}^{\perp}$, we may assume $a_{2}\geq 0$.

We now bound $a_{0}$ from above by considering $f(\pi/4)=a_{0}+a_{2}-a_{3}$
and $f(\pi_{2})=a_{0}-a_{1}+a_{3}$. We consider two cases.

  1. (1)

$a_{1}\geq 3a_{3}$, in which case $-a_{1}\leq 3a_{3}$ and so
$a_{2}-a_{3}\geq-\frac{a_{1}+a_{3}}{4}$

  2. (2)

$a_{1}\leq 3a_{3}$, in which case $-a_{1}+a_{3}\geq-\frac{a_{1}+a_{3}}{2}$.

Since $f$ is non-positive, in either case we have that
$a_{0}\leq\frac{a_{1}+a_{3}}{4}$. However, the average holomorphic sectional
curvature is simply $a_{0}$, which shows that the average holomorphic
sectional curvature in the $(\mathcal{X},\mathcal{X}^{\perp})$-plane is at
most $1/5$ of the minimum holomorphic sectional curvature. Furthermore, since

(75)
$a_{0}+\frac{\sqrt{2}}{2}a_{1}-\left(1-\frac{\sqrt{2}}{2}\right)a_{2}=f(-\pi/8)\geq
f(0)=a_{0}+a_{1}+a_{3},$

it necessary follows that

(76)
$-\left(1-\frac{\sqrt{2}}{2}\right)a_{1}-a_{3}\geq\left(1-\frac{\sqrt{2}}{2}\right)a_{2}.$

By combining (73) with (76) and doing some straightforward numeric
calculations, which find that for $|\theta|<\frac{\pi}{12}$, we have that

(77) $\displaystyle f(\theta)$ $\displaystyle\leq$ $\displaystyle
f\left(\frac{\pi}{12}\right)$ $\displaystyle=$ $\displaystyle
a_{0}+\frac{\sqrt{3}}{2}a_{1}+\frac{a_{3}}{2}+\frac{2-\sqrt{3}}{4}a_{2}$
$\displaystyle\leq$ $\displaystyle
a_{0}+\frac{\sqrt{3}}{2}a_{1}+\frac{a_{3}}{2}+\frac{1}{14}a_{2}$
$\displaystyle\leq$ $\displaystyle a_{0}.$

As we established earlier, $a_{0}\leq\frac{H(\mathcal{X})}{5}$. As such, this
shows that for any unit holomorphic tangent vector $\mathcal{Y}$ whose angle
with $\mathcal{X}$ is less than $\frac{\pi}{12}$, the holomorphic sectional
curvature satisfies

(78) $H(\mathcal{X})\leq\frac{H(\mathcal{X})}{5}.$

The set of holomorphic vectors whose angle with $\mathcal{X}$ is at most
$\pi/12$ has positive measure in $\mathbb{S}^{n-1}$. Combining Equation (69)
with Inequality (78), we establish Corollary 8. ∎

Before moving on, let us make several remarks about this result. First, for a
metric with negative anti-bisectional curvature, it is possible to control the
entire curvature tensor in terms of the polarized holomorphic sectional
curvature and $\mathfrak{O}$. As such, this estimate shows that we can control
the flow at positive times solely in terms of the function $\mathfrak{O}$.

Second, Lemma 8 can be adapted to arbitrary Kähler metrics with either non-
negative or non-positive holomorphic sectional curvature. For example, we can
adapt the proof to obtain the following result.

######  Proposition.

Suppose $\mathbb{M}$ is a Kahler manifold with non-negative holomorphic
sectional curvature. Then, there exists a constant $C(n)$ depending only on
the dimension so that for all unit $(1,0)$ tangent vectors $\mathcal{X}$, the
scalar curvature satisfies the inequality

$S\geq C(n)H(\mathcal{X}).$

Furthermore, for metrics with non-positive holomorphic sectional curvature, we
obtain a similar result with the final inequality reversed.

###  4.2.  Kähler-Einstein metrics with negative cost-curvature

We can also use Theorem 1 in combination with Tong’s result on negatively
curved spaces to find tube domains whose Kähler-Einstein metrics have negative
anti-bisectional curvature. Corollary 1 (restated here for convenience)
follows immediately from these two results.

######  Corollary.

Given a tube domain which admits a complete metric which of strongly negative
holomorphic sectional curvature and anti-bisectional curvature, the unique
complete Kähler-Einstein metric with scalar curvature -2 also has non-positive
anti-bisectional curvature.

When $\Omega$ is a smooth, bounded and strongly convex domain, it is possible
to strengthen this result to show that the polarized holomorphic sectional
curvature is negatively pinched.

######  Corollary.

Suppose that $\Omega$ is a smooth, bounded, and strongly convex domain and
that $T\Omega$ admits a complete Kähler metric with negative anti-bisectional
curvature and strongly negative holomorphic sectional curvature. Then the
complete Kähler-Einstein metric also has negative cost-curvature.

###### Proof.

From Corollary 1, the Kähler-Einstein metric has non-positive anti-bisectional
curvature. To show that it has negative cost-curvature, we must show that the
polarized holomorphic sectional curvature is strongly negative.

To do so, we note that for tube domains, the Kähler-Einstein metric
$\omega_{K.E.}$ has negatively pinched holomorphic sectional curvature in a
neighborhood of the strictly pseudoconvex boundary points (see [57]
Proposition 3.5 for details [57]). When $\Omega$ is smooth and strongly
convex, the entire boundary is strictly pseudoconvex, which shows that the
Kähler-Einstein metric has negative cost curvature near its boundary.

As a result, when $\Omega$ is bounded, we can find a relatively compact open
set $\Omega^{\lambda}\subset\Omega$ such that the holomorphic sectional
curvature has a uniform lower bound on $T(\Omega\backslash\Omega^{\lambda})$.
For the sake of contradiction, suppose that $\omega_{K.E.}$ does not have
negative cost-curvature. Since the holomorphic sectional curvature is strongly
negative away from $T\Omega^{\lambda}$, $\Omega^{\lambda}$ is relatively
compact and the metric is translation-invariant in the fibers, there must be a
point $z\in T\Omega^{\lambda}$ and a polarized vector at $z$ (which we assume
without loss of generality is $\mathcal{E}_{1}$) such that
$H(\mathcal{E}_{1})=0$.

