/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Let $$ \mathcal{C}:=\mathcal... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 20

Let $$ \mathcal{C}:=\mathcal{H} \times \mathcal{H} $$ Consider the following composition law ("product") $$ \begin{aligned} \mathcal{C} \times \mathcal{C} & \longrightarrow \mathcal{C} \\ \left(\left(H_{1}, H_{2}\right),\left(K_{1}, K_{2}\right)\right) & \mapsto\left(H_{1} K_{1}-\bar{K}_{2}^{\prime} H_{2}, H_{2} \bar{K}_{1}^{\prime}+K_{2} H_{1}\right) \end{aligned} $$ Here \(\bar{H}^{\prime}\) denotes the adjoint (conjugate transpose) matrix of \(H \in \mathcal{H} \subset M(2 \times\) \(2 ; \mathbb{C})\) Show that this defines on \(\mathcal{C}\) an \(\mathbb{R}\)-bilinear map, which has no divisors of zero (or is non-degenerated), i.e. the "product" of two elements in \(\mathcal{C}\) is zero, iff one of the two factors vanishes. This "CAYLEY multiplication" is, in general, neither commutative nor associative. A deep theorem (M. A. KERVAIRE (1958), J. MILNOR (1958), J. BOTT (1958)) says that on an \(n\)-dimensional \((n<\infty)\) real vector space \(V\) a bilinear form free of divisors of zero can only exist when \(n=1,2,4\) or \(8 .\) Examples of such structures are the "real numbers", the "complex numbers", the "HAMILTONian quaternions" and the "CAYLEY numbers". Compare with the article of F. HIRZEBRUCH, [Hi].

Short Answer

Expert verified
Verify bilinearity and zero divisor conditions separately and validate these properties of the Cayley multiplication.

Step by step solution

01

Define the Structure

We start by defining \( \mathcal{C} = \mathcal{H} \times \mathcal{H} \), where \( \mathcal{H} \) is a subset of \( M(2 \times 2; \mathbb{C}) \). \( \mathcal{C} \) would then be a 4-dimensional real vector space since each element is composed of two 2x2 matrices.
02

Understand the Composition Law

The composition law is defined such that if we take two pairs of elements \( (H_1, H_2) \) and \( (K_1, K_2) \) in \( \mathcal{C} \), the product is defined by \( (H_1 K_1-\bar{K}_2^{\prime} H_2, H_2 \bar{K}_1^{\prime}+K_2 H_1) \). This operation serves as the 'Cayley multiplication' we need to investigate.
03

\\(\mathbb{R} \\)-Bilinearity Verification

To show \( \mathbb{R} \)-bilinearity, we need to verify linearity in both components separately. Checking linearity in the first component, we evaluate \( ((aH_1+bG_1, H_2), (K_1, K_2)) = (a(H_1K_1-\bar{K}_2^{\prime}H_2) + b(G_1K_1-\bar{K}_2^{\prime}H_2), H_2\bar{K}_1^{\prime}+K_2(aH_1+bG_1)) \). Similarly for the second, evaluate appropriately.
04

