Lemma 16.15.11EFC27557FD59

Proof. (1) In the coordinates $w$ the metric and its inverse have the components

By these rules and the chain rule $\partial _{w_ c}(F\circ z)=J_{kc}\, (\partial _{z_ k}F)\circ z$, which holds because $\partial _{w_ c}\bar z_ k=0$, we have $\partial _{w_ c}g'_{a\bar b}=(\partial _{w_ c}J_{ia})g_{i\bar j}\overline{J_{jb}}+J_{ia}J_{kc}\, \partial _ kg_{i\bar j}\, \overline{J_{jb}}$, and hence

Let $T$ be a tensor, and differentiate its transformation rule by $\partial _{w_ c}$. By the product rule, this gives the transformed array $J_{kc}\partial _ kT$, plus one term for each holomorphic index, in which $\partial _{w_ c}$ falls on the Jacobian factor of that index; the factors of the antiholomorphic indices are antiholomorphic and are not differentiated. For an upper holomorphic index, the term contains $\partial _{w_ c}(J^{-1})_{dm}=-(J^{-1})_{dm'}(\partial _{w_ c}J_{m'b})(J^{-1})_{bm}$, and it cancels against the contribution of the second term of (16.15.12) to $\Gamma '^ d_{cb}T'^{\cdots b\cdots }$; for a lower holomorphic index $a$, the term contains $\partial _{w_ c}J_{ia}$, and it cancels against the contribution of the second term of (16.15.12) to $-\Gamma '^ b_{ca}T'_{\cdots b\cdots }$. The contributions of the first term of (16.15.12) are the transforms of the connection terms of $T$. Hence $\nabla _ cT'$ is the transform of $\nabla _ kT$, contracted with $J_{kc}$, so $\nabla _ pT$ is a tensor; by conjugation, so is $\nabla _{\bar q}T$.

The Leibniz rule follows from the product rule for $\partial _ p$ and the additivity of the connection terms over the indices. If an upper and a lower holomorphic index of $T$ are contracted, their connection terms $\Gamma ^ m_{pa}T^{a\cdots }{}_{\cdots m}$ and $-\Gamma ^ a_{pm}T^{m\cdots }{}_{\cdots a}$ cancel, so $\nabla $ commutes with contractions. Next,

because $g^{a\bar l}g_{a\bar j}=\delta _{lj}$, and $\nabla _{\bar q}g_{i\bar j}=\overline{\nabla _ qg_{j\bar i}}=0$, since $\overline{g_{i\bar j}}=g_{j\bar i}$. The connection terms of the identity tensor cancel, so $\nabla \delta =0$, and the Leibniz rule applied to $g^{i\bar j}g_{l\bar j}=\delta ^ i_ l$ gives $(\nabla g^{i\bar j})g_{l\bar j}=0$, hence $\nabla g^{-1}=0$. For a complete contraction $S$, a function, $\nabla _ pS=\partial _ pS$, and the last assertion follows from the Leibniz rule and the commutation with contractions.

(2) By (16.15.12) for $g$ and for $h$, the inhomogeneous terms of $\Gamma '$ and $\widehat\Gamma '$ coincide, so the difference transforms as a tensor. Moreover $\widehat\nabla _ ig_{j\bar l}=\partial _ ig_{j\bar l}-\widehat\Gamma ^ p_{ij}g_{p\bar l}$, and contracting with $g^{m\bar l}$ gives $\Gamma ^ m_{ij}-\widehat\Gamma ^ m_{ij}$.

(3) The second term of (16.15.12) is holomorphic in $w$, so it is annihilated by $\partial _{\bar w_ e}=\overline{J_{le}}\, \partial _{\bar z_ l}$, and $R$ transforms as a tensor; it is symmetric in $i$ and $j$ because $\Gamma $ is. Since $\Gamma ^ m_{im}=g^{m\bar l}\partial _ ig_{m\bar l}=\partial _ i\log \det g$, we get $R_{i\bar j}=-\partial _{\bar j}\partial _ i\log \det g$. Both sides of each of the two identities are tensors, so it suffices to prove them at a point $x$ in coordinates in which $g_{i\bar j}(x)=\delta _{ij}$ and $\mathrm dg(x)=0$, which exist by lemma 16.15.3. There $\Gamma =0$, so covariant derivatives at $x$ are ordinary derivatives, and $\partial g^{-1}=-g^{-1}(\partial g)g^{-1}$ and $\bar\partial g^{-1}$ vanish at $x$. Using a local Kähler potential $\varphi $, $g_{i\bar j}=\partial _ i\partial _{\bar j}\varphi $, whose derivatives commute, we find at $x$

and

in each case all terms containing a first derivative of $g$ or of $g^{-1}$ vanish at $x$.

(4) A tensor with only holomorphic indices receives no connection term from $\nabla _{\bar q}$, so $\nabla _{\bar q}T=\partial _{\bar q}T$. The tensor $\nabla _ pT$ again has only holomorphic indices, so $\nabla _{\bar q}\nabla _ pT^ m_{ij}$ is the $\partial _{\bar q}$-derivative of $\partial _ pT^ m_{ij}+\Gamma ^ m_{pa}T^ a_{ij}-\Gamma ^ a_{pi}T^ m_{aj}-\Gamma ^ a_{pj}T^ m_{ia}$. On the other hand, $\nabla _{\bar q}T=\partial _{\bar q}T$ has the additional antiholomorphic index $\bar q$, which receives no term from $\nabla _ p$, so $\nabla _ p\nabla _{\bar q}T^ m_{ij}=\partial _ p\partial _{\bar q}T^ m_{ij}+\Gamma ^ m_{pa}\partial _{\bar q}T^ a_{ij}-\Gamma ^ a_{pi}\partial _{\bar q}T^ m_{aj}-\Gamma ^ a_{pj}\partial _{\bar q}T^ m_{ia}$. The difference consists of the terms in which $\partial _{\bar q}$ falls on $\Gamma $; since $\partial _{\bar q}\Gamma ^ m_{pa}=-R^ m{}_{a\bar qp}$ by the symmetry of $R$, this is the asserted formula. $\square$