40 votes

What is the best way to draw curvature?

The best way I know to illustrate the notion of curvature is via Toponogov's theorem. We can compare any (geodesic) triangle in a Riemannian manifold $M$ with one with the same edge lengths in ...
Mohammad Ghomi's user avatar
32 votes
Accepted

Is the minimal volume a topological invariant?

Minimal volume is not a homeomorphism invariant. It is shown in [L. Bessières, Un théorème de rigidité différentielle, Comm. Math. Helv. 73 443-479 (1998)] that the minimal volume of the connected sum ...
Igor Belegradek's user avatar
19 votes
Accepted

Techniques to solve a non-linear differential equation related to curvature

Well the standard techniques would take advantage of the fact that the equation doesn't explicitly involve the independent variable $x$ to integrate the equation once, thereby leading to the ...
Robert Bryant's user avatar
18 votes
Accepted

Geometric interpretation of the Weyl tensor?

There is such an interpretation, with a few caveats. Essentially, there is a canonical connection on a certain vector bundle for which the "principal part" of the curvature is the Weyl ...
Jeffrey Case's user avatar
  • 1,483
17 votes
Accepted

Constant Gaussian curvature surfaces in 3-space containing lines

Given any point $p$ on a surface $S$ of Gauss curvature -1, there exists an open neighborhood $U\subset S$ and $p$-centered coordinates $(x,y):U\to\mathbb{R}^2$, whose image is a domain $R = (x,y)(U)\...
Robert Bryant's user avatar
17 votes

What is the best way to draw curvature?

With advances in discrete differential geometry, it is now nearly routine to compute curvature on meshed surfaces. Here are two of many possible color-coded examples.       ...
Joseph O'Rourke's user avatar
16 votes
Accepted

Are there some intrinsic invariants of surfaces other than Gaussian curvature?

As others have pointed out, it's not hard to show that any function $F(\kappa_1,\kappa_2)$ that is intrinsic to the surface metric must be a function of $K = \kappa_1\kappa_2$, so that settles what ...
Robert Bryant's user avatar
16 votes

What is the best way to draw curvature?

This is not what you're looking for, but I always remember Milnor's diagram in Chapter 9 of his book on Morse Theory describing the symmetries of the curvature tensor.
15 votes
Accepted

Locally Riemannian Connection

Perhaps I can offer some information and comment on this problem. An essential part of the problem is how to interpret terms such as 'observe', 'accessible', 'identify', as the OP wants to know how ...
Robert Bryant's user avatar
15 votes

Does every ‘curvature’ tensor induce a metric?

Here's a quick summary. The answers provided in the link cited by @RBega2 have more details. Given a curvature-like tensor $R$ at a point, there always exists a metric whose curvature tensor at that ...
Deane Yang's user avatar
14 votes
Accepted

Taylor expansion of the metric tensor in the normal coordinates

Using the reference https://arxiv.org/pdf/0903.2087.pdf, which agrees with https://arxiv.org/pdf/hep-th/0001078v1.pdf, which agrees with the reference U. Müller, C. Schubert and Anton M. E. van de Ven,...
Robert Bryant's user avatar
14 votes
Accepted

Gauss-Bonnet Theorem: Neither Gauss nor Bonnet

Shiing-Shen Chern's Historical remarks on Gauss-Bonnet seems an authoritative source. The formula goes back to Gauss (1827), Bonnet (1848), and Binet (unpublished). Gauss considered a triangle, Bonnet ...
Carlo Beenakker's user avatar
13 votes

Riemannian vs Non-Riemannian curvature

NB: In what follows, to save typing, I will be working on a manifold $M$, but I will write $T$, $T^*$, etc. to denote the bundles $TM$, $T^*M$, etc. and let $M$ be understood. It seems that the OP ...
Robert Bryant's user avatar
13 votes

What is the best way to draw curvature?

Mohammed Ghomi's answer reminds me of a related picture that Cedric Villani drew to depict Ricci curvature ([1] Chapter 14). Similar to the $\operatorname{CAT}(\kappa)$ inequality, this idea can be ...
12 votes
Accepted

Principal curvatures of $\mathbb{R}^{n^2}$-embedded SO(n)

What you are asking about is the second fundamental form of the embedding. Since this is a Lie group, it's enough to know what the second fundamental form is at the identity matrix $I_n=e$. Since ...
Robert Bryant's user avatar
11 votes

Differential geometric interpretation of cohomology

For geometric interpretations of cup product and Poincaré duality let us assume that in the following (dual) homology classes are representable by smooth submanifolds (and everything is orientable). ...
Daniel Valenzuela's user avatar
11 votes

$S^3 \setminus S^1$ doesn't have hyperbolic structure

You probably mean $M$ does not admit complete hyperbolic metrics of finite volume. Since $M$ is topologically the interior of a solid torus, a complete hyperbolic structure just identifies $M$ as the ...
Xin Nie's user avatar
  • 1,764
11 votes

Curvature of nonsymmetric metric tensors?

