Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 17 / 4

The unheaded definitions use the complex structure $J$ on the real tangent bundle. Compatibility means $g(Ju,Jv)=g(u,v)$. Its fundamental Hermitian form is $\omega (u,v)=g(Ju,v)$; the compatibility identity makes this real, alternating and of type $(1,1)$. The metric is a Kähler metric precisely when $d\omega =0$. In complex dimension one a real three-form is zero, so every compatible metric is a Kähler metric.

Now assume closedness and prove the Kähler normal holomorphic coordinates condition. Start with holomorphic coordinates $w$ centered at the point and make a complex-linear change so that the positive Hermitian coefficient matrix satisfies $h(0)=I$. Closedness of the $(1,1)$ form givesDefine $a_{i\bar{j},k}={\partial }_{w_k}{h}_{i\bar{j}}(0)$; it is symmetric in $i,k$. Choose new coordinates implicitly byThe derivative of this holomorphic map at zero is the identity, so the holomorphic inverse function theorem makes $z$ valid local coordinates. In the new coordinates,At zero, $h^{\prime }=I$, and differentiating givesHermitian symmetry makes all antiholomorphic first derivatives zero as well. Taylor's theorem for a smooth function now yieldsThus condition (a) implies condition (b), with the exact factor $i/2$ in the fundamental-form convention retained.

Conversely, choose the stated normal holomorphic coordinates at an arbitrary point. The coefficient matrix equals $I+O(|z{|}^2)$, so all of its first derivatives at the center vanish. Since the coordinate differentials themselves are closed,Every point can serve as the center, hence $d\omega =0$ on all of $X$. Combined with the preceding construction, this proves the equivalence of (a) and (b); it is a first-jet characterization rather than a claim that the metric is flat on a neighborhood.

To prove (a) implies (c), shrink to a contractible coordinate neighborhood. The real Poincare lemma gives a real one-form $\eta$ with $\omega =d\eta$. Write $\eta ={\eta }^{1,0}+{\eta }^{0,1}$. Because $\omega$ has type $(1,1)$, $\bar{\partial }{\eta }^{0,1}=0$ and $\partial {\eta }^{1,0}=0$. The Dolbeault-Poincaré lemma supplies a smooth function $\psi$ with ${\eta }^{0,1}=\bar{\partial }\psi$. Reality of $\eta$ gives ${\eta }^{1,0}=\partial \overline{\psi }$. ThusTaking the real function $f=2\mathrm{Im}\psi$ gives the local real potential for a closed (1,1)-form. Conversely, $d(i\partial \bar{\partial }f)=0$, by ${\partial }^2={\bar{\partial }}^2=0$ and anticommutation. Hence (a) and (c) are equivalent, completing all three conditions.

For the unheaded radial continuation, take the real potential $f$ as in (c). Rotation invariance makes $f$ constant on every circle of radius $r>0$, sois well defined and smooth by composition. This avoids treating a smooth function as if it had a convergent Taylor series. At nonzero $z$, put $t=\log |z{|}^2$. The identities ${\partial }_{z}t=1/z$ and ${\partial }_{\bar{z}}t=1/\bar{z}$ giveLet $q(z)=f_{z\bar{z}}(z)$, which is smooth on the whole plane. ThenAt the origin, rotation invariance gives the finite Taylor expansion $f(z)=f(0)+b|z{|}^2+O(|z{|}^4)$, with $b=q(0)$. Alternatively continuity of $q$ directly givesThe metric is positive away from the origin exactly when $u^{\prime \prime }(t)>0$ for all finite $t$, and it is positive at the origin exactly when this limit is positive. Smoothness at the origin is already supplied by the original smooth potential. Consequently the positivity criterion for a radial Kähler potential isClosedness is automatic from the potential, and positivity completes the Kähler condition. No completeness of this metric is asserted. The rotation-invariant Kähler potential on the complex plane is understood as real; if a complex radial potential is initially allowed with real $\omega$, its smooth radial imaginary part is harmonic and hence constant, so that constant can be removed.