We now consider the Kähler-Ricci flow with initial condition $\omega_{K.E.}$.
Since $\omega_{K.E.}$ is Kähler-Einstein, we must have that

(79) $\frac{\partial}{\partial t}H(\mathcal{E}_{1})\equiv 0.$

On the other hand, since the anti-bisectional curvature of $\omega_{K.E.}$ is
non-positive, by Equation (30), we have that

(81) $\displaystyle\frac{\partial}{\partial t}\widetilde{H}(\mathcal{E}_{1})$
$\displaystyle=$
$\displaystyle\Delta\widetilde{H}(\mathcal{E}_{1})+4\widetilde{R}_{1\overline{1}p\overline{q}}\widetilde{R}_{q\overline{p}1\overline{1}}-2\widetilde{R}_{1\overline{p}1\overline{q}}\widetilde{R}_{p\overline{1}q\overline{1}}$
$\displaystyle=$
$\displaystyle\Delta\widetilde{H}(\mathcal{E}_{1})-\widetilde{R}_{1\overline{p}1\overline{q}}\widetilde{R}_{p\overline{1}q\overline{1}}$
$\displaystyle+(4\widetilde{R}_{1\overline{1}p\overline{q}}\widetilde{R}_{q\overline{p}1\overline{1}}-\widetilde{R}_{1\overline{p}1\overline{q}}\widetilde{R}_{p\overline{1}q\overline{1}}).$

From Inequality (56), (81) is non-positive. As such, we have that

(82) $\displaystyle\frac{\partial}{\partial t}\widetilde{H}(\mathcal{E}_{1})$
$\displaystyle\leq$
$\displaystyle\Delta\widetilde{H}(\mathcal{E}_{1})-\widetilde{R}_{1\overline{p}1\overline{q}}\widetilde{R}_{p\overline{1}q\overline{1}}.$

Since $\Delta\widetilde{H}(\mathcal{E}_{1})\leq 0$, we must have that
$\widetilde{R}_{1\overline{p}1\overline{q}}\widetilde{R}_{p\overline{1}q\overline{1}}\leq
0$ in order for $\frac{\partial}{\partial t}\widetilde{H}(\mathcal{E}_{1})$ to
vanish. Again appealing to Inequality (56), this implies that

(83) $\displaystyle
0\geq\widetilde{R}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1},\mathcal{E}_{i},\overline{\mathcal{E}}_{j})\widetilde{R}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1},\mathcal{E}_{j},\overline{\mathcal{E}}_{i})$

However, due to the fact that the metric is generated by a convex function on
$\Omega$,
$\widetilde{R}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1},\mathcal{E}_{i},\overline{\mathcal{E}}_{j})$
is symmetric in $i$ and $j$. As such, Inequality (83) shows that

(84)
$0\geq\sum_{i,j}|\widetilde{R}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1},\mathcal{E}_{i},\overline{\mathcal{E}}_{j})|^{2},$

which implies that for all $i$ and $j$,

(85)
$\widetilde{R}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1},\mathcal{E}_{i},\overline{\mathcal{E}}_{j})\equiv
0.$

As such, the holomorphic bisectional curvature
$R(\mathcal{E}_{1},\overline{\mathcal{E}}_{1},\mathcal{Z},\overline{\mathcal{Z}})$
vanishes for all $(1,0)$-vectors $\mathcal{Z}$. We can then use the following
modified version of Berger’s lemma to calculation the Ricci curvature
$\textrm{Ric}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1})$.

######  Lemma 9.

For any (1,0) vector $\mathcal{X}$, the Ricci curvature
$Ric(\mathcal{X},\overline{\mathcal{X}})$ is given by the integral

(86)
$\operatorname{Ric}(\mathcal{X},\overline{\mathcal{X}})=\frac{n-1}{Vol(\mathbb{S}^{2n-3})}\int_{|\mathcal{Z}|=1,\mathcal{Z}\in
T_{z}^{(1,0)}\mathbb{M}}R(\mathcal{X},\overline{\mathcal{X}},\mathcal{Z},\overline{\mathcal{Z}})\,d\theta(\mathcal{Z}).$

Combining (85) and (86), it follows that
$\textrm{Ric}(\mathcal{E}_{1},\overline{\mathcal{E}}_{1})=0$, which
contradicts the assumption that $\omega_{K.E.}$ is Kähler-Einstein with
negative scalar curvature. ∎

As we noted in the introduction, it is possible to prove Corollary 2 when the
base is a compact Hessian manifold, in which case we can use the compactness
of $M$ to use the standard tensor maximum principle. It is of interest to
determine whether this result can be sharpened to obtain quantitative
estimates on the pinching of the polarized holomorphic sectional curvature.
However, at present we are not able to do so.

###  4.3. Towards a differential Harnack inequality

Although negativity of the anti-bisectional curvature is a strong assumption,
its relation to other curvature conditions is not straight-forward. For
instance, it does not imply negative Ricci curvature, even in the special case
where the potential is $O(n)$-symmetric (which were studied previously by the
authors [27]). The lack of straightforward relationship between the anti-
bisectional curvature and the Ricci curvature has an important consequence
which pertains to differential Harnack estimates.

Differential Harnack estimates are some of the most important results for
Ricci flow. The original version of this inequality was proved by Hamilton and
gives $C^{1}$ control over the scalar curvature for positive time whenever the
initial metric has positive curvature operator [22]. Cao proved another
version for Kähler-Ricci flow under the assumption that the bisectional
curvature is positive [7]. For metrics of mixed curvature, Perelman found a
related inequality [43] which involves coupling the Ricci flow with a
conjugate heat equation.

