Example 12.34.14.706929790BE5 Let $E=\mathbb C/(\mathbb Z+\mathrm i\mathbb Z)$, let $X=E^ n$ with $n\geq 2$, and let $\eta _ j=(\mathrm i/2)\mathrm dz_ j\wedge \mathrm d\bar z_ j$ be the flat forms pulled back from the factors. Put $\omega _0=\sum _{j=1}^ n\eta _ j$. All the flows below start at $\omega _0$, with potential zero and initial density one relative to $\omega _0^ n$. In particular, hypotheses on the initial data expressed through a decomposition $\mathrm e^{\psi ^+-\psi ^-}$ of the density, with quasi-plurisubharmonic functions $\psi ^\pm $ subject to fixed $L^ p$ bounds, hold uniformly for all these flows, with $\psi ^+=\psi ^-=0$.
For each $M>0$, the twist $\zeta _ M=M\eta _1-\eta _2$ gives the explicit solution
\[ \omega _ M(t)=(1+Mt)\eta _1+(1-t)\eta _2+ \sum _{j=3}^ n\eta _ j, \qquad 0\leq t<1, \]
of $\partial _ t\omega =-\operatorname {Ric}(\omega )+\zeta _ M$. Its maximal smooth existence time is exactly one, while
\[ \operatorname {diam}(X,\omega _ M(1/2)) \geq \sqrt{1+M/2}\, \operatorname {diam}(E,\eta _1). \]
For the single fixed twist $\zeta =\omega _0$, the solution is $(1+t)\omega _0$ for all $t\geq 0$, and its diameter grows as $\sqrt{1+t}\, \operatorname {diam}(X,\omega _0)$.
Proof.
Each of these metrics has constant coefficients, so its Chern–Ricci form vanishes; differentiation in $t$ proves the flow equations, and positivity of the coefficients gives the stated existence intervals. In the first family, pair a hypothetical positive representative at a time $t\geq 1$ with the closed positive form $\eta _1\wedge \eta _3\wedge \cdots \wedge \eta _ n$, where the empty product is $1$ when $n=2$. The potential term integrates to zero, and the pairing is a positive fixed multiple of $1-t$, so it cannot be strictly positive. Integrating the flow equation shows that every smooth solution differs from the affine background by $\mathrm{dd}^{\mathrm{c}}$ of a smooth function, so no smooth positive continuation beyond $t=1$ is possible.
Projection to the first elliptic factor does not increase lengths once the metric of that factor is scaled by $1+Mt$. Taking two points that differ only in the first factor proves the lower bound for the diameter. For the second flow, every path length is multiplied by $\sqrt{1+t}$, which proves the exact formula for the diameter.
$\square$
Pome
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