Search Results

The answer is 'no'. In fact, even for a Ricci-flat Kähler manifold $(M^4,J,g)$, the map $W_-(x):\Lambda^2_-(T_xM)\to\Lambda^2_-(T_xM)$ can be any symmetric traceless linear map. Hence, the only cons …

No. $\Omega = A + i B$ is closed, so $A$ and $B$ (the real and imaginary parts) are closed. Moreover, with respect to any Hermitian metric $g$, we will have $*_gA = B$ and $*_g B = -A$, so the real …

Well, I'm at the next airport and have a little time. While I don't have a complete answer worked out about the necessary and sufficient conditions on the curvature, I can definitely say that you don' …

To answer the first question: Only when $G$ is abelian. In fact, if $G$ is not abelian and has dimension $2d$, then there is no element of $H^2(G,\mathbb{R})$ whose $d$-th power is nonzero in $H^{2d …

Here is a partial answer: Let $(M^{2n},g,J)$ be a compact Riemannian manifold endowed with a $g$-orthogonal almost complex structure $J$ with the property that, for every nonzero $v\in T_pM$ there ex …

You asked for references to original papers. In Kobayashi and Nomizu, Vol. 2, pp. 170–171, they give a proof of this result and then write, "This has been proved independently by Hawley [1] and Igusa …

The answer is 'yes, we can'. Since we are in the simply-connected case, by the deRham Theorem, we can assume that $(M_i,g_i)$ for $i=1,2$ are isometric to $$ (\mathbb{C}^m,h_0)\times (N_1,h_1)\times\c …

There is a more enlightening proof of this statement than Berger's calculation and, in fact, it proves something a bit more general. First, a definition: Let $(M,g)$ be a Riemannian $n$-manifold wit …

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) …

The answer to $(1)$ is 'yes'. It is a Lie group. This is true for any compact complex manifold. Basically, this is because the equations for a holomorphic vector field on a compact manifold always …

The answer to the newly edited question is now "no". The reason is that $\Omega^{(\bullet,\bullet)}(M)$ is not an irreducible $\frak{sl}_2$-module, even when you fix $\rho(h) = H$. For example, yo …

Suppose that the metric on $M^n$ has irreducible holonomy, is simply connected (or, slightly more generally, that the restricted holonomy $H^0$ acts irreducibly), and that there exist two independent …

It's not a silly question, but there's a standard answer, and it's a purely local result: If the Kähler metric is $C^2$ and has constant scalar curvature, then it is real-analytic with respect to the …

YangMills' answer shows that it is not always possible to represent a real $(1,1)$-form $\phi$ in the desired form globally on a compact complex manifold but doesn't answer the question of how to tell …

NB: New evidence has changed my conclusions. In the first nontrivial case where this question makes sense, i.e., when $X$ has dimension $16$ and $S\subset X$ has dimension $4$, a preliminary calculat …