In order to find such inequalities, the general strategy is to consider a
gradient Ricci soliton and find tensor expressions which vanish. Doing so
provides candidate tensors which may have a sign along the flow. The soliton
calculations done by Cao and Hamilton use the fact the Ricci curvature has a
sign to ensure that their expression is positive for small times, so cannot be
used for the anti-bisectional curvature. It would be of interest to find a
similar inequality for metrics with negative anti-bisectional curvature,
especially since doing so seems to require a different approach.

######  Question 3.

Can we establish a differential Harnack inequality for Kähler-Ricci flow for
metrics of negative anti-bisectional curvature?

##  5\. Orthogonal anti-bisectional curvature and Kähler Ricci Flow

We now study the behavior of orthogonal anti-bisectional curvature under
Kähler-Ricci flow. Recall that in optimal transport, the cost $\Psi(x-y)$ is
weakly regular whenever the orthogonal anti-bisectional curvature is non-
negative.

######  Theorem.

Suppose that $\Omega\subset\mathbb{R}^{2}$ is a convex domain and
$\Psi:\Omega\to\mathbb{R}$ is a strongly convex function so that the Kähler
manifold $(T\Omega,\omega_{\Psi})$

  1. (1)

is complete,

  2. (2)

has bounded curvature, and

  3. (3)

has non-negative orthogonal anti-bisectional curvature.

Then the orthogonal anti-bisectional curvature remains non-negative along
Kähler-Ricci flow.

###### Proof.

The proof is nearly identical to the proof that non-positive anti-bisectional
curvature is preserved, except for the verification of the null-vector
condition. As such, we will only provide the details for that step.

As before, we use Uhlenbeck’s trick so that the curvature evolves by the
equation.

(87) $\displaystyle\frac{\partial}{\partial
t}\widetilde{R}_{i\overline{j}k\overline{\ell}}$ $\displaystyle=$
$\displaystyle\Delta\widetilde{R}_{i\overline{j}k\overline{\ell}}+2\widetilde{R}_{i\overline{j}p\overline{q}}\widetilde{R}_{q\overline{p}k\overline{\ell}}+2\widetilde{R}_{i\overline{\ell}p\overline{q}}\widetilde{R}_{q\overline{p}k\overline{j}}-2\widetilde{R}_{i\overline{p}k\overline{q}}\widetilde{R}_{p\overline{j}q\overline{\ell}}$

We then define a function

$\displaystyle\varphi(z,t)$ $\displaystyle=$
$\displaystyle\inf\left\\{\theta\in\mathbb{R}~{}|~{}A_{\mathcal{V}\overline{\mathcal{W}}\mathcal{V}\overline{\mathcal{W}}}(\theta,z,t)\geq
0\textrm{ for }\mathcal{V}=v^{pol.},\mathcal{W}=w^{pol.}\textrm{ with
}\tilde{g}(v,w)=0\right\\}$

where the tensor $A(\theta,z,t)$ is defined as

(88)
$A(\theta)_{i\overline{j}k\overline{\ell}}=\widetilde{R}_{i\overline{j}k\overline{\ell}}+\theta\widetilde{g}_{ik}\widetilde{g}_{\overline{j}\overline{\ell}}.$

As before, we define the tensor

(89) $A(z,t):=A(\varphi(z,t),z,t),$

which is now positive definite when restricted to lifts of pairs of orthogonal
vectors.

######  Lemma 10.

Suppose that $\\{A_{i\overline{j}k\overline{\ell}}\\}$ is a tensor with the
same symmetries as $\\{\widetilde{R}_{i\overline{j}k\overline{\ell}}\\}$.
Suppose that $\tilde{g}$ is the identity at one point and that
$\mathcal{V}=v^{pol.},\mathcal{W}=w^{pol.}$ are vectors with
$\tilde{g}(v,w)=0$.

Suppose that

(90)
$A(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})=0,$

and that for all other orthogonal pairs of vectors $\xi$ and $\eta$,

(91)
$A(\xi^{pol.},\overline{\eta}^{pol.},\xi^{pol.},\overline{\eta}^{pol.})\geq
0.$

Then, if we let $Q$ be as in (31)

(92)
$Q(A)(\mathcal{V},\overline{\mathcal{W}},\mathcal{V},\overline{\mathcal{W}})\geq
0.$

By an affine change of coordinates, we can assume $\mathcal{V}=e_{1}^{pol.}$
and $\mathcal{W}=e_{2}^{pol.}$. Note that $e_{1}+se_{2}\perp
e_{2}-\overline{s}e_{1}$. As such, we have that

(93)
$\frac{d}{ds}A((e_{1}+se_{2})^{pol.},\overline{(e_{2}-se_{1})^{pol.}},(e_{1}+se_{2})^{pol.},\overline{(e_{2}-se_{1})^{pol.}})|_{s=0}=0$

Simplifying the expression, we find the following.

(94) $A_{1\overline{2}1\overline{1}}=A_{1\overline{2}2\overline{2}}$

Computing $Q(A)$, we find that

(95) $\displaystyle Q(A)_{1\overline{2}1\overline{2}}$ $\displaystyle=$
$\displaystyle
2A_{1\overline{2}p\overline{q}}A_{q\overline{p}1\overline{2}}+2A_{1\overline{2}p\overline{q}}A_{q\overline{p}1\overline{2}}-2A_{1\overline{p}1\overline{q}}A_{p\overline{2}q\overline{2}}$
$\displaystyle=$ $\displaystyle
4A_{1\overline{2}p\overline{q}}A_{q\overline{p}1\overline{2}}-2A_{1\overline{p}1\overline{q}}A_{p\overline{2}q\overline{2}}$
$\displaystyle=$ $\displaystyle
4A_{1\overline{2}1\overline{1}}A_{1\overline{1}1\overline{2}}+4A_{1\overline{2}2\overline{2}}A_{2\overline{2}1\overline{2}}\textrm{
when }p=q\textrm{ (the other terms vanish)}$
$\displaystyle-2A_{1\overline{1}1\overline{2}}A_{1\overline{2}2\overline{2}}-2A_{1\overline{2}1\overline{1}}A_{2\overline{2}1\overline{2}}\textrm{
when }p\neq q\textrm{ (the other terms vanish)}$ $\displaystyle=$
$\displaystyle 4(A_{1\overline{1}1\overline{2}})^{2}.$

Identity (95) uses Equation (94) to simplify the expression. This is non-
negative, which completes the proof.

