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2 votes

Existence of smooth functions decaying fast near the boundary

The following proof results from my discussion with Prof. John Lee. His comments and suggestions are very helpful. Step 1. First suppose that $U\subset\mathbb{R}^n$. Cover $U$ by countably many balls ...
Sardines's user avatar
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1 vote

Let $G$ a lie group and $\varphi:G\times G\to G$ given by $(g,h)\to ghg^{-1}$. what is $(d\varphi)_{(g,h)}(X_g,X_h)$.

Consider the composition $\varphi = \theta \circ \psi$, where $\psi:G\times G\to G\times G$ is defined by $$\psi(g,h) = (gh, g^{-1}).$$ Let $i:G\to G$ be the inversion $i(g) = g^{-1}$. Then $$d\psi_{(...
Pengin's user avatar
  • 375
1 vote

A centre of a simple closed planar curve

Centroid & Barycenter & Center of gravity & Center of Mass are very commonly used terms for what you want. Using the techniques involved there & taking the curve to have "Constant ...
Prem's user avatar
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0 votes

Reference for Surface area measure

As Behram Esmayli mentions, one reasonable notion of "surface area" for sets whose boundaries are "worse" than $C^1$ comes from BV theory. Note however that once your "surface&...
pseudocydonia's user avatar
4 votes

Given $F: \mathbb R^n \to\mathbb R^m,$ $n < m,$ if $F$ has no critical points in a subset $U,$ then is F necessarily injective in $U?$

If $f\colon \mathbb R \to \mathbb R^2$ is given by $f(t)= (\cos(t),\sin(t))^\intercal$, then $f'(t) = (-\sin(t),\cos(t))^\intercal$ never vanishes, and so $f$ has no critical points (and indeed has ...
krm2233's user avatar
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0 votes

$\frac {\partial }{\partial z^{j}}$ and $\frac {\partial }{\partial {\bar {z}}^{j}}$ form bases for complex tangent bundles

Well note the bundle $TM \otimes \mathbb{C}$ has as its fiber over $p \in M$ simply $T_p M \otimes \mathbb{C}$. This is a vector space with basis being the tensor products $\frac{\partial}{\partial x^...
Integral fan's user avatar
0 votes

About Differentiability in geometric definition of tangent space

The concept of smoothness was originally introduced for maps between open subsets of Euclidean spaces. This is the subject of multivariable calculus. The concept was extended to maps between smooth ...
Paul Frost's user avatar
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3 votes
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Kähler manifolds properties: parallel transp induces a complex linear map & a Riemannian manifold is Kähler iff $Hol_\nabla(p) \subseteq U(T_pM)$?

I am not sure why you haven't given the definition of a Kähler manifold that you would like to use. This is particularly relevant, since in my opinion the conditions 1. - 4. that you list are mainly ...
Andreas Cap's user avatar
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3 votes
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When is $h=d\omega$

Probably this question is at least in spirit a duplicate, but I couldn't find an appropriate post to point to. Suppose $h = d\omega$ for some $\omega$ (that is, that $h$ is exact); then $d h = d^2 \...
Travis Willse's user avatar
1 vote
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Submanifold of matrix space

So far your approach looks good, though n.b. we only need $D\phi$ to be surjective on $S$. Hint First restrict $\phi$ to the open subset $\operatorname{GL}(n, \Bbb R) := M_{\Bbb R}(n, n) \setminus \...
Travis Willse's user avatar
1 vote
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Length of a regular arc

As requested by @SammyBlack , here is the reason why everything goes out as desired. $\forall x \in [a,b], h'(x)>0$ so $h$ is increasing. Furthermore, $h$ is surjective since it is a diffeomorphism....
julio_es_sui_glace's user avatar
3 votes

Concatenation of $f$-related vector fields/tangent vectors

So I found the problem. The concatenation encountered in the Lie bracket, is ,in fact, not defined as I did it in my question, but as follows. Let $h \in \mathcal{C}^\infty (M)$. Then $$[X^1,X^2]h: p ...
welahi's user avatar
  • 283
1 vote
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Möbius strip is a smooth submanifold

You could see it as a sub manifold of the filled torus, which would be $$X = \left\{\left( \left(1+\frac{\lambda}{2}\cos \beta \right)\cos \alpha, \left(1+\frac{\lambda}{2}\cos \beta \right)\sin \...
julio_es_sui_glace's user avatar
0 votes

Where is the mistake in this proof that 2-form is non-vanishing?

