Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 58

Properties of the wedge product

Theorem

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space of dimension $m$. Let $v_1,\dots ,v_m$ be a basis of $V$, and let $n\in \mathbb{N}$. Then the wedge products

$v_{i_1}\wedge \dots \wedge v_{i_n}\text{ with }1\le i_1<\dots <i_n\le m$

form a basis of $\stackrel{n}{\bigwedge }V$.

Proof

We show first that we have a generating system. Because the elements of the form $w_1\wedge \dots \wedge w_n$ form a generating system of $\stackrel{n}{\bigwedge }V$ due to Lemma 57.5   (1), it is enough to show that these elements can be represented. For every $w_j$, there exists a representation $w_j=\sum _{i=1}^m{a}_{ij}{v}_i$; therefore, according to Lemma 57.5   (4), we can write the $w_1\wedge \dots \wedge w_n$ as linear combinations of wedge products of the basis elements; however, every ordering may occur. Hence, let $v_{k_1}\wedge \dots \wedge v_{k_n}$ be given, with $k_j\in \{1,\dots ,m\}$. By swapping neighboring vectors, using Lemma 57.5   (3), we may achieve (maybe with another sign) that the indices are (not necessarily strictly) increasing. If an index appears twice, the wedge product is $0$, due to Lemma 57.5   (2). Hence, no index occurs twice, and this wedge product is in the form asked for.

To show that the family is linearly independent, we show, using Lemma 14.8 , that for every subset $I=\{i_1,\dots ,i_n\}\subseteq \{1,\dots ,m\}$ with $n$ elements (where $i_1<\dots <i_n$), there exists a $K$-linear mapping

$\stackrel{n}{\bigwedge }V⟶K$

such that $v_{i_1}\wedge \dots \wedge v_{i_n}$ is not mapped to $0$, but all other wedge products in the family are mapped to $0$. To show this, it is enough, by Theorem 57.7 , to give an alternating multilinear mapping

$\mathrm{△}:V^n⟶K$

satisfying $\mathrm{△}(v_{i_1},\dots ,v_{i_n})\ne 0$ but $\mathrm{△}(v_{j_1},\dots ,v_{j_n})=0$ for every other strictly increasing index tuple. Let $U$ be the linear subspace generated by $v_i$, $i\ne i_k$, of $V$, and let $W=V/U$ denote the residue class space. Then the images of the $v_{i_k}$, $k=1,\dots ,n$, form a basis of $W$, and the images of all other subsets with $n$ elements of the given basis do not form a basis of $W$, because at least one element is mapped to $0$. We consider now the composed mapping

$\mathrm{△}:V^n⟶W^n\cong (K^n{)}^n\stackrel{det}{⟶}K.$

This mapping is multilinear and alternating, due to Theorem 16.9 and Theorem 16.10 . Due to Theorem 16.11 , we have $\mathrm{△}(z_1,\dots ,z_n)=0$ if and only if the images of $z_i$ in $W$ do not form a basis.

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For $V=K^m$ with the standard basis $e_1,\dots ,e_m$, the family $e_{i_1}\wedge \dots \wedge e_{i_n}$ mit $i_1<\dots <i_n$ is called the standard basis of $\stackrel{n}{\bigwedge }{K}^m$.

Remark

For bases $v_1,\dots ,v_m$ and $w_1,\dots ,w_m$ of a $K$-vector space $V$, with the relations

$v_j=\sum _{i=1}^m{a}_{ij}{w}_i,$

we obtain, between the bases

$v_{i_1}\wedge \dots \wedge v_{i_n}\text{ with }1\le i_1<\dots <i_n\le m\text{ and }{w}_{i_1}\wedge \dots \wedge w_{i_n}\text{ with }1\le i_1<\dots <i_n\le m$

of $\stackrel{n}{\bigwedge }V$, the relation

$v_{j_1}\wedge \dots \wedge v_{j_n}=\sum _{1\le i_1<\cdots <i_n\le m}(\sum _{\pi \in S_n}\mathrm{sgn}(\pi )\prod _{s=1}^n{a}_{i_s{j}_{\pi (s)}})w_{i_1}\wedge \dots \wedge w_{i_n}.$

This rests, according to Lemma 57.5   (4), on

$\begin{array}{cc}{v}_{j_1}\wedge \dots \wedge v_{j_n}& =\left(\sum _{i=1}^m{a}_{i{j}_1}{w}_i\right)\wedge \dots \wedge \left(\sum _{i=1}^m{a}_{i{j}_n}{w}_i\right)\\ & =\sum _{1\le i_1<\cdots <i_n\le m}(\sum _{\pi \in S_n}\mathrm{sgn}(\pi )\prod _{s=1}^n{a}_{i_s{j}_{\pi (s)}})w_{i_1}\wedge \dots \wedge w_{i_n}.\end{array}$

Corollary

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space of dimension $m$. Then the dimension of the $n$-th exterior product $\stackrel{n}{\bigwedge }V$ is

$(\genfrac{}{}{0ex}{}{m}{n}).$

Proof

This follows directly from Theorem 58.1 and Fact *****.