∎

For $n\geq 3$, the quadratic $Q$ has terms which are negative definite, and so
$Q$ need not have a sign. As such, this result is special to complex dimension
two.

###  5.1. Applications

From a geometric standpoint, the assumption of non-negative orthogonal anti-
bisectional curvature alone does not provide strong control over the geometry
of a tube domain. For instance, it is possible to find metrics with non-
negative orthogonal anti-bisectional curvature which do not converge to a
Kähler-Einstein metric under the flow. As such, the natural field of
applications for Theorem 2 is optimal transport rather than complex geometry.

####  5.1.1. Monge’s cost $c(x,y)=|x-y|$

Since Kähler-Ricci flow tends to smoothen metrics, one natural application of
the flow is to prove regularity estimates for for optimal transport when the
cost function is rough. In order to motivate this question and give a concrete
example of such a function, consider the cost

(96) $c_{0}(x,y)=\|x-y\|,$

which was used in Monge’s original paper on optimal transport [40].

From the perspective of the modern regularity theorem, this function is quite
pathological. In particular, it fails to be $C^{1}$ at the origin and is
degenerate elsewhere (in that the kernel of $c_{i,j}(x,y)$ is non-trivial).
These properties affects the behavior of optimal transport. For instance, the
optimal transport with respect to the cost $c_{0}$ generally fails to be
unique. Even more strongly, when we consider the Kantorovich relaxation, it is
possible to find optimal couplings which are non-deterministic, in that they
are not supported on the graph of a function $T:X\to Y$.

Nevertheless, Monge’s cost can be uniformly approximated (in $C^{\alpha}$) by
the costs

(97) $c_{\epsilon}=\sqrt{\epsilon+\|x-y\|^{2}},$

which are smooth, non-degenerate, and have non-negative MTW tensor. As a
result, it is possible to develop a type of regularity theory for Monge’s cost
in terms of the ray-monotone optimal transport map (see Chapter 3 of
Santambrogio [46] for details). As such, Monge’s cost is much better behaved
than a generic Lipschitz cost function. However, the deformation from $c_{0}$
to $c_{\epsilon}$ is somewhat ad hoc from a conceptual standpoint, and it
would be preferable to have a canonical way to deform the cost function. For
this purpose, Kähler-Ricci flow seems to be a natural candidate.

At present, it is not possible to do this, since Monge’s cost corresponds to a
“Kähler metric” which is singular at the origin and degenerate elsewhere.
Furthermore, this “metric” is not complete. However, if we can find a Kähler-
Ricci flow whose initial condition corresponds to the Monge’s cost in some
weak sense, we would be able to fit Monge’s cost into the existing regularity
theory.

####  5.1.2. Weak regularity using the Kähler-Ricci flow

At first, it seems that in order to satisfy the MTW(0) condition, a cost must
be $C^{4}$ (in that $\frac{\partial^{2}}{\partial x^{i}\partial
x^{j}}\frac{\partial^{2}}{\partial y^{k}\partial y^{l}}c(x,y)$ is well
defined). However, there are results to suggest that it is possible to obtain
regularity for less smooth cost functions. For instance, Villani showed that
the MTW(0) condition for the squared-distance cost is stable under Gromov-
Hausdorff convergence [59]. Furthermore, Guillen and Kitagawa [20] found a
synthetic version of the $MTW(0)$ condition known as quantitative quasi-
convexity which only requires that the potential be $C^{3}$. For complete
$\Psi$-costs in two dimensions, KR weak regularity (Definition 2) provides
another version of $MTW(0)$, which is meaningful for even less regular costs.

Theorem 2 shows that for cost functions which are sufficiently smooth, this
definition is equivalent to the normal MTW(0) condition. However, it has the
advantage of being meaningful whenever the parabolic flow (7) exists and
smoothens the convex potential.

It is natural to ask whether KR weak regularity provides a meaningful
generalization of weak regularity. In other words, these two definitions are
equivalent for costs which are sufficiently smooth. However, unless we can say
something about the optimal transport when the cost function is less smooth
than $C^{4}$, there is no reason to use the Kähler-Ricci flow to define weak
regularity. To address this question, we provide the following result.

######  Corollary 6.

Let $(X,\mu)$ and $(Y,\nu)$ be probability spaces and
$\Psi_{0}:\Omega\to\mathbb{R}$ be a strongly convex $W^{2,p}$ function (with
$p>2$) such that following assumptions hold.

  1. (1)

$X,~{}Y$ and $\Omega$ are subsets of $\mathbb{R}^{2}$.

  2. (2)

$X-Y\subset\Omega$.

  3. (3)

$\Psi$ induces a complete Hessian metric on $\Omega$.

  4. (4)

$X$ and $Y$ are smooth and bounded.

  5. (5)

There exists $\epsilon>0$ such that whenever $t<\epsilon$, $X$ and $Y$ are
uniformly strongly relatively $c$-convex with respect to the cost
$c(x,y)=\Psi_{t}(x-y)$ where $\Psi_{t}$ solves the flow (7). For a definition
of relative $c$-convexity, see [35] Definition 2.3.

  6. (6)

$\Psi_{0}$ is KR weakly regular (in the sense of Definition 2).

  7. (7)

The measures $\mu$ and $\nu$ are absolutely continuous with respect to the
Lebesgue measure and $\log\left(\frac{d\mu(x)}{d\nu(y)}\right)\in
L^{\infty}(X\times Y)$.

