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?
Best Answer
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.