Section II.6.3

Exterior algebra

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Problem II.6.3.1

Let V=RnV=\mathbb {R}^{n} be the real space spanned by the column vectors A(1),…,A(n)A^{(1)}, \ldots , A^{(n)}, and let BB be an arbitrary column vector. Show that the components λk\lambda_{k} of the solution of the vector equation

∑kλkA(k)=B \sum _{k} \lambda _{k} A^{(k)}=B

are given by the relation

(A(1)∧…∧A(n))λk=A(1)∧…∧A(k−1)∧B∧A(k+1)∧…∧A(n) \left(A^{(1)} \wedge \ldots \wedge A^{(n)}\right) \lambda _{k}=A^{(1)} \wedge \ldots \wedge A^{(k-1)} \wedge B \wedge A^{(k+1)} \wedge \ldots \wedge A^{(n)}

Deduce from this Cramer's formulas [ВА I, Ch. 3, § 3].

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Problem II.6.3.2

Let VV be an nn-dimensional vector space over a field K\mathfrak {K} of characteristic zero,

Z=∑k=0nZk(Zk∈Λk(V)) Z=\sum _{k=0}^{n} Z_{k} \quad \left(Z_{k} \in \Lambda ^{k}(V)\right)

an element of the exterior algebra Λ(V)\Lambda (V),

Z(Λ(V))={Z∣Z∧X=X∧Z∀X∈Λ(V)} \mathcal{Z}(\Lambda (V))=\left\{ Z \mid Z \wedge X=X \wedge Z \quad \forall X \in \Lambda (V)\right\}

the center of the exterior algebra.

Show:

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(a)

for invertibility of ZZ, i.e., for the existence of Z−1Z^{-1}, it is necessary and sufficient that Z(0)≠0Z_{(0)} \neq 0;

(b)

Z∈Z(Λ(V))⟺{Z2i−1=0 for i=1,…,m, if n=2m,Z2i−1=0 for i=1,…,m−1, if n=2m−1.Z \in \mathcal{Z}(\Lambda (V)) \Longleftrightarrow \begin{cases} Z_{2 i-1}=0 \text{ for } i=1, \ldots , m, \text{ if } n=2 m, \\ Z_{2 i-1}=0 \text{ for } i=1, \ldots , m-1, \text{ if } n=2 m-1.\end{cases}

How should the indicated conditions be reformulated in the case of an infinite-dimensional space VV?

Problem II.6.3.3

Let A:V→V\mathcal{A}: V \rightarrow V be a linear operator. Its pp-th exterior power is the linear operator

ΛpA:Λp(V)→Λp(V) \Lambda ^{p} \mathcal{A}: \Lambda ^{p}(V) \rightarrow \Lambda ^{p}(V)

defined on decomposable pp-vectors by the formula

(∧pA)(x1∧…∧xp)=A(x1)∧…∧A(xp) \left(\wedge ^{p} \mathcal{A}\right)\left(\mathbf{x}_{1} \wedge \ldots \wedge \mathbf{x}_{p}\right)=\mathcal{A}\left(\mathbf{x}_{1}\right) \wedge \ldots \wedge \mathcal{A}\left(\mathbf{x}_{p}\right)

This is obviously an alternating multilinear map. For a fixed basis of the space VV, one may speak of the matrix AA of the operator A\mathcal{A} and of its pp-th exterior power ∧pA\wedge^{p} A. Show that

det⁡∧pA⋅det⁡∧n−pA=(det⁡A)(np). \operatorname {det} \wedge ^{p} A \cdot \operatorname {det} \wedge ^{n-p} A=(\operatorname {det} A)^{\binom {n}{p}}.
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Problem II.6.3.4

Let a nonsingular quadratic form qq be given on a vector space VV. Let ff be the bilinear form polar to qq. Then on Λp(V)\Lambda^{p}(V) one can introduce a quadratic form q∧pq^{\wedge p} by the formulas

q∧0=1,q∧p(x1∧…∧xp)=∣f(x1,x1)⋯f(x1,xp)⋯⋯⋯f(xp,x1)⋯f(xp,xp)∣ q^{\wedge 0}=1, \qquad q^{\wedge p}\left(\mathbf{x}_1 \wedge \ldots \wedge \mathbf{x}_p\right)=\begin{vmatrix} f(\mathbf{x}_1,\mathbf{x}_1) & \cdots & f(\mathbf{x}_1,\mathbf{x}_p) \\ \cdots & \cdots & \cdots \\ f(\mathbf{x}_p,\mathbf{x}_1) & \cdots & f(\mathbf{x}_p,\mathbf{x}_p) \end{vmatrix}

Prove that the resulting extension of the form qq to the algebra Λ(V)\Lambda (V) is a nonsingular quadratic form.

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Problem II.6.3.5

An orientation of an nn-dimensional Euclidean vector space (V,(∗∣∗))(V,(* \mid *)) is an element d∈Λn(V)d \in \Lambda^{n}(V) for which (d∣d)∧n=1(d \mid d)^{\wedge n}=1.

Let, further, (e1,…,en),(e1′,…,en′)\left(\mathbf{e}_{1}, \ldots , \mathbf{e}_{n}\right),\left(\mathbf{e}_{1}^{\prime }, \ldots , \mathbf{e}_{n}^{\prime }\right) be two bases in VV. These bases are said to be equally oriented if the determinant of the transition matrix from one of them to the other is positive. It is clear that the order of the basis vectors matters here. It is also clear that the set of ordered bases of the space VV splits into exactly two classes consisting of equally oriented bases, while bases from different classes are oriented differently (or oppositely). The choice of one of these classes is called an orientation of the space VV.

Is there any connection between the two definitions of orientation?

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