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Results tagged with alexandrov-geometry
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user 1441
Alexandrov geometry studies non smooth analogues of Riemannian manifolds with curvature bounded from below or above. It includes spaces with curvature bounded below (briefly $\mathrm{CBB}[\kappa]$) and spaces with curvature bounded above (briefly $\mathrm{CAT}[\kappa]$).
2
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Doubling theorem for Alexandrov spaces
You may check our draft:
"A journey into Alexandrov geometry: curvature bounded below" by Vitali Kapovitch and me.
It is based on a graduate course.
We do not pove Perelman’s theorem about conic neigh …
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1
vote
Accepted
Connectedness of fibers of almost Riemannian submersions
In the definition of submersion you should assume that $d_xf\colon \textrm{T}_x\to \textrm{T}_{f(x)}$ is onto.
In this case there is an almost horizontal lift $\phi_x\colon\textrm{T}_{f(x)}\to\textrm{ …
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1
vote
Dimension of Alexandrov space which is homeomorphic to a manifold
If $n$ is defined, then the statement has already been proved in the paper by Burago, Gromov, and Perelman.
However, there might be no such $n$; in other words, the space has infinite dimension in thi …
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1
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Cusp points in Alexandrov spaces
Theorem 8 on page 166 in "Intrinsic geometry of surfaces" by Alexandrov and Zalgaller implies that if $\omega(\{p\})=0$, then the total angle around $p$ vanishes. Therefore, the space of directions at …
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4
votes
Accepted
Does the Alexandrov angle define convex functions along geodesics in CAT(0) spaces?
The answer is "no" even for 3-dimensional Hadamard manifolds.
Moreover, implication
$$\measuredangle[a\,^b_c]<\tfrac\pi2 \ \&\ \measuredangle[a\,^b_d]<\tfrac\pi2\ \Rightarrow\ \measuredangle[a\,^b_x] …
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2
votes
Accepted
Lower estimate on sectional curvature of the boundary
I assume that convex surfaces have nonnegative second fundamental forms.
If $\lambda\geqslant 0$ then the answer is yes, and it follows from the Gauss formula.
On the other hand, if $\lambda<0$, then …
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1
vote
Question on G. Perelman's paper "Elements of Morse theory on Aleksandrov spaces"
Indeed, openness does not solely imply regularity.
(Say, double branching covering of the plane is open, but not regular.)
But the map is admissible and that is the key word.
The definition of admissi …
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2
votes
Source for: Geodesics in CAT(0) spaces
You may try our book An invitation to Alexandrov geometry: CAT(0) spaces.
It contains a chapter about cubical complexes.
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3
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Accepted
Topological spaces admitting CAT(1) metrics
There are examples of that type.
I will construct a compact metrizable contractible and locally contractible space that does not admit a CAT(1) length metric.
Let $H$ be Bing's house (or the dunce hat …
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1
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isoperimetric problems on Alexandrov spaces
The existence follows from theory of currents the same way as for Riemannian manifolds.
As far as I know, there are no regularity results for $\partial D$.
But look at the proof of Levy--Gromov isoper …
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4
votes
Accepted
Length and curvature for closed curves in negatively curved spaces
The Reshetnyak majorization theorem (see 9.56) states that any closed rectifiable curve $\alpha$ in a CAT(0) length space $U$ can be majorized by a convex plane figure $F$; that is, there is short (= …
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2
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Accepted
Intersection of conical neighbourhoods on a polyhedral space
You say "The same is true for a tubular neighborhood of the edge".
This is not correct, but it is true if you stay away from the endpoints.
So $U$ should be defined as a tubular neighborhood of subarc …
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1
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If M times circle admits a locally CAT(0)-metric, then M also carries a locally CAT(0)-metric?
Here is a partial answer, I will prove the following:
If $M\times \mathbb{S}^1$ equipped with locally $\mathrm{CAT}(0)$ length metric, then there is isometric $\mathbb{S}^1$-action on $M\times \mathb …
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2
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Tangent cone of a proper CAT(0) is a proper CAT(0) space
The answer is "yes" for geodesically complete CAT(к) spaces.
It follows directly from the comparison.
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2
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When is the angular metric on the space of directions intrinsic?
One may ask if locally compact Alexandrov spaces have intrinsic spaces of directions --- I do not know the answer.
See 13.40 in our book.
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