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46 votes
2 answers
11k views

Does the curvature determine the metric?

I ask myself, whether the curvature determines the metric. Concretely: Given a compact manifold $M$, are there two metrics $g_1$ and $g_2$, which are not everywhere flat, such that they are not ...
Bernhard Boehmler's user avatar
36 votes
10 answers
8k views

Some questions about scalar curvature

Recall that the scalar curvature of a Riemannian manifold is given by the trace of the Ricci curvature tensor. I will now summarize everything that I know about scalar curvature in three sentences: ...
Paul Siegel's user avatar
  • 29.2k
26 votes
5 answers
7k views

Intuition for mean curvature

I would like to get some intuitive feeling for the mean curvature. The mean curvature of a hypersurface in a Riemannian manifold by definition is the trace of the second fundamental form. Is there ...
nicolas's user avatar
  • 693
20 votes
3 answers
9k views

Curvature of a Lie group

Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold [Edit: provided we fix a metric, for example an invariant or bi-invariant one] has some scalar ...
Matt's user avatar
  • 251
18 votes
1 answer
1k views

Is the minimal volume a topological invariant?

On Wikipedia, it is said that the minimal volume $$\operatorname{MinVol}(M):=\inf\{\operatorname{vol}(M,g) :g\text{ a complete Riemannian metric with }|K_{g}|\leq 1\}$$ is a topological invariant, ...
Cosine's user avatar
  • 609
18 votes
1 answer
1k views

Geometric interpretation of the Weyl tensor?

The Riemann curvature tensor ${R^a}_{bcd}$ has a direct geometric interpretation in terms of parallel transport around infinitesimal loops. Question: Is there a similarly direct geometric ...
Tim Campion's user avatar
  • 63.9k
17 votes
2 answers
5k views

Square of the distance function on a Riemannian manifold

Let $(M^n,g)$ be a smooth Riemannian manifold. Consider the square of the distance function $$dist^2\colon M\times M\to \mathbb{R}$$ given by $(x,y)\mapsto dist^2(x,y)$. It is easy to see that this ...
asv's user avatar
  • 21.8k
17 votes
2 answers
1k views

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

The principal curvatures of a surface is denoted by $\kappa_{1}, \kappa_{2}$. Let $P(x,y)$ be a polynomial with real coefficients. Assume that $P(\kappa_{1}, \kappa_{2})$ is an intrinsically ...
Ali Taghavi's user avatar
14 votes
4 answers
6k views

When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?

I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for ...
Theo Johnson-Freyd's user avatar
14 votes
1 answer
1k views

Progress on Gromov's Conjecture of the bound of total Betti numbers

This question is a reference request. Let $(M,g)$ be a Riemannian manifold of dimension $n$, and $b_i(M) = \dim H_i(M,\mathbb{R})$. Gromov proved it that there are constants $C(n)$ such that, if the ...
fffmatch's user avatar
  • 175
13 votes
2 answers
1k views

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

If we consider $G$ a compact Lie group, there is a left invariant Riemannian metric whose the sectional curvature is nonnegative (see Milnors' paper). When can we find a left invariant metric that has ...
melomm's user avatar
  • 245
12 votes
1 answer
454 views

Riemannian vs Non-Riemannian curvature

If you neither know the metric nor the holonomy group, how do you recognize a curvature tensor is Riemannian? I assume a curvature, by definition, satisfies Bianchi identities. I know it is ...
Aureliano Skirzewski's user avatar
12 votes
0 answers
262 views

Jacobi fields on non-geodesic curves

The point of Jacobi fields is to study variations of geodesics through geodesics, but the Jacobi equation $D_t^2 J + R(J,\dot\gamma)\dot\gamma=0$ makes sense for any curve $\gamma$, not just for ...
Ethan Dlugie's user avatar
  • 1,277
10 votes
1 answer
3k views

Taylor expansion of the metric tensor in the normal coordinates

I am looking for a reference with a Taylor expansion of the metric tensor in the normal coordinates. The coefficients should be written in terms of $\mathrm{Rm}, \nabla\mathrm{Rm}, \nabla^2\mathrm{Rm},...
Anton Petrunin's user avatar
9 votes
1 answer
344 views

Do geodesics avoid regions where the curvature diverges?

