Problem 19
Question
Consider the Clifford algebra over \(\mathbf{R}\). The standard notation is \(C_{n}\) if \(E=\mathbf{R}^{n}\) with the negative definite form, and \(C_{n}^{\prime}\) if \(E=\mathbf{R}^{n}\) with the positive definite form. Thus \(\operatorname{dim} C_{n}=\operatorname{dim} C_{n}^{\prime}=2^{n}\) (a) Show that \(C_{1}=\mathbf{C} \quad C_{2}=\mathbf{H}\) (the division ring of quaternions) $$ C_{1}^{\prime}=\mathbf{R} \times \mathbf{R} \quad C_{2}^{\prime}=M_{2}(\mathbf{R})(2 \times 2 \text { matrices over } \mathbf{R}) $$
Step-by-Step Solution
Verified Answer
In summary:
\(C_1\) is isomorphic to the complex numbers \(\mathbf{C}\) since they share the same multiplication rules, with \(i = e_1\). \(C_2\) is isomorphic to the quaternion algebra \(\mathbf{H}\) as both have the same basis and relations. \(C_1'\) is isomorphic to \(\mathbf{R} \times \mathbf{R}\) due to the same product structure, and finally, \(C_2'\) is isomorphic to the matrix algebra \(M_2(\mathbf{R})\) since the elements of \(C_2'\) can be represented as \(2 \times 2\) matrices over \(\mathbf{R}\), and their multiplication follows the relations of \(C_2'\).
1Step 1: Obtain the properties of Clifford algebras
A Clifford algebra over a vector space \(E\) is a unital associative algebra formed by taking the tensor algebra of \(E\) and applying a certain quotient process. More concretely, given a quadratic form \(Q\) on \(E\), we impose the relation \(v^2 = Q(v)\) for all \(v \in E\). In this exercise, we are given the negative definite form for \(C_n\) and the positive definite form for \(C_n'\).
2Step 2: Analyze \(C_1\) and show that it is isomorphic to the complex numbers \(\mathbf{C}\)
In this case, \(E = \mathbf{R}^1\), so \(C_1\) has dimension 2. Let \(e_1\) be a basis element for \(\mathbf{R}^1\) and define \(C_1 = \mathbf{R} \oplus \mathbf{R} e_1\). The multiplication rule is given by \(e_1^2 = -1\), which is identical to the complex numbers' multiplication rule with \(i = e_1\). Therefore, \(C_1\) is isomorphic to the complex numbers \(\mathbf{C}\).
3Step 3: Analyze \(C_2\) and show that it is isomorphic to the quaternion algebra \(\mathbf{H}\)
Let \(E = \mathbf{R}^2\) and choose \(e_1\) and \(e_2\) as basis elements. Then, \(C_2 = \mathbf{R} \oplus \mathbf{R} e_1 \oplus \mathbf{R} e_2 \oplus \mathbf{R} e_1 e_2\). We have the relations \(e_1^2 = e_2^2 = -1\) and \(e_1 e_2 = -e_2 e_1\). The algebra \(C_2\) is isomorphic to the quaternion algebra \(\mathbf{H}\), which has the basis \(1, i, j, k\) and the relations \(i^2 = j^2 = -1\) and \(ij = -ji = k\).
4Step 4: Analyze \(C_1'\) and show that it is isomorphic to \(\mathbf{R} \times \mathbf{R}\)
In this case, \(E = \mathbf{R}^1\) and \(Q\) is the positive definite form. The multiplication rule is \(e_1^2 = 1\). The algebra \(C_1'\) is given by \(\mathbf{R} \oplus \mathbf{R} e_1\), and it is isomorphic to the direct product of two copies of the real numbers \(\mathbf{R} \times \mathbf{R}\) with the product \((a, b) \cdot (c, d) = (ac + bd, ad + bc)\).
5Step 5: Analyze \(C_2'\) and show that it is isomorphic to the matrix algebra \(M_2(\mathbf{R})\)
For \(C_2'\), we have \(E = \mathbf{R}^2\) and \(Q\) is the positive definite form. The multiplication rules are \(e_1^2 = e_2^2 = 1\) and \(e_1 e_2 = e_2 e_1\). The algebra \(C_2'\) is given by \(\mathbf{R} \oplus \mathbf{R} e_1 \oplus \mathbf{R} e_2 \oplus \mathbf{R} e_1 e_2\). We can represent elements of \(C_2'\) as \(2 \times 2\) matrices over \(\mathbf{R}\):
$$
1 \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad
e_1 \to \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad
e_2 \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad
e_1 e_2 \to \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$
Matrix multiplication of these matrices follows the relations of \(C_2'\). Therefore, \(C_2'\) is isomorphic to the matrix algebra \(M_2(\mathbf{R})\).
