Subjects: Differential Geometry, Linear Algebra, Clifford Algebra
Links: Dual Vector Spaces, Exterior Algebra of Vector Spaces, Multilinear Transformations, Tensor Product of Modules, Graded Ring, Tensor Product of Linear Functions, Symmetric Group

Let $K$ be characteristic $0$ field. We are going to drop, the fact that the codomain is $K$ for the rest of the note, since this notation is a bit cumbersome.

Def: A covariant $k$-tensor on $V$ is said to be alternating if its value changes sign whenever two arguments are interchanged, or equivalently if any permutation of the arguments causes its value by the sign of the permutation. Alternating covariant $k$-tensors are also called exterior forms, multicovectors or $k$-covectors. The space of all $k$-covectors on $V$ is denoted by $\stackrel{k}{\bigwedge }(V^{\ast })$. All $0$-tensors and $1$-tensors are alternating.

Lemma: Let $\alpha$ be a covariant $k$-tensor on a finite-dimensional vector space $V$. The following are equivalent.

• $\alpha$ is alternating.
• $\alpha (v_1,\dots ,v_k)=0$ whenever the $k$-tuple $(v_1,\dots ,v_k)$ is linearly independent.
• $\alpha$ gives the value zero whenever two of its arguments are equal: $$\alpha(v_1,\dots, w,\dots, w, \dots v_k) = 0.$$

Def: We define a projection $\text{Alt}:T^k(V^{\ast })\to \stackrel{k}{\bigwedge }(V^{\ast })$, called alternation as follows: $$\text{Alt } \alpha = \frac1{k!} \sum_{\sigma \in S_k} \text{sgn }\sigma (\sigma \alpha),$$where $S_k$ is the symmetric group on $k$ elements. More explicitly, this means $$(\text{Alt } \alpha)(v_1,\dots, v_k) = \frac1{k!} \sum_{\sigma \in S_k} \text{sgn }\sigma \alpha(v_{\sigma(1)},\dots, v_{\sigma(k)}). $$
Properties of Alternation: Let $\alpha$ be a covariant tensor on a finite-dimension $K$-vector space.

• $\text{Alt }\alpha$ is alternating.
• $\text{Alt }\alpha =\alpha$ iff $\alpha$ is alternating.

Elementary Alternating Tensors

Def: Given a positive integer $k$, an ordered $k$-tuple $I=(i_1,\dots ,i_k)$ of positive integers is called a multi-index of length $k$. If $I$ is such a multi-index and $\sigma \in S_k$, we write $I_{\sigma }$ for the following multi-index: $$I_\sigma := (i_{\sigma(1)},\dots, i_{\sigma(k)}) $$Note that $I_{\sigma \tau }:=(I_{\sigma }{)}_{\tau }$ for $\sigma ,\tau \in S_k$.

Def: Let $V$ be a finite-dimensional $K$-vector space, and suppose $({\epsilon }^1,\dots ,{\epsilon }^n)$ is any basis for $V^{\ast }$. We know define a collection of $k$-covectors on $V$ that generalise the determinant function on ${\mathbb{R}}^n$. For each multi-index $I=(i_1,\dots ,i_k)$ of length $k$ such that $1\le i_1\dots ,i_k\le n$, define a covariant $k$-tensor ${\epsilon }^I={\epsilon }^{i_1,\dots ,i_k}$ by $$\varepsilon^I(v_1,\dots, v_k) = \det \begin{pmatrix}
\varepsilon^{i_1}(v_1)& \cdots & \varepsilon^{i_1}(v_k) \
\vdots & \ddots & \vdots \ \varepsilon^{i_k}(v_1) & \cdots & \varepsilon^{i_k}(v_k)
\end{pmatrix} = \det \begin{pmatrix}
v_1^{i_1}& \cdots & v_k^{i_1} \
\vdots & \ddots & \vdots \ v_1^{i_k} & \cdots & v_k^{i_k}
\end{pmatrix}.$$In other words, if $\mathbb{v}$ is denotes the $n\times k$ matrix whose columns are the components of the vectors $v_1,\dots ,v_k$ with respect to the basis $(E_i)$ dual to $({\epsilon }^i)$, then ${\epsilon }^I(v_1,\dots ,v_k)$ is the determinant of the $k\times k$ submatrix consisting of the rows $i_1,\dots ,i_k$ of $\mathbb{v}$ . We call ${\epsilon }^I$ an elementary alternating tensor or elementary $k$-covector.

