PomeProposition 17.16.22.2B828BD01DBB The unique action minimizer between two points of $\mathcal B_2$ is their constant-speed Perron segment, and its action is half their squared distance. On every smooth path in $\mathcal B_\infty $, $\mathcal A_ B$ equals the raw action $\mathscr S$. Every smooth basic Perron segment with strictly positive spatial forms is stationary for $\mathscr S$ under all smooth fixed-endpoint variations on $X$.
Proposition 17.16.222B828BD01DBB
Proof. The length inequality and Cauchy–Schwarz give
\[ d_ B(\gamma _0,\gamma _1)^2 \leq \left(\int _0^1|\dot\gamma |_{d_ B}\, \mathrm dt\right)^2 \leq 2\mathcal A_ B(\gamma ). \]
Equality holds for the constant-speed geodesic. Equality for a connecting curve forces constant speed and equality between length and endpoint distance, hence forces a geodesic; uniqueness follows from the CAT$(0)$ property.
For a smooth basic path $\pi ^*v_ s$, put $h=v_{s+\delta }-v_ s$. On its short base Perron segment, parametrized by $0\leq \tau \leq 1$, the chord bounds above, and
\[ v_ s+\tau h-C\delta ^2\tau (1-\tau ) \]
bounds below. Indeed, the spatial forms along the chord have a uniform positive lower bound, $\mathrm dh=O(\delta )$, and proposition 17.16.9 makes the displayed path a subgeodesic for a fixed sufficiently large $C$. The right initial velocity of the Perron segment is therefore $h+O(\delta ^2)$ uniformly. The classical smooth-endpoint Kähler distance formula gives
\[ \lim _{\delta \to 0}\frac{d_ B(\pi ^*v_ s,\pi ^*v_{s+\delta })^2}{\delta ^2} =2n\int _ Y\dot v_ s^{\, 2}(\kappa +\mathrm{dd}^{\mathrm{c}}v_ s)^ d. \]
Integration proves $\mathcal A_ B=\mathscr S$ along this path.
For stationarity, let $L(w)$ be the first variation at a smooth basic Perron path. Uniform spatial positivity allows every smooth endpoint-zero variation. Group invariance of the action and of the path gives $L(g^*w)=L(w)$ for $g\in E$. Average over Haar probability measure to obtain
\[ L(w)=L\left(\int _ Eg^*w\, \mathrm dg\right)=L(\pi ^*q) \]
for a smooth endpoint-zero base variation $q$. Fiber integration reduces this to $2n$ times the closed-base first variation. With $\eta _ t=\kappa +\mathrm{dd}^{\mathrm{c}}v_ t$, integration by parts gives that variation as
\[ \int _0^1\int _ Yq_ t\left[-\ddot v_ t\eta _ t^ d +d\, \mathrm d\dot v_ t\wedge \mathrm d^{\mathrm c}\dot v_ t\wedge \eta _ t^{d-1}\right]\mathrm dt. \]
The bracket vanishes by the smooth base homogeneous equation. Thus $L(w)=0$ for every variation on $X$. $\square$
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