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The reason the complex Hessian (actually, it ought to be called the 'Hermitian Hessian', since it defines an Hermitian form at every point, but 'the complex Hessian' is entrenched in the literature) i …

Actually, the correct notion of maps preserving harmonicity is that of 'harmonic morphisms': A map $f:(M,g)\to (N,h)$ between Riemannian manifolds is a harmonic morphism if it pulls back $h$-harmonic …

The answer is 'no', there is no holomorphic curve in $\mathbb{C}^n$ (for any $n$) such that the induced metric has constant negative curvature. To my knowledge, this was first proved by E. Calabi man …

Here is a revised and somewhat expanded version of my answer, with a preparatory 'toy version' to help orient the reader. A simple warmup problem: Before discussing a quantitative variant of the Sch …

Let me point out a somewhat different answer. Rbega explicitly mentions $$ Q = \left(\frac1{z^3} + \frac1{z^2}\right)\ (dz)^2 $$ as an example of the sort of meromorphic quadratic differential that i …

1.If $U$ satisfies LAP then there exists a $V$ such that $(U,V)$ satisfies CR. In fact, $V$ is unique up to the addition of a term of the form $a + bx + cy + d xy$, where $a$, $b$, $c$, and $d$ are c …

It is not a finite dimensional Lie group. For example, all of the maps $$ \Phi\bigl(z,[a,b]\bigr) = \bigl(z, [a+p(z)b,\ b]\bigr), $$ where $p:\mathbb{C}^\ast\to \mathbb{C}$ is holomorphic, belong to …

You can prove by induction that, if $f$ satisfies your equation, then there exist holomorphic functions $f_0,\ldots,f_{n-1}$ on the domain of $f$ such that $$ f = f_0(z) + f_1(z)\ \bar z + \cdots + f_ …

By your last sentence, this is clearly not true in general. Let $X$ be a $K3$ surface and let $\omega$ be a Kähler form on $X$ whose associated metric is Ricci flat. Let $H$ denote the space of th …

Yes, this is true. In fact, a more precise statement holds: Unless $f$ vanishes identically on $\mathbb{D}$, there is an integer $n$ and a nonzero complex number $a$ such that $f(z) = a\,z^n + f_{n+ …

No such conformal map exists. Conformal mapping in dimensions above 2 is very different from conformal mapping in dimension 2. In dimensions above 2, any conformal mapping is a (finite) composition …

A cultural remark to begin: The comments asking for clarification of your question may have sounded a bit rough, but please understand that they weren't meant personally. In the culture of mathematic …

The group $H$ acts transitively and primitively on $\mathbb{C}=\mathbb{R}^2$. ('Primitive' means that $H$ preserves no nontrivial foliation.) It's a consequence of the classification of transitive pr …

I assume that you are assuming $\sigma > 3\delta$, otherwise, the problem doesn't make sense. Can't you just apply the Argument Principle? Fix a $z_0\in S_{\sigma-3\delta}$ and an $R>>0$ so big th …

As Alexandre Eremenko points out, there's no PDE that $n$ would have to satisfy, at least for local solvability, which is believable when you think of it in heuristic terms: There are $3$ unknowns in …