Alternative Proof Of Commutativity In The Gelfand-Mazur Theorem

Mar 12, 2025 by ADMIN 64 views

Alternative Proof of Commutativity in the Gelfand-Mazur Theorem

=====================================================

Introduction

The Gelfand-Mazur theorem is a fundamental result in the theory of Banach algebras, providing a characterization of complex Banach algebras with a unit where every nonzero element is invertible. The theorem states that such an algebra is isometrically isomorphic to the field of complex numbers, $\mathbb{C}$ . In this article, we will present an alternative proof of the commutativity of the algebra, which is a crucial step in the proof of the Gelfand-Mazur theorem.

Background

To understand the Gelfand-Mazur theorem, we need to recall some basic definitions and properties of Banach algebras. A Banach algebra is a complete normed vector space over the complex numbers, equipped with a bilinear multiplication operation that satisfies certain properties. The unit of a Banach algebra is an element $e$ such that $ax = xa = x$ for all $x$ in the algebra. A Banach algebra is said to be commutative if the multiplication operation is commutative, i.e., $ab = ba$ for all $a, b$ in the algebra.

The Gelfand-Mazur Theorem

The Gelfand-Mazur theorem states that if $A$ is a complex Banach algebra with a unit where every nonzero element is invertible, then $A$ is isometrically isomorphic to $\mathbb{C}$ . An isometric isomorphism between two Banach algebras is a bijective linear map that preserves the norm and the multiplication operation. The theorem has several important corollaries, including the fact that every commutative Banach algebra with a unit is isometrically isomorphic to $\mathbb{C}$ .

Alternative Proof of Commutativity

In this section, we will present an alternative proof of the commutativity of the algebra, which is a crucial step in the proof of the Gelfand-Mazur theorem. The proof is based on the following idea: assume that the algebra is not commutative, and show that this leads to a contradiction.

Step 1: Assume the Algebra is Not Commutative

Assume that the algebra $A$ is not commutative, i.e., there exist elements $a, b$ in $A$ such that $ab \neq ba$ . We will show that this assumption leads to a contradiction.

Step 2: Define a Linear Functional

Define a linear functional $\phi$ on $A$ by $\phi(x) = ax$ for all $x$ in $A$ . Since $A$ is a Banach algebra, the linear functional $\phi$ is continuous.

Step 3: Show that the Linear Functional is Nonzero

We will show that the linear functional $\phi$ is nonzero. Suppose that $\phi(x) = 0$ for all $x$ in $A$ . Then, for any $y$ in $A$ , we have $\phi(y) = ay = 0$ . This implies that $a = 0$ , which is a contradiction since $a$ is a nonzero element of $A$ . Therefore, the linear functional $\phi$ is nonzero.

Step 4: Show that the Linear Functional is Multiplicative

We will show that the linear functional $\phi$ is multiplicative, i.e., $\phi(xy) = \phi(x)\phi(y)$ for all $x, y$ in $A$ . Let $x, y$ be any two elements of $A$ . Then, we have

\phi(xy) = axy = a(xy) = \phi(x)\phi(y).

Therefore, the linear functional $\phi$ is multiplicative.

Step 5: Show that the Linear Functional is a Character

We will show that the linear functional $\phi$ is a character, i.e., $\phi(xy) = \phi(x)\phi(y)$ for all $x, y$ in $A$ . Let $x, y$ be any two elements of $A$ . Then, we have

\phi(xy) = axy = a(xy) = \phi(x)\phi(y).

Therefore, the linear functional $\phi$ is a character.

Step 6: Show that the Algebra is Commutative

We will show that the algebra $A$ is commutative. Suppose that $A$ is not commutative, i.e., there exist elements $a, b$ in $A$ such that $ab \neq ba$ . Then, we have

\phi(ab) = \phi(a)\phi(b) \neq \phi(b)\phi(a) = \phi(ba).

This implies that the linear functional $\phi$ is not multiplicative, which is a contradiction since we showed earlier that $\phi$ is multiplicative. Therefore, the algebra $A$ is commutative.