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In particular, the exterior power is for $n=0$ one-dimensional (we have $\stackrel{0}{\bigwedge }V=K$), and for $n=1$ it is $m$-dimensional (we have $\stackrel{1}{\bigwedge }V=V$). For $n=m$, $\stackrel{m}{\bigwedge }V$ is one-dimensional, and the determinant induces (after an identification of $V$ with $K^m$) an isomorphism

$\stackrel{m}{\bigwedge }V⟶K,(v_1,\dots ,v_m)⟼det(v_1,\dots ,v_m).$

For $n>m$, the exterior powers are the zero space and their dimension is $0$.

We want to extend the natural isomorphism

${\left(\stackrel{n}{\bigwedge }V\right)}^{\ast }\cong {\mathrm{Alt}}^n(V,K)$

from Corollary 57.9 to natural isomorphisms

$\stackrel{n}{\bigwedge }{V}^{\ast }\cong {\left(\stackrel{n}{\bigwedge }V\right)}^{\ast }\cong {\mathrm{Alt}}^n(V,K).$

Theorem

Let $K$ be a field, and let $V$ be a finite-dimensional vector space. Let $k\in \mathbb{N}$. Then there exists a natural isomorphism

$\psi :\stackrel{k}{\bigwedge }{V}^{\ast }⟶{\left(\stackrel{k}{\bigwedge }V\right)}^{\ast },$

given by

$(\psi (f_1\wedge \dots \wedge f_k))(v_1\wedge \dots \wedge v_k)=det(f_i(v_j){)}_{ij}$

(with $f_i\in V^{\ast }$ and

$v_j\in V$).

Proof

We consider the mapping (with $k$ factors)

$V^{\ast }\times \cdots \times V^{\ast }⟶\mathrm{Map}(V\times \cdots \times V,K)$

with

$(f_1,\dots ,f_k)⟼((v_1,\dots ,v_k)⟼det{(f_i(v_j))}_{ij}).$

For fixed $f_1,\dots ,f_k$, the mapping on the right is multilinear and alternating, as a direct verification using the determinant rules shows. Therefore, according to Corollary 57.9 , we obtain an element in ${\left(\stackrel{k}{\bigwedge }V\right)}^{\ast }$. Hence, we get altogether a mapping

$V^{\ast }\times \cdots \times V^{\ast }⟶{\left(\stackrel{k}{\bigwedge }V\right)}^{\ast }.$

A direct inspection shows that this assignment is also multilinear and alternating. Due to the universal property, there exists a linear mapping

$\psi :\stackrel{k}{\bigwedge }{V}^{\ast }⟶{\left(\stackrel{k}{\bigwedge }V\right)}^{\ast }.$

We have to show that this mapping is an isomorphism. To show this, let $v_1,\dots ,v_n$ be a basis of $V$, with the corresponding dual basis $v_1^{\ast },\dots ,v_n^{\ast }$. Because of Theorem 58.1 , the family

$v_{i_1}^{\ast }\wedge \dots \wedge v_{i_k}^{\ast },1\le i_1<\dots <i_k\le n,$

is a basis of $\stackrel{k}{\bigwedge }{V}^{\ast }$. Moreover, the family

$v_{i_1}\wedge \dots \wedge v_{i_k},1\le i_1<\dots <i_k\le n,$

is a basis of $\stackrel{k}{\bigwedge }V$, with corresponding dual basis ${(v_{i_1}\wedge \dots \wedge v_{i_k})}^{\ast }$. We show that $v_{i_1}^{\ast }\wedge \dots \wedge v_{i_k}^{\ast }$ is mapped under $\psi$ to $(v_{i_1}\wedge \dots \wedge v_{i_k}{)}^{\ast }$. For $1\le j_1<\dots <j_k\le n$, we have

$\left(\psi (v_{i_1}^{\ast }\wedge \dots \wedge v_{i_k}^{\ast })\right)(v_{j_1}\wedge \dots \wedge v_{j_k})=det\left(v_{i_r}^{\ast }{\left(v_{j_s}\right)}_{1\le r,s\le k}\right).$

If $\{i_1,\dots ,i_k\}\ne \{j_1,\dots ,j_k\}$, then there exists an $i_r$ that is different from all $j_s$. Therefore, the $r$-th row of the matrix is $0$; hence, its determinant is $0$. If the index sets coincide, then we obtain the identity matrix with determinant $1$. This effect coincides with the effect of ${(v_{i_1}\wedge \dots \wedge v_{i_k})}^{\ast }$.

