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Chapter 3: Problem 20

(a) Consider the space obtained from \([0,1] \times \mathbb{R}^{n}\) by identifying \((0, v)\) with \((1, T v)\), where \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) is a vector space isomorphism. Show that this can be made into the total space of a vector bundle over \(S^{1}\) (a generalized M枚bius strip). (b) Show that the resulting bundle is orientable if and only if \(T\) is orientation preserving.

Short Answer

Expert verified
The space is a vector bundle over \( S^{1} \) with fiber \( \mathbb{R}^{n} \. \) The bundle is orientable if and only if \( T \) is orientation-preserving.

Step by step solution

01

Understand the Space Definition

The space provided is \( [0,1] \times \mathbb{R}^{n} \) with the points \( (0,v) \) and \( (1, Tv) \) being identified. This suggests we are dealing with a fiber bundle over a base space which, in this case, is \( S^{1} \) obtained by identifying the ends of the interval \( [0,1] \).
02

Recognize the Base Space

Since the top and bottom of \( [0,1] \) are identified, it rolls up into a circle, making \( S^{1} \) the base space of the bundle. Consider this interval as the base of the generalized M枚bius strip.
03

Define the Fiber

The fiber over any point in the interval \( [0,1] \) is \( \mathbb{R}^{n} \.\) This means that for every \( t \in [0, 1] \), we attach a vector space \( \mathbb{R}^{n}\) at that point.
04

Bundle Construction

To construct the bundle, observe that moving from \( t = 0 \) to \( t = 1 \) results in mapping points \( (0, v) \) to \( (1, Tv) \.\) Therefore, the fibers at \( t = 0 \) and \( t = 1 \) are connected through the isomorphism \( T \.\)
05

Verification of Vector Bundle

The total space of the vector bundle is \( ([0,1] \times \mathbb{R}^{n}) /\sim\) where \( (0,v) \sim (1,Tv) \.\) We verify that this setup satisfies the structure of a vector bundle over \( S^{1} \) with fibers being \( \mathbb{R}^{n} \.\)
06

Argument for Orientability

A bundle is orientable if there exists a consistent choice of orientation in the fibers throughout the entire base space. Here, \( T \) must preserve orientation to ensure that applying \( T \) across the identification \( (0,v) \sim (1,Tv) \) does not reverse orientation.
07

Conclusion on Orientability

Therefore, the bundle is orientable if and only if \( T \) is orientation-preserving.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vector bundle
A vector bundle is a collection of vector spaces parametrized by another space, called the base space. Each point of the base space has a vector space attached to it, called the fiber. You can think of it like having a vector space hanging over every point in the base. For example, in the exercise provided, we have a vector bundle over the circle \( S^1 \), where each point on the circle has an \( \mathbb{R}^n \) vector space attached. The construction makes sure that the space smoothly varies as you move along the base.
orientation preserving
A transformation is orientation preserving if it maintains the handedness of the basis vectors. In simpler words, it doesn't flip the space inside out. For a space to be orientable, there must be a consistent way to choose an orientation for all its fibers, in every little neighborhood. In the context of the given problem, the vector space isomorphism \( T \) must be orientation-preserving for the whole bundle over \( S^1 \) to be orientable. If \( T \) reverses the orientation, the resulting bundle would not uniformly maintain that same 'handedness' throughout.
fiber bundle
A fiber bundle is a more general concept than a vector bundle. It consists of a base space, a total space, and a projection map, all glued together in a way that locally looks like a product space. The given exercise involves fiber bundles because the interval \([0,1]\) with its ends identified as a circle \( S^1 \) serves as the base space. The total space is the product \([0,1] \times \mathbb{R}^n\), and the map projects points in the total space down to the base space \( S^1 \). This bundle, where fibers over each point are vector spaces, qualifies as a vector bundle.
M枚bius strip
A M枚bius strip is a classic example of a non-orientable surface created by taking a rectangular strip of paper, giving it a half-twist, and then joining the ends together. In the exercise, a generalized M枚bius strip structure is suggested by the identification \((0,v) \sim (1,Tv)\). But, this forms a vector bundle over \( S^1 \) rather than a simple M枚bius strip. The twist in the usual M枚bius strip makes it non-orientable. By ensuring \( T \) is orientation preserving, we avoid such reversals that make the bundle non-orientable.
vector space isomorphism
A vector space isomorphism is a linear mapping between two vector spaces that preserves the operations of vector addition and scalar multiplication. It essentially means you can transform the vector space without distorting its structure. In the exercise, the transformation \( T: \mathbb{R}^n \to \mathbb{R}^n\) is a vector space isomorphism, indicating that \( T \) reshuffles \( \mathbb{R}^n \) in a fully reversible way. For the bundle to be orientable, \( T \) must preserve orientations, meaning it must not flip the base in ways that could alter handedness.

