4

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 ...


3

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 ...


3

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 $det(X^TX)=det(D)\geq 0$.


3

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 ...


2

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 ...


2

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 ...


2

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, $f$ ...


2

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 ...


2

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


2

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, ...


2

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 $...


1

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 $i, j \in \{1, 2, \dots, n\}$. ...


1

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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