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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
1
vote
Hessian of distance function from a Jordan curve
Yes, there are some formulas. Let's assume you are close-enough to the curve that the geometric tubular neighbourhood theorem applies, i.e. there is a unique closest point to $x$ in the curve $\gamma …
3
votes
Accepted
Algorithm for finding inverse images of a local diffeomorphism
With a Lipschitz bound on the derivative, Newton's method gives an algorithm to efficiently approximate $F^{-1}$. Hubbard's calculus textbook has a write-up using this perspective, viewing the result …
13
votes
Accepted
Interpretation of Riemann tensor antisymmetry
This is a re-flavouring of Alexander's answer but in a language I prefer.
Take two vectors $v,w \in T_p N$, and consider the `rectangle' $exp(xv+yw)$ where $0 \leq x \leq a$ and $0 \leq y \leq b$. …
8
votes
Following curves on S^n
No, your solutions do not need to be closed curves. For example, take the vector field "multiplication by $i$" on the $3$-sphere, thought of as the sphere of unit quaternions. That has closed soluti …
3
votes
Accepted
Examples of Lie group actions on manifolds with singular quotients
A general nonsense answer to your question would be the orbit decomposition theorem. Given a compact lie group $G$ acting on a manifold $M$, it describes a stratification of $M$ by orbit types.
The …
3
votes
Embedding a pair of manifolds, torus knots
As stated, yes $d(M,N)$ is always finite -- it's bounded above by $2n+1$ by the strong Whitney embedding theorem. The idea is to use the smooth Urysohn Lemma to write $M$ as $M=f^{-1}(0)$ for some sm …
5
votes
Holonomy of compact manifolds
The holonomy group need not be compact. For example, take $S^1$, trivialize its tangent bundle and let $\Gamma_{1,1}^1 = 1$, constant on $S^1$. If you parallel transport any vector around $S^1$, the …
3
votes
Accepted
Extending Smooth Embeddings of S^n into S^{n+k}
Your condition can be stated equivalently as the knot longitude, as an embedded $S^n$ in the knot complement $C_h = S^{n+k} \setminus h(S^n)$ is null homotopic.
Generally the answer is no when $k=2$ …
1
vote
Geodesics on spheres are great circles
Here is another argument.
Given a great circle in $S^n$, there is an isometry of the sphere $S^n$ whose fixed points are precisely that great circle, and on the normal bundle (fibers) the symmetry is …
3
votes
Poincare-Hopf theorem for polytopes?
There are certainly analogous theorems but the most direct analogy you might not find particularly useful. For example, manifolds with boundary could be considered a step towards a general Poincare-H …
12
votes
Is the space of immersions of $S^n$ into $\mathbb R^{n+1}$ simply connected?
Let me just fill-in the gap in j.c.'s exposition. Smale-Hirsch states that the derivative from the space of immersions $Imm(S^n, \mathbb R^{n+1})$ to the space of bundle monomorphisms $Mono(TS^n, T\ma …
4
votes
Accepted
Singular Homology Groups from Compact Oriented Submanifolds?
Expanded version of my comment:
Please google "the steenrod realization problem". This is a foundational problem in algebraic topology. It has a nice answer -- Glen Bredon's book "Topology and Geomet …
6
votes
Accepted
space of simple loops in the plane
The space of smooth embeddings $S^1 \to \mathbb R^2$ has the homotopy-type of $O_2$. Denote this embedding space by $Emb(S^1,\mathbb R^2)$.
The proof goes like this. Let $Emb(D^2, \mathbb R^2)$ be …
4
votes
Accepted
What kinds of manifolds admit concave boundary?
I think the answer is yes and in a strong way.
Precisely, let $M$ be a compact manifold, and put any Riemann metric on $\partial M$ Then I claim there is a Riemann metric on $M$, extending the Riem …
3
votes
Accepted
Circle bundles over $CP^1$ and self-intersection number of $CP^1$ embeddings
You can find $\mathbb CP^1$ in a wide variety of $4$-manifolds having any Euler class you like. One really simple way is to take the connect-sum of $k$ copies of $\mathbb CP^2$. The idea is to embed …