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11 votes
2 answers
2k views

Retraction of a Riemannian manifold with boundary to its cut locus

This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus. Recall that the cut locus of a set $S$ in a ...
Dmitri Panov's user avatar
  • 28.9k
16 votes
3 answers
1k views

SO(3) action on (simply connected) 6 manifold with discrete fixed point

If a 6-dimensional orientable smooth manifold $M$ admits a smooth effective $SO(3)$ action with discrete fixed point set, what can we say about the topology of $M$? What if we assume that in addition ...
Yuhang Liu's user avatar
14 votes
1 answer
3k views

How metric is Riemannian geometry

Let $(M, g)$ be a finite-dimensional Riemannian manifold. It is well-known, that the Riemannian metric induce a metric on the manifold by $$d(x, y) = \text{inf} \int_a^b \| \dot\gamma(t) \| \, dt\,,$$...
Tobias Diez's user avatar
  • 5,824
8 votes
1 answer
599 views

Exact condition for smooth homogeneous to imply Riemannian homogeneous for compact manifolds

Let $ (M,g) $ be a homogeneous Riemannian manifold. That is, the isometry group $ Iso(M,g) $ acts transitively on $ M $. Let $ \pi_1(M) $ be the fundamental group of $ M $. Then $ \pi_1(M) $ has ...
Ian Gershon Teixeira's user avatar
5 votes
1 answer
1k views

Does every smooth manifold admit a metric with bounded geometry and uniform growth?

Let $M$ be a smooth manifold, $g_M$ a Riemannian metric, and consider for $x\in M$ the volume growth function, $gr_x$ that maps $r>0$ to the volume $vol_{g_M}(B(x,r))$. My interest is to see ...
user44172's user avatar
  • 541
3 votes
2 answers
265 views

For a closed Riemannian manifold $M$, must the set of points with non-unique closest points to a closed submanifold $S$ of $M$ be of 0 volume measure?

Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of finite dimension, with the volume measure $\mu:= \mu(E):=\int_{E}dvol_g \forall E \in \mathcal{B}(M),$ the Borel sigma algebra ...
Learning math's user avatar
2 votes
1 answer
138 views

noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous

Is it true that a manifold $ E $ admits a metric with respect to which the isometry group is transitive ($ E $ is Riemannian homogeneous) if and only if $ E $ is the total space of a $ K $ equivariant ...
Ian Gershon Teixeira's user avatar
2 votes
1 answer
484 views

Mapping torus of orientation reversing isometry of the sphere

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$ Let $ f_n $ be an orientation reversing isometry of the round ...
Ian Gershon Teixeira's user avatar