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If one interprets the OP's question literally, the answer is 'yes', but I imagine that the OP didn't literally mean what the OP wrote. First, interpret everything literally: Assume that $(M,g,\omega) …

If the Hermitian metric is not Kähler, the Laplacian won't respect bi-degree. In fact, a more general result is true: If an almost Hermitian metric is not Kähler, then its Laplacian will not respect …

In general, even if the fibers of $\phi$ are connected and the map is proper, there need not be an almost complex structure on $B^{2k}$ such that the differential of $\phi$ is complex linear. (I assu …

The answer to the question as asked is 'No', at least when $2n=4$. In that case, the conditions (1) and (2) only involve the Weyl curvature of the underlying metric $g$, so, in particular, when $g$ i …

The answer is "Yes, there are many solutions with $u$ and $v$ non-constant." Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex sy …

Yes, this can happen. For a simple example, consider $M = \mathrm{SL}(2,\mathbb{C})/\Lambda$ where $\Lambda\subset \mathrm{SL}(2,\mathbb{C})$ is a discrete, co-compact lattice. Then $M$ is a compact …

It is an old theorem of Samelson that any compact Lie group $G$ of even rank has an integrable complex structure, which, in particular applies to the case of $\mathrm{SU}(3)$. Basically, one chooses …

I think that this follows from basic properties of the moduli space $D(Z)$, since one knows that, for each $x\in M$, the normal bundle $\nu_x$ in $Z$ of each fiber $q(x) = p^{-1}(x)\simeq \mathbb{CP}^ …

Update: I have had a little time this weekend to think more deeply about the construction I outlined below and I have concluded that, unfortunately, the construction that I thought would work to produ …

First, the definition: A Riemannian $n$-manifold $(M^n,g)$ is of Hessian type if there exist $(n{+}1)$ functions $x^1,\ldots,x^n, u$ on $M$ such that $dx^1\wedge\cdots\wedge dx^n\not=0$ and such that …

Yes, it is well-known that a $6$-manifold has an $\mathrm{SU}(3)$-structure if and only if it is orientable and spinnable (i.e., it has a spin structure). The necessity of these two conditions is c …

I assume that you mean 'positive definite symmetric matrices'. Here's one proof, though it's certainly not the most clean. A linear automorphism of the space of all symmetric matrices that preserves …