We then fix an open set $X^{\lambda}$ which is relatively compact in the
interior of $X$. Then there is an optimal transport $T_{0}$ with respect to
the cost $c_{0}(x,y)=\Psi_{0}(x-y)$. Furthermore, there exists constants
$\alpha^{\prime}\in(0,1)$ and $C>0$ so that

(98) $\|T_{0}\|_{C^{\alpha^{\prime}}(X^{\lambda})}<C.$

The constant $\alpha^{\prime}\in(0,1)$ depends on $p$, the subset
$X^{\lambda}\subset X$ and the $L^{\infty}$-norm of
$\log\left(\frac{d\mu}{d\nu}\right)$. The Hölder norm depends the $W^{2,p}$
norm of $\Psi_{0}$ on $X-Y$.

###### Proof.

Consider the transport $T_{t}$ which is optimal with respect to the cost
$c_{t}(x,y)=\Psi_{t}(x-y)$ for $t>0$. When $\Psi_{0}$ is $C^{2,\alpha}$,
$\Psi_{t}$ will be $C^{\infty}$ for $t>0$ and bounded in $C^{2,\alpha}$ for
small time. By Theorem 2, $c_{t}$ will also satisfy the MTW(0) condition.

Using the fifth assumption, it is possible to obtain a Hölder estimate for the
transport $T_{t}$ on $X^{\lambda}$ for small times $t$. To do so, we first
note that the $W^{2,p}$ estimate on the cost implies a $C^{1,\alpha}$-estimate
via Morrey’s inequality for some $\alpha>0$. Using this, we appeal to Theorem
9.5 of Figalli, Kim, McCann [17] to obtain a uniform Hölder estimate on
$T_{t}$ for $t>0$. It is worth noting that the original proof of these result
uses a $C^{2}$-estimate on $c$. However, the $C^{2}$-estimate only appears as
an error term, so it is straightforward to adapt the proof to only require a
$C^{1,\alpha}$-estimate, at the possible expense of weakening the final Hölder
exponent999To apply the Figalli-Kim-McCann argument, we can replace the error
term $\left\|D_{xx}^{2}c\right\|_{L^{\infty}(U\times V)}s^{2}|v|^{2}$ by
$\left\|D_{x}c\right\|_{C^{\alpha}}(X\times Y)s^{1+\alpha}|v|^{1+\alpha}$. To
apply the rest of the argument, we must have that $\alpha$ satisfies
$\alpha>1/K$, where $K$ is the constant determined by the dimension $n$ and
the $L^{\infty}$ bound on $\log\left(\frac{d\mu(x)}{d\nu(y)}\right)$. When
$\alpha\leq 1/K$, we can instead set $K^{\prime}=2/\alpha$ and repeat the rest
of the argument in terms of $K^{\prime}$. This will decrease the final Hölder
exponent, but otherwise leaves the argument unchanged. We can find a sequence
$t_{k}\to 0$ so that the transport maps $T_{t_{k}}$ converge on $X^{\lambda}$
in $C^{\alpha^{\prime}}$ for $\alpha^{\prime}<\alpha$ to a limiting function
$T_{0}$.

By the Kantorovich theorem, we know that an optimal $c_{0}$-coupling between
$(X,\mu)$ and $(Y,\nu)$ exists (i.e., a coupling satisfying (17)). This
coupling is unique and deterministic, by Theorem 10.28 of Villani [58]. As a
coupling, it depends continuously on the cost function in the weak-∗ topology
(see Corollary 5.20 for a precise statement [58]). As such, we can pick a sub-
sequence so that the transport maps $T_{t_{k}}$ converge to an optimal
coupling $\pi_{0}$ with respect to $c_{0}$. However, the support of a
$\pi_{0}$ is contained within the Kuratowski limit inferior of the supports of
$T_{t_{k}}$, so in particular is contained within $T_{0}$. As such, $T_{0}$ is
the optimal transport with respect to $c_{0}$ restricted to $X^{\lambda}$, and
is bounded in $C^{\alpha^{\prime}}$. ∎

Corollary 6 gives a continuity result for optimal transport when the cost
function is $W^{2,p}$ and the flow (7) exists. It seems likely that this flow
is well defined for convex functions $\Psi_{0}$ which are $W^{2,p}$, which
would allow us to remove the second assumption. For the purpose of comparison,
it is worth discussing the corresponding situation for compact Kähler
manifolds. We cannot directly apply these results to our situation, but this
does give some indication of what might be in our setting. Song and Tian [52]
showed that if $\mathbb{M}$ is a compact Kähler manifold with Kähler form
$\omega_{0}$ and volume form $\Omega$, the Kähler-Ricci flow exists, is unique
and immediately smoothens the metric whenever the initial data is in the space
$PSH(\mathbb{M},\omega_{0},\Omega)_{p}$ for some $p>1$. Here,
$PSH(\mathbb{M},\omega_{0})$ is the space of all quasi-plurisubharmonic
functions, which are upper semi-continuous functions on $\mathbb{M}$
satisfying $\omega_{0}+\sqrt{-1}\partial\bar{\partial}\psi\geq 0$. The space
$PSH(\mathbb{M},\omega_{0},\Omega)_{p}$ is defined as

$\displaystyle PSH(\mathbb{M},\omega_{0},\Omega)_{p}$ (99) $\displaystyle=$
$\displaystyle\left\\{\psi\in PSH(\mathbb{M},\omega_{0})\cap
L^{\infty}(\mathbb{M})~{}|~{}\frac{\left(\omega_{0}+\sqrt{-1}\partial\bar{\partial}\psi\right)^{n}}{\Omega}\in
L^{p}(\mathbb{M})\right\\}.$

In our context, this is roughly equivalent to the Monge-Ampére measure
associated with $\Psi_{0}$ having a Radon-Nikodym derivative with respect to
the Lebesgue measure which is $L^{p}$ for some $p>1$. However, showing that
the flow exists and smoothens the metric in the non-compact case requires
careful analysis, since it is not possible to use the maximum principle.

