In 1971 Phillip.A.Griffith proved that

Let $B_n^\ast=\{z\in\mathbb{C}^n:0<\|z\|\le 1\}$ be the punctured ball in $\mathbb{C}^n$,and $f:B_n^\ast\rightarrow M$ a holomorphic mapping into a compact K\"ahler manifold $M$.Then $f$ extends to a meromorphic mapping from the ball $B_n=\{z:\|z\|\le 1\}$ into $M$,provided $n\ge 3$.

We recall the definition of a meromorphic mapping.Let $M$ and $N$ be connected complex manifolds of complex dimension $m$ and $n$ respectively and where $M$ is assumed to be compact.A holomorphic mapping \begin{equation*} f:N\rightarrow M \end{equation*} which is defined on the complement of a proper subvariety $S\subset N$ will be said to be meromorphic if the closure $\overline{\Gamma}_f$ in $N\times M$ of the graph $\Gamma_f\subset (N-S)\times M$ is an analytic subvariety of $N\times M$.

Griffith also examined the example of the Hopf manifold $M$ (which is not K\"ahler) and showed that $f:B_n^\ast\rightarrow M$ CANNOT extend meromorphically across the origin.

Note that there is a gap between K\"ahler manifolds and the Hopf manifold.Say,balanced manifolds (a balanced manifold $M^m$ is a Hermitian manifold with the K\"ahler form $\omega$ satisfying $d\omega^{m-1}=0$) are not good enough to be K\"ahler but are not as bad as the Hopf manifold.So I want to consider the problem whether $f:B_n^\ast\rightarrow M$ extends meromorphically across the origin when $M$ is a compact balanced manifold.To begin with,I want to check an example when $M$ is an Iwasawa manifold,which is a compact balanced but non-K\"ahler manifold.


Let $B_n^\ast=\{z\in\mathbb{C}^n:0<\|z\|\le 1\}$ be the punctured ball in $\mathbb{C}^n$,and $f:B_n^\ast\rightarrow M$ a holomorphic mapping into the Iwasawa manifold $M$.Then does $f$ extend to a meromorphic mapping from the ball $B_n=\{z:\|z\|\le 1\}$ into $M$?

Even for this concrete example,I don't know whether the above statement is true.Can you help me to support or reject my viewpoint?Thanks in advance!

  • $\begingroup$ For $n\geq 2$, the punctured ball is simply connected. Thus you can lift the holomorphic map to the universal cover of the Iwasawa manifold. Isn't this Stein? If so, then you can apply Hartog. $\endgroup$ Jul 31, 2015 at 16:56
  • 1
    $\begingroup$ *Hartog --> Hartogs. Sorry for the misspelling. $\endgroup$ Jul 31, 2015 at 23:19

1 Answer 1


First of all, if you are speaking of Phillip Augustus Griffiths, his surname has an "s" at the end. For your specific question about the Iwasawa manifold: the Iwasawa manifold $M$ is the quotient of the complex Heisenberg group $\mathbf{H}_{3,\mathbb{C}}$ by a cocompact, discrete subgroup $\Gamma$. In particular, the quotient holomorphic map, $$q:\mathbf{H}_{3,\mathbb{C}} \to M,$$ is the universal covering.

Let $n\geq 1$ be an integer. The punctured ball $B_n^*$ deformation retracts onto the sphere $\mathbf{S}^{2n-1}$. If $n\geq 3$, $B_n^*$ is simply connected. Thus every continuous map $$f: B_n^* \to M,$$ factors through a continuous map to the universal cover, $$\widetilde{f}:B_n^* \to \mathbf{H}_{3,\mathbb{C}}.$$ Since $q$ is a local biholomorphism, the continuous map $f$ is holomorphic if and only if the continuous map $\widetilde{f}$ is holomorphic. Finally, as a complex manifold $\mathbf{H}_{3,\mathbb{C}}$ is biholomorphic to $\mathbb{C}^3$. Therefore, by the extension theorem of Hartogs (sorry for misspelling his name above!), the holomorphic map $\widetilde{f}$ is the restriction to $B_n^*\subset B_n$ of a holomorphic map $$\widetilde{g}:B_n \to \mathbf{H}_{3,\mathbb{C}}.$$ Defining $g = q\circ f$, then $$g:B_n \to M,$$ is a holomorphic extension of $f$.


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