Proposition 23.2.24.32084430C8AD The Gauduchon form $\omega =\alpha +\beta $ of section 23.1 is Vaisman, with Lee form $\lambda =-\mathrm{d}s$, and
\[ \kappa _\lambda =\alpha , \qquad D_\lambda 1=\omega . \]
In particular, the bounded mass property of definition 13.14.14 fails for the ordinary potential theory of a background that is simultaneously Gauduchon and Vaisman.
Proof.
Using $\alpha =\mathrm{dd}^{\mathrm{c}}s$ and $\beta =\mathrm{d}s\wedge \mathrm d^{\mathrm c}s$, direct differentiation gives
\[ \mathrm{d}\omega =-\mathrm{d}s\wedge \alpha =-\mathrm{d}s\wedge \omega , \qquad \mathrm{e}^ s\omega =\mathrm{dd}^{\mathrm{c}}(\mathrm{e}^ s)=\mathrm{dd}^{\mathrm{c}}|z|^2. \]
Although $s$ is not a function on the quotient, $\mathrm{d}s$ descends to it. Its integral along the loop induced by $z\mapsto \mathrm{e}^{-1}z$ is $-2$, so the Lee form is not exact.
Let $g_{S^5}$ be the round metric on the unit sphere. Up to a constant factor, the Riemannian metric associated with $\omega $ on the cover is
\[ \frac{|\mathrm{d}z|^2}{|z|^2} =\mathrm{d}t^2+g_{S^5},\qquad t=\log |z|. \]
The one-form $\mathrm{d}t$ is parallel for this product metric, which proves the Vaisman condition. Finally, $\kappa _\lambda =\mathrm{dd}^{\mathrm{c}}s=\alpha $ and $\lambda \wedge \lambda ^{\mathrm c}=\beta $, so $D_\lambda 1=\omega $ by (23.2.22). The unbounded masses are given by theorem 23.1.1.
$\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.