####  5.1.3. General cost functions

There are several limitations to Definition 2. For instance, we are restricted
to considering costs whose associated Kähler metric is complete. This
restriction does not seem to be fundamental and can be solved by finding
boundary conditions for Kähler-Ricci flow so that Theorem 2 still holds.

The other major restriction is that we can only consider costs which are
induced by a convex function. At present, this seems to be a fundamental
obstacle; the associated geometry for a general cost function is pseudo-
Riemannian, not Kählerian [28, 29]. For pseudo-Riemannian metrics, the Ricci
flow is no longer weakly parabolic. As a result, we do not have a canonical
way to evolve general cost functions.

##  6\. Examples

Initially, one might be concerned that non-positive anti-bisectional curvature
is a very strong assumption, and that Theorem 1 has limited application. In
this section, we provide various examples of metrics satisfying this
assumption, which suggests that such metrics exist in abundance.

###  6.1. Simple examples

We start by providing two simple examples.

####  6.1.1. An example in one dimension

Consider the one-dimensional domain $\Omega=\mathbb{R}_{>0}$ with the
potential

(100) $\Psi(x)=-\log(x).$

The tube domain $T\mathbb{R}_{>0}$ is the half space $\mathbb{H}$ and the lift
$\Psi^{h}$ induces its standard hyperbolic metric. This is a metric of
constant negative holomorphic sectional curvature, so the anti-bisectional
curvature is negative (as a Riemann surface, there is no orthogonal anti-
bisectional curvature).

We can also use this example to obtain a compact Hessian manifold. Namely, if
we consider the $\mathbb{Z}$-action $\psi$ on $\mathbb{R}_{>0}$ given by
$\psi(k,x)=2^{k}x$ for $k\in\mathbb{Z},x\in\mathbb{R}_{>0}$, we can consider
the quotient manifold, which gives an affine structure on the circle which is
distinct from the standard one.

####  6.1.2. Two versions of the bidisk

Using the previous example, we can find a Hessian manifold whose tube domain
is biholomorphically isometric to the bidisk (with its product metric) by
considering the domain

(101) $\Omega=\\{x\in\mathbb{R}^{2}~{}|~{}x_{1},x_{2}>0\\}$

with Hessian metric

(102) $\Psi(x)=-\log(x_{1})-\log(x_{2})$

However, this potential is not unique. For instance, the potential

(103) $\Psi=-\log(\cos(x_{1})+\cos(x_{2})),$

defined on the domain

(104) $\Omega=\\{|x_{1}|+|x_{2}|<\pi/2\\}$

also lifts to a metric which is biholomorphically isometric to the bidisk.

This metric has negative holomorphic sectional curvature and non-positive
anti-bisectional curvature (so is negatively cost-curved).

###  6.2. Other examples

We will now provide examples which require more computation, and do not just
follow from the basic properties of the anti-bisectional curvature. Since
computing the anti-bisectional curvature is fairly involved, we have written a
Mathematica notebook which can be used to do the calculation and numerically
check for negativity [25].

####  6.2.1. Calabi’s example

In [6], Calabi studied the tube domain $TB$ where $B$ is a unit ball in
$\mathbb{R}^{n}$. In particular, he showed that this space admits a complete
Kähler-Einstein metric which is not homogeneous. He also found a semi-explicit
expression for this metric in terms of the solution to an ordinary
differential equation.

Using Corollary 2, we can show that this metric has negative cost-curvature.
In order to do so, it is sufficient to find a single metric on this space
which has non-positive anti-bisectional curvature and negative holomorphic
sectional curvature. For this purpose, we consider the potential

(105) $\Psi(x)=-\log\left(1-\sum_{i=1}^{n}(x_{i})^{2}\right).$

The metric $\omega_{\Psi}$ becomes asymptotically hyperbolic near the
boundary, and there are orthogonal vectors whose anti-bisectional curvature
vanishes to third-order as one approaches $\partial T\mathbb{B}$. However, the
metric $\omega_{\Psi}$ is negatively cost-curved. Furthermore, it also has
negative holomorphic sectional curvature. We postpone a derivation of this
example to the appendix.

####  6.2.2. A cone in $\mathbb{R}^{2}$

There are tube domains whose base is non-compact which admit negatively cost-
curved metrics. For example, on the cone

(106) $C=\\{(x_{1},x_{2})~{}|~{}|x_{1}|>|x_{2}|\\},$

the convex potential

(107) $\Psi=-\log(x_{1}^{2}-x_{2}^{2})$

lifts to a complete metric whose anti-bisectional curvature satisfies the
identity

(108)
$2\mathfrak{A}(v^{pol.},(w^{\sharp})^{pol.})=-1+\sin[2\theta]\sin[2\phi],$

where

(109) $\displaystyle v$ $\displaystyle=$
$\displaystyle\cos[\theta]\frac{\partial}{\partial{x_{1}}}+\sin[\theta]\frac{\partial}{\partial{x_{2}}},$
(110) $\displaystyle w$ $\displaystyle=$
$\displaystyle\sin[\phi]dx_{1}-\cos[\phi]dx_{2}.$

Hessian metrics on convex cones have been previously studied (see, e.g.,
Chapter 4 of [50] and [60]). It would be of interest to find other cones whose
tangent bundles admit negatively cost-curved metrics.

###  6.3. Surfaces with non-negative orthogonal anti-bisectional curvature

In recent work of the authors and J. Zhang [27], we studied the problem of
finding $O(n)$-symmetric metrics on tube domains whose orthogonal anti-
bisectional curvature is non-negative. The motivation for this was to find
weakly regular cost functions which only depended on the Euclidean distance
between points. In that paper, we showed that for such a metric to be
complete, the underlying Hessian manifold must be $\mathbb{R}^{n}$. We also
found several examples, such as the lift of the convex potential

(111) $\Psi(x)=\|x\|-C\log(\|x\|+C)\textrm{ for }C>0.$

###  6.4. Solitons and Statistical Mirror Symmetry

In recent work of Zhang and the first named author [65], we study T-duality in
the context of tube domains. We refer to this phenomena as “statistical mirror
symmetry,” due to its roots in information geometry.

