Proposition 23.2.43.503628CC8BE5 Let $\omega $ be a metric as in theorem 23.2.41, with Lee form $\lambda $. There is a positive smooth function $U$ on $H_ n$ with $\omega =D_\lambda U$; in particular, the class of $\omega $ in $H^{1,1}_{\mathrm{BC},\lambda }(H_ n,\mathbb R)$ is zero. If $\omega _1,\omega _2$ are two such metrics, with positive potentials $U_1,U_2$ and cohomologous Lee forms $\lambda _1$ and $\lambda _2=\lambda _1+\mathrm da$, then for every choice of the additive constant in $a$,
\[ \mathrm e^{-a}\omega _2-\omega _1=D_{\lambda _1}(\mathrm e^{-a}U_2-U_1). \]
In particular, among metrics of the form (23.2.36), a solution of (23.2.26) is determined by its volume form and Lee class, even if the potential $u$ in (23.2.26) is not assumed radial.
Proof.
Write $l=kC^{-m}<0$ and $f(r)=\int _0^ rl(s)\, \mathrm ds$ for real $r$. Then $f(r+2)=f(r)+L_0$. The solution of $U'-lU=C$ given by
\[ \begin{aligned} U(0)& =\frac{\mathrm e^{L_0}}{1-\mathrm e^{L_0}} \int _0^2\mathrm e^{-f(s)}C(s)\, \mathrm ds,\\ U(r)& =\mathrm e^{f(r)} \left(U(0)+\int _0^ r\mathrm e^{-f(s)}C(s)\, \mathrm ds\right) \end{aligned} \]
is positive on $[0,2]$ and satisfies $U(2)=U(0)$. Since the equation is periodic, all derivatives also match at the endpoints, so $U$ defines a positive smooth function on $H_ n$. By (23.2.19), its twisted Hessian is
\[ \begin{split} D_\lambda U & =(U'-lU)\alpha + \bigl [U''-2lU'+(l^2-l')U\bigr ]\beta \\ & =C\alpha +(C'-lC)\beta =\omega . \end{split} \]
This proves that the twisted class is zero. The identity $\mathrm e^{-a}D_{\lambda _2}U_2=D_{\lambda _1}(\mathrm e^{-a}U_2)$ of subsection 23.2.25 gives the formula for the difference. For two metrics of the form (23.2.36), equal volume forms give the same function $F$, and equal Lee classes give the same period, so theorem 23.2.41 applies. This uniqueness concerns the metric, not the potentials representing it.
$\square$
Pome
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