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Each Euclidean manifold $M$ admits a flat Riemannian metric (obtained by pull-back of the flat Riemannian metric on $E^n$ via coordinate charts of the Euclidean atlas). If $M$ is compact, the metric is complete (Hopf-Rinow theorem) , so is its lift to the universal covering space $X$ of $M$ (see this question). By Cartan-Killing-Hopf theorem, each simply ...

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The idea for this sort of thing (where it's still low-dimensional enough) is to try and draw it, and see where the problem is on the drawing, and what that problem corresponds to. Once you see that, you can do a formal, precise argument by basing what you're looking for on the drawing (although of course the drawing itself isn't enough !) To draw it, notice ...

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Note that $X^TX$ is self-adjoint, hence it is diagonalizable. Also note that $(x,X^TXx)=(Xx,Xx)=||Xx||^2\geq 0$ for any vector $x$. Hence, all eigenvalues are nonnegative. Hence we can write $X^TX=VDV^*$ where $D$ is a diagonal matrix with nonnegative entries and $V$ is unitary. Hence $det(X^TX)=det(D)\geq 0$.

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They talk about this in example 8.4. Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ be smooth and $p\in \mathbb{R}^n$. Take standard coordinates of $\mathbb{R}^n$ and $\mathbb{R}^m$ via $(x^1,\cdots, x^n)$ and $(y^1\cdots, y^m)$, respectively. Then, the map $F_*:T_p\mathbb{R}^n\rightarrow T_{F(p)}\mathbb{R}^m$ is a linear map. The entries of the matrix ...

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Definitely not. First recall the definition of a regular domain: it's a properly embedded (hence closed) codimension-$0$ smooth submanifold with boundary in $M$. The basic problem is that wherever the boundary of a sublevel set intersects the boundary of $M$, you're likely to get a corner or worse. A simple counterexample is to take $M$ to be the closed ...

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Sets of the form $\mathcal U$ form a neighborhood basis of the point $(p,0) \in TU$, so you are free to choose $\mathcal U$ of that form. And regarding the notation $|v| < \epsilon_1$, what's being used here is not the full panoply of Riemannian geometry, but just an arbitrary smoothly varying norm on the tangent spaces at points of $V$, chosen for the ...

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Yes, there is always a regular domain (i.e., a smooth, codimension-$0$, closed, embedded submanifold with boundary) that contains $A$. Here's a proof. References are to my Introduction to Smooth Manifolds (2nd ed.). First of all, Proposition 2.28 shows that there is a smooth positive exhaustion function $f\colon M\to (0,\infty)$. Because $A$ is compact, $f$ ...

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It's hard to give a hint to this problem without giving the whole thing away. By definition, $\text{Int} M$ is the subset of all $x \in M$ for which there exists a chart $(U_i,\phi_i)$ such that $x \in U_i$ and such that $\phi_i(U_i) \subset \mathbb R^{n-1} \times (0,\infty)$. This existence property is clearly also true for every $y \in U_i$, using the ...

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$Z\setminus\{(0,0,0)\}$ is not connected - that can only happen with 1-dimensional manifolds. But a neighbourhood of e.g. $(1,0,1)$ clearly looks 2-dimensionsl.

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Let's just work in $\mathbb{R}^n$ to try to get some geometric intuition. I will directly connect the intuitive definition with the abstract one. First, let's actually start with $n=3$. In $\mathbb{R}^3$, we can visually a vector at a point $p$ as an arrow starting at $p$, where the direction of the arrow is based on the coordinates of the vector. Visually, ...

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Or another way I look at it; since $S$ is defined to be the image $f(N)$, that means that for every point $n\in N$, $f(n)\in S$. This fact, combined with the fact that $f$ is $C^{\infty}$, seems like a sufficient condition for $\tilde{f}$ to be smooth. But that's not the definition of what it means for $\tilde{f}$ to be smooth. By definition, $\tilde{f}:N\... 2 The other answer covers it, but I think it is worthwhile to remark that the very definition of differential encodes how$(F_*)_p$transforms tangent basis vectors, and so how it acts on$T_pN:$if$p\in U\subseteq N$and$(U,\phi)$is a chart, then$\phi_*:T_pU\cong T_pM\to T_p \mathbb R^n$is an isomorphism (because$\phi$is a diffeomorphism) and so it ... 1 Two approaches: First, it suffices to note that if$X$has linearly-independent rows, then$XX^T$is positive definite. It follows that the eigenvalues of$XX^T$are positive, so that$XX^T$has positive determinant. Another approach is to use the Cauchy-Binet formula to find that $$\det(AA^T) = \sum_{S \subset \{1,\dots,n\}, |S| = m} \det(A_S)^2$$ ... 1 Define$v$by $$v(t) = ((2t-1)^2, (2t-1)^3).$$ Then$v(1) - v(0) = (0, 2)$, But it is easy to check that$v'(t)$is never a scalar multiple of$(0, 2)$. This can be easily seen from the following plot. The statement can be made true with a bit of correction, which results in a version of Cauchy's mean value theorem. Let$v : \mathbb{R} \to \mathbb{R}^2$... 1 If you have a map$ f : U \to V$between open subsets$U \subset \mathbb R^m, V \subset \mathbb R^n$, then the derivative at$p \in U$is the best linear approximation of$f$in$p$. This is the (unique) linear map$df(p) : \mathbb R^m \to \mathbb R^n$such that $$\lim_{h \to 0} \dfrac{\lVert f(p+h) - (f(p) + df(p)(h)) \rVert}{\lVert h \rVert} = 0.$$ Note ... 1 For the first thing you do not get: That$\exp_p$is defined at$v\in T_pM$, means that there is some$\delta>0$, with$|v|<\delta$such that$\exp_p$is defined on$B(0_p,\delta)\subset T_pM$. Just check the domain of the generalized$\exp$in Proposition 2.7. So, the sphere$S$of all$w$s with$|w|=|v|$is contained in the domain of$\exp_p$. Now, ... 1 Yes: if$M$is a smooth manifold, then whenever you talk about a chart in$M$(or on$M$, or of$M$, etc.) that always refers to a chart in the atlas of$M$unless specified otherwise. 1 No. For a really simple example, consider rectangular and polar coordinates on$\mathbb{R}^2$. Starting at$(1,0)$(in rectangular coordinates), the curve given in rectangular coordinates given by a vertical velocity vector is a vertical line. But in polar coordinates, it is a circle about the origin (since a vertical velocity vector points in the$\theta$... 1 The notion of convexity you are using is unnatural for functions defined on compact (connected) Riemannian manifolds: With this definition every convex function is constant. Indeed, suppose that$f: M\to {\mathbb R}$is a convex function on a compact connected Riemannian manifold. Then$f$is necessarily convex and, hence, attains its maximum at some point$...

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I provide a partial answer below, but I would be interested to see what people think. Claim: If $A$ is not a multiple of the identity, then $f$ is not convex. Let $\lambda_i$, $e_i$ denote the (real) eigenvalue-eigenvector pairs for $A$, for $i = 1, 2, \dots, n$. Suppose that $\lambda_i < \lambda_j$ for some distinct $i, j \in \{1, 2, \dots, n\}$. ...

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Be careful with how you apply the Proper s-cobordism Theorem. The obstruction to an inclusion being an infinite simple homotopy equivalence is more complicated than you indicate. (See Section 3 of Siebenmann's "Infinite Simple Homotopy Types".) It involves not just the fundamental group of the space, but also its fundamenatal group system at infinity. In ...

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