As a particular example, we show that the Siegel-Jacobi space
$\mathbb{H}\times\mathbb{C}$ with its special-affine invariant metric is
mirror to the Siegel half-space, which is a space of constant negative
holomorphic sectional curvature on the tube domain $T\Omega$ where

(112) $\Omega=\\{(x_{1},x_{2})~{}|~{}x_{1}>x_{2}^{2}\\}.$

In forthcoming work, we solve the Kähler-Ricci flow explicitly on both spaces
and show that they remain coupled eternally. Furthermore, the Siegel-Jacobi
space is a complex two-dimensional Kähler-Ricci soliton with non-negative
orthogonal bisectional curvature and the Siegel half-space is a negatively
cost-curved Kähler-Einstein metric (see [65], Subsubsection 4.4.4). As such,
these two flows provide examples which satisfy the curvature conditions
studied in this paper.

##  7\. Acknowledgements

The first named author would like to thank Robert McCann, Mizan Khan, and Jun
Zhang for some helpful suggestions. He was partially supported by AFOSR Grant
FA9550-19-1-0213 (“Brain-Inspired Networks for Multifunctional Intelligent
Systems and Aerial Vehicles”, MURI subcontract from UCLA). The second named
author is partially supported by a research grant from NSFC with grant number
12071050.

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##  Appendix A Verification of Calabi’s example

In this section, we verify that the potential
$\Psi=\log\left(1-\sum_{i=1}^{n}x_{i}^{2}\right)$ lifts to a negatively cost-
curved metric. We start by considering the case in complex dimension two. Let
$\Omega$ be the unit ball in $\mathbb{R}^{2}$ and consider the point
$(x^{1},x^{2})\in\Omega$. Let $(v,w)$ be the vector-covector pair

$v=\cos(\theta)\frac{\partial}{\partial
x_{1}}+\sin(\theta)\frac{\partial}{\partial x_{2}}$
$w=\sin(\phi)dx_{1}-\cos(\phi)dx_{2}.$

Since the potential is rotationally symmetric, we can rotate the space so that
$x_{2}=0$. Calculating the anti-bisectional curvature, we find the following.

$\displaystyle\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle=$
$\displaystyle\cos[2(\theta-\phi)]-\frac{2(1+x_{1}^{2}+2x_{1}^{4})}{(1+x_{1}^{2})^{3}}+\frac{2(-x_{1}^{2}+x_{1}^{4})\cos[2\phi]]}{(1+x_{1}^{2})^{3}}$

We now want to show that this expression is negative. To do so, note that it
will be maximized in $\theta$ whenever $\phi=\theta$ (i.e.,
$g(v,w^{\sharp})=0$). As such, we have that

$\displaystyle\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle\leq$ $\displaystyle
1-\frac{2(1+x_{1}^{2}+2x_{1}^{4})}{(1+x_{1}^{2})^{3}}+\frac{2(-x_{1}^{2}+x_{1}^{4})\cos[2\phi]]}{(1+x_{1}^{2})^{3}}$

Furthermore, the coefficient in front of the last term is non-positive
definite, so the final term is maximized (in $\phi$) when
$\phi=\frac{\pi}{2}$. As such, we have that

$\displaystyle\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle\leq$ $\displaystyle
1+\frac{-2(1+x_{1}^{2}+2x_{1}^{4})+2(x_{1}^{2}-x_{1}^{4})}{(1+x_{1}^{2})^{3}}$
$\displaystyle=$ $\displaystyle 1+\frac{-2(1+3x_{1}^{4})}{(1+x_{1}^{2})^{3}}.$

To show that this is negative, we let $s=x_{1}^{2}$ and consider the quantity

$\displaystyle(1+s)^{3}\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle\leq$ $\displaystyle(1+s)^{3}-2(1+3x_{1}^{4})$ $\displaystyle=$
$\displaystyle(s-1)^{3}$ $\displaystyle<$ $\displaystyle 0\textrm{ (for
}s<1).$

This shows that the anti-bisectional curvature is negative. To see that the
metric is negatively cost-curved, note that when $v=w^{\sharp}$,

$\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)=-2+\frac{(-1+s)(1+s^{2}-2s\cos[2\theta])}{(1+s)^{3}}$

which forces the expression to be strongly negative.

To show that the holomorphic sectional curvature is strongly negative, we note
that the metric becomes asymptotically hyperbolic near the boundary, so it
suffices to verify this property when $s$ is strictly smaller than $1$. We
then consider the non-polarized vector $\mathcal{X}=\partial
z_{1}+(a+\sqrt{-1}b)\partial z_{2}$ and compute its holomorphic sectional
curvature.

(113)
$(1-s)^{4}(1+s)R(\mathcal{V},\overline{\mathcal{V}},\mathcal{V},\overline{\mathcal{V}})=-2\left(\begin{aligned}
&3+3s+9s^{2}+s^{3}-2b^{2}(1-s)^{3}\\\
&+a^{4}(1-s)^{2}(3+s)+b^{4}(1-s)^{2}(3+s)\\\
&+2a^{2}(1-s)\left(-3-2s-3s^{2}+b^{2}(-3+2s+s^{2})\right)\end{aligned}\right)$

We use Mathematica to find the maximum for this in terms of $a$ and $b$. Doing
so we find that

$R(\mathcal{V},\overline{\mathcal{V}},\mathcal{V},\overline{\mathcal{V}})\leq-\frac{2(3+3s+9s^{2}+s^{3})}{(s-1)^{4}(s+1)}.$

This is strictly negative for $s<1$, so we find that the holomorphic sectional
curvature is negative.

