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14

First, I think you have some misconceptions about what it means to be a "topological property." A topological property is a property that is preserved by homeomorphisms. Norms and metrics are definitely not topological properties. For example, the unit ball in $\mathbb R^n$ is homeomorphic (in fact, diffeomorphic) to $\mathbb R^n$ itself, but the two spaces ...


8

"Why is there no natural metric on manifolds?" Think of three physical variables, like $p$, $V$, $T$, dependent on each other via some natural law $$W(p, V, T)=0\ ,\tag{1}$$ where $W$ is an explicit expression in three variables (say, van der Waal's law) chosen by a professor of theoretical physics. The equation $(1)$ defines a two-dimensional manifold ...


7

Any two nonintersecting curves are transversal. Given $f:X\to Y$ and $Z\subset Y$ such that the image of $f$ doesn't intersect $Z$, the preimage of $Z$ under $f$ is empty. Then the condition "$Image(df_x)+T_{f(x)}(Z) = T_{f(x)}Y$ for all $x\in f^{-1}(Z)$" is trivially satisfied, because $f^{-1}(Z)$ is empty.


4

I think that you are missing an important point about differentiation (but don't feel ashamed, it's not an obvious point at all if you don't have enough background). The important thing here is that differentiation is a local concept, i.e. we have the concept of differentiation at a point, and this depends only on a(n arbitrarily small) open neighborhood of ...


4

The key is to find a set that is the image of two different injective immersions that induce different topologies on it. Here's a hint: $\qquad$


4

The two-holed torus is a topological two-manifold (i.e. a surface), so each point has a neighbourhood homeomorphic to an open subset of $\mathbb{R}^2$; this property is often called locally Euclidean of dimension two. If the two circles meeting at a point (often called the wedge sum of two circles, written $S^1\vee S^1$) were homeomorphic to the two-holed ...


4

No. For example, if $M=(0,1)$, then $M$ is closed in itself and embeds boundedly in $\Bbb{R}$, but is not compact. If $M$ were embedded as a closed submanifold of $\Bbb{R}^N$, the result you want would follow immediately. But there's no guarantee that $M$ has a closed embedding in $\Bbb{R}^N$ just because it has an embedding in $\Bbb{R}^N$.


3

Imagine a disc with a countably infinite number of parallel lines shading it. How to get from one of these lines to another?


3

If $M$ is $\ne S^2$ we can realize it as a "domain" in the plane ${\mathbb R}^2$. Both loops then appear as smooth closed curves in the plane, intersecting transversally. Since $\alpha$ is supposed simple, by Jordan's curve theorem it separates the plane into two open regions, one of them, call it $\Omega$, compact, and $\alpha=\pm\partial\Omega$. The sign ...


3

No it is not two-sided because the complement of $RP^n$ in $RP^{n+1}$ is an $(n+1)$-cell.


3

I found this series of video lectures on youtube that looks interesting. From the description: Note: Some of you may have studied point-set topology (metric and topological spaces, continuous maps, compactness, etc.). The content of this course is different: it is usually called algebraic and differential topology.


3

COMPLETE EDIT: Now that you have removed cohomology and the question has been clarified in my mind, I believe you are asking the following question. Suppose $\omega$ is a $k$-form with compact support contained in $U$ and there is a smooth form $\beta$ on U with $d\beta=\omega$ on $U$; is $\omega$ globally exact on $M$? The answer is NO, not ...


3

Here's one way to prove it. If you write $d\Phi_p(v) = a^i \partial/\partial x^i$ (using the summation convention), you can compute each coefficient $a^k$ by letting $d\Phi_p(v)$ act on the $k$-th coordinate function $x^k\colon \mathbb R^n\to \mathbb R$: $$ a^k = d\Phi_p(v)(x^k) = v (x^k \circ \Phi) = v(\Phi^k). $$ Part of the reason you had a hard time ...


3

As @InTransit suggested in a comment, O'Neill's phrase "and defined on an open subset of $\mathbb R^m$" seems to take care of the problem. Note that the subset of $\mathbb R^m$ on which $\psi\circ f \circ \varphi^{-1}$ is defined is $(f\circ\varphi^{-1})^{-1}(V) = \varphi(f^{-1}(V)\cap U)$. O'Neill's definition stipulates that this set is open in $\mathbb ...


2

Given a Riemannian metric, giving a compatible almost complex structure is equivalent to giving a reduction of the holonomy group from $O(n)$ to $U(n)$. An orientable surface must have holonomy group in $SO(2)$, which is just $U(1)$, so the parallel transport automatically preserves a compatible almost complex structure chosen at some point. And in two ...


2

I think there is something simpler. Let $x,y,z$ denote the coordinates in $\mathbb{R}^3$. By compactness, there is $p\in f(S^2)$ such that $z(p)$ is the maximal value of $z$ along $f(S^2).$ Any path $\alpha:(-\epsilon,\epsilon)\to f(S^2),\;t\mapsto(x(t),y(t),z(t))$ with $\alpha(0)=p$, satisfies $\dot{z}(0)=0$. It follows that ...


