Proposition 12.6.227294934ED49C

Proof. Since the forms $\mathrm d\alpha _ j$ are holomorphic, we have

Indeed $\bar\partial (\alpha _ j\wedge \bar\alpha _ k) =-\alpha _ j\wedge \mathrm d\bar\alpha _ k$, and applying $\partial $ and multiplying by $(\mathrm i/(2\pi ))^2$ gives the identity in (12.6.24). Diagonalizing the positive coefficient matrix expresses the result as a sum of nonnegative multiples of $\beta \wedge \bar\beta $ for invariant $(2,0)$-forms $\beta $. In dimension three these are nonnegative $(2,2)$-forms: the wedge product with $\mathrm i\xi \wedge \bar\xi $, for a $(1,0)$-form $\xi $, has the squared modulus of $\xi \wedge \beta $ as coefficient.

By proposition 12.6.5, $\theta _ A$ is balanced, hence Gauduchon. The maximum case of theorem 12.16.3 therefore gives $v_+(\theta _ A)=\int _ X\theta _ A^3=H\det A$. The supremum of $\mathcal V_\nu (\theta _ A)$ over $\nu \in \mathscr M$ equals this upper mass by theorem 12.15.4, and (12.6.6) shows that it is attained at $\mu _0$. Finally, proposition 13.14.15 makes the finiteness of the upper mass a property of the manifold. $\square$