Let $X$ be a connected compact Kähler manifold of dimension $n$ and $\theta $ be a smooth closed real $(1,1)$-form on $X$ representing a big cohomology class.
Definition 17.10.2.8699FCF63094 Let us fix $\varphi _0,\varphi _1\in \mathrm{PSH}(X,\theta )$. A subgeodesic from $\varphi _0$ to $\varphi _1$ is a family $(\varphi _ t)_{t\in (0,1)}$ in $\mathrm{PSH}(X,\theta )$ such that
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if we define
\[ \Phi \colon X\times \Bigl \{ z\in \mathbb {C}:\operatorname{Re}z\in (0,1) \Bigr \} \rightarrow [-\infty ,\infty ),\quad (x,z)\mapsto \varphi _{\operatorname{Re}z}(x), \]
then $\Phi $ is $p_1^*\theta $-psh, where $p_1\colon X\times \{ z\in \mathbb {C}:\operatorname{Re}z\in (0,1)\} \rightarrow X$ is the natural projection;
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when $t\to 0+$ (resp. $t\to 1-$), $\varphi _ t$ converges to $\varphi _0$ (resp. $\varphi _1$) with respect to the $L^1$-topology.
We also say $(\varphi _ t)_{t\in [0,1]}$ is a subgeodesic.
We call $\Phi $ the complexification of the subgeodesic $(\varphi _ t)_ t$.
When we do not want to specify $\varphi _0$ and $\varphi _1$, we shall say $(\varphi _ t)_{t\in (0,1)}$ is a subgeodesic.
More generally, a family $(\psi _ t)_{t\in [a,b]}$ (resp. $(\psi _ t)_{t\in (a,b)}$) in $\mathrm{PSH}(X,\theta )$ for some $a\leq b$ is called a subgeodesic if $(\psi _{tb+(1-t)a})_{t\in [0,1]}$ (resp. $(\psi _{tb+(1-t)a})_{t\in (0,1)}$) is a subgeodesic.
In general, there are no subgeodesics from $\varphi _0$ to $\varphi _1$. In fact, by proposition 17.10.7 below, the existence of a subgeodesic from $\varphi _0$ to $\varphi _1$ forces $\varphi _0\land \varphi _1\not\equiv -\infty $, a condition that may fail (see example 21.12.22).
We first note that the subgeodesics are well-behaved under the change of $\theta $:
Proposition 17.10.5.2042C299A40E Let $g$ be a smooth real function on $X$. Let $\theta '=\theta +\mathrm{dd}^{\mathrm{c}}g$. Suppose that $(\varphi _ t)_{t\in [0,1]}$ is a subgeodesic in $\mathrm{PSH}(X,\theta )$. Then $(\varphi _ t-g)_{t\in [0,1]}$ is a subgeodesic in $\mathrm{PSH}(X,\theta ')$.
Proof.
This follows directly from the definition of the complexification.
$\square$
Example 17.10.6.2F494BEED8F5 Let $\varphi _0\in \mathrm{PSH}(X,\theta )$, $C\in \mathbb {R}$. Let
\[ \varphi _ t=\varphi _0+tC,\quad t\in [0,1]. \]
Then $(\varphi _ t)_{t\in [0,1]}$ is a subgeodesic. Indeed, since $\operatorname{Re}z$ is harmonic in $z$, the complexification $\Phi (x,z)=\varphi _0(x)+C\, \operatorname{Re}z$ is $p_1^*\theta $-psh.
As a consequence, the constant $(\varphi _0)_{t\in [0,1]}$ is a subgeodesic, called the constant subgeodesic at $\varphi _0$.
A more general version is as follows: Suppose that $(\varphi _ t)_{t\in [0,1]}$ is a subgeodesic in $\mathrm{PSH}(X,\theta )$, $C_1,C_2\in \mathbb {R}$. Then $(\varphi _ t+C_1t+C_2)_{t\in [0,1]}$ is also a subgeodesic.
Proposition 17.10.7.1880C1FBD88F Let $\varphi _0,\varphi _1\in \mathrm{PSH}(X,\theta )$ and $(\varphi _ t)_{t\in (0,1)}$ be a subgeodesic from $\varphi _0$ to $\varphi _1$. Then for each $x\in X$, $[0,1] \ni t\mapsto \varphi _ t(x)$ is a convex function. Furthermore,
\[ \inf _{t\in (0,1)}\varphi _ t\in \mathrm{PSH}(X,\theta ),\quad \inf _{t\in (0,1)}\varphi _ t\leq \varphi _0\land \varphi _1. \]
Proof.