###  A.1. Negative anti-bisectional curvature in higher dimensions

In higher dimensions, we can again take advantage of symmetry to rotate so
that all the coordinates except $x_{1}$ vanish,

$\displaystyle v$ $\displaystyle=$ $\displaystyle\cos(\theta)\partial
x_{1}+\sin(\theta)\partial x_{2},\textrm{ and }$ $\displaystyle w$
$\displaystyle=$
$\displaystyle\sin(\phi)dx_{1}-\cos(\phi)\cos(\alpha)dx_{2}-\cos(\phi)\sin(\alpha)dx_{3}.$

In this case, again denoting $x_{1}^{2}=s$ we can express the anti-bisectional
curvature in the following way.

$\displaystyle(1+s)^{3}\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle=$
$\displaystyle-(1-s)(1+2s+s^{2})\cos[\theta]^{2}\cos[\phi]^{2}\cos[\alpha]^{2}$
$\displaystyle-(3+7s+5s^{2}+s^{3})\cos[\phi]^{2}\cos[\alpha]^{2}\sin[\theta]^{2}$
$\displaystyle-(3+3s+9s^{2}+s^{3})\cos[\theta]^{2}\sin[\phi]^{2}$
$\displaystyle-(1-s)(1-2s+s^{2})\sin[\theta]^{2}\sin[\phi]^{2}$
$\displaystyle+(1+s)^{3}\cos[\alpha]\sin[2\theta]\sin[2\phi]$
$\displaystyle-(1-s)(1+2s+s^{2})\cos[\theta]^{2}\cos[\phi]^{2}\sin[\alpha]^{2}$
$\displaystyle-(1-s)(1+2s+s^{2})\cos[\phi]^{2}\sin[\theta]^{2}\sin[\alpha]^{2}.$

We can simplify this expression using a routine but tedious calculation to
find the following.

$\displaystyle(1+s)^{3}\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle=$ $\displaystyle-(1-s)(1+s)^{2}\cos[\phi]^{2}$
$\displaystyle-(1-s)^{3}\sin[\phi]^{2}$
$\displaystyle-2(1+s)^{3}\left(\cos[\theta]\sin[\phi]-\cos[\phi]\cos[\alpha]\sin[\theta]\right)^{2}.$

Doing so, we find that

$\displaystyle(1+s)^{3}\mathfrak{A}\left(v^{pol.},\left(w^{\sharp}\right)^{pol.}\right)$
$\displaystyle\leq$ $\displaystyle-(1-s)^{3}.$

This shows that the metric has negative anti-bisectional curvature. We can
then use the same calculation as in the two-dimensional case to show that the
holomorphic sectional curvature is strongly negative.

##  Appendix B The polarized holomorphic sectional curvature and
$\mathfrak{O}$ control the Riemann curvature tensor for metrics with negative
anti-bisectional curvature

Here we show that for a metric with negative anti-bisectional curvature, we
can control the Riemann curvature tensor in terms of the polarized holomorphic
sectional curvatures and the orthogonal anti-bisectional trace curvature.

As before, we work in a polarized unitary frame. Let $s$ be a free parameter
and consider

$\displaystyle\mathfrak{A}(\mathcal{E}^{i},\mathcal{E}^{j}+t\mathcal{E}^{k})$
$\displaystyle=$
$\displaystyle\mathfrak{A}(\mathcal{E}^{i},\mathcal{E}^{j})+2sR(\mathcal{E}^{i},\overline{\mathcal{E}}^{j},\mathcal{E}^{i},\overline{\mathcal{E}}^{k})+s^{2}\mathfrak{A}(\mathcal{E}^{i},\mathcal{E}^{k}).$

The left hand side is non-positive. By our assumptions, we have bounds on the
first and third terms, which allows us to obtain two sided bounds on the
middle term as well.

Now we consider the quantity

$\displaystyle H(\mathcal{E}^{i}+\mathcal{E}^{j})$ $\displaystyle=$
$\displaystyle
H(\mathcal{E}^{i})+4R(\mathcal{E}^{i},\overline{\mathcal{E}}^{j},\mathcal{E}^{i},\overline{\mathcal{E}}^{i})+4R(\mathcal{E}^{i},\overline{\mathcal{E}}^{i},\mathcal{E}^{j},\overline{\mathcal{E}}^{j})$
$\displaystyle+2\mathfrak{A}(\mathcal{E}^{i},\mathcal{E}^{j})+4R(\mathcal{E}^{i},\overline{\mathcal{E}}^{j},\mathcal{E}^{j},\overline{\mathcal{E}}^{j})+H(\mathcal{E}^{i}).$

Again, the left hand side is negative and bounded from below. From the
previous estimate (setting $i=j$) and our assumption, we have two sided bounds
on all the terms except for
$4R(\mathcal{E}^{i},\overline{\mathcal{E}}^{i},\mathcal{E}^{j},\overline{\mathcal{E}}^{j})$.
Rearranging the expression, we obtain two-sided bounds on this term.

Now we consider the expression

$\displaystyle\mathfrak{A}(\mathcal{E}^{i}+t\mathcal{E}^{j},\mathcal{E}^{i}+s\mathcal{E}^{k})$
$\displaystyle=$ $\displaystyle
2tsR(\mathcal{E}^{i},\overline{\mathcal{E}}^{i},\mathcal{E}^{j},\overline{\mathcal{E}}^{k})+\textrm{bounded
terms},$

where both $s$ and $t$ are free parameters. The left hand side is negative, so
we derive a two-sided bound on the first term by choosing the parameters
appropriately.

Finally, we consider

$\displaystyle\mathfrak{A}(\mathcal{E}^{i}+t\mathcal{E}^{k},\mathcal{E}^{j}+s\mathcal{E}^{\ell})$
$\displaystyle=$ $\displaystyle
2tsR(\mathcal{E}^{i},\overline{\mathcal{E}}^{j},\mathcal{E}^{k},\overline{\mathcal{E}}^{\ell})+\textrm{terms
with a repeated index}.$

Since all the terms with a repeated index have been bounded and the left hand
side is non-positive, we can choose $t$ and $s$ to obtain two-sided bounds on
the first term on the right hand side. This gives a bound on the curvature in
our polarized unitary frame, which shows that the curvature is bounded.