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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 …

There are probably too many such $(M,g,I_+,I_-)$ to really expect a 'classification'. For instance, consider the case when a complex manifold $(M,I_+)$ has real dimension $4$, and the $I_+$-holomorphi …

The answer is 'yes' there do exist such functions that are non-constant with singularities only along surfaces $\Sigma\subset\mathbb{C}^2$, and here is how one can understand them: First, it helps to …

The answer is that the only solutions have the form $$ f = (f_1,f_2) = \bigl(c, h(\,\overline{z}_1, z_2)\bigr) $$ where $h:\mathbb{C}^2\to\mathbb{C}$ is holomorphic and $c$ is a constant, which must e …

Note that, if you take $\phi=0$, then the equation reduces to $w_y =0$, i.e., if $D\subset C$ is the domain of $w$ and $x:D\to\mathbb{R}$ is the projection on the $x$-axis and has connected fibers, t …

There is a trivial construction that shows that the answer is 'yes' for all complex manifolds, not just those that admit an anti-holomorphic involution. Let $(M,J)$ be a (finite-dimensional) complex $ …

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 …

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+ …

I think that there is a smooth (or analytic) result of the kind that you are seeking: Let $M^m$ be a connected smooth (or analytic) manifold, and let $P:M\times\mathbb{R}\to\mathbb{R}$ be a smooth …

Edit: (21 May 2017) I have modified my answer to cover the case that the OP meant to ask, i.e., the assumption is that the closure of an orbit has nonempty interior. Now that you have added the assu …

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 …

It's easy to derive a third-order (nonlinear) differential equation for $u(x,y)$ that satisfies your conditions (1) and (2): Namely, set $\theta(x,y) = \arctan\bigl(u_y(x,y)/u_x(x,y)\bigr)$ and then …

The answer to (1) in the space case is a definite 'no'. The actual image 'surface' $h(\Delta)$ in $\mathbb{R}^3$ can actually change when you reparametrize with $\phi:S^1\to S^1$, in which case, ther …

I think that the answer to your question(s), when everything is sorted out, is 'no'. What is actually going on in the complex case that you are concerned with is that there are two Hermitian forms in …

The answer is basically 'no', there is no 'elementary method' involving elementary operations and quadrature (i.e., finding antiderivatives of known holomorphic functions) that will give you a solutio …