Let $\Phi $ be the complexification of $(\varphi _ t)_{t\in (0,1)}$.
For each $x\in X$, the map
\[ \Bigl \{ z\in \mathbb {C}: \operatorname{Re}z\in (0,1) \Bigr \} \rightarrow [-\infty ,\infty ),\quad z\mapsto \Phi (x,z) \]
is either subharmonic or constantly $-\infty $, as follows from definition 17.10.2 1 and proposition 5.13.17. In the latter case, the convexity of $[0,1] \ni t\mapsto \varphi _ t(x)$ is trivial. In the former case, the convexity on the interval $(0,1)$ follows from proposition 5.13.12.
In order to verify the convexity at the boundary, let us fix $s\in (0,1)$. We need to show that
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17.10.8
\begin{equation} \varphi _ s(x)\leq s\varphi _1(x)+(1-s)\varphi _0(x) \end{equation}
for all $x\in X$. Thanks to proposition 5.14.6, it suffices to prove this for almost all $x$.
Choose sequences $a_ j\searrow 0$ and $b_ j\nearrow 1$ such that $\varphi _{a_ j}\to \varphi _0$ and $\varphi _{b_ j}\to \varphi _1$ almost everywhere. Outside a set $Z\subseteq X$ of zero Lebesgue measure, all these convergences hold and all the values involved are finite. For $x\in X\setminus Z$ and $j$ large enough that $a_ j<s<b_ j$, the convexity on $(0,1)$ gives
\[ \varphi _ s(x)\leq \frac{b_ j-s}{b_ j-a_ j}\varphi _{a_ j}(x)+\frac{s-a_ j}{b_ j-a_ j}\varphi _{b_ j}(x). \]
Letting $j\to \infty $ gives (17.10.8) almost everywhere.
Let us prove the last assertion. Let
\[ \varphi \coloneqq \inf _{t\in (0,1)}\varphi _ t. \]
By Kiselman's principle proposition 5.14.15, we know that $\varphi \in \mathrm{PSH}(X,\theta )\cup \{ -\infty \} $. Since the complexification $\Phi $ is not identically $-\infty $, Fubini's theorem shows that for almost every $x\in X$, the function $t\mapsto \varphi _ t(x)$ is finite for almost every $t\in (0,1)$. For such an $x$, the convexity just proved implies that $t\mapsto \varphi _ t(x)$ is a finite convex function on $(0,1)$, hence it is bounded from below on $(0,1)$. Thus $\varphi \not\equiv -\infty $, and so $\varphi \in \mathrm{PSH}(X,\theta )$.
For any $t\in (0,1)$, using the convexity established above, we have
\[ \varphi \leq (1-t)\varphi _0+t\varphi _1. \]
It follows that $\varphi \leq \varphi _0$ and $\varphi \leq \varphi _1$ almost everywhere and hence everywhere by proposition 5.14.6. Our assertion follows.
$\square$
Corollary 17.10.9.C645EE3517E1 Let $\varphi _0,\varphi _1\in \mathrm{PSH}(X,\theta )$ and $(\varphi _ t)_{t\in (0,1)}$ be a subgeodesic from $\varphi _0$ to $\varphi _1$. Then
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17.10.10
\begin{equation} \lim _{t\to 0+}\varphi _ t\leq \varphi _0,\quad \lim _{t\to 1-}\varphi _ t\leq \varphi _1. \end{equation}
Moreover,
\[ \lim _{t\to 0+}\varphi _ t=\varphi _0,\quad \lim _{t\to 1-}\varphi _ t=\varphi _1\quad \textup{quasi-everywhere}. \]
The inequalities in (17.10.10) are not equalities everywhere in general. For example, consider an arbitrary subgeodesic from an unbounded $\varphi _0\in \mathcal{E}(X,\omega )$ to a bounded $\varphi _1\in \mathrm{PSH}(X,\omega )$ for some Kähler form $\omega $ on $X$.
Proof.
