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20 votes
Accepted

Are all maps $\mathbb{R}^2 \to \mathbb{R}^2$ with fixed singular values affine?

Answer modified on 1 February 2020: It's not true 'locally' in the sense that non-affine $f$'s satisfying this system of PDE can be constructed on some open sets in $\mathbb{R}^2$. This first order,...
Robert Bryant's user avatar
9 votes
Accepted

Alexandrov's generalization of Cauchy's rigidity theorem

The following is Theorem 27.2 of Igor Pak's book Lectures on Discrete and Polyhedral Geometry (which in general is a very nice resource for these sorts of questions): Let $P,Q\subset\mathbb{R}^d$ (...
j.c.'s user avatar
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9 votes

Are all maps $\mathbb{R}^2 \to \mathbb{R}^2$ with fixed singular values affine?

I would like to propose a simple local example: Consider the map in polar coordinates, $\mathbb C\to \mathbb C$ that takes a complex number $z=e^{2\pi i \theta}r$ to $e^{(\sigma_1/\sigma_2)\cdot 2\...
Dmitri Panov's user avatar
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6 votes
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Cartesian monoidal star-autonomous categories

[I'm going to assume $S' \cong S$, which holds in every symmetric monoidal $*$-autonomous category. (See e.g. Lemma 5.6 of this paper.) This applies here since cartesianness implies symmetry. Part of ...
Tobias Fritz's user avatar
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5 votes

Alexandrov's generalization of Cauchy's rigidity theorem

Wikipedia is correct. This is discussed in Alexandrov's book "Convex polyhedra" in Section 3.6.5.
Ivan Izmestiev's user avatar
5 votes

Alexandrov's generalization of Cauchy's rigidity theorem

This may help: Bauer, C. "Infinitesimal Rigidity of Convex Polytopes." Discrete Comput Geom (1999) 22: 177. https://doi.org/10.1007/PL00009453 "Aleksandrov [1] proved that a simple convex $d$-...
Joseph O'Rourke's user avatar
5 votes
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Cocycle superrigidity

Yes. Instead of writing down explicitly such a cocycle (which is not hard), let me take this opportunity to explain how to think of such objects in a "cooridnate free" manner. Given a cocycle $c:\...
Uri Bader's user avatar
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5 votes
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Constant Gaussian curvature disks

This is an addendum to the proofs by Anton and Deane, completing the missing part of the argument. Lemma. Let $D$ be the closed unit disk and $f: D\to S^2$ an immersion such that $f(\partial D)$ is a ...
Moishe Kohan's user avatar
  • 12.3k
3 votes
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Shrinking a disk with fixed differential

Here are a few comments that you might find useful, though they don't completely solve the problem. First, using symmetries of the problem, you can easily reduce to the case that $f$ is mapping the ...
Robert Bryant's user avatar
3 votes

Counterexample to mostow rigidity theorem

Yes, I believe there are many. For example, if you think of hyperbolic $2$-space as a geodesic subspace of hyperbolic $3$-space, any group of hyperbolic isometries of hyperbolic $2$-space extends ...
Ryan Budney's user avatar
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3 votes

How to correctly state Cauchy's rigidity theorem?

Yes, formally speaking the statement is incorrect. The correct assumption is: there is a continuous bijection between the surfaces of two polyhedra that maps each facet to a facet by isometry. ...
Anton Petrunin's user avatar
3 votes
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Forbidden minors of a graph with treewidth at most 4

I have a copy of Sander's PhD thesis. Counting $K_6$, there are actually $76$ excluded minors for treewidth at most $4$ (found by computer) in the thesis, but it is unknown if this list is complete (...
Tony Huynh's user avatar
  • 32.1k
3 votes

Forbidden minors of a graph with treewidth at most 4

This is not a complete solution in the sense of a complete list, just a description to get an easy example of a graph of treewidth $5$ that does not have $K_6$ as a minor. As far as I know, the ...
karlheuer's user avatar
  • 105
1 vote

Constant Gaussian curvature disks

The answer is yes. Since curvature is 1, there is an isometric immersion $\iota\colon D\looparrowright \mathbb{S}^2$. Note that the curve $\iota(\partial D)$ has constant curvature, therefore $\iota(\...
Anton Petrunin's user avatar
1 vote

Show that duality functor is anti-monoidal

First of all, by Mac Lane Coherence Theorem we may assume that $\mathcal{C}$ is strict. Therefore we may omit associativity and unit constraints and we are left to check that $$J_{U,W \otimes V} \circ ...
Ender Wiggins's user avatar

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