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Is geometric algebra (Clifford algebra) isomorphic to tensor algebra? If so, how then would one relate a unique 2-vector (this is what I'm going to call a multivector that is the sum of a scalar, vector, and bivector) to every 2nd rank tensor?


Edit by the OP, converted from "answer"

Okay. Well I'm still curious if there's a way to represent any 2nd-rank tensor by a bivector, vector, and scalar. Or in particular, can any $3 \times 3$ matrix be represented by a 2-vector in 3D.

It seems to me that they can't because I would guess the matrix representation of a bivector (grade 2 element of a 2-vector) would be exactly the same as the $3 \times 3$ matrix representation of the cross product (i.e. $[a \times b]_{ij} = a_j b_i - a_i b_j$) which only uniquely identifies 3 components.
I would also assume that the scalar part of the 2-vector would be represented by a scalar times the $3 \times 3$ identity matrix. This would fill in 3 numbers, but really only uniquely gives 1 component.
I don't know how to represent the vector component of the 2-vector as a $3 \times 3$ matrix but I don't see how it could identity the remaining 5 components by itself.

Am I right then in assuming that there is a canonical matrix representation of a general 2-vector, but that there are matrices that cannot be represented by any 2-vector?

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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

$$\partial_a \underline A(a) = \partial_a \cdot \underline A(a) + \partial_a \wedge \underline A(a) = T + B$$

where $T$ is a scalar, the trace, and $B$ is a bivector. The linear map $\underline A$ can then be written as

$$\underline A(a) = \frac{T}{n} a + \frac{1}{2} a \cdot B + \underline S(a)$$

where $\underline S$ is some traceless, symmetric map. While the scalar can be turned into a multiple of the identity, in $T \underline I/n$, and the bivector can be directly turned into an antisymmetric map in $a \cdot B$, the map $\underline S$ is very much part of $\underline A$, yet not representable in general through a single algebraic element of the geometric algebra. This is just one example of such an object.

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  • $\begingroup$ I'm not super familiar with the idea of representing a tensor by a multilinear map, but could we say that a tensor is a mapping $T$, such that $T(I \wedge J)=\alpha$ (or possibly $T(I,J)=\alpha$ if $I \wedge J$ doesn't make sense), where $I$ is a pseudoscalar (k-blade) in our tangent space, $J$ is a pseudoscalar (j-blade) in our cotangent space, and $\alpha$ is a scalar? $\endgroup$ – user137731 Jul 12 '14 at 14:52
  • $\begingroup$ @Bye_World Yeah, I think that wedge wouldn't make sense, but as separate arguments, sure. That said, not every tensor can be extended through outermorphism. The Riemann tensor is a good example, as it fundamentally takes bivector inputs. $\endgroup$ – Muphrid Jul 12 '14 at 15:28
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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 without regard to component indices. Go to GA and look at bookGA.pdf in "GA Notes". Note that any tensor has a definite rank (number of multi-linear vector arguments), but a multivector can be the sum of different pure grade multivectors. An even mulitvector (contains only even grade components) can represent a spinor, but a tensor cannot. In that sense a multivector is more general than a tensor. However, pure grade multivectors can only represent completely antisymmetric tensors. In that sense tensors are more general than pure grade multivectors. The rank of a tensor is not restricted by the dimension of the base vector space like the grade of a multivector.

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  • $\begingroup$ In first line, isn't the multilinear function T called a tensorproduct rather than a tensor, while an element $T(a_1,\dots , a_r)=a_1\otimes\cdots\otimes a_r$ is called a tensor? $\endgroup$ – Lehs Jul 2 '16 at 5:22
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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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Similar question: what's the relationship of tensor and multivector

The short answer is that all multivectors are tensors, but not all tensors are multivectors, so geometric algebra is not and cannot be isomorphic to tensor algebra.

However, we can try to embed geometric algebra inside of tensor algebra. These resources give some guidelines for doing so (in the special case where the vector space is $\mathbb{R}^3$ and the inner product is the "regular" inner product, as opposed to say the Minkowski metric on $\mathbb{R}^4$):

http://www2.ic.uff.br/~laffernandes/teaching/2013.1/topicos_ag/lecture_18%20-%20Tensor%20Representation.pdf

https://www.docdroid.net/uwfvUxE/tensor-representation-of-geometric-algebra.pdf.html

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