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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.

30 votes
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When is a closed differential form harmonic relative to some metric?

A closed $k$-form is called intrinsically harmonic if there is some Riemannian metric with respect to which it is harmonic. E. Calabi (Calabi, Eugenio, An intrinsic characterization of harmonic one-fo …
Dan Fox's user avatar
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22 votes

Why is there no symplectic version of spectral geometry?

From a certain point of view the premise of the question is wrong. The study of sympletic manifolds with no additional structure is akin to differential topology rather than differential geometry. Fro …
Dan Fox's user avatar
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12 votes
Accepted

"Famous" 2d Riemannian manifolds with non-constant curvature

A Killing field is preserved by the Ricci flow. By a theorem of Daskalopoulos, Hamilton and Sesum (arXiv:0902.1158), on a compact surface an ancient (defined for all negative time) solution to the Ric …
Dan Fox's user avatar
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10 votes

"Noncommutative heat equation" -- a strange generalization of Killing vectors for a flat metric

Here is an observation which does not answer the question, but does at least tell where not to look for examples. It is shown that for compact Riemannian manifolds with non-positive Ricci curvature th …
Dan Fox's user avatar
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9 votes
Accepted

Why symmetric spaces?

The algebra of invariant differential operators on a symmetric space is commutative, and this is certainly not true for an arbitrary homogeneous space. While it is not true that the commutativity of t …
Dan Fox's user avatar
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6 votes
Accepted

Upper bound on the sectional curvature of the orthogonal group

The $n = 3$ case is a straightforward computation using the identification of the cross product on $\mathbb{R}^{3}$ with the Lie bracket on $\mathfrak{so}(3)$. Namely, defined $\hat{x}:\mathbb{R}^{3} …
Dan Fox's user avatar
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4 votes

Covariant derivative of determinant of the metric tensor

The determinant of a metric makes perfectly good sense, but it is not a function, rather a $2$-density. Formally, this means that it transforms as a section of the bundle associated with the frame bun …
Dan Fox's user avatar
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