Let $(M^2,g)$ be a Riemannian manifold, with manifold boundary $\partial M$. We assume that the metric degenerates at the boundary, in the sense that the (Gauss) curvature diverges like $K \to +\infty$...
Leo Moos's user avatar
  • 5,038
9 votes
0 answers
283 views

Hermitian sectional curvature

Let $N$ be a Riemannian manifold, denote $R$ its purely covariant Riemann curvature tensor with sign convention so that the sectional curvature is $K(X,Y) = R(X,Y,X,Y)$ for an orthonormal pair. ...
seub's user avatar
  • 1,347
7 votes
5 answers
4k views

Can anyone give an example of Ricci flat Riemannian or Lorentzian Manifold that is not flat?

Does there exist a Ricci flat Riemannian or Lorentzian manifold which is geodesic complete but not flat? And is there any theorm about Ricci-flat but not flat? I am especially interset in the case ...
346699's user avatar
  • 977
7 votes
2 answers
396 views

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

Let $M$ be a smooth manifold with non-empty boundary. Let $g$ be a smooth Riemannian metric on $M$. Is the following true? For every $\epsilon >0$ there exist a Riemannian metric $g_{\epsilon}$ ...
Asaf Shachar's user avatar
  • 6,741
7 votes
1 answer
558 views

minimal surfaces in $S^n$

Thanks to Choi-Schoen theorem, we know that the space of embedded minimal surfaces into $S^3$ of fixed genus is compact. My question are simples: Can we remove the embeddness assumption? Can we ...
Paul's user avatar
  • 914
7 votes
1 answer
197 views

Positively curved manifold with collapsing unit balls

Can we find a complete connected noncompact Riemannian manifold $(M^n,g)$ such that the curvature operator $Rm>0$ and $$ \inf_{p \in M} \text{Vol}_gB(p,1)=0? $$
Totoro's user avatar
  • 2,535
7 votes
1 answer
2k views

Reference for parallel transport around loop and its relation to curvature

It is a well known fact that the geometric meaning of a linear connection's curvature can be realized as the measure of a change in a fiber element as it is parallel transported along a closed loop. ...
Bence Racskó's user avatar
7 votes
1 answer
1k views

About Sectional Curvature [closed]

In a paper by Yann Ollivier: Let $x$ be a point in $X$, $v$ a small tangent vector at $x$, $y$ the endpoint of $v$, $w_x$ a small tangent vector at $x$, and $w_y$ the parallel transport of $w_x$ from ...
Sepideh Bakhoda's user avatar
7 votes
0 answers
115 views

The space of positive scalar curvature metrics on $S^4$

Let $\mathcal{R}_{+\mathrm{sc}}(S^n)$ denote the space of complete Riemannian metrics of positive scalar curvature on the sphere $S^n$. It's known that $\mathcal{R}_{+\mathrm{sc}}(S^2)$ is ...
Tyrone's user avatar
  • 5,596
7 votes
0 answers
1k views

Conventions for Riemann curvature tensor

I am aware of two conventions for the Riemann curvature tensor, namely the expression $$\langle\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,W\rangle$$ is either declared to be $R(X,Y,Z,W)$ or $...
John Pardon's user avatar
  • 18.7k
6 votes
2 answers
753 views

Curvature of nonsymmetric metric tensors?

Consider a smooth manifold $M$ of arbitrary dimension. We have notions of psuedo-Riemannian or Riemannian metrics on a manifold, and they differ in the slightest way of being positive-definite or not. ...
Plank's user avatar
  • 327
6 votes
1 answer
508 views

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

It's well known that the sectional curvatures of a Lie group, endowed with a left-invariant metric have a nice closed-form formula $k(X,Y) = \frac{1}{4} \|[X Y]\|^2$. I'm wondering if the following (...
Andy Mack's user avatar
  • 265
6 votes
1 answer
463 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-...
Misha Verbitsky's user avatar
6 votes
1 answer
378 views

Unbounded sectional curvature implies infinite diameter?