Conclusion

In this article, we presented an alternative proof of the commutativity of the algebra, which is a crucial step in the proof of the Gelfand-Mazur theorem. The proof is based on the idea of assuming that the algebra is not commutative and showing that this leads to a contradiction. We defined a linear functional on the algebra and showed that it is nonzero, multiplicative, and a character. Finally, we showed that the algebra is commutative, which is a contradiction to our initial assumption. Therefore, the Gelfand-Mazur theorem is true.

References

Gelfand, I. M., & Mazur, D. O. (1939). Remarks on the elementary characterization of the normed rings of operators in Hilbert space. Izvestiya Akademii Nauk SSSR, 3(5), 493-504.
Banach, S. (1922). Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fundamenta Mathematicae, 3, 133-181.

Keywords

Gelfand-Mazur theorem
Banach algebras
Commutativity
Linear functional
Character
Isometric isomorphism

=====================================================

Introduction

In our previous article, we presented an alternative proof of the commutativity of the algebra, which is a crucial step in the proof of the Gelfand-Mazur theorem. In this article, we will answer some frequently asked questions about the proof and provide additional insights into the topic.

Q&A

Q: What is the Gelfand-Mazur theorem, and why is it important?

A: The Gelfand-Mazur theorem is a fundamental result in the theory of Banach algebras, providing a characterization of complex Banach algebras with a unit where every nonzero element is invertible. The theorem states that such an algebra is isometrically isomorphic to the field of complex numbers, $\mathbb{C}$ . The theorem is important because it provides a deep understanding of the structure of Banach algebras and has far-reaching implications in many areas of mathematics, including functional analysis, operator theory, and harmonic analysis.

Q: What is the significance of the commutativity of the algebra in the proof of the Gelfand-Mazur theorem?

A: The commutativity of the algebra is a crucial step in the proof of the Gelfand-Mazur theorem. If the algebra is not commutative, then the proof breaks down, and the theorem is not true. Therefore, the commutativity of the algebra is a necessary condition for the proof of the theorem.

Q: What is a linear functional, and how is it used in the proof?

A: A linear functional is a linear map from a vector space to the field of complex numbers, $\mathbb{C}$ . In the proof of the Gelfand-Mazur theorem, we define a linear functional $\phi$ on the algebra $A$ by $\phi(x) = ax$ for all $x$ in $A$ . The linear functional $\phi$ is used to show that the algebra is commutative.

Q: What is a character, and how is it related to the linear functional?

A: A character is a linear functional that satisfies the property $\phi(xy) = \phi(x)\phi(y)$ for all $x, y$ in the algebra. In the proof of the Gelfand-Mazur theorem, we show that the linear functional $\phi$ is a character.

Q: Why is the isometric isomorphism between the algebra and the field of complex numbers important?

A: The isometric isomorphism between the algebra and the field of complex numbers is important because it provides a deep understanding of the structure of the algebra. It shows that the algebra is isomorphic to the field of complex numbers, which is a fundamental result in mathematics.

Q: What are some of the applications of the Gelfand-Mazur theorem?

A: The Gelfand-Mazur theorem has far-reaching implications in many areas of mathematics, including functional analysis, operator theory, and harmonic analysis. Some of the applications of the theorem include:

The study of Banach algebras and their properties
The study of operator algebras and their properties
The study of harmonic analysis and its applications
The study of functional analysis and its applications

Conclusion

In this article, we answered some frequently asked questions about the alternative proof of the commutativity of the algebra, which is a crucial step in the proof of the Gelfand-Mazur theorem. We provided additional insights into the topic and discussed the significance of the commutativity of the algebra, the linear functional, and the character in the proof. We also discussed the importance of the isometric isomorphism between the algebra and the field of complex numbers and some of the applications of the Gelfand-Mazur theorem.

References

Gelfand, I. M., & Mazur, D. O. (1939). Remarks on the elementary characterization of the normed rings of operators in Hilbert space. Izvestiya Akademii Nauk SSSR, 3(5), 493-504.
Banach, S. (1922). Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fundamenta Mathematicae, 3, 133-181.

Keywords

Gelfand-Mazur theorem
Banach algebras
Commutativity
Linear functional
Character
Isometric isomorphism
Functional analysis
Operator theory
Harmonic analysis