The inequalities in (17.10.10) and the existence of the limits follow from the convexity in proposition 17.10.7. Let us prove the quasi-everywhere equality at $t=0$. Take a decreasing sequence $t_ j\searrow 0$ with $t_ j>0$. Then $\varphi _{t_ j}\to \varphi _0$ in $L^1(X)$ as $j\to \infty $. So by corollary 5.14.8, we have
\[ \varlimsup _{j\to \infty }\varphi _{t_ j}= \varphi _0\quad \textup{quasi-everywhere}. \]
Therefore, we have
\[ \varphi _0=\varlimsup _{j\to \infty }\varphi _{t_ j}=\lim _{t\to 0+}\varphi _ t\leq \varphi _0\quad \textup{quasi-everywhere}. \]
The assertion at $t=1$ is similar.
$\square$
Corollary 17.10.11.E31D1E7DDF87 Let $(\varphi _ t)_{t\in [0,1]}$ be a subgeodesic in $\mathrm{PSH}(X,\theta )$. Then
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17.10.12
\begin{equation} \adjustlimits \sup _ X \inf _{s\in (0,1)}\varphi _ s \leq \sup _ X \varphi _ t\leq \sup _ X \varphi _0\lor \sup _ X \varphi _1 \end{equation}
for all $t\in (0,1)$.
Furthermore, $[0,1] \ni t\mapsto \varphi _ t$ is continuous with respect to the $L^1$-topology.
In particular, in view of proposition 5.17.3, the family $\{ \varphi _ t\} _{t\in [0,1]}$ is relatively compact with respect to the $L^1$-topology on $\mathrm{PSH}(X,\theta )$.
Proof.
(17.10.12) follows immediately from proposition 17.10.7. Let us verify the continuity of $t\mapsto \varphi _ t$. The continuity at $t=0$ or $t=1$ follows from the definition of subgeodesic. While the continuity on $(0,1)$ follows from the convexity of $t\mapsto \varphi _ t$, and the fact that $\varphi _ t\in L^1(X)$ for each $t\in (0,1)$.
$\square$
Proposition 17.10.13.5ECD747A7B0B Let $(\varphi ^ i_0)_{i\in I}$, $(\varphi ^ i_1)_{i\in I}$ be two non-empty uniformly bounded from above families in $\mathrm{PSH}(X,\theta )$. Let $(\varphi ^ i_ t)_{t\in (0,1)}$ be subgeodesics from $\varphi _0^ i$ to $\varphi _1^ i$ for each $i\in I$. Then
\[ \left(\operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi ^ i_ t\right)_{t\in (0,1)} \]
is a subgeodesic from $\operatorname *{sup}_{i}{}\! {}^{*} \varphi ^ i_0$ to $\operatorname *{sup}_{i}{}\! {}^{*} \varphi ^ i_1$.
Proof.
We may assume that $\varphi _0^ i,\varphi _1^ i\leq 0$ for all $i\in I$. Then it follows that $\varphi _ t^ i\leq 0$ for all $t\in (0,1)$ and all $i\in I$ by proposition 17.10.7.
We define
\[ \varphi _ t\coloneqq \operatorname *{sup}_{i\in I}{}\! {}^{*} \varphi ^ i_ t\in \mathrm{PSH}(X,\theta ) \]
for all $t\in [0,1]$. Observe that $[0,1] \ni t\mapsto \varphi _ t$ is convex by the same argument leading to (17.10.8).
Let $(\psi _ t)_{t\in (0,1)}$ be the subgeodesic whose complexification $\Phi _{\psi }$ corresponds to $\operatorname *{sup}_{i}{}\! {}^{*} \Phi _{\varphi ^ i}$, where $\Phi _{\varphi ^ i}$ is the complexification of $(\varphi ^ i_ t)_{t\in (0,1)}$. Then clearly, $\varphi _ t\leq \psi _ t$ for each $t\in (0,1)$. On the other hand, by proposition 5.14.5,
\[ \psi _ t=\sup _{i\in I}\varphi ^ i_ t=\varphi _ t\quad \textup{almost everywhere} \]
for almost all $t\in (0,1)$. Therefore, using proposition 5.14.6, we find $\psi _ t=\varphi _ t$ for almost all $t\in (0,1)$. Since both functions are convex in $t$, we conclude that $\psi _ t=\varphi _ t$ for all $t\in (0,1)$.
It remains to argue that $\varphi _ t\xrightarrow {L^1}\varphi _0$ as $t\to 0+$ and $\varphi _ t\xrightarrow {L^1}\varphi _1$ as $t\to 1-$. By symmetry, it suffices to argue the former.