Suppose that $v=(x_1,y_1,z_1)$ and $w=(x_2,y_2,z_2)$ both lie in $T_pF^{-1}(0)\subset \mathbb R^3$ such that at $p$ we have $dF/dz\neq0$. Then we have that: \begin{align} (dF/dx)x_1+(dF/dy)y_1+(dF/dz)...
Chris's user avatar
  • 2,946
4 votes

Geometry of origami saddle surfaces made of five or six square paper sheets connected around a point

If you conceptualize the surface as being composed of some number of quarter unit circle sectors, it's clear that the total arclength is $n\pi/2$ for some integer $n \ge 4$, and that the boundary of ...
heropup's user avatar
  • 137k
2 votes
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Dot product of two exterior products and associativity of geometric product?

I am not aware of a simple formula for the product of higher graded pieces of the Clifford algebra (a.k.a. geometric algebra). The formula $xy = x\cdot y + x\wedge y$ only works if $x,y$ are vectors. ...
Callum's user avatar
  • 4,386
2 votes
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Confusion about one pushforward calculation on $\mathbb{S}^{n-1}$

Your calculation is correct and results in a tangent vector at the appropriate base point. In more detail: First let $M:= \{ (p, r, X) \colon \ p \in \mathbb{S}^{n-1}, \ r > 0, \ X \in T_p \mathbb{...
Sven-Ole Behrend's user avatar
2 votes

derive Lie derivative of volume form by using Cartan's magic formula

$$ \begin{aligned} \mathcal{L}_X \Omega &=\iota_X d \Omega+d \iota_X \Omega =d \iota_X \Omega \\ &=\sum ^4 _{i=1} d (X^i \sqrt{-g}dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots \wedge dx^4)...
kookie's user avatar
  • 31
1 vote

Prove the directional derivative is a linear map

The mapping is linear if transforming a linear combination of the inputs is the same as transforming the inputs independently and then combining them linearly, so considering the vectors $h,l \in \...
Juan Martinez's user avatar
1 vote
Accepted

Self duality of a connection is invariant under a gauge transformation

Elements of $\Omega^2(M,E)$ are just sections of the vector bundle $E\otimes \wedge^2T^*M$. Your bundle morphism is really $\theta^{-1}\otimes \operatorname{id}_{\wedge^2T^*M}$ (well-defined since ...
user17945's user avatar
  • 4,018
1 vote
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Two $SO(2)$ connections on a smooth $SO(2)$-vector bundle

We have that by the Leibniz rule for covariant derivatives: \begin{align} \exp(if)\cdot \nabla'_X(\exp(-if)\cdot s))=&(\exp(if)\cdot s)\cdot d(\exp(-if)(X)+\exp(if)\cdot \exp(-if)\cdot \nabla'_Xs\\...
Chris's user avatar
  • 2,946
1 vote
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Understanding of connections differential geometry

Re 1.: Locally, when can define a connection $\nabla$ by picking a smooth local frame $E_{1}, E_{2}, \ldots , E_{n}$ on an open set $U\subseteq M$ and defining the connection on the frame. That is, ...
THW's user avatar
  • 1,380
0 votes

The relation between the third fundamental forms and the covariant derivative of the second fundamental form?

I think this is impossible for very silly reasons. The normal indices do not line up. For the second fundamental form, $n+1\le\alpha\le n+p_1$, and for the third fundamental form, $n+p_1+1\le \alpha\...
Ted Shifrin's user avatar
0 votes

Why specifying values on a local section is enough to determine the local values of a tensorial form on a principal $G$-bundle?

Staring at it again, it is not only possible to show that the monstrosity is vertical, it is in fact trivial to show that, just by applying $\pi_*$ to it and using $\pi \circ R_g = \pi$.
rosecabbage's user avatar
  • 1,655
1 vote
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Why $\phi_{*0}(\frac{d}{dt}|_0)=\frac{d\phi^{\mu}(t)}{dt}|_0\frac{\partial}{\partial g^{\mu}}|_e$?

It's just the chain rule and the definition of the pushforward. Things are easier to see if you give the resulting vector in $T_eG$ some function $f$ in $C^k(G)$ to act on. We will have; $$\phi_{*}\...
Volk's user avatar
  • 1,857
0 votes

Orthographic Projection and Concentric Circles

The answer is obviously no in the general case. Consider the case $n = 90$. Then the circle with radius $\sqrt{\dfrac1\pi}$ is the projection of the line of latitude at $89$ degrees latitude (one ...
David K's user avatar
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1 vote
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When does the map from Fundamental Group to Holonomy Group Injective?

Here is the construction. Let $F$ be your field (say, complex or real numbers), $\rho: \pi_1(M)\to G= GL(n,F)$ a representation. Let $V=F^n$, the $n$-dimensional vector space over $F$. Let $\tilde M\...
Moishe Kohan's user avatar
  • 97.9k
0 votes

Is ellipse the only planar regular curve with exactly four vertices?

Generally, it is not true. One should consider a semi-ellipse and smoothen its "flat side": there are still 4 points of extremum of the curvature function.
Denis D. Bavrin's user avatar
0 votes
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Orthographic Projection and Concentric Circles

We will derive the radius of the circles of equally spaced latitude under orthographic projection along the polar axis of the unit sphere. Since by construction, consecutive equally spaced circles of ...
heropup's user avatar
  • 137k
2 votes
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About Differentiability in geometric definition of tangent space

The definition of the tangent space $T_x M$ at a point $x$ on a differentiable manifold $M$ can indeed be conceptually challenging. Let me break it down for you: Smooth Curves on $M$: Consider a ...
Zoe's user avatar
  • 132
0 votes

Small question regarding left invariant vectorfields

Here my attempt to make sense of the definition but using a different notation. Notice that in the Lie group $G$ it holds that for any $x\in G$, $L_x:G\to G$ is a diffeomorphism. For any $p\in G$, the ...
Daniel B.'s user avatar
  • 181
2 votes
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Integral over a form

Welcome, Tanizaki! Keep in mind that the functions $ \theta _ i $ don't take only the values $ 0 $ and $ 1 $. The satisfy $ \sum _ i \theta _ i = 1 $ everywhere, even on an overlap $ U _ i \cap U _ j ...
Toby Bartels's user avatar
  • 4,714
0 votes

A property of the exponential map on flat torus

The problem is that you are using a presentation of the torus that obscures the simplicity of the issue. I would suggest that you use the (equivalent) presentation of the torus as a suitable ...
Mikhail Katz's user avatar
  • 42.5k
1 vote
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Principal eigenfunction of $\Delta$ on rectangular patch of $S^d$

As a not so trivial example take hyperspherical coordinates in $\mathbb R^4$. I use Mathematica for short $$\text{SeparatedLaplacian}(1)=\text{Expand}\left[ \frac{1}{R(r)\ \Theta (\theta ) R(r) \Psi (\...
Roland F's user avatar
  • 2,212
1 vote
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Parallel lines tangent to a strictly convex smooth plane body dividing its border in two arcs of equal length

Yes, it is true in general. Consider two arbitrary parallel lines $L1$ and $L2$ touching arbitrary strictly convex plane body $B$ with smooth border at points $P1$ and $P2$. Assume that arc $A1$, ...
Vladimir_U's user avatar
0 votes

Flat metric defined by an abelian differential form.

In short, $f(z)$ has zeros(and poles), which make conic singularities, the metric is flat outside those singularities. Imagine a cube, which is flat except 8 vertices. You can search with quadratic ...
Peter Wu's user avatar
  • 877
0 votes

Is this Proof on 1-Form with Compact Support Correct?

Your proof is not correct in either the one dimensional case nor the higher dimensional case unfortunately. Here is what goes wrong in your proof (and alternative proofs) Your proof of the $1-$...
Pelota's user avatar
  • 548
0 votes

Understanding an integrable almost complex structure

An almost complex structure is simply a smooth choice of complex structure $J$ on each tangent bundle. This is a weaker statement than the manifold being a complex manifold itself since we haven't ...
Callum's user avatar
  • 4,386
0 votes

How do I translate the description of geodesics of hyperbolic space in the hyperboloid model to the Poincaré ball and half-space models?