In order to streamline the computations with the elementary $k$-covectors, we can extend the Kronecker delta notation in the following way. If $I$ and $J$ are multi-indices of length $k$, we define$$\delta^I_J := \det \begin{pmatrix}
\delta_{j_1}^{i_1}& \cdots & \delta_{j_k}^{i_1} \
\vdots & \ddots & \vdots \ \delta_{j_1}^{i_k} & \cdots & \delta_{j_k}^{i_k}
\end{pmatrix} .$$
Obs: We see that $$\delta^I_J = \begin{cases}\text{sgn }\sigma & \text{if neither } I \text{ nor } J \text{ has repeated index and } J = I_\sigma \text{ for some }\sigma\in S_k, \
0 & \text{if }I \text{ or }J \text{ has a repeated index or } J \text{ is not a permutation of }I.
\end{cases} $$
Properties of Elementary $k$-covectors: Let $(E_i)$ be a basis for $V$ and let $({\epsilon }^i)$ be the dual basis for $V^{\ast }$, and let $I$ be a multi-index.

• If $I$ has a repeated index, then ${\epsilon }^I=0$.
• If $J=I_{\sigma }$, for some $\sigma \in S_k$, then ${\epsilon }^J=(\text{sgn }\sigma ){\epsilon }^J.$
• The result of evaluating ${\epsilon }^I$ on a sequence of basis vectors is $$\varepsilon^I(E_{j_1},\dots, E_{j_k}) = \delta^I_J. $$
  Def: A multi-index $I=(i_1,\dots ,i_k)$ is said to be increasing if $i_1<\cdots <i_k$.

A Basis for $\stackrel{k}{\bigwedge }(V^{\ast })$: Let $V$ be a $n$-dimensional $K$-vector space. If $({\epsilon }^i)$ is any basis for $V^{\ast }$, then for each positive $k\le n$, the collection of $k$-covectors $$\mathcal E:= {\varepsilon^I\mid I \text{ is an increasing multi-index of length }k} $$is a basis for $\stackrel{k}{\bigwedge }(V^{\ast })$. Therefore $$\dim {\displaystyle \bigwedge}^{!k}(V^*) = {n\choose k}. $$If $k>n$, then $\dim \stackrel{k}{\bigwedge }(V^{\ast })=0$.

Prop: Suppose $V$ is an $n$-dimensional $K$-vector space and $\omega \in \stackrel{k}{\bigwedge }(V^{\ast })$. If $T:V\to V$ is any linear map and $v_1,\dots ,v_n$ are arbitrary vectors in $V$, then $$\omega(Tv_1,\dots, Tv_n) = (\det T) \omega(v_1,\dots, v_n). $$

The Wedge Product

Determinant Convention: Given $\omega \in \stackrel{k}{\bigwedge }(V^{\ast })$ and $\eta \in \stackrel{l}{\bigwedge }(V^{\ast })$, we define their wedge product or exterior product to be the following $(k+k)$-covector: $$\omega \wedge \eta := \frac{(k+l)!}{k!l!}\text{Alt }(\omega\otimes \eta).$$
Alt Convention: Given $\omega \in \stackrel{k}{\bigwedge }(V^{\ast })$ and $\eta \in \stackrel{l}{\bigwedge }(V^{\ast })$, we define their wedge product or exterior product to be the following $(k+k)$-covector: $$\omega \wedge \eta := \text{Alt }(\omega\otimes \eta).$$

Lemma: Let $V$ be a $n$-dimensional $K$-vector space and let $({\epsilon }^1,\dots ,{\epsilon }^n)$ be a basis for $V^{\ast }$. For any multi-indices $I=(i_1,\dots ,i_k)$ and $J=(j_1,\dots ,j_l)$, $$\varepsilon^I \wedge \varepsilon^J = \varepsilon^{IJ}, $$where $IJ:=(i_1,\dots ,i_k,j_1,\dots ,j_l)$ is obtained by concatenating $I$ and $J$. (under the determinant convention). $$\varepsilon^I \wedge \varepsilon^J =\frac{k!l!}{(k+l)!} \varepsilon^{IJ}, $$under the Alt convention.

Properties of the Wedge Product: Suppose $\omega$, ${\omega }^{\prime }$, $\eta$, ${\eta }^{\prime }$ and $\xi$ are multicovectors on a finite-dimensional $K$-vector space $V$.