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Most popular questions from this chapter

(a) Show that for any bundle \(\pi: E \rightarrow B\), the map \(s: B \rightarrow E\) with \(s(p)\) the 0 vector of \(\pi^{-1}(p)\) is a section. (b) Show that an \(n\)-plane bundle \(\xi\) is trivial if and only if there are \(h\) sections \(s_{1}, \ldots, s_{n}\) which are everywhere linearly independent, i.e., \(s_{1}(p), \ldots, s_{n}(p) \in\) \(\pi^{-1}(p)\) are linearly independent for all \(p \in B\). (c) Show that locally every \(n\)-plane bundle has \(n\) linearly independent sections.

(a) If \(M\) and \(N\) are \(C^{\infty}\) manifolds, and \(\pi_{M}\) [or \(\left.\pi_{N}\right]: M \times N \rightarrow M\) [or \(\left.N\right]\) is the projection on \(M\) [or \(N]\), then \(T(M \times N) \simeq \pi_{M}^{*}(T M) \oplus \pi_{N}^{*}(T N)\). (b) If \(M\) and \(N\) are orientable, then \(M \times N\) is orientable. (c) If \(M \times N\) is orientable, then both \(M\) and \(N\) are orientable.

Let \(M^{n} \subset \mathbb{R}^{N}\) be a \(C^{\infty} n\)-dimensional submanifold. By a chord of \(M\) we mean a point of \(\mathbb{R}^{N}\) of the form \(p-q\) for \(p, q \in M\). (a) Prove that if \(N>2 n+1\), then there is a vector \(v \in S^{N-1}\) such that (i) no chord of \(M\) is parallel to \(v\), (ii) no tangent plane \(M_{p}\) contains \(v\).(b) Let \(\mathbb{R}^{N-1} \subset \mathbb{R}^{N}\) be the subspace perpendicular to \(v\), and \(\pi: \mathbb{R}^{N} \rightarrow \mathbb{R}^{N-1}\) the corresponding projection. Show that \(\pi \mid M\) is a one-one immersion. In particular, if \(M\) is compact, then \(\pi \mid M\) is an imbedding. (c) Every compact \(C^{\infty} n\)-dimensional manifold can be imbedded in \(\mathbb{R}^{2 n+1}\). Note: This is the easy case of Whitney's classical theorem, which gives the same result even for non-compact manifolds (H. Whitney, Differentiable manifolds, Ann. of Math. \(37(1935), 645-680)\). Proofs may be found in Auslander and MacKenzie, Introduction to Differentiable Manifolds and Sternberg, Lectures on Differential Geometry. In Munkres, Elementary Differential Topology, there is a different sort of argument to prove that a not-necessarily-compact \(n\)-manifold \(M\) can be imbedded in some \(\mathbb{R}^{N}\) (in fact, with \(\left.N=(n+1)^{2}\right) .\) Then we may show that \(M\) imbeds in \(\mathbb{R}^{2 n+1}\) using essentially the argument above, together with the existence of a proper map \(f: M \rightarrow \mathbb{R}\), given by Problem '2-30 (compare Guillemin and Pollack, Differential Topology). A much harder result of Whitney shows that \(M^{n}\) can actually be imbedded in \(\mathbb{R}^{2 n}\) (H. Whitney, The self-intersections of a smooth n-manifold in \(2 n\)-space, An?. of Math. \(45(1944), 220-246) .\) Hint: Consider certain naps from appropriate open subsets of \(M \times M\) and \(T M\) to \(S^{N-1}\)

(a) If \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) is a linear transformation, \(T^{*}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\), the adjoint of \(T\), is defined by \(\left(T^{*} v, w\right\rangle=(v, T w\rangle\) (for each \(v\), the map \(w \mapsto(v, T w\rangle\) is linear, so it is \(w \mapsto\left(T^{*} v, w\right\rangle\) for a unique \(\left.T^{*} v\right)\). If \(A\) is the matrix of \(T\) with respect to the usual basis, show that the matrix of \(T^{*}\) is the transpose \(A^{\mathrm{t}}\). (b) A linear transformation \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) is self-adjoint if \(T=T^{*}\), so that \((T v, w\rangle=(v, T w\rangle\) for all \(v, w \in \mathbb{R}^{n} .\) If \(A\) is the matrix of \(T\) with respect to the standard basis, then \(T\) is self-adjoint if and only if \(A\) is symmetric, \(A^{\mathbf{t}}=A\). It is a standard theorem that a symmetric \(A\) can be written as \(C D C^{-1}\) for some diagonal matrix \(D\) (for an analytic proof, see Calculus on Manifolds, pg. 122). Show that \(C\) can be chosen orthogonal, by showing that eigenvectors for distinct eigenvalues are orthogonal. (c) A self-adjoint \(T\) (or the corresponding symmetric \(A\) ) is called positive semidefinite if \((T v, v\rangle \geq 0\) for all \(v \in \mathbb{R}^{n}\), and positive definite if \((T v, v\rangle>0\) for all \(v \neq 0\). Show that a positive definite \(A\) is non-singular. Hint: Use the Schwarz inequality. (d) Show that \(A^{\mathbf{t}} \cdot A\) is always positive semi-definite. (e) Show that a positive semi-definite \(A\) can be written as \(A=B^{2}\) for some \(B\). (Remember that \(A\) is symmetric.) (f) Show that every \(A \in \mathrm{GL}(n, \mathbb{R})\) can be written uniquely as \(A=A_{1} \cdot A_{2}\) where \(A_{1} \in \mathrm{O}(n)\) and \(A_{2}\) is positive definite. Hint: Consider \(A^{\mathrm{t}} \cdot A\), and use part (e). (g) The matrices \(A_{1}\) and \(A_{2}\) are continuous functions of \(A\). Hint: If \(A^{(n)} \rightarrow A\) and \(A^{(n)}=A^{(n)} 1 \cdot A^{(n)} 2\), then some subsequence of \(\left\\{A^{(n)}\right\\}\) converges. (h) \(\mathrm{GL}(n, \mathbb{R})\) is homeomorplic to \(\mathrm{O}(n) \times \mathbb{R}^{n(n+1) / 2}\) and has exactly two components, \(\\{A: \operatorname{det} A>0\\}\) and \(\\{A: \operatorname{det} A<0\\}\). (Notice that this also gives us another way of finding the dimension of \(\mathrm{O}(n) .\) )

