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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.

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.) …
Harry Richman's user avatar
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 …
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 …
Misha Verbitsky's user avatar
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 …
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 …
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 …
Misha Verbitsky's user avatar
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 …
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar