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No, the dual quaternions contain zero divisors, one of which is a nonzero element $\epsilon$ with square zero. Such elements cannot have inverses. It turns out that the units are exactly the $a+\epsilon b$ where $a$ is nonzero. To see this, first notice that $1+\epsilon b$ has the obvious inverse $1-\epsilon b$. Suppose now $a\neq0$. Then $1+\epsilon ...


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You might want to look at GA and the documentation GA.pdf in the "LaTeX docs" directory. Also in the examples directory both in the terminal_check.py and the latex_check.py examples are examples of conformal geometric algebra - def F(x): global n, nbar Fx = ((x * x) * n + 2 * x - nbar) / 2 return(Fx) def make_vector(a, n=3, ga=None): if ...


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A tensor (not an abstract tensor in the sense of Penrose) is a multilinear function from the repeated Cartesian product of a vector space the real numbers thus the multi-linear function $T(a_1,...,a_r)$ where $a_1$ through $a_r$ are vectors in the same vector space. The rank of this tensor would be $r$. All the standard tensor operations can be defined ...


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Every element of a geometric algebra can be identified with a tensor, but not every tensor can be identified with an element of a geometric algebra. It's helpful to consider the vector derivative of a linear operator, of a map from vectors from vectors. Call such a map $\underline A$. The vector derivative is then $$\partial_a \underline A(a) = ...


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Tensor algebras are (in general) infinite dimensional, but the most common Clifford algebras (over finite dimensional vector spaces) are finite dimensional, so they aren't isomorphic. Clifford algebras can be constructed as quotients of tensor algebras, so there is a relationship between them.


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Okay, the answer is No, and the reason is quite simple: It can be shown that the Dirac matrices $\gamma_i:=\Phi(e_i)$ are traceless, so the same is true for all $\Phi(v)$ with $v\in\mathbb{R}^{p,q}$. Now, if $\Phi(v)=1$ we can take traces and find $0=\dim S$, a contradiction.



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