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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, determined PDE system is hyperbolic, so there are many local solutions. However, it turns out (see below) that all $C^3$ solutions with domain equal to $\... 6 I would like to propose a simple local example: Consider the map in polar coordinates,$\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\pi i \theta}r\sigma_2$. (apologies for the previous wrong example, I confused in it singular values with eigenvalues...) 5 Let$n\geq 3.$Let$\Omega$be an open connected subset of$\mathbb R^n,$and let$f:\Omega\to\mathbb R^n$be a function having a pointwise derivative$Df(x)$everywhere satisfying$(Df)^T(Df)=g(x)I$with$g(x)>0.$Then$f$is continuously differentiable. By the inverse function theorem,$f$is a local homeomorphism. By shrinking$\Omega$we can assume$...

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In general, having constant mean curvature is not preserved. This can be seen by a direct computation, or by looking at examples: The round metric on the 3-sphere without a point is conformal to flat 3-space. The Clifford torus in the 3-sphere (even without a point) is minimal, hence of constant mean curvature, and embedded. After stereographic projection ...

4

$\newcommand{\R}{\mathbb{R}}\newcommand{\pa}{\partial}$Edit: The answer is now LaTeXified. Below are my notes on this. I reworked the proof: Proof. Let $S:=\big\{T\in[0,\infty):$ the equation has a solution in $C^{2+\alpha,1+\alpha/2}(M\times[0,T],N)\big\}$. Let $T_0:=\sup S$. By existence of local solution, $T_0>0$. We claim that $T_0=\infty$. Suppose ...

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This answers the first (simple) half of the question, asking just about a smooth map. In fact, you've already given an answer to it, in some sense. Apply the map $f: re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$ to a unit disk that doesn't contain $(0,0)$, say radius $1$ disk $D$, centred at $(2,0)$. Then, the image $f(D)$ is contained in the ...

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Yes, this is true. It is shown (either in the paper you cite or the other McShane-Rivin paper) that the length of a simple closed geodesic is quasi-the-same as the combinatorial length ($m+n$) (this is easy, because a simple geodesic stays away from the cusp), and that, in turn, is easily seen to be quasi-the-same as the Euclidean length.

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Charles R. Johnson and I gave a proof of the Gauss–Lucas theorem that utilizes the field of values (or numerical range) of a matrix, which can be viewed as geometric in nature (the field itself is an uncountably infinite intersection of half-planes). The proof can be found in the article Matricial Proofs of Some Classical Results about Critical Point ...

3

Yes, it is true. Note that $M$ is isometric to a quotient of the Euclidean space $\mathbb{E}^n$ by a totally discontinuous free isometric action of a group $\Gamma$. Your condition implies that the soul of $M$ is a single point. It follows that $\Gamma$ fixes a point in $\mathbb{E}^n$. Since the action is free, the group $\Gamma$ is trivial.

3

As in my previous solution in the 3-dimensional case (discussed here), we can set $f=-(p/h)x$ for a function $x$ that satisfies $$\Delta x + p\,x^2 = |\nabla x|^2 - x^3 = 0.\tag 1$$ Conversely, if $x$ satisfies this system for a metric $g$, then $f = -(p/h)x$ will satisfy the original equations. Again, the same argument as before shows that the $1$-form $... 2 The answer is negative if$\dim M=2$and positive otherwise, as shown in the paper: Croke, C. B., & Karcher, H. (1988). VOLUMES OF SMALL BALLS ON OPEN MANIFOLDS: LOWER BOUNDS AND EXAMPLES. AMERICAN MATHEMATICAL SOCIETY (Vol. 309). https://www.ams.org/journals/tran/1988-309-02/S0002-9947-1988-0961611-7/S0002-9947-1988-0961611-7.pdf The negative result ... 2 Yes, this is possible. Note that a strictly convex hypersuface in$\mathbb R^{n+1}$has positive$Rm$. To get an example, consider the following graph$H\subset \mathbb R^{n+1}$over the open unit$n$-disk in$\mathbb R^n$: $$H:=\left(x_{n+1}=\frac{1}{1-\sum_{i=1}^n{x_i^2}}\right).$$ Clearly,$H$is convex. Moreover, the scalar curvature of$H$tends to ... 2$\newcommand{\pa}{\partial}Edit: The answer is now LaTeXified. I happen to have written notes on this. :) \begin{aligned} \frac{\pa\kappa(u_t)}{\pa t}&=h_{\alpha\beta}(u_t)\frac{\pa^2 u_t^\alpha}{\pa t^2}\frac{\pa u_t^\beta}{\pa t}\\&=\Big\langle\frac{\pa^2u_t}{\pa t^2},\frac{\pa u_t}{\pa t}\Big\rangle. \end{aligned} By Ricci's identity,$\... 2 There are a number of ways to see this. One way is to take the covariant derivative of the isometric embedding equation$\partial_iu\cdot\partial_ju = g_{ij}$and "differentiate by parts". The calculation below is with respect to local coordinates, and$u$is treated as a$q$-tuple of scalar real-valued functions. Therefore,$\nabla^2_{ij}u = \nabla^2_{ji}u$.... 2 The argument here is due to Mahan Mj: If$A, B$are compact and not isometric, then the isometry group is the product. Here is the point: Let$X=A \times B$be the product manifold with the product metric. Let$f: X \to X$be an isometry. Then each$f(A \times \{b\})$is totally geodesic for every b. so that$X$fibers over$B$with fibers$f(A).\$ Further ...

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Ian's argument of mean curvature is wonderfully simple. Here is another one. Rotate your surface to put it in generic position with respect to the heigth function z; then, the preimage of z is a Morse function f on RP^2, which has no critical point of index 1 (saddle point) since the surface is locally convex. Hence, every critical point of f has index 0 or ...

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