Checking Zero Divisors

Assume \( (H_1, H_2) \cdot (K_1, K_2) = (0, 0) \). From the first component, \( H_1K_1 - \bar{K}_2^{\prime}H_2 = 0 \) and the second \( H_2 \bar{K}_1^{\prime} + K_2H_1 = 0 \). Check if either \( (H_1, H_2) = (0, 0) \) or \( (K_1, K_2) = (0, 0) \) must necessarily be zero, confirming no zero divisors.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Real Vector Space
A real vector space is a fundamental concept in linear algebra. It is essentially a collection of vectors where you can perform vector addition and scalar multiplication. In this exercise, the set \(\mathcal{C} = \mathcal{H} \times \mathcal{H}\) forms a real vector space. To understand why, consider the following:
  • Each element in \(\mathcal{C}\) consists of a pair of matrices from \(\mathcal{H}\), which is a subset of 2x2 complex matrices \(M(2 \times 2; \mathbb{C})\).
  • This structure allows us to perform operations like addition, where two pairs \((H_1, H_2)\) and \((K_1, K_2)\) add to \((H_1 + K_1, H_2 + K_2)\).
  • Scalar multiplication in this space means multiplying each matrix in the pair by a real number \(a\), resulting in \((aH_1, aH_2)\).
Thus, \(\mathcal{C}\) satisfies the properties of a vector space over the field of real numbers, making it a 4-dimensional real vector space.
Bilinear Map
A bilinear map is a type of function where linearity is preserved separately in each of two variables. The Cayley multiplication here defines such a map on \(\mathcal{C}\). To understand how this map works, note the following:
  • The map combines two pairs of matrices \((H_1, H_2)\) and \((K_1, K_2)\) into another pair of matrices \((H_1 K_1 - \bar{K}_2^{\prime} H_2, H_2 \bar{K}_1^{\prime} + K_2 H_1)\).
  • For \(\mathbb{R}\)-bilinearity, checking linearity in one component means ensuring that the operation respects scaling and addition. For instance, when scaling by \(a\) or adding another element, the result respects the same properties.
  • Since this holds for both inputs separately, the map is bilinear.
This bilinear feature allows the Cayley multiplication to be consistent over its domain in a way that blends well mathematically with real vector spaces.
Zero Divisors Free
The term 'zero divisors free' means that if the product of two elements is zero, then at least one of the elements must be zero. This property is crucial as it implies that the space is non-degenerate. Here's how it applies here:
  • Given two pairs \((H_1, H_2)\) and \((K_1, K_2)\), their product being zero is defined as \((H_1 K_1 - \bar{K}_2^{\prime} H_2, H_2 \bar{K}_1^{\prime} + K_2 H_1) = (0, 0)\).
  • For this equation to hold, calculations would reveal that it requires either \((H_1, H_2)\) or \((K_1, K_2)\) to be \((0, 0)\).
    • Checking mathematical properties affirm that neither part of the operation contributes its effect if one factor is zero.
Thus, possessing no zero divisors means that the set \(\mathcal{C}\) does not allow trivial solutions without involving genuine zero factors. This helps maintain its mathematical integrity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(f: D \rightarrow \mathbb{C}\) is analytic, \(D \subseteq \mathbb{C}\) is open, and one of the following conditions holds: (a) Re \(f=\) constant, (b) \(\operatorname{Im} f=\) constant, (c) \(|f|=\) constant, then \(f\) is locally constant.

Let \(\widetilde{\mathbb{C}}\) be another field of complex numbers. Determine all mappings \(\varphi: \mathbb{C} \rightarrow \widetilde{\mathbb{C}}\) with the following properties; \(\begin{aligned}(a) & \varphi(z+w) &=\varphi(z)+\varphi(w) & & \text { for all } z, w \in \mathbb{C} \\\\(b) & \varphi(z w) &=\varphi(z) \varphi(w) & & \text { for all } z, w \in \mathbb{C} \\\\(c) & \varphi(x) &=x & & \text { for all } x \in \mathbb{R} \end{aligned}\) Remark. It turns out that such mappings exist, and they are automatically bijective; thus they give isomorphisms \(\mathbb{C} \rightarrow \widetilde{\mathbb{C}}\) that leave \(\mathbb{R}\) fixed element by element. The field of complex numbers is therefore essentially uniquely determined. In the special case \(\mathbb{C}=\widetilde{\mathbb{C}}\) we get automorphisms of \(\mathbb{C}\) with the fixed field \(\mathbb{R}\). Remark: What automorphisms (i.e. isomorphisms with itself) does the real field \(\mathbb{R}\) have ? Hint: Such an automorphism of \(\mathbb{R}\) must preserve the ordering of \(\mathbb{R} !\)

Let \(\left(x_{n}\right)_{n \geq 0}\) be a sequence in \(\mathbb{R}^{p} . a \in \mathbb{R}^{p}\) is called an accumulation value of the sequence \(\left(x_{n}\right)\) if for each \(\varepsilon\)-disk \(U_{\varepsilon}(a)\) there are infinitely many indices \(n\) such that \(x_{n} \in U_{\varepsilon}(a)\). Show (BOLZANO-WEIERSTRASS Theorem): Any bounded sequence \(\left(x_{n}\right), x_{n} \in\) \(\mathbb{R}^{p}\) has an accumulation point. A subset \(K \subseteq \mathbb{R}^{p}\) is called sequentially compact if each sequence \(\left(x_{n}\right)_{n \geq 0}\) with \(x_{n} \in K\) has (at least) one accumulation point in \(K\) Show: For a subset \(K \subseteq \mathbb{R}^{p}\) the following are equivalent: (a) \(K\) is compact, (b) \(K\) is sequentially compact. Remark. These equivalences hold for any metric space.

There is no continuous function \(l: \mathbb{C}^{\bullet} \rightarrow \mathbb{C}\) such that $$ \exp (l(z))=z \text { for all } z \in \mathbb{C}^{\bullet} $$

Which is the maximal open set \(D \subseteq \mathbb{C}\), such that \(f: D \rightarrow \mathbb{C}, f(z):=\log \left(z^{5}+\right.\) 1), is well defined and analytic.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.