Exterior Algebra

The exterior algebra (or Grassmann algebra) $\bigwedge V^{\ast }$ is the algebraic structure formed by equipping the vector space of skew-symmetric tensors with the wedge product. It is a sibling space to the full Tensor Algebra.

Corollary (Dimension of Exterior Powers)

If $n$ is the dimension of the underlying vector space $V$ (i.e., the size of its basis set, $n=\dim (V)$), then the dimension of the $k$-th exterior power is:

$$\dim \left(\stackrel{k}{\bigwedge }{V}^{\ast }\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{(n-k)!k!}$$

Example (Exterior Powers in 3D Space)

In standard 3D space ($n=3$) with coordinates $(x,y,z)$, the dimension formula $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ dictates the size of the basis for each exterior power. There are exactly four meaningful spaces:

1. 0-forms ($k=0$): Dimension $\left(\genfrac{}{}{0ex}{}{3}{0}\right)=1$. A 0-form is simply a scalar. $$f\in \stackrel{0}{\bigwedge }{V}^{\ast }=\mathbb{R}$$
2. 1-forms ($k=1$): Dimension $\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3$. The basis is $\{dx,dy,dz\}$. A 1-form is a covector. $$\stackrel{1}{\bigwedge }{V}^{\ast }=\{\underset{{\mathbf{u}}_{1-\text{form}}}{\underbrace{u_{1}dx+u_{2}dy+u_{3}dz}}\mid \mathbf{u}\in {\mathbb{R}}^3\}$$
3. 2-forms ($k=2$): Dimension $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$. The basis consists of pairs of wedge products $\{dy\wedge dz,dz\wedge dx,dx\wedge dy\}$. $$\stackrel{2}{\bigwedge }{V}^{\ast }=\{\underset{{\mathbf{w}}_{2-\text{form}}}{\underbrace{w_{1}dy\wedge dz+w_{2}dz\wedge dx+w_{3}dx\wedge dy}}\mid \mathbf{w}\in {\mathbb{R}}^3\}$$
4. 3-forms ($k=3$): Dimension $\left(\genfrac{}{}{0ex}{}{3}{3}\right)=1$. The basis is a single volume element $\{dx\wedge dy\wedge dz\}$. $$\stackrel{3}{\bigwedge }{V}^{\ast }=\{\underset{f_{3-\text{form}}}{\underbrace{fdx\wedge dy\wedge dz}}\mid f\in \mathbb{R}\}$$

(Note: Any exterior power where $k>3$ is strictly $0$ in 3D space, because you would inevitably repeat a basis element, and by skew-symmetry $dx\wedge dx=0$.)

Definition (Wedge Product / Exterior Product)

The wedge product extends the linear combination operator on differential forms, making the space of all differential forms into an exterior algebra.

$$\wedge :\stackrel{k}{\bigwedge }{V}^{\ast }\times \stackrel{\ell }{\bigwedge }{V}^{\ast }\stackrel{\text{bilinear}}{\to }\stackrel{k+\ell }{\bigwedge }{V}^{\ast }$$

This is kind of like a natural multiplication on Differential Forms. The wedge product is defined axiomatically by two rules:

1. Associativity: $(\alpha \wedge \beta )\wedge \gamma =\alpha \wedge (\beta \wedge \gamma )$
2. Skew on 1-forms: If $\alpha \in {\bigwedge }^1{V}^{\ast }$, then $\alpha \wedge \alpha =0$

Consequently, for any $k$-form $\alpha$ and $\ell$-form $\beta$:

$$\alpha \wedge \beta =(-1{)}^{k\ell }\beta \wedge \alpha$$

Example (Wedge Product in 3D Space)

1. ${\mathbf{u}}_{1-\text{form}}\wedge {\mathbf{v}}_{1-\text{form}}=(\mathbf{u}\times \mathbf{v}{)}_{2-\text{form}}$
2. ${\mathbf{u}}_{1-\text{form}}\wedge {\mathbf{w}}_{2-\text{form}}=(\mathbf{u}\cdot \mathbf{w}{)}_{3-\text{form}}$

Definition (Pairing with Vectors)

Just like a single covector pairs with a vector to produce a scalar, a $k$-form can be paired with $k$ vectors. We denote this pairing as:

$$\omega [[X_1,\dots ,X_k]]$$

and it will satisfy skew symmetry:

$$\omega [[X_1,\dots ,X_i,\dots ,X_j,\dots ,X_k]]=-\omega [[X_1,\dots ,X_j,\dots ,X_i,\dots ,X_k]]$$

