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

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.