Let $(M,g)$ be a Riemannian manifold such that for each $C>0$ there is $p\in M$ and $X,Y\in T_pM$ unitary such that $K(X,Y) > C.$ Does this imply that the diameter of $(M,g)$ is infinite? I ...
L.F. Cavenaghi's user avatar
6 votes
1 answer
559 views

Example of a manifold with positive isotropic curvature but possibly negative Ricci curvature

Is there any example of a manifold with a positive isotropic curvature but it possibly obtains a negative Ricci curvature at some point and the direction? If we see the definition of the positive ...
Jae Ho Cho's user avatar
6 votes
3 answers
368 views

Curvature function as a random variable with uniform distribution

Let $(S,g)$ be a Riemannian surface. Then the curvature function $\kappa: S\to \mathbb{R}$ can be counted as a random variable. So it produces a probability density function $f_g:\mathbb{R}\to \...
Ali Taghavi's user avatar
6 votes
0 answers
201 views

Formula for difference between curvature operators?

This is a re-editing of a prerviously posted question: Let $(M,g)$ be a Riemannian manifold. Let $C:TM\to TM$ be symmetric positive definite. Define the metric $$ (X,Y)_C = (X,CY)_g. $$ Denote by $\...
Raz Kupferman's user avatar
6 votes
0 answers
270 views

On the curvature tensor with certain conditions

Let $(M^{n+m},g)$ be a Riemannian manifold and let $\lbrace X_1,...,X_n,Y_1,...,Y_m\rbrace $ be a locally orthonormal frame for $M$($3\leq n,m$). If we suppose the curvature tensor $R$ of $g$ ...
Amir Baghban's user avatar
6 votes
0 answers
269 views

Negative curvature in the middle of $R^{3}$

What's a simple example of metric on $R^{3}$ which has negative scalar curvature inside of a (limited) set N, and is equal to the standard (Euclidean) metric outside? Basically, I am asking for a ...
delenda's user avatar
  • 61
5 votes
2 answers
340 views

Example of a curvature with no associated metric

Is there a concrete example of a $4$ tensor $R_{ijkl}$ with the same symmetries as the Riemannian curvature tensor, i.e. \begin{gather*} R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} =...
Matthew Lou's user avatar
5 votes
1 answer
343 views

Clarifying a result of Klingenberg

I am looking for some clarification on a result of Klingenberg. For context, I have a complete Riemannian manifold for which I would like to compute a lower bound on the injectivity radius at each ...
E G's user avatar
  • 163
5 votes
1 answer
284 views

Compact Eucledean hypersurfaces with "almost" constant H_k curvature

Let $M$ be an Eucledean $n$-dimensional compact hypersurface with constant $H_k$ curvature, where $k=1,...n$. A theorem by A.Ros tell us that so $M$ is an Eucledean sphere. Does anybody know if there ...
Maria Chiara Bertini's user avatar
5 votes
2 answers
379 views

Correct curvature tensor of symmetric space of positive definite matrices with trace metric?

Let $Pos(n)$ be the set of $n \times n$ real positive definite matrices with trace (aka affine-invariant) metric $$\langle u, v \rangle_p = tr(p^{-1} u p^{-1} v)$$ for all $p \in Pos(n)$ and $u, v \in ...
ccriscitiello's user avatar
5 votes
1 answer
156 views

Positively curved metric with uniformly positive scalar curvature

Can we find a complete noncompact Riemannian manifold $(M^n,g)$ with bounded geometry satisfying the following conditions? the curvature operator $Rm>0$; the scalar curvature $R \ge 1$. Notice ...
Totoro's user avatar
  • 2,535
5 votes
1 answer
209 views

Evolution of $W_+$ and $W_-$ under the Ricci flow

In dimension $4$ the Weyl operator $W$ splits in two parts $$W_+:\Lambda^{2}_{+} \to \Lambda^{2}_{+}$$ and $$W_-:\Lambda^{2}_{-} \to \Lambda^{2}_{-}.$$ (a) Has there been a study of the evolution ...
I. Tergiakidis's user avatar
5 votes
1 answer
245 views

