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Results tagged with riemannian-geometry
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user 3377
Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.
11
votes
1
answer
599
views
Flat manifolds and irreducible representations
Let $M$ be a compact Riemannian manifold with vanishing curvature of Levi-Civita connection. Such manifolds were classified by Bieberbach; sometimes they are called Bieberbach manifolds. According to …
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10
votes
Accepted
Infinite dimensional Riemannian geometry
Lempert, László The Dolbeault complex in infinite dimensions. III. Sheaf cohomology in Banach spaces. Invent. Math. 142 (2000), no. 3, 579-603.
Lempert, László The Dolbeault complex in infinite dimen …
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9
votes
2
answers
724
views
Bieberbach theorem for compact, flat Riemannian orbifolds
In his thesis, Bieberbach solved Hilbert 18 problem and
proved that any compact, flat Riemannian manifold is a
quotient of a torus. I need a reference to an orbifold version
of this result: any comp …
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9
votes
Roadmap to learning about Ricci Flow?
Here is a list of literature which I compiled when I taught the course
on Ricci flow.
Basic differential geometry:
Einstein Manifolds (Besse).
Riemannian geometry (Gallot S., Hulin D., Lafontaine J.) …
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8
votes
1
answer
332
views
reference to a theorem about a product of harmonic and parallel forms
Let $\alpha$ be an exterior product of a harmonic and a parallel form on a Riemannian manifold. Then $\alpha$ is known to be harmonic. I have heard that this is an old result due to R. Bott, but I cou …
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7
votes
Is the group of isometries of a homogeneous Riemannian manifold maximal?
Let $G$ be a compact Lie group with left-invariant metric $h$. For general $h$, the group of isometries is $G$. However, when $h$ is bi-invariant, the group of isometries is $G\times G$. I think this …
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6
votes
The Chern connection on a Hermitian symmetric domain
If I understood your question correctly, you define a holomorphic
connection as one which satisfies $\bar\partial (\nabla_x y)=0$
for any holomorphic vector fields $x, y$. Then the holomorphicity
is …
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6
votes
1
answer
441
views
Holonomy bounded in terms of area and the curvature
I suppose the following result follows
from Ambrose-Singer theorem, but I cannot
find a reference, and the arguments I found
in the literature are usually weaker. The idea
is that holonomy over a null …
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4
votes
Accepted
Flat manifolds and irreducible representations
Here is the solution (with thanks to Misha Kapovich
who pointed this out).
Gerhard Hiss, Andrzej Szczepanski,
"On torsion free crystallographic groups",
Journal of Pure and Applied Algebra
Volume 7 …
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4
votes
Non simply connected HyperKähler 4-manifolds without ALE metrics
"Peter Kronheimer showed that each simply connected HyperKähler 4-manifold possesses an ALE metric."
This is false as stated. K3 surface, or the Taub-NUT space, are simply connected 4-manifolds which …
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3
votes
Accepted
Nonpositive curvature of Stein manifolds
Complete, simply connected manifolds of non-positive sectional curvature are diffeomorphic to $R^n$, by Cartan-Hadamard theorem. Conversely, if it's $R^n$, you can put the metric of negative curvature …
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2
votes
Is a linear vector field a geodesible vector field?
This observation seems to be very easy,
but it takes care of many examples.
Suppose that $A$ corresponds to a contraction
(that is, all eigenvalues are $< 1$ in absolute value).
Decompose ${\Bbb R}^n …
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1
vote
The trace of a wedge product of matrices
This is valid for any vector bundle $B$.
One considers $R$ as a 2-form with coefficients in endomorphisms of a bundle $B$.
Then $R\wedge R$ is a 4-form with coefficients in endomorphisms, and the trac …
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