= Exterior power
{title2=$\Lambda^kV$}
The $k$th <exterior power> $\Lambda^kV$ is the <quotient vector space> of the <tensor power> $V^{\otimes k}$ that makes every elementary tensor with two equal factors zero. More precisely, take the <quotient vector space> by the <linear span> of all such tensors. Its elementary products $v_1\wedge\cdots\wedge v_k$ turn alternating <multilinear maps> into <linear maps>: every alternating $k$-linear map on $V$ factors uniquely through $\Lambda^kV$.<Exterior Powers by Keith Conrad>[Exterior Powers, §3, Definition 3.1 and Theorem 3.3: quotient construction and universal property]{ref}
If $e_1,\ldots,e_n$ is an <ordered basis>, the increasing products
$$
e_{i_1}\wedge\cdots\wedge e_{i_k},\qquad
1\leq i_1<\cdots<i_k\leq n
$$
form a <basis> of $\Lambda^kV$. There is one for each $k$-element choice of basis directions, so
$$
\boxed{\dim\Lambda^kV=\binom nk}.
$$
Repeated directions vanish; products with a different order can be sorted using their sign. Coordinate <determinants> distinguish the remaining increasing products, establishing <linear independence>.<Exterior Powers by Keith Conrad>[Exterior Powers, §4, Theorem 4.2: basis and rank of an exterior power]{ref}
For $V=\mathbb R^3$, a <basis> of $\Lambda^2V$ is $e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3$. This explains the three <bivector> components in the worked example. The top <exterior power> $\Lambda^3V$ has one basis element, $e_1\wedge e_2\wedge e_3$, while $\Lambda^4V=0$: four independent directions cannot fit into that <vector space>. By convention $\Lambda^0V$ is the scalar <field>, even when $V$ has <dimension> zero.
Back to article page