Consider a bilinear form $b \in \mathcal{C}^\infty (T^*M\otimes T^*M, \mathbb{R})$ and an affine connection $\nabla \colon \mathcal{\Gamma}^\infty(TM) \to \mathcal{\Gamma}^\infty(T^*M\otimes TM)$ ...
Vít Tuček's user avatar
  • 8,157
11 votes

Taylor expansion of the square of the distance function on a Riemannian manifold

This is just a remark about an alternative, more `low tech', derivation of this famous formula by using Taylor series. It relies on this property of a Riemannian metric: If $(M,g)$ is a Riemannian ...
Robert Bryant's user avatar
10 votes
Accepted

Who first proved that a vanishing Riemann tensor is sufficient for the existence of Euclidean coordinates?

Presumably it was Riemann himself who proved the sufficiency of $R=0$ for the flatness. Spivak's Chapter 4 in Volume II of "A comprehensive introduction to differential geometry" is a good source on ...
Ivan Izmestiev's user avatar
10 votes
Accepted

What is known about Lie groups with (strictly) positive curvature?

The following result is, for example, exercise 3 on pg. 104 of Do Carmo's Riemannian Geometry book. Suppose $X$ is a Killing field on a compact even dimensional Riemannian manifold of positive ...
Jason DeVito - on hiatus's user avatar
10 votes

$S^3 \setminus S^1$ doesn't have hyperbolic structure

The following contribution comes from conversations with Bill Goldman, any mistakes however are mine alone. Any (geodesically complete) geometric 3-manifold $N=\mathbb{M}/G$ with infinite order ...
Sean Lawton's user avatar
  • 8,394
10 votes
Accepted

Is every metric uniformly close to a metric with negative scalar curvature?

Jochen Lohkamp answers your first question in "Curvature h-principles", Ann. of Math, vol 142, p 457–498. If $\dim M\ge 3$, then the space of negative scalar curvature metrics is dense in the $C^0$ ...
Sebastian Goette's user avatar
9 votes
Accepted

minimal surfaces in $S^n$

Without embeddedness, the Choi--Schoen theorem is false. For example, there is a huge family of rotationally symmetric immersed tori in $\mathbb{S}^3$ (the only embedded one is the Clifford torus, ...
Otis Chodosh's user avatar
  • 7,087
9 votes
Accepted

Positive scalar curvature on the double of a manifold

The answer is negative at least if you do not add some sort of convexity hypothesis for the boundary, and at least in dimension $2$. Take a round $2$-sphere with $h\ge 2$ round holes. It has positive ...
Benoît Kloeckner's user avatar
9 votes
Accepted

What should a meaningful notion of curvature satisfy, in the absence of a smooth structure?

A curvature bound for a non-smooth metric space is often known as a "synthetic curvature bound" or a "coarse curvature." Here's one possible definition for the concept. Definition:...
Gabe K's user avatar
  • 5,384
8 votes

Do curvature differences obstruct a.e orientation-preserving isometries?

There is a discontinuous map $f\colon\mathbb{S}^2\to\mathbb{R}^2$ such that $d_xf$ is defined and isometric for almost all $x$. (If you want a continuous one then I am sure the answer is "no") To ...
Anton Petrunin's user avatar
8 votes
Accepted

Holonomy of a Ricci-flat affine connection

The answer depends on the dimension. When $n=2$, Ricci-flatness of a connection implies that it is flat, so, in that case, yes, you get holonomy reduction locally. However, when $n>2$, Ricci-...
Robert Bryant's user avatar
8 votes
Accepted

Realizing the cross product of $\mathbb{R}^3$ as the curvature tensor of a Riemannian metric on $\mathbb{R}^3$

There are two interpretations of your question. Metric cross product Assuming that you are looking for a Riemannian metric $g$ such that (by the triple-product formula) $$ R_g(X,Y)Z = g(Z,X)Y - g(...
Willie Wong's user avatar
  • 37.7k
8 votes
Accepted

Positive scalar curvature on the total space of a circle bundle

It is a theorem of Gromov and Lawson, also Schoen and Yau, that no closed orientable three-manifold which contains an aspherical factor in its prime decomposition can admit a metric of positive scalar ...
Michael Albanese's user avatar

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