Key Concepts
Vector SpaceQuadratic FormQuaternion AlgebraMatrix Algebra
Vector Space
Vector spaces form the foundational framework in many mathematical theories, including Clifford algebras. A vector space, often denoted as \(E\), is a collection of vectors. These vectors can be added together and multiplied by scalars, forming a core component of linear algebra.
For example, when considering the Clifford algebra \(C_n\) over a vector space \(E = \mathbf{R}^n\), each vector can be represented in \(\mathbf{R}^n\). This concept allows us to apply operations specific to vector spaces, like linear combinations.
For example, when considering the Clifford algebra \(C_n\) over a vector space \(E = \mathbf{R}^n\), each vector can be represented in \(\mathbf{R}^n\). This concept allows us to apply operations specific to vector spaces, like linear combinations.
- **Vectors:** Basic elements of a vector space that follow specific algebraic rules.
- **Scalars:** Real numbers from the field over which the vector space is defined, typically \(\mathbf{R}\).
- **Vector Addition and Scalar Multiplication:** The operations that define a vector space, following distributive, associative, and commutative properties.
Quadratic Form
A quadratic form is a crucial concept when working with Clifford algebras, as it influences the relations within the algebra. In simple terms, a quadratic form \(Q\) on a vector space \(E\) is a special type of polynomial that associates numbers to vectors based on their components.
Quadratic forms can be either positive definite or negative definite, which fundamentally impacts Clifford algebras:
Quadratic forms can be either positive definite or negative definite, which fundamentally impacts Clifford algebras:
- **Positive Definite Quadratic Form:** Here, \(Q(v) > 0\) for any non-zero vector \(v\). This type was used in the exercise for \(C_n'\).
- **Negative Definite Quadratic Form:** In this case, \(Q(v) < 0\) for any non-zero vector \(v\). This was the form used for \(C_n\).
Quaternion Algebra
Quaternion algebra, represented as \(\mathbf{H}\), is an extension of the complex numbers and is crucial in the study of rotations in three dimensions. It is a four-dimensional normed division algebra over the real numbers, which means it forms a vector space of dimension four.
The algebra \(C_2\) in the exercise showed isomorphism to \(\mathbf{H}\), which consists of:
The algebra \(C_2\) in the exercise showed isomorphism to \(\mathbf{H}\), which consists of:
- **Basis Elements:** \(1, i, j, k\), where \(i^2 = j^2 = k^2 = -1\)
- **Multiplication Rules:** Crucially, \(ij = k,\; ji = -k,\; jk = i,\; kj = -i,\; ki = j,\; ik = -j\). These rules illustrate the non-commutative nature of quaternions.
Matrix Algebra
Matrix algebra is a rich area of study involving arrays of numbers that can represent linear transformations. Matrices can be manipulated using operations like addition, subtraction, and multiplication to solve systems of linear equations or transform geometric objects.
In the exercise, \(C_2'\) was shown to be isomorphic to the matrix algebra \(M_2(\mathbf{R})\), which consists of \(2 \times 2\) matrices with real number entries. This connection is crucial because:
In the exercise, \(C_2'\) was shown to be isomorphic to the matrix algebra \(M_2(\mathbf{R})\), which consists of \(2 \times 2\) matrices with real number entries. This connection is crucial because:
- **Linear Transformations:** Matrices can describe linear functions mapping vectors to other vectors.
- **Compositions and Inversions:** Matrix multiplication allows for the composition of these transformations, much like function composition.
- **Determinants and Trace:** These are properties of matrices that provide insights into their behavior. For example, invertibility is determined by a non-zero determinant.
Other exercises in this chapter
Problem 17
Suppose that \(k=\mathbf{R}\), so \(E\) is a real vector space, which we now assume of even dimension \(2 m\). We also assume that \(g\) is non-degenerate. We o
View solution Problem 18
Still assuming \(g\) non-degenerate, let \(J\) be an automorphism of \((E, g)\) (i.e. \(g(J x, J y)=g(x, y)\) for all \(x, y \in E\) ) such that \(J^{2}=-\) id.
View solution Problem 20
Establish isomorphisms: $$ \mathrm{C} \otimes_{\mathrm{k}} \mathrm{C}=\mathbf{C} \times \mathrm{C} ; \quad \mathrm{C} \otimes_{\mathrm{R}} \mathrm{H}=M_{2}(\mat
View solution Problem 21
(a) Establish isomorphisms $$ C_{n+2}=C_{n}^{\prime} \otimes C_{2} \quad \text { and } \quad C_{n+2}^{\prime}=C_{n} \otimes C_{2}^{\prime} $$ [Hint: Let \(\left
View solution