• Bilinearity: For $a,b\in \mathbb{R}$, $$\begin{align*}
  (a\omega+ b\omega')\wedge \eta &= a(\omega\wedge \eta) + b(\omega' \wedge \eta), \ \eta\wedge (a\omega + b\omega') &= a(\eta \wedge \omega)+ b(\eta\wedge \omega').
  \end{align*} $$
• Associativity: $$\omega \wedge (\eta \wedge \xi) = (\omega \wedge \eta ) \wedge\xi. $$
• Anticommutativity: $$\omega \wedge\eta = (-1)^{kl}\eta\wedge\omega. $$
• If $({\epsilon }^i)$ is any basis for $V^{\ast }$ and $I=(i_1,\dots ,i_k)$ is any multi-index, then $$\varepsilon^{i_1}\wedge\dots \wedge\varepsilon^{i_k} = \varepsilon^I. $$
• For any covectors ${\omega }^1,\dots ,{\omega }^k$ and vectors $v_1,\dots ,v_k$, $$\omega^1\wedge\dots \wedge \omega^k(v_1,\dots, v_k) = \det(\omega^j(v_i)), $$under the determinant convention. We also have $$\omega^1\wedge\dots \wedge \omega^k(v_1,\dots, v_k) =\frac{1}{k!} \det(\omega^j(v_i)). $$
  Def: A $k$-covector is decomposable if it can be expressed in the form $\eta ={\omega }^1\wedge \cdots \wedge {\omega }^k$, where ${\omega }^1,\dots ,{\omega }^k$ are covectors.

Def: For any $n$-dimensional $K$-vector space $V$, define a vector space $\bigwedge (V^{\ast })$ by $$\bigwedge(V^*) := \bigoplus_{k = 0}^n {\textstyle \bigwedge}^{!k} (V).$$We see that $\bigwedge (V)$ is a $K$-vector space with dimension $2^n$.

Obs: We see that $(\bigwedge (V),\wedge )$ is an anticommutative graded algebra, and it is called the exterior algebra or Grassmann algebra of $V$.

Interior Multiplication

Def: Let $V$ be a finite-dimensional $K$-vector space. For each $v\in V$, we define a linear map $i_v:\stackrel{k}{\bigwedge }(V^{\ast })\to \stackrel{k-1}{\bigwedge }(V^{\ast })$, called interior multiplication by $v$, as follows: $$i_v\omega(w_1,\dots, w_{k-1}) := \omega(v, \omega_1,\dots, w_{k-1}). $$In other words, $i_v\omega$ is obtained from $\omega$ by inserting $v$ into the first slot. By convention, we interpret $i_v\omega$ to be zero when $\omega$ is a $0$-covector. Another common notation is $$v;\lrcorner ;\omega := i_v \omega .$$This is often read as '$v$ into $w$'.

Lemma: Let $V$ be a finite-dimensional $K$-vector space and $v\in V$.

• $i_v\circ i_v=0$.
• If $\omega \in \stackrel{k}{\bigwedge }(V^{\ast })$ and $\eta \in \stackrel{l}{\bigwedge }(V^{\ast })$, $$i_v(\omega\wedge\eta) = (i_v \omega)\wedge \eta + (-1)^k\omega\wedge(i_v \eta). $$

when the wedge product is defined using the Alt convention, interior multiplication of a vector with a $k$-form has to be defined with an extra $k$: $$\bar i_v \omega(w_1,\dots, w_{k-1}) = k\omega(v, w_1,\dots, w_{k-1}). $$This definition ensures that the interior multiplication ${\overline{i}}_v$ still satisfies $$\bar i_v(\omega\wedge\eta) = (\bar i_v \omega)\wedge \eta + (-1)^k\omega\wedge(\bar i_v \eta). $$

Duality

Let $V$ be a space with inner product.

For each natural $k$, we can define an interior product on ${\bigwedge }^k(V)$. Let ${\omega }^1\wedge \dots {\omega }^k,{\eta }^1\wedge \dots {\eta }^k\in {\bigwedge }^k(V)$, then $$\langle \omega^1\wedge\dots \wedge\omega^k,\eta^1\wedge\dots\wedge\eta^k\rangle := \det(\langle(\omega^i)^\sharp, (\eta^j)^\sharp\rangle). $$So basically, we are calculating pulling the covectors in the original vector space using the musical isomorphisms, and then calculating the inner product there.