Exterior algebra

Definition

Let $F$ be a field and let $V$ be a vector space over $F$. Let $T(V) = \bigoplus_{k \ge 0} V^{\otimes k}$ be the tensor algebra of $V$, where $V^{\otimes 0} = F$ and the product is the tensor product. Let $I$ be the two-sided ideal of $T(V)$ generated by the elements $v \otimes v$ for $v \in V$. The exterior algebra of $V$ is the quotient algebra $\Lambda(V) = T(V) / I$. The product in $\Lambda(V)$ is written $\wedge$.

Remarks

For all $u, v \in V$, $v \wedge v = 0$ and $u \wedge v = -v \wedge u$. The image of $V^{\otimes k}$ is the $k$-th exterior power $\Lambda^k(V)$; if $\dim V = n$, then $\dim \Lambda^k(V) = \binom{n}{k}$ and $\dim \Lambda(V) = 2^n$. If $(e_1, \dots, e_n)$ is a basis of $V$, then $\Lambda^n(V)$ is spanned by $e_1 \wedge \cdots \wedge e_n$. For $v_j = \sum_i a_{ij} e_i$, $v_1 \wedge \cdots \wedge v_n = \det(a_{ij})\, e_1 \wedge \cdots \wedge e_n$.

Explanations

L1 human

The exterior algebra multiplies vectors with the wedge product, which changes sign when we swap two factors. In $\R^3$, $u \wedge v$ describes an oriented parallelogram, and $u \wedge v \wedge w$ an oriented volume. The top wedge product of $n$ vectors gives the determinant.