# Tag Info

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 ...

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

The definition of a manifold as such does not take metric structures into account. I just requires local diffeomorphism (homeomorphic in case of topological manifolds) with some $\mathbb{R}^n$, which sometimes (sloppily) is rephrased as locally diffeomorphic to a Euclidean space. This is still correct, cause a diffeomorphism does not have to preserve scalar ...

4

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 ...

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 ...

3

I'm going to assume that you by neighbourhood mean an open neighbourhood, else substitute that in the proof. Then observe that since $M$ is compact all closed subsets are compact. Thus since $N$ is hausdorff and f is continous the image of a closed set must be compact hence closed, thus you get $f(M-U_1-\dots-U_n)$ is closed. But an open set minus a closed ...

3

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

Let me summarize my comments in an answer. I don't have answers to all the questions but I don't expect (myself) to really come up with any more, so here goes. First, I am mostly thinking about this space as the algebraic variety over $\mathbb{R}$ (complex conjugation is not $\mathbb{C}$-scalar) cut out by the $n^2$ (in the $\mathbb{R}$ case) or $2n^2$ (in ...

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 ... 2 There is a similar post a few days ago asking why T(\mathbb{S}^{2}\times \mathbb{S}^{1}) $$is trivial. The idea is to use the extra dimension coming from T(\mathbb{S}^{1}) to fill in the place of the normal bundle. For your question I think it is similar but more subtle, as you have to consider carefully how the pull back bundle splits and how to place ... 2 I just want to point out that this is elementary. Consider S^n \times [1,2] \subseteq \mathbb{R}^{n+1} let the vector fields be defined by \mathbf{e}_i the constant fields in the coordinate directions. Now identify the points (x,1) and (x,2) this gives S^n \times S^1 the vector fields are equal at the identified points so this gives n+1 fields on ... 2 No, not in general. For example, the image of a circle in a plane under a homeomorphism of athe plane can be a curve of positive area. Indeed, here you can see the construction of a simple curve of positive area. It can be made closed by adding an arc, and then, by the Schoenflies theorem, it is the image of a circle under a homeomorphism of \mathbb R^2. ... 2 To show that g^*(U\times \mathbb R^m) is open in TM, it suffices to show that$$g^*(U\times \mathbb R^m) = TM \cap (W\times\mathbb R^k),$$because W\times\mathbb R^k is open in M\times\mathbb R^k, and TM has the subspace topology it inherits as a subset of M\times\mathbb R^k. It follows from your definitions that g^*(U\times \mathbb R^m) ... 2 Sure: w(\xi_1) \cdots w(\xi_n) \overline{w}(\xi_1) \cdots \overline{w}(\xi_n) = w(\xi_1) \overline{w}(\xi_1) \cdots w(\xi_n)\overline{w}(\xi_n) = 1\in H^*(M, \mathbb{Z}_2). 2 You don't need to define the Taylor development for functions M \to N since you're working in the coordinate charts he mentions in the definition. For example if the coordinate chart about p is \phi : U \subset M \to \mathbb R^m and likewise \psi : V \subset N \to \mathbb R^n about f(p) then he is talking about the Taylor polynomials of the maps ... 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 ... 1 First, in the theorem, you have to assume that manifolds are connected. Next, it is a theorem (Rado and Caratheodory, I think), that every topological surface has a smooth structure and that two surfaces are homeomorphic if and only if they are diffeomorphic. See here for more references than you probably wanted. As for Riemann surfaces: if you want to ... 1 HINT: One approach is to apply the Local Immersion Theorem in Guillemin/Pollack twice: first to X and, then, working locally in the appropriate parametrization of X, to the inclusion Z\hookrightarrow X. EDIT: So the Local Immersion Theorem tells us that we can choose local coordinates (x_1,\dots,x_k) on X and Z so that the inclusion map f\colon ... 1 MJD, since you refer back to the original question, and given your previous comments, I assume you are trying to be provocative. However, your question does address a very common misconception. In a sense, the same one addressed by the original question. It is common to think that the Mobius strip that can be modeled by giving a strip of paper a half-twist ... 1 I'm assuming that your end goal is the computation of w_2(TS(E)), rather than that of the Wu class. If so, here's how I'd proceed. First, I'll abuse notation and let p:S(E)\rightarrow \mathbb{T}^2 denote the projection map. First note that \pi_1(\mathbb{T}^2)\cong\mathbb{Z}^2 acts trivially on H^\ast(S^3). This follows because the bundle is ... 1 To simplify, denote$$X=\sum_{i=1}^nx^i\frac{\partial}{\partial x^i} and $\Omega=dx^1\wedge dx^2 \cdots \wedge dx^n$, then $\sigma=i(X)\Omega$, where $i$ is the Interior Product operator, that is, $\sigma(\cdots)=\Omega(X,\cdots)$. $\forall p\in \mathbb{R}^n$, let $q=r(p)$. Note that $X$ has the following property: ...

1

My favorite reference for this (Poincare duality via differential forms) is Bott and Tu "Differential Forms in Algebraic Topology", Chapter 1, section 5.

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