Let $V$ be an $n$-dimensional vector space.
The exterior algebra $\Lambda V$ of $V$ is the direct sum of the exterior powers $\Lambda^kV$. It comes with a product (called the exterior product) which is bilinear, alternating and anticommutative. The dimension of $\Lambda V$ is $2^n$.
(For physics buffs, the exterior algebra is an example of a supersymmetric and supercommutative algebra.)
The exterior algebra is a geometric algebra with trivial quadratic form (I think). It is a quotient of the tensor algebra of $V$ (by the two-sided ideal generated by set $\{v \otimes v\ |\ v\in V\}$).
The exterior algebra $\Lambda\textbf{R}^1$ of $\textbf{R}^1$ is isomorphic to the dual numbers.
What is the exterior algebra $\Lambda\textbf{R}^2$ of $\textbf{R}^2$ (as in, through which isomorphic objects can I understand it, specially for the purpose of doing explicit computations)? Is it related to $\Lambda\textbf{C}$? Does it admit a matrix representation, like $\Lambda\textbf{R}^1$?
(If this question is too low-powered please do migrate it / close it.)
Best Answer
The exterior algebra $\Lambda \mathbb{R}^2$ is a real vector space of dimension 4 with basis $1, e_1, e_2, e_1 \wedge e_2$. So its every element is a unique linear combination of these basis elements, say $a_1 \cdot 1 + a_2 e_1 + a_3 e_2 + a_4 e_1 \wedge e_2$, for real numbers $a_1, a_2, a_3, a_4$, which can be chosen arbitrarily. The multiplication operation is written $\wedge$, is associative and has $1$ as identity, and has $e_1 \wedge e_1 = 0$ and $e_2 \wedge e_2=0$, $e_1 \wedge e_2$ is the basis element by that name, and $e_2 \wedge e_1=-e_1 \wedge e_2$. That gives you the complete multiplication table, by associativity. Please comment if this answer is not sufficient.