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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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  • $\begingroup$ If you want to do linear maps on geometric algebra, define a k-blade, which is a k-vector formed by the exterior product of k 1-vectors. Then proyect a vector or space into the k-blade and you are done. K-blade are k-vectors but also a subspace. Note that a multivector of grade n can be decomposed or not in 1-vector. If not, they are not blades and if you project into them, you should get rubbish. $\endgroup$ Dec 24, 2022 at 11:12

5 Answers 5

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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, 2014 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, 2014 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, 2016 at 5:22
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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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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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No.

Firstly, the geometric product is not the same as the tensor product.

Secondly, in geometric algebra, we can add up 2 elements of different grades, and the result is a mixed-graded element, while in tensor algebra, we can only add up elements of the same grades, and that means all tensors are pure-graded.

Lastly, if you want to confirm any associations between geometric algebra and tensor algebra, the answer is:

Pure-graded elements (scalar, vectors, bivectors, trivectors...) in geometric algebra are differential forms in tensor analysis, in short, they are just tensors $ω_{a_1 a_2 ... a_n}$satisfied with: $$ ω_{a_1 a_2 ... a_n} = ω_{[a_1 a_2 ... a_n]} $$ where the $[...]$ operator is defined by $$ ω_{[a_1 a_2 ... a_n]}:=\frac{1}{n!} \sum(-1)^ε ω_{a_{k_1} a_{k_2}...a_{k_n}} $$ where $k_1 k_2 ... k_n$ iterates through all permutations of $a_1 a_2...a_n$ and ε is even-or-odd of corresponding permutation.

So the elements of geometric algebra is only a subset of tensors. But that's the most association. Notice that since these two algebras are built on totally different products (geometric product and tensor product), both can not contains each other as a sub-algebra. Their relationship and associations are very weak.

Plus:

we often define exterior product of vectors (1-form) by formula: $$ a_1 \wedge a_2 \wedge ... \wedge a_n:=\frac{1}{n!} \sum(-1)^ε a_{k_1} a_{k_2}...a_{k_n} $$

now, if we specify the vector products in the right side of equation as geometric product, we'll get the geometric algebra version of exterior product, similarly, we can get the tensor algebra version of exterior product. The point is that these two versions of exterior product are the same. So geometric algebra and tensor algebra contain the same sub-algebra of exterior product, but geometric product and tensor product are not the same.

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