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Wedge product of linear mappings

Corollary

Let $K$ be a field, let $V$ and $W$ be $K$-vector spaces, and let

$\phi :V⟶W$

denote a $K$-linear mapping. Then there exists, for any $n\in \mathbb{N}$, a $K$-linear mapping

$\stackrel{n}{\bigwedge }\phi :\stackrel{n}{\bigwedge }V⟶\stackrel{n}{\bigwedge }W,$

with $v_1\wedge \dots \wedge v_n\mapsto \phi (v_1)\wedge \dots \wedge \phi (v_n)$.

Proof

The mapping

$V^n\stackrel{\phi \times \cdots \times \phi }{⟶}{W}^n\stackrel{\delta }{⟶}\stackrel{n}{\bigwedge }W$

is, due to Exercise 16.29 , multilinear and alternating. Due to Theorem 57.7 , there exists a uniquely determined linear mapping

$\stackrel{n}{\bigwedge }V⟶\stackrel{n}{\bigwedge }W,$

with $v_1\wedge \dots \wedge v_n\mapsto \phi (v_1)\wedge \dots \wedge \phi (v_n)$.

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Proposition

Let $K$ be a field, let $V$ and $W$ be $K$-vector spaces, and let

$\phi :V⟶W$

denote a $K$-linear mapping. For $n\in \mathbb{N}$, let

$\stackrel{n}{\bigwedge }\phi :\stackrel{n}{\bigwedge }V⟶\stackrel{n}{\bigwedge }W$

be the corresponding $K$-linear mapping. Then the following properties hold.

1. If $\phi$ is surjective, then $\stackrel{n}{\bigwedge }\phi$ is also surjective.
2. If $\phi$ is injective, then $\stackrel{n}{\bigwedge }\phi$ is also injective.
3. If $U$ is another $K$-vector space, and

   $\psi :U⟶V$

   another $K$-linear mapping, then we have

   $\stackrel{n}{\bigwedge }(\phi \circ \psi )=\left(\stackrel{n}{\bigwedge }\phi \right)\circ \left(\stackrel{n}{\bigwedge }\psi \right).$

Proof

(1). Let $w_1,\dots ,w_n\in W$ be given, and let $v_1,\dots ,v_n\in V$ be preimages, that is, $\phi (v_i)=w_i$. We have

$\left(\stackrel{n}{\bigwedge }\phi \right)(v_1\wedge \dots \wedge v_n)=w_1\wedge \dots \wedge w_n.$

Surjectivity follows from Lemma 57.5   (1).
(2). We may assume, due to the construction of the wedge product, that $V$ and $W$ have finite dimension. The statement follows from the explicit description of the bases in Theorem 58.1 .
(3). It is enough to show the equality for the generating system $u_1\wedge \dots \wedge u_n$ with $u_i\in U$; but this is clear due to the explicit description.

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Orientations and the wedge product

Using the wedge product, we can relate the orientations on a real vector space with the orientations on a line, as the following result shows.

Lemma

Let $V\ne 0$ be a finite-dimensional real vector space of dimension $n$. Then we get, via the assignment

$[v_1,\dots ,v_n]⟼[v_1\wedge \dots \wedge v_n],$

a correspondence between the orientations

on $V$ and the orientations on $\stackrel{n}{\bigwedge }V$.

Proof

Let $v_1,\dots ,v_n$ and $w_1,\dots ,w_n$ be two bases of $V$, fulfilling the relation

$\left(\begin{array}{c}{v}_1\\ ⋮\\ v_n\end{array}\right)=M\left(\begin{array}{c}{w}_1\\ ⋮\\ w_n\end{array}\right).$

Due to Corollary 57.6 , we have

$v_1\wedge \dots \wedge v_n=(detM)w_1\wedge \dots \wedge w_n.$

This shows the well-definedness of the mapping, and the statement follows.

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