11.4 Conformal Gauduchon metrics4ECC3E68F7D5

Proof. For a smooth real function $v$ on $X$, define

This real second-order operator has positive definite principal symbol and no zeroth-order term. By the maximum principle, its kernel consists of the constants. Its index is zero, since the linear path joining it to a Laplace operator consists of elliptic operators. Fredholm theory and elliptic regularity therefore show that the formal adjoint $L^*$ of $L$ with respect to $\eta ^ n$ has a one-dimensional kernel, generated by a nonzero smooth real function $\rho $.

We first prove that $\int _ X\rho \eta ^ n\ne 0$. Otherwise the constant function $1$ would be orthogonal to the kernel of $L^*$, and the Fredholm alternative would solve $Lv=1$, which is impossible at a maximum point of $v$. We may therefore rescale so that $\int _ X\rho \eta ^ n=1$.

Next we show that $\rho \geq 0$. Suppose instead that $\rho $ is negative somewhere. Choose a nonzero smooth function $h\geq 0$ supported where $\rho <0$, and put $c=\int _ Xh\rho \eta ^ n<0$. The function $h-c$ is strictly positive and orthogonal to the kernel of $L^*$, so the Fredholm alternative would solve $Lv=h-c>0$, which is again impossible at a maximum point of $v$.

In real local coordinates $(x_1,\ldots ,x_{2n})$, write $L^*=A^{ij}\partial _ i\partial _ j+B^ i\partial _ i+C$, where $\partial _ i=\partial /\partial x_ i$, the coefficients are smooth, and repeated indices are summed from $1$ to $2n$. The leading symbol $(A^{ij})$ is positive definite. Since $\rho \geq 0$,

and the operator $L^*-\max (C,0)$ has nonpositive zeroth-order term. By the strong minimum principle, a zero of $\rho $ would force $\rho $ to vanish identically on the connected manifold $X$. Hence $\rho >0$.

Finally, integration by parts gives

Thus the equation $L^*\rho =0$ is precisely $\mathrm{dd}^{\mathrm{c}}(\rho \eta ^{n-1})=0$, and the one-dimensional adjoint kernel gives uniqueness up to positive scaling. Since $\gamma ^{n-1}=\rho \eta ^{n-1}$, the form $\gamma $ is Gauduchon. $\square$