Pome

17.16.28 Explicit basic geodesics and singular potentials1DFB5A5992AD

Diagonal projective automorphisms give smooth nonconstant geodesics. For real $a_0,\ldots ,a_{n-1}$, set

\[ v_ t([z])=\log \frac{\sum _{j=0}^{n-1}\mathrm e^{ta_ j}|z_ j|^2}{\sum _{j=0}^{n-1}|z_ j|^2}. \]

The family is spatially nonconstant when the $a_ j$ are not all equal.

Proposition 17.16.29.A3E893F6A601 The pullback $\pi ^*v_ t$ is the full Hermitian Perron segment and is stationary under all smooth fixed-endpoint variations. Its distance and action are

\[ \begin{aligned} d_ B(0,\pi ^*v_1)^2& =\frac{2[(\sum _ ja_ j)^2+\sum _ ja_ j^2]}{n+1},\\ \mathscr S(\pi ^*v)& =\mathcal A_ B(\pi ^*v) =\frac{(\sum _ ja_ j)^2+\sum _ ja_ j^2}{n+1}. \end{aligned} \]

Proof. On the strip, the holomorphic map $F([z],s)=[\mathrm e^{a_0s/2}z_0:\cdots :\mathrm e^{a_{n-1}s/2}z_{n-1}]$ satisfies $\operatorname {pr}_ Y^*\kappa +\mathrm{dd}^{\mathrm{c}}V=F^*\kappa $. The imaginary-time phases are unitary, so $V$ depends only on $\operatorname{Re}s$. This form has rank $d$, strictly positive spatial slices, and zero $(d+1)$st power. The classical bounded Kähler Dirichlet uniqueness identifies it with the base Perron segment; proposition 17.16.16, proposition 17.16.22 give the full segment and its stationarity.

For $x_ j=|z_ j|^2/\sum |z_ j|^2$, the initial velocity is $\sum _ ja_ jx_ j$. A standard complex Gaussian vector has independent exponential squared coordinates, and its projective direction has distribution $\kappa ^ d$. Dividing those coordinates by their sum gives the uniform simplex distribution, with moments

\[ \int _ Yx_ i^2\kappa ^ d=\frac2{n(n+1)}, \qquad \int _ Yx_ ix_ j\kappa ^ d=\frac1{n(n+1)}\quad (i\ne j). \]

The smooth distance formula and (17.16.20) give the claimed squared distance; the action is half of it. $\square$

For instance, $n=3$ and $(a_0,a_1,a_2)=(1,-1,0)$ give squared distance one and action $1/2$.

The completed space also contains potentials with different ordinary singularity types. Put $o=[1:0:\cdots :0]\in Y$ and

\[ \psi =\log \frac{\sum _{j=1}^ d|z_ j|^2}{\sum _{j=0}^ d|z_ j|^2}, \qquad v_ a=1-(1-\psi )^ a,\qquad 0<a<1, \]

with the canonical value $v_ a(o)=-\infty $.

Proposition 17.16.30.9355B610E2C3 Every $v_ a$ has non-pluripolar mass one, and

\[ v_ a\in \mathcal E^2(Y,\kappa ) \quad \Longleftrightarrow \quad a<\frac d{d+2}. \]

At equality its squared energy diverges logarithmically. The pullback has mass $2n$ and pole set $\pi ^{-1}(o)$. If $0<a<b<d/(d+2)$, the two pullbacks have different ordinary singularity types and finite positive $d_ B$ distance.

Proof. In the chart $z_0=1$, write $s=\log |z'|^2$, $L=\log (1+\mathrm e^ s)$, $P=\mathrm e^ s/(1+\mathrm e^ s)$, and $T=1+L-s\geq 1$. The local potential $\Phi =L+1-T^ a$ has derivative

\[ A=P+a(1-P)T^{a-1},\qquad A'=P'(1-aT^{a-1})+a(1-a)(1-P)^2T^{a-2}>0. \]

Both $A$ and $A'$ are positive, so the radial complex Hessian is positive. Its bounded-above local potential extends plurisubharmonically across $o$ by lemma 5.24.7. The function extends smoothly across the hyperplane at infinity. Its bounded plurisubharmonic truncations decrease to a nontrivial canonical potential: near $o$, $|v_ a|=O((1-s)^ a)$ is integrable against radial smooth volume $O(\mathrm e^{ds})\, \mathrm ds$.

The radial mass flux is $Q(s)=A(s)^ d$. Its sphere coefficient is one, since $\mathrm d^{\mathrm c}\log |z'|^2$ integrates to one on the Hopf circle and $\mathrm{dd}^{\mathrm{c}}\log |z'|^2$ induces mass-one Fubini–Study on $\mathbb P^{d-1}$. Stokes on an annulus gives mass $Q(s_2)-Q(s_1)$. Since $Q(-\infty )=0$ and $Q(+\infty )=1$, finite-level locality and annular exhaustion give non-pluripolar mass one. The pole has no atom, and smoothness gives no mass on the hyperplane at infinity.

As $s\to -\infty $,

\[ |v_ a|^2\sim T^{2a},\qquad Q'(s)\sim d a^ d(1-a)T^{-d(1-a)-1},\qquad T\sim 1-s. \]

The squared energy therefore converges exactly when $2a-d(1-a)<0$. At equality the integrand has order $T^{-1}$, giving logarithmic divergence. All other compact regions have smooth potentials and measures. Fiber integration gives the pullback assertions. Finally, $v_ a-v_ b=T^ b-T^ a\to +\infty $ at $o$, so mutual bounded comparison fails. Both points lie in the completed metric space and are distinct, giving finite positive distance. $\square$

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