By Choquet's lemma proposition 5.14.3, applied to the uniformly bounded above family of complexifications $(\Phi _{\varphi ^ i})_{i\in I}$ on the strip, there is a countable subset $I'\subset I$ such that
\[ \operatorname *{sup}_{i\in I}{}\! {}^{*}\Phi _{\varphi ^ i}=\operatorname *{sup}_{i\in I'}{}\! {}^{*}\Phi _{\varphi ^ i}. \]
Applying Choquet's lemma proposition 5.14.3 once more to the endpoint families if necessary, and enlarging $I'$, we may also assume
\[ \operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _0^ i=\operatorname *{sup}_{i\in I'}{}\! {}^{*}\varphi _0^ i,\quad \operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _1^ i=\operatorname *{sup}_{i\in I'}{}\! {}^{*}\varphi _1^ i. \]
Thus, after replacing $I$ by $I'$, we may assume that $I$ is countable.
We know that for any $t\in (0,1)$ and any $j\in I$,
\[ \varphi ^ j_ t\leq \varphi _ t\leq t\varphi _1+(1-t)\varphi _0. \]
Letting $t\to 0+$, we find that
\[ \varphi _0^ j\leq \varlimsup _{t\to 0+}\varphi _ t\leq \varphi _0 \]
almost everywhere. Since $I$ is countable, we conclude that
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17.10.14
\begin{equation} \varphi _0=\varlimsup _{t\to 0+}\varphi _ t \end{equation}
almost everywhere.
Fix $i_0\in I$. Recall that by proposition 17.10.7, for each $t\in (0,1)$, we have
\[ \adjustlimits \inf _{t\in (0,1)}\sup _ X \varphi _ t\geq \adjustlimits \inf _{t\in (0,1)}\sup _ X \varphi ^{i_0}_ t\geq \adjustlimits \sup _ X \inf _{t\in (0,1)}\varphi ^{i_0}_ t>-\infty , \]
so the set $\{ \varphi _ t\} _{t\in (0,1)}$ is relatively compact with respect to the $L^1$-topology by proposition 5.17.3. Let $\psi $ be a cluster point as $t\to 0+$. It suffices to show that $\psi =\varphi _0$. By corollary 5.14.8 and (17.10.14), this holds almost everywhere. Therefore, it holds everywhere by proposition 5.14.6.
$\square$
Proposition 17.10.15.E5064A497AB8 Let $(\varphi _ t)_{t\in [0,1]}$ be a subgeodesic in $\mathrm{PSH}(X,\theta )$. Then for any $0\leq a\leq b\leq 1$, the segment $(\varphi _{t})_{t\in [a,b]}$ is a subgeodesic.
Proof.
It suffices to show that
\[ \varphi _{tb+(1-t)a}\xrightarrow {L^1}\varphi _ a,\quad \varphi _{tb+(1-t)a}\xrightarrow {L^1}\varphi _ b \]
as $t\to 0+$ and $t\to 1-$ respectively. In other words, we need to show that for any $c\in (0,1)$, we have
\[ \varphi _ t\xrightarrow {L^1}\varphi _ c \]
as $t\to c$. For this purpose, observe that by proposition 17.10.7,
\[ \adjustlimits \sup _ X \inf _{s\in (0,1)}\varphi _ s\leq \sup _{X}\varphi _ t\leq \left( \sup _ X \varphi _0 \right) \lor \left( \sup _ X\varphi _1 \right) \]
for any $t\in (0,1)$. Therefore, $\{ \varphi _ t\} _{t\in (0,1)}$ is a relatively compact family with respect to the $L^1$-topology on $\mathrm{PSH}(X,\theta )$ by proposition 5.17.3. It suffices to show that any cluster point $\psi $ of $\varphi _ t$ as $t\to c$ is equal to $\varphi _ c$. By corollary 5.14.8 and the convexity proposition 17.10.7, we have $\varphi _ c=\psi $ almost everywhere and hence everywhere by proposition 5.14.6.
