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Results tagged with complex-geometry
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user 13972
Complex geometry is the study of complex manifolds, complex algebraic varieties, complex analytic spaces, and, by extension, of almost complex structures. It is a part of differential geometry, algebraic geometry and analytic geometry.
6
votes
Accepted
Nontrivial extension of the action of complex hyperbolic group $H$ on $\mathbb{C}$
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 …
- 108k
9
votes
Is the Gödel universe Wick rotatable?
I may be misreading the sources that you list for the definition of Wick-rotatable, but, I believe that the following construction does fit that definition: According to the Wikipedia page that the O …
- 108k
12
votes
Accepted
Determine whether a (1,2) tensor is Nijenhuis tensor
Yes, there are pointwise algebraic conditions on a section $N$ of $T\otimes\Lambda^2(T^*)$ in order for $N$ to equal $N_J$ for some almost complex structure $J$, but there are differential conditions …
- 108k
13
votes
Kähler metric with two compatible complex structures
No, you cannot prove this because it is not true. For example, consider $M$ to be the product of two oriented, complete Riemannian surfaces $M=\Sigma_1\times\Sigma_2$ where $g$ is the product metric …
- 108k
6
votes
Accepted
Does $F_{A}^{0,2}=0$ for a connection $A$ on $TM$ almost complex give a complex structure?
Here is an example to think about: Let $S^6 = \mathrm{G}_2/\mathrm{SU}(3)$ be the $6$-sphere endowed with its $\mathrm{G}_2$-invariant almost Hermitian structure. There is a $\mathrm{G}_2$-invariant …
- 108k
9
votes
Accepted
Are holomorphic Lagrangians locally graphs?
The answer is 'yes'. Specifically, the holomorphic version of the Darboux-Weinstein theorem holds, just as it does in the smooth category. In particular, if $L\subset M$ is a holomorphic Lagrangian …
- 108k
15
votes
Accepted
Algebraic atlas on smooth manifolds
The answers to your question in the case $n=1$ are well-known. In higher dimensions, the answers are less complete, but something is known.
For example, in the real case when $n=1$, there is only one …
- 108k
4
votes
Example of usual Laplacian does not respect bidegree for general hermitian manifolds
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 …
- 108k
8
votes
Are the quaternionic Grassmannians quaternionic Kaehler manifolds?
Perhaps the OP really wants to know why quaternionic Grassmannians other than the quaternionic projective spaces are not considered to be 'quaternion-Kähler'.
The reason goes back to Berger's classifi …
- 108k
6
votes
Accepted
What is the definition of a Calabi-Yau metric on a non-compact manifold?
There are two slightly different definitions. The first is that it is a Kähler metric that is Ricci-flat, and the second is that it is a Kähler metric on a (usually connected) complex $n$-manifold wi …
- 108k
4
votes
Accepted
almost complex $\mathbb{Z}^{6}$-action
Here is a construction of a family of such examples that will work, but you will have to choose a particular map to get an explicit example.
Let's use coordinates $v_1,v_2,v_3,v_4,x,y$ (each periodic …
- 108k
8
votes
Accepted
Non-calibrated area-minimising surface
Actually, a better example along the lines Otis suggests would be the geodesic $\mathbb{RP}^1\subset\mathbb{RP}^2$. Of course, $\mathbb{RP}^1$ is orientable and it is homologically mass-minimizing, …
- 108k
14
votes
Unique almost complex structure up to diffeomorphism
Dmitri's answer is fine, but there is a different argument that is purely local that is worth bearing in mind as well:
On a $2n$-manifold $M$, the set of almost complex structures on $M$ are the secti …
- 108k
3
votes
Complex manifold defined over $\mathbb{Q}$
Note that this is a correct answer to the original question, so I will leave it here, even though the question has now been changed. (The original question is recoverable by going back to the previous …
- 108k
3
votes
Complex manifold defined over $\mathbb{R}$
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 $ …
- 108k