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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
11
votes
1
answer
510
views
A very torsioned closed curve in space
Is there a simple smooth closed curve $\gamma$ in $\mathbb{R}^{3}$ such that for all $x,y\in \gamma$ with $x \neq y $, $l_{x}$ and $l_{y}$ are skew lines, where $l_{x} $ and $l_{y}$ are s …
10
votes
2
answers
1k
views
Obstructions for a metric to be conformally equivalent to a product metric
Is there a Riemannian metric on $\mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$ which is not conformally equivalent to a product metric?
More generally, assume that $M$ and $N$ are two manifolds. Wha …
0
votes
Does for every vector field there always exist a volume form for which the vector field is a...
If a vector field $V$ has a semi stable limit cycle then there is no any volum form $\Omega$ such that $Div_{\Omega} X=C$, a constant.
Explanation:
A semi stable limit cycle is an isolated perio …
3
votes
when constant scalar curvature implies Einstein?
To have a wide familly of counter example lets consider the Yamabe problem which says every compact manifold admite a metric of constant scalar curvature. So every manifold which does not …
3
votes
2
answers
329
views
Triangles in rigid Riemann surfaces
Edit: We thank Vladimir Matveev for his comment on this post which leeds us to revise the question as follows:
Assume that $M_{g}$ is a compact Riemann surface with constant negative cuvature (That …
2
votes
0
answers
161
views
Shape-related vector fields
Assume that $M$ is a surface in $\mathbb{R}^{3}$. We denote its shape operator by $S$. A vector field $X$ is shape related to $Y$ if $S(X)=Y$.
(of course it is not an equivalent relation).
Ass …
2
votes
2
answers
505
views
A strongly non-integrable distribution
What is an example of a three-dimensional smooth distribution $D$ of $\mathbb{R}^4$ with this property:
Not only $D$ is not integrable but also there is no a two-dimensional foli …
3
votes
1
answer
311
views
Is there a $2$ dimensional foliation tangent to this particular $3$ dimensional distribution?
Is there a $2$- dimensional foliation of $\mathbb{R}^4\setminus \{0\}$ whose tangent space is contained in $\ker \alpha$ where $\alpha$ is the following non integrable $1$-form?
$$\alpha=(x^2+y^2)dx+ …
7
votes
2
answers
536
views
A non integrable distribution which is totally geodesic
Is there a non integrable $2$ dimensional distribution $D$ of a $3$ dimensional Riemannian manifold such that the distribution is totally geodesic in the following sense:
Every geodesic whose tang …
8
votes
1
answer
442
views
The differential of the Gauss normal map from a Lie algebraic view point
Let $S\subset \mathbb{R}^3$ be a smooth surface with the Gauss normal map $N:S\to S^2$.
Then for every $x\in S$, the differential $(dN)_x:T_xS\to T_{N(x)}S^2$ can be considered as an endom …
2
votes
1
answer
321
views
An orientable compact even dimensional manifolds whose all even cohomologies do not vanish b...
What is an example of an orientable compact $2n$ dimensional manifold $M$ whose all even dimensional De Rham cohomology groups $H_{\mathrm{DeR}}^{2i}(M)$ are nonzero, but $M$ does not admit any symple …
4
votes
0
answers
278
views
Orthonormal vector fields on a Riemannian manifold
Let $(M,g)$ be a Riemannian manifold. We equip the tangent bundle $TM$ with the Sasaki metric $g_s$.
Assume that $X: M \to TM$ is a vector field on $M$.
We say that $X$ is an orthonormal vector field …
1
vote
0
answers
229
views
A new Lie group associated to a given Lie group
Edit: According to the comment of Ycor we remove the phrase "Naturally arises from a left invariant metric'
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. We fix an orientation for $G$. We d …
3
votes
1
answer
567
views
A surface on which all regular curves have nowhere vanishing curvature
Let $S$ be a surface in $\mathbb{R}^{3}$ such that every regular curve $\gamma\subset S$ has nowhere vanishing curvature, that is $\kappa(z)\neq 0$ for all $z\in \gamma$. Does this imply that …
1
vote
1
answer
144
views
An special isometric embedding
Let $M$ be a Riemannian manifold and $\gamma$ be a non closed geodesic.
Is there an isometric embedding of $M$ into some $\mathbb{R}^{n}$ which send $\gamma$ into an straight line?
The second questio …