Lemma 14.5.1BC4847C0CD7D

Proof. The identity $\mathrm{d}\omega _ u=\mathrm{d}\omega $ is equivalent to

together with its complex conjugate; this can also be checked directly, since the third derivatives of $u$ cancel.

For tensors with two holomorphic and one antiholomorphic lower index, we use the norm

Set $A_{ik\bar j}=\nabla _ i h_{k\bar j}$ and $D_{ik\bar j}=T_{ik}^ p h_{p\bar j}$. We first claim

To prove this, start with $\Delta _ hS=h^{i\bar j}g^{k\bar\ell }\nabla _ i\nabla _{\bar j}h_{k\bar\ell }$. Switch $\bar j,\bar\ell $ by the conjugate of (14.5.3), commute $\nabla _ i,\nabla _{\bar\ell }$, and then switch $i,k$ by (14.5.3). The terms with a derivative of $h$ are

The first switch also produces the term $-h^{i\bar j}g^{k\bar\ell }\overline{T_{j\ell }^ q}\nabla _ i h_{k\bar q}$. Applying (14.5.3) to it gives one of the two conjugate derivative terms displayed above, together with

The part of this expression containing $h_{p\bar q}$ is $\| D\| ^2$. Differentiating the logarithmic determinant twice gives

By the antisymmetry of $T$, the two remaining derivative terms combine to $2\operatorname {Re}\langle A,D\rangle $, where $\langle \cdot ,\cdot \rangle $ is the inner product of the norm above. These identities give (14.5.4).

To bound the remainder $E$ explicitly, let

The remainder is the contraction by $h^{i\bar j}g^{k\bar\ell }$ of

The commutator $\mathcal C$ is the curvature of $\nabla $ acting linearly on $h$, as one sees by inserting $\nabla =\partial +\Gamma $ and canceling the derivatives of $h$. In a $g$-unitary frame diagonalizing $h$, every coefficient in this remainder is bounded by $C_ g(1+S)$, so the contraction is bounded by $C_ g(1+S)\sum _ i h^{i\bar i}=C_ g(1+S)B$, as asserted. In particular, no derivative of $h$ remains in $E$.

It remains to control the gradient term introduced by the logarithm. Define

By (14.5.3),

Moreover $\| H\| ^2=|\partial S|_ h^2/S$. Expanding the square therefore gives

Combining this with (14.5.4) and with $\Delta _ h\log S=\Delta _ hS/S-|\partial S|_ h^2/S^2$, we obtain the exact identity

Since $K\geq 0$ and $S\geq 1$, and since $E/S\geq -2C_ gB$, enlarging the constant proves the lemma. $\square$