Corollary 21.13.68.551D529C92A2 Let $\varphi _0,\varphi _1\in \mathrm{PSH}_{\mathrm{tor}}(X,\theta )\cap \mathrm{PSH}(X,\theta )_{>0}$. Then the following are equivalent:
8E1A729499A6
$\varphi _0\sim _ P \varphi _1$;
39AFF113945C
there is a subgeodesic from $\varphi _0$ to $\varphi _1$.
If these conditions are satisfied, let $(\varphi _ t)_{t\in (0,1)}$ be the geodesic from $\varphi _0$ to $\varphi _1$. Then $\varphi _ t\in \mathrm{PSH}_{\mathrm{tor}}(X,\theta )$ for all $t\in (0,1)$ and
8A0A9FDE2DF5
21.13.69
\begin{equation} G_{\varphi _ t}=(1-t)G_{\varphi _0}+tG_{\varphi _1}. \end{equation}
Proof.
The equivalence between 1 and 2 follows from a general fact theorem 16.8.32. In the toric case, 2 $\implies $ 1 can also be argued more directly.
Assume that these conditions are satisfied. Let $(\varphi _ t)_{t\in (0,1)}$ be the geodesic from $\varphi _0$ to $\varphi _1$. Since the geodesic is the upper envelope of toric subgeodesics, $\varphi _ t\in \mathrm{PSH}_{\mathrm{tor}}(X,\theta )$ for all $t\in (0,1)$. Let $\Psi '$ be the proper convex function on $M_{\mathbb {R}}\times \mathbb {R}$ defined by corollary 21.13.66. Then $\Psi '$ is the minimum of all $\Psi $ satisfying (21.13.67). We claim that
C4D285074403
21.13.70
\begin{equation} \Psi '(m,s)=G_{\varphi _0}(m)\lor \left(G_{\varphi _1}(m)+s\right). \end{equation}
It suffices to show that the right-hand side is proper, namely, $G_{\varphi _0}\lor G_{\varphi _1}$ is not identically $\infty $. But recall that by proposition 17.10.7, we have $\varphi _0\land \varphi _1\in \mathrm{PSH}(X,\theta )$. Therefore, by proposition 21.13.56,
\[ G_{\varphi _0}\lor G_{\varphi _1}=G_{\varphi _0\land \varphi _1}\not\equiv \infty . \]
In particular, (21.13.70) follows.
Now by construction,
\[ G_{\varphi _ t}(m)=\inf _{s\in \mathbb {R}} \left( \Psi '(m,s)-st\right)=(1-t)G_{\varphi _0}(m)+tG_{\varphi _1}(m) \]
for all $t\in (0,1)$. So (21.13.69) follows.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.