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Results tagged with riemannian-geometry
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user 18050
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.
6
votes
Accepted
Non-negatively curved manifolds and the volume of balls
It is certainly not true that every complete nonflat open manifold of nonnegative curvature has Euclidean volume growth. Counterexamples are trivial to construct. Say, a capped cylinder. More generall …
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12
votes
Accepted
Geodesic circles on riemannian manifolds
I'm not sure but I believe you are asking if there always exists a closed geodesic such that it gives a distance preserving embedding of $S^1$ with respect to the length metric on the circle and th …
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4
votes
Example for Busemann function is not an exhaustion when Ricci $\ge 0$
This is an interesting question. I don't have an answer but I want to clarify what is being asked as there seems to be some confusion on the issue.
Given a point $p$ and a ray $\gamma$ starting at $p …
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10
votes
Accepted
Dimension of certain subgroup of isometry group of positively curved manifold
Forgetting positive curvature, if $\dim M^n/G=k$ then by looking at the transitive action of $G$ on the principal orbit one gets a trivial bound $\dim G\le \dim O(n-k)=\frac{(n-k+1)(n-k)}{2}$. This bo …
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16
votes
Accepted
Behavior of sectional curvature under metric deformations
Formula 2) is the correct one in general except it's the derivative of the sectional curvature i.e of $\frac{k_t(X,Y)}{|X\wedge Y|^2_t}$ (and not just of $k_t(X,Y)$) for an orthonormal frame $X,Y$ wi …
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12
votes
Manifold with all geodesics of Morse index zero but no negatively curved metric?
As mentioned by Rbega the question should be amended to ask whether it's true that a closed manifold $M$ without conjugate points admits a metric of non-positive (rather than negative) curvature (othe …
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9
votes
Accepted
Is the bundle map of the Eguchi-Hanson metric a Riemannian submersion?
No. The reason is basically the same why if you take flat $\mathbb R^4\setminus 0$ and quotient by the standard Hopf $\mathbb S^1$ action you get a punctured cone over $\mathbb S^2$ but the projecti …
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8
votes
Accepted
Is it known whether a closed simply-connected manifold of non-negative curvature admits posi...
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 kn …
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11
votes
Accepted
Riemann surfaces with bounded curvature
You need to specify what limit you are talking about as the question makes no sense otherwise. The weakest natural topology to consider in this setting is pointed Gromov-Hausdorff topology.
Gromov-Ha …
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5
votes
Explanation that Twistor Space of $S^4$ is $\mathbb{C}P^3$?
I am sure there are many ways to see this but here is a quick one. The oriented frame bundle of $\mathbb S^4$ is $SO(4)\to SO(5)\to \mathbb S^4$. The spin cover of this is $Spin(4)\to Spin(5)\to S^4$ …
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16
votes
Accepted
Consequences of Gromov's Conjecture
As Igor mentioned knowing the optimal bound is always better than knowing a non-optimal one such as the bound provided by Gromov's proof. It rules out a lot more examples. A proof of the sharp bound w …
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10
votes
Accepted
Conformally-flat
I'm not quite sure what you mean by always non-positively curved. If you are asking if this metric is non-positively curved for any $f$ then this is false. If you are asking for conditions on $f$ ensu …
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14
votes
Accepted
Metric deformations from non-negative to positive curvature
As Benoît Kloeckner points out this is false for non simply connected manifolds with $RP^2\times RP^2$ being a counterexample (by Synge's theorem). For simply connected manifolds this is a well known …
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3
votes
Isometry groups of Riemannian submersions with totally geodesic fibers
I'm not aware of any natural conditions that would ensure existence of the lifts (extensions) as you want. But I want to point out that your statement in (ii) is wrong. There is no reason to expect th …
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2
votes
Is every homogeneous space Riemannian homogeneous?
It immediately follows from the long exact homotopy sequence of the bundle $H\to G\to G/H$ that $\pi_1$ of any Riemannian homogeneous space $M$ is virtually abelian. So if you have a homogeneous space …
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