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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 …
Ryan Budney's user avatar
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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 …
Ryan Budney's user avatar
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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 $0 \leq y \leq b$. …
Ryan Budney's user avatar
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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 …
Ryan Budney's user avatar
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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 …
Ryan Budney's user avatar
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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 …
Ryan Budney's user avatar
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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 …
Ryan Budney's user avatar
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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 $k=2$ …
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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 …
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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 …
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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 …
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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 …
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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 …
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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 …
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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 …
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