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 ...
- 7,149
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 ...
- 29.1k
23
votes
Accepted
Progress on Gromov's Conjecture of the bound of total Betti numbers
The $2^n$ bound is true for all rationally elliptic spaces, which (I think) includes all known closed simply-connected nonnegatively curved manifolds. Simply-connected symmetric (or more generally, ...
- 29.1k
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 ...
- 1,713
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)\...
- 108k
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.
...
- 151k
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.
16
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 ...
- 27.5k
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 ...
- 108k
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,...
- 108k
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 ...
- 188k
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 ...
- 108k
13
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 ...
- 108k
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 ...
11
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 ...
- 6,546
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 ...
- 1,804
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 $\nabla \colon \mathcal{\Gamma}^\infty(TM) \to \mathcal{\Gamma}^\infty(T^*M\otimes TM)$ ...
- 8,597
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 ...
- 108k
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 ...
- 6,337
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 ...
- 8,529
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 $C^0$ ...
- 6,813
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 ...
- 14.4k
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:...
- 6,001
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 ...
- 45k
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(...
- 39k
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 ...
- 19.3k
8
votes
Sectional curvature of the manifold of symmetric positive definite matrices
Dolcetti and Pertici give an explicit expression for the Riemannian curvature tensor on the symmetric space $\mathcal{P}_n$ of $n \times n$ positive definite matrices with affine-invariant metric. ...
- 637
8
votes
Fáry-Milnor theorem for positively curved metrics on $S^3$?
In fact, a knot in a positively curved metric can have zero curvature.
Consider a 2-bridge knot, then one can make it into a $\pi$-orbifold (or bifold in the terminology of Kronheimer-Mrowka) so that ...
- 68.8k
8
votes
Accepted
Is it known whether a closed simply-connected manifold of non-negative curvature admits positive Ricci?
No there are no such examples known. Most known examples of manifolds of nonnegative sectional curvature come from biquotients or cohomogeneity one manifolds. If these are simply connected they are ...
- 7,842
8
votes
Accepted
Compact complex non-Kähler manifolds with nef canonical bundle
Let $X$ and $Y$ be compact complex manifolds. Note that $K_{X\times Y} \cong \pi_1^*K_X\otimes \pi_2^*K_Y$. If $Y$ has trivial canonical bundle, then $K_{X\times Y} \cong \pi_1^*K_X$. Now the pullback ...
- 19.3k
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