(a) If \(M\) is a manifold-with-boundary, the tangent bundle \(T M\) is defined exactly as for \(M\); elements of \(M_{p}\) are \(\tilde{p}\) equivalence classes of pairs \((x, v)\). Although \(x\) takes a neighborhood of \(p \in \partial M\) onto \(\mathbb{H}^{n}\), rather than \(\mathbb{R}^{n}\), the vectors \(v\) still run through \(\mathbb{R}^{n}\), so \(M_{p}\) still has tangent vectors "pointing in all directions". If \(p \in \partial M\) and \(x: U \rightarrow \mathbb{H}^{n}\) is a coordinate system around \(p\), then \(x_{*}^{-1}\left(\mathbb{R}^{n-1} x(p)\right) \subset M_{p}\) is a subspace. Show that this subspace does not depend on the choice of \(x\); in fact, it is \(I_{*}(\partial M)_{p}\), where \(i: \partial M \rightarrow M\) is the inclusion. (b) Let \(a \in \mathbb{R}^{n-1} \times\\{0\\} \subset \mathbb{M}^{n} .\) A tangent vector in \(\mathbb{M}^{n}{ }_{a}\) is said to point "invard" if, under the identification of \(T \mathbb{M}^{n}\) with \(\varepsilon^{n}\left(\mathbb{M}^{n}\right)\), the vector is \((a, v)\) where \(v^{n}>0 .\) A vector \(v \in M_{p}\) which is not in \(i_{*}(\partial M)_{p}\) is said to point "inward" if \(x_{*}(v) \in \mathbb{M}^{n} x(p)\) points inward. Show that this definition does not depend on the coordinate system \(x\). (c) Show that if \(M\) has an orientation \(\mu\), then \(\partial M\) has a unique orientation \(\partial \mu\) such that \(\left[v_{1}, \ldots, v_{n-1}\right]=(\partial \mu)_{p}\) if and only if \(\left[w, i_{*} v_{1}, \ldots, i_{*} v_{n-1}\right]=\mu_{p}\) for every outward pointing \(w \in M_{p}\) (d) If \(\mu\) is the usual orientation of \(\mathbb{H}^{n}\), show that \(\partial \mu\) is \((-1)^{n}\) times the usual orientation of \(\mathbb{R}^{n-1}=\partial \mathbb{H}^{n} .\) (The reason for this choice will become clear in Chapter 8.) (e) Suppose we are in the setup of Problem 2-14. Define \(g: \partial M \times[0,1) \rightarrow\) \(\partial N \times[0,1)\) by \(g(p, t)=(f(p), t) .\) Show that \(T P\) is obtained from \(T M \cup T N\) $$ v \in(\partial M)_{p} \quad \text { with }\left(\beta^{-1}\right)_{*} g_{*} \alpha_{*}(v) \in(\partial N)_{f(p)} $$ (f) If \(M\) and \(N\) have orientations \(\mu\) and \(\nu\) and \(f:(\partial M, \partial \mu) \rightarrow(\partial N, \partial \nu)\) is orientation-reversing, show that \(P\) has an orientation which agrees with \(\mu\) and \(\nu\) on \(M \subset P\) and \(N \subset P\). (g) Suppose \(M\) is \(S^{2}\) with two holes cut out, and \(N\) is \([0,1] \times S^{1}\). Let \(f\) be a diffeomorphism from \(M\) to \(N\) which is orientation preserving on one copy of \(S^{1}\) alnd orientation reversing on the other. What is the resulting manifold \(P ?\)
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