Let $X$ be a smooth irreducible projective complex variety of dimension $n$. Let $X=Y_0\supseteq\cdots\supseteq Y_n$ be an admissible flag. Consider $n$ line bundles $L_1,\ldots,L_n$ on $X$. Let $\Delta(L_1),\ldots,\Delta(L_n)$ be the corresponding Newton–Okounkov bodies. By a theorem of Jow, the numerical class of $L_i$ is determined by the convex body $\Delta(L_i)$. In particular, the movable intersection product $\langle L_1,\ldots,L_n\rangle$ is determined by the collection $\Delta(L_1),\ldots,\Delta(L_n)$ with respect to all admissible flags. I'm wondering if there is an explicit formula in general?

  • $\begingroup$ I thought that in Jow's theorem you only have that if $\Delta(L)=\Delta(M)$ for all admissible flags, then $L\equiv M$, but that only having this equality for just one fixed flag (like in your question) may not be enough $\endgroup$
    – YangMills
    Nov 12, 2021 at 14:50
  • $\begingroup$ @YangMills Of course. By a formula, I have in mind a formula that involves all admissible flags. In fact, I have a more precise conjecture: $\langle L_1,\ldots,L_n \rangle=n!\sup \mathrm{vol} (\Delta(L_1),\ldots,\Delta(L_n))$, where the sup is taken over all admissible flags. $\endgroup$ Nov 12, 2021 at 16:27


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.