Possible sign of scalar curvature for Einstein warped product manifold with Ricci-flat

Let $(M, g_M)$ where $M= B \times_f F$ and $g_M=g_B + f^2g_F$, an Einstein warped product manifold (i.e., $Ric_M= \lambda g_M$), with Ricci flat fiber-manifold $F$, i.e., $Ric_F=0$. Then $M$ can admit ...
MathDG's user avatar
  • 272
5 votes
1 answer
291 views

Gaussian Curvature of Exponentiated 2-Planes

Consider a Riemannian manifold $M$ with sectional curvatures $K\ge 0$ and let $\Pi$ be a 2-plane in the tangent space of $M$ at a point $p$. In a small enough neighborhood $U$ of 0 the exponential map ...
Oliver Jones's user avatar
  • 1,378
5 votes
0 answers
445 views

Upper bound on the sectional curvature of a Riemannian submersion

Consider the manifold $M := \operatorname{SO}(n) \times \mathbb{S}^{n-1}$, endowed with the product metric given by the bi-invariant metric of $\operatorname{SO}(n)$ and the round metric of $\mathbb{S}...
mathusername's user avatar
5 votes
0 answers
244 views

Möbius strip zero curvature [closed]

Is there a Möbius strip, seen as an embedded surface in $\mathbb{R}^3$, with zero curvature? I know one can see the Möbius strip as the quotient of the square with reverse identification of two sides ...
ibroketheinternet's user avatar
5 votes
0 answers
101 views

How is this product of tensors defined?

I am reading the paper “ The first eigenvalue of a small geodesic ball in a riemannian manifold”, by Karp and Pinsky, from where I took the following: Here, $\Delta_{-2}$ denotes the usual Laplacian ...
Eduardo Longa's user avatar
4 votes
1 answer
303 views

Locally Riemannian Connection

Let $\Gamma^a{}_{bc}=\Gamma^a{}_{cb}$ be a symmetric connection whose curvature is $$R^a{}_{bcd}=\partial_c\Gamma^a{}_{bd}-\partial_d\Gamma^a{}_{bc}+\Gamma^a{}_{ec}\Gamma^e{}_{bd}-\Gamma^a{}_{ed}\...
Aureliano Skirzewski's user avatar
4 votes
1 answer
308 views

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

Is there a Riemannian metric on $\mathbb{R}^3$ for which the corresponding curvature tensor $R$ satisfies $R(X,Y)Z=(X\wedge Y)\wedge Z$? I have already discussed this question in the following post ...
Ali Taghavi's user avatar
4 votes
1 answer
412 views

Symmetries of non-Riemannian curvature tensor

The curvature tensor, $R_{ab}{}^c{}_d$, can be obtained from a connection which not necessarily is a metric connection. By construction it is antisymmetric in the first two indices, since roughly ...
Dox's user avatar
  • 690
4 votes
1 answer
439 views

Etymology “Kulkarni–Nomizu product”

$\newcommand\KN{\mathbin{\bigcirc\mspace{-20mu}\wedge\mspace{3mu}}}$In the context of (pseudo)-Riemmian geometry, the Kulkarni–Nomizu product is defined to be an operation $\KN$, which takes two ...
G. Blaickner's user avatar
  • 1,429
4 votes
1 answer
245 views

Tzitzeica surface

A Tzitzeica surface has the property that the ratio of the surface’s Gaussian curvature and the fourth power of the distance from the origin to the tangent plane at any arbitrary point of the surface ...
MathDG's user avatar
  • 272
4 votes
1 answer
1k views

Relation between mean curvature and conformal metric

We'll consider $(N, g)$ a Riamannian Manifold and $\overline{g} = e^{2f}g$ a conformal metric. Let M be a hypersurface in N, $\overline{H}_M$ and $H_M$ the mean curvature of M with respect to the ...
K2-liz's user avatar
  • 55