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It's also easily computed from the structure theorem for Clifford algebras. In your particular case, the algebra is isomorphic to $M_2(\Bbb R)\times M_2(\Bbb R)$ whose center is $\Bbb R\times \Bbb R$.


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You ask: "Why is the grade of the wedge product of two arbitrary blades the sum of the two blades grades independently?" The answer is that this is the definition of the wedge product. Why is this the definition? Because it is useful. About the grade of 0. If the grade-k multivectors are to be a vector space (highly desirable), then the grade-k ...


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I don't think there's any issue here. $(e_1 \wedge e_2) \wedge (e_2 \wedge e_3)$ is the zero 4-vector. It should not be confused with the zero scalar, although in geometric algebra, we can and often do use the same symbol (0) to denote any zero $k$-vector, or whole linear combinations of these zero $k$-vectors. I would say the grade of the zero 4-vector ...


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You're right that your book is wrong: The correct statement should be $$\operatorname{grade}(\mathbf A \wedge \mathbf B) = \begin{cases} \operatorname{grade}(\mathbf A) + \operatorname{grade}(\mathbf B), & \text{for } \mathbf A\ \bot\ \mathbf B \\ 0, & \text{otherwise}\end{cases}$$ This is trivially shown from the definition of the wedge product of ...


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This is indeed a Clifford algebra. A reference for the proof would be e.g. the first chapter of Friedrich's "Dirac operators in Riemannian geometry."


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Ok if nobody else is going to address this one let me give it a try. In complex numbers the real and imaginary components operate by different rules. Multiplication by the real component scales a vector in or out from the origin, whereas multiplication by the imaginary component rotates it about the origin. In Clifford Algebra there is no distinction ...


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The gamma matrices are four $\Bbb C$-linearly independent matrices of $M_2(\Bbb C)$, a space with $\Bbb C$-dimension $4$, so the gamma matrices form a basis. Since the image of $\gamma$ contains this basis, it is the entire space. Thus $\gamma$ is surjective and hence injective.



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