$\square$
For the purpose of this chapter, it is useful to reformulate the gluing lemma corollary 5.14.20 in the following way, which allows us to glue subgeodesics directly:
Lemma 17.10.16.3BEBB86C7F79 Let $(\varphi _ t)_{t\in [0,1]}$ be a subgeodesic in $\mathrm{PSH}(X,\theta )$. Let $0\leq a<b\leq 1$ and let $(\psi _ t)_{t\in [a,b]}$ be a subgeodesic such that $\psi _ a=\varphi _ a$, $\psi _ b=\varphi _ b$, and
\[ \psi _ t\geq \varphi _ t,\quad t\in (a,b). \]
Define
\[ \gamma _ t= \begin{cases} \psi _ t,& \textup{if }t\in (a,b),\\ \varphi _ t,& \textup{if }t\in [0,1]\setminus (a,b). \end{cases} \]
Then $(\gamma _ t)_{t\in [0,1]}$ is a subgeodesic.
Proof.
The $L^1$-convergence of $\gamma _ t$ to $\varphi _0$ at $0$ and $\varphi _1$ at $1$ follows immediately from the corresponding property for $(\varphi _ t)_ t$ and $(\psi _ t)_ t$. It remains to prove that the complexification of $(\gamma _ t)_ t$ is $p_1^*\theta $-psh.
Let
\[ S=\{ z\in \mathbb {C}:0<\operatorname{Re}z<1\} ,\quad S'=\{ z\in \mathbb {C}:a<\operatorname{Re}z<b\} . \]
Let $\Phi $ and $\Psi $ be the complexifications of $(\varphi _ t)_ t$ and $(\psi _ t)_ t$ on the corresponding strips. Since $\Psi \geq \Phi $ on $X\times S'$, the complexification of $(\gamma _ t)_ t$ is equal to $\Phi \lor \Psi $ on $X\times S'$ and to $\Phi $ on $X\times (S\setminus S')$. By the gluing lemma corollary 5.14.20, it is enough to verify that
\[ \varlimsup _{\substack {S'\ni z'\to z\\ x'\to x}}\Psi (x',z')\leq \Phi (x,z) \]
at every point $(x,z)\in X\times S$ with $\operatorname{Re}z=a$ or $\operatorname{Re}z=b$.
Suppose for instance that $a>0$, and let $(x_ j,z_ j)$ tend to $(x,z)$ with $\operatorname{Re}z=a$ and $a<\operatorname{Re}z_ j<b$. Set $t_ j=\operatorname{Re}z_ j$. Since $(\psi _ t)_ t$ is a subgeodesic from $\varphi _ a$ to $\varphi _ b$, the sequence $\psi _{t_ j}$ converges to $\varphi _ a$ in $L^1$. Applying corollary 5.14.8 to this sequence gives
\[ \varphi _ a=\inf _{m>0}\left(\operatorname *{sup}_{j\geq m}{}\! {}^{*}\psi _{t_ j}\right). \]
For fixed $m$, the upper semicontinuity of $\operatorname *{sup}_{j\geq m}{}\! {}^{*}\psi _{t_ j}$ gives
\[ \varlimsup _{j\to \infty }\psi _{t_ j}(x_ j)\leq \left(\operatorname *{sup}_{j\geq m}{}\! {}^{*}\psi _{t_ j}\right)(x). \]
Letting $m\to \infty $, we get
\[ \varlimsup _{j\to \infty }\psi _{t_ j}(x_ j)\leq \varphi _ a(x). \]
Since $\Phi (x,z)=\varphi _ a(x)$ when $\operatorname{Re}z=a$, this is the desired boundary inequality along $\{ \operatorname{Re}z=a\} $. The same argument applies along $\{ \operatorname{Re}z=b\} $ when $b<1$. Hence corollary 5.14.20 shows that the pasted complexification is $p_1^*\theta $-psh.
$\square$
Definition 17.10.17.872270015CE7 A ray $\ell =(\ell _ t)_{t\geq 0}$ is a subgeodesic ray in $\mathrm{PSH}(X,\theta )$ if for any $0\leq a\leq b$, the segment $(\ell _{t})_{t\in [a,b]}$ is a subgeodesic in $\mathrm{PSH}(X,\theta )$. We say $\ell $ emanates from $\ell _0$.
The complexification of a subgeodesic ray $\ell $ is defined as the potential
\[ \Phi \colon X\times \{ z\in \mathbb {C}:\operatorname{Re}z>0\} \rightarrow [-\infty ,\infty ),\quad (x,z)\mapsto \ell _{\operatorname{Re}z}(x). \]
Note that $\Phi $ is $p_1^*\theta $-psh, where $p_1\colon X\times \{ z\in \mathbb {C}:\operatorname{Re}z>0\} \rightarrow X$ is the natural projection.
Pome
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