PomeProposition 19.6.4.29A7DDA38CA2 Any Okounkov test curve $(\Delta _{\tau })_{\tau < \Delta _{\max }}$ relative to $\Delta $ is continuous in $\tau $. Moreover, $\operatorname{vol}\Delta _{\tau }>0$ for all $\tau <\Delta _{\max }$.
Proposition 19.6.429A7DDA38CA2
Proof. We first claim that $\operatorname{vol}\Delta _{\tau '}>0$ for all $\tau '<\Delta _{\max }$. By Condition (2b) in definition 19.6.1 and theorem 19.7.5, we know that $\operatorname{vol}\Delta _{\tau ''}>0$ when $\tau ''$ is small enough. Fix one such $\tau ''$. We may assume that $\tau ''\leq \tau '$ since otherwise there is nothing to prove. Next take $\tau '''\in (\tau ',\Delta _{\max })$. Take $t\in (0,1)$ such that $\tau '=t\tau '''+(1-t)\tau ''$. It follows that
\[ \operatorname{vol}\Delta _{\tau '}\geq \operatorname{vol}\left( t\Delta _{\tau '''}+(1-t)\Delta _{\tau ''}\right)\geq (1-t)^ n\operatorname{vol}\Delta _{\tau ''}>0. \]
Next we claim that $\operatorname{vol}\Delta _{\tau }$ is continuous for $\tau < \Delta _{\max }$. In fact, it follows from theorem 19.7.14 that $(-\infty ,\Delta _{\max }) \ni \tau \mapsto \log \operatorname{vol}\Delta _{\tau }$ is concave, but we already know that it is finite, hence the continuity follows.
Next we show that
\[ \Delta _{\tau }=\bigcap _{\tau '<\tau }\Delta _{\tau '}. \]
The $\subseteq $ direction is obvious. By the continuity of the volume, both sides have the same volume and the volume is positive, we therefore obtain the equality.
Similarly, we have
\[ \Delta _{\tau }=\overline{\bigcup _{\tau '>\tau }\Delta _{\tau '}}. \]
The continuity of $\Delta _{\tau }$ at $\tau <\Delta _{\max }$ is proved. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.