2

There is a similar post a few days ago asking why $$ T(\mathbb{S}^{2}\times \mathbb{S}^{1}) $$ is trivial. The idea is to use the extra dimension coming from $T(\mathbb{S}^{1})$ to fill in the place of the normal bundle. For your question I think it is similar but more subtle, as you have to consider carefully how the pull back bundle splits and how to place ...


2

I just want to point out that this is elementary. Consider $S^n \times [1,2] \subseteq \mathbb{R}^{n+1}$ let the vector fields be defined by $\mathbf{e}_i$ the constant fields in the coordinate directions. Now identify the points $(x,1)$ and $(x,2)$ this gives $S^n \times S^1$ the vector fields are equal at the identified points so this gives $n+1$ fields on ...


2

Since they both are smooth convex sets, the "central projection" maps: $$\pi:C\to S,\qquad \pi(x,y,z)=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}},$$ $$\pi^{-1}:S\to C,\qquad \pi^{-1}(x,y,z)=\sqrt{\frac{2}{(1-x^2)+\sqrt{1-2x^2+5x^4}}}\;(x,y,z)\tag{1}$$ are differentiable as a consequence of the Dini's theorem, since the "scale ratio" function is the inverse of a ...


2

The problem is that $U \times V$ won't just contain the arc that $(e^{ip}, e^{i \alpha p})$ lives on. Because $f(\Bbb{R})$ is dense in the torus, $U \times V$ contains lots of other arcs too, each of which forms a (local) path-component.


2

Metric is not a topological structure. You can only say a metric defines a metric topology, or a topological manifold is metrizable (the metrization theorems). 1) 2) 3) do not ensure a differentiable manifold structure, you also need the second countability axiom. You will benefit from local Euclidean property the following: locall connectedness: which ...


2

Yes, your curve is the image $\gamma (\mathbb R)$ of the global immersion $$ \gamma:\mathbb R \to \mathbb R^2:t\mapsto (t^2,t^3-t)$$ Edit The idea for the parametrization comes from algebraic geometry: Consider the one parameter family of lines $L_t=\{y=t(x-1)\}$ through $(1,0)$ and look at the other point of intersection of that line $L_t$ with the ...


2

This follows from the fact that pulling back by homotopic maps produces isomorphic bundles. Let $ p:E\to B $ and $ s:B\to E $ denote the projection and zero section, respectively. These are homotopy inverses. In particular $ sp $ is homotopic to the identity, and so $(sp)^*F=p^*s^*F =E\times_B F_B $ is isomorphic to $ F$.


2

As a follow up to the comment placed by the OP after the answer of Mark Grant, I would like to add some details. I hope the OP will find them an helpful complement to the response by Mark Grant. Let $\pi:E\longrightarrow M$ be a vector bundle, and $H$ an homotopy from $N$ to $M$ $$\begin{align}H:N\times[0,1]&\longrightarrow M\\(x,t)&\longmapsto ...


1

Let $c$ denote the curve in question, i.e. $c=\{y^2=x(x-1)^2\}\subset\mathbb{R}^2$. It is easy to verify that $c\setminus\{(1,0)\}$ is a smooth curve (around every point there is a neighborhood with a smooth parametrization). Furthermore, each one of the two "branches" at $(1,0)$ has a smooth parametrization. For example: $$t\mapsto (t,\sqrt{t}(t-1))$$ and ...


1

No, $C^0$ denotes the ring of continuous functions, as opposed to $C^k$, $k\ge 1$, which denotes functions with up to $k$th order derivatives continuous.


1

You've got 99% of the proof already. Hint: Suppose $d\pi_M|_S$ is not surjective and think about the isomorphisms $T_{(p,q)} S+ T_{(p,q)}(\{p\} \times N) \cong T_{(p,q)}(M \times N) \cong T_p M \times T_q N$. Click/rollover for full answer below:


1

Another way to map $S^2$ diffeomorphically to $W^2=\{x^4+y^2+z^2=1\}\subset\mathbb{R}^3$ is by $$(x,y,z)\mapsto(x,\sqrt{1+x^2}\cdot y,\sqrt{1+x^2}\cdot z).$$ This map is obviously smooth with a smooth inverse, and it is well defined as shown immediately.


1

It's probably more accurate to say that inward-pointing vectors are those with positive $x^n$ component. In the given boundary chart, each element of $T_pM$ can be written $v = v^i\partial/\partial x^i$ (using the summation convention), and $v$ is inward-pointing if and only if $v^i>0$.


1

A tangent vector $v\in T_pM$ does not have any coordinates. However, if we have a smooth boundary chart $\mathbf{X}:U\subset H^n\to M$ with $\mathbf{X}(0)=p$, then the derivative $d\mathbf{X}$ is an isomorphism $\mathbb{R}^n\to T_pM$. Now, in $\mathbb{R}^n$ it is clear enough whether a given tangent vector points up, down, or none of them. In other words, a ...



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