Here's how I would approach this: First, using the half-space and ball models only: Verify that in the half-space model vertical lines are geodesics and in the ball model lines through the origin are ...
Deane's user avatar
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0 votes

Using parallel transport to construct geodesically convex regions

One natural way of defining Minkowski "addition" on a complete Riemannian manifold $M$ would, I think, be the following: Let $\Omega \subset M$ be a compact domain with Lipschitz boundary ...
Deane's user avatar
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0 votes
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Do we need to use wedge product or tensor product for Curvature form in local frame?

$e^\alpha \otimes e_\beta$ pops up because $\Theta\in C^\infty(M, \Omega^2_M\otimes \text{End } E)$ is a 2-form which takes values in $\text{End } E \cong E^* \otimes E$ . $\Theta = \Theta^ \beta_{\...
Pengin's user avatar
  • 375
0 votes

$[a,b]$ has smooth boundary???

Yes, it is smooth. To see this, you can use one of the classical characterizations of smooth sub-manifolds of $\mathbb R^n$. Then, the set $\{a,b\}$ is a smooth ($C^\infty$) manifold of dimension $0$.
S. Maths's user avatar
  • 2,238
2 votes
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Generic intersection of hyperplanes

$\Rightarrow$: Existence of decomposition Take $L = V_1 \cap \cdots \cap V_n$. For each $i = 1, \cdots, n$, take $L_i$ such that $$L \oplus L_i = V_1 \cap \cdots \widehat{V_i} \cap \cdots \cap V_n$$ (...
dummy's user avatar
  • 500
3 votes

integral curves and Lie group actions

Suppose $X$ is a smooth vector field on $M$ with complete flow, meaning the flow map $\Phi:\Bbb{R}\times M\to M$ is defined everywhere. The meaning of the point $\Phi(t,x)\in M$ is that it is the ...
peek-a-boo's user avatar
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1 vote
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Will combining smooth functions by a partition of unity always give me a smooth function?

Let me fix some notation. Let $f_j: U_j \rightarrow \mathbb{R}$ be smooth functions and let $$ g_j: \mathbb{R}^2 \rightarrow \mathbb{R}, g_j(x)= \begin{cases} f_j(x),& x\in U_j, \\ 0,& x\in \...
Severin Schraven's user avatar
4 votes
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Why are geodesics between two points in hyperbolic space unique?

The real hyperbolic $n$-space is $$\mathbb H^n:=\{x \in \mathbb R^{n+1}: \langle x,x \rangle =-1 \text{ and }x_0>0\}$$ where $x=(x_0,x_1,\ldots,x_n)$ and the bilinear form is $$\langle x,y \rangle =...
Hugo's user avatar
  • 3,795
2 votes

Is a vector field a mathematical field?

No. Vector field and field (math) are quite different objects. And it quickly becomes a little trouble if we venture into physics, because in physics, a vector field is simply called a field, creating ...
user1286767's user avatar
13 votes
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Why do we need to prove extension lemmas?

These extension lemmas are a bit special to working with smooth things over $\mathbb{R}$. If you work with analytic things over $\mathbb{C}$, then for instance all global analytic functions on the ...
Daniël Apol's user avatar
  • 4,122
20 votes

Why do we need to prove extension lemmas?

The function $$\begin{array}{ccc} \mathbb R \setminus \{0\} & \longrightarrow & \mathbb R \\ x & \longmapsto & \dfrac 1x \end{array}$$ cannot be the restriction of a continuous ...
Héhéhé's user avatar
  • 1,431
1 vote

One technicality in proving partition and integral gives same definition of arclength

First, we restrict ourselves to a neighborhood in which $\exp$ is a diffeomorphism. Then let's define $\rho:[0, \delta)\to T_{\alpha(s_0)} M $ such that $\alpha(s_0 + t) = \exp_{\alpha(s_0)} (\rho(t))$...
Pengin's user avatar
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