Theorem (Determinant Formula for Pairing)

If we have a decomposable $k$-form ${\alpha }^1\wedge \cdots \wedge {\alpha }^k$ (where each ${\alpha }^i$ is a covector), its pairing with $k$ vectors $X_1,\dots ,X_k$ is computed via the determinant of their individual dual pairings:

$$({\alpha }^1\wedge \cdots \wedge {\alpha }^k)[[X_1,\dots ,X_k]]=\det \left[\begin{array}{ccc}⟨{\alpha }^1|X_1⟩& \dots & ⟨{\alpha }^1|X_k⟩\\ ⋮& \ddots & ⋮\\ ⟨{\alpha }^k|X_1⟩& \dots & ⟨{\alpha }^k|X_k⟩\end{array}\right]$$

This evaluation is skew-symmetric in both the vector arguments and the covector arguments.

Remark (Determinant as a Skew-Symmetry Engine)

The determinant formula mathematically guarantees this skew-symmetry because of how the inputs are arranged in the matrix:

• Columns (Vectors): All pairings with $X_1$ are in the first column, $X_2$ in the second column, and so on. Swapping two input vectors physically swaps two columns in the matrix.
• Rows (Covectors): All pairings with ${\alpha }^1$ are in the first row, ${\alpha }^2$ in the second row, and so on. Swapping two covectors in the wedge product physically swaps two rows.

Because a fundamental axiom of the determinant is that swapping any two rows or columns multiplies the result by $-1$, the pairing inherently satisfies the skew-symmetry requirement for both its vector and covector arguments.

Example (Pairings in 3D Space)

In 3D Euclidean space, the determinant pairing formula reproduces standard operations from vector calculus. If we represent 1-forms as vectors $\mathbf{u}$, 2-forms as vectors $\mathbf{w}$, and 3-forms as scalars $f$, their pairings with standard vectors (${\mathbf{u}}_{\text{vec}},{\mathbf{v}}_{\text{vec}},{\mathbf{w}}_{\text{vec}}$) are exactly:

1. 1-form Pairing (Dot Product): $${\mathbf{u}}_{\text{1-form}}[[{\mathbf{v}}_{\text{vec}}]]=\mathbf{u}\cdot {\mathbf{v}}_{\text{vec}}=\left[\begin{array}{ccc}{u}_1& u_2& u_3\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]$$
2. 2-form Pairing (Scalar Triple Product): $${\mathbf{w}}_{\text{2-form}}[[{\mathbf{u}}_{\text{vec}},{\mathbf{v}}_{\text{vec}}]]=\det \left[\begin{array}{ccc}|& |& |\\ \mathbf{w}& {\mathbf{u}}_{\text{vec}}& {\mathbf{v}}_{\text{vec}}\\ |& |& |\end{array}\right]$$ We can represent ${\mathbf{w}}_{\text{2-form}}$ as a 3D vector $\mathbf{w}$. This is computing the volume of the parallelpiped formed by all 3 vectors, precisely $\mathbf{w}\cdot (\mathbf{u}\times \mathbf{v})$. But this is precisely the Determinant of all three vectors.
3. 3-form Pairing (Scaled Triple Product): $$f_{\text{3-form}}[[{\mathbf{u}}_{\text{vec}},{\mathbf{v}}_{\text{vec}},{\mathbf{w}}_{\text{vec}}]]=f\det \left[\begin{array}{ccc}|& |& |\\ {\mathbf{u}}_{\text{vec}}& {\mathbf{v}}_{\text{vec}}& {\mathbf{w}}_{\text{vec}}\\ |& |& |\end{array}\right]$$

In particular, these are special cases because $\left(\genfrac{}{}{0ex}{}{3}{x}\right)=3$ for $x=1,2$. This is why we can map the results to vectors for 1-forms and 2-forms.

Definition (Interior Product / Contraction)

Another fundamental operation on exterior powers is the interior product (or contraction), which extends the standard covector-vector pairing. It “inserts” a vector $X\in V$ into a $k$-form, returning a $(k-1)$-form:

$$i_X:V\times \stackrel{k}{\bigwedge }{V}^{\ast }\stackrel{\text{bilinear}}{\to }\stackrel{k-1}{\bigwedge }{V}^{\ast }$$

It’s like as if we fed $1$ out of the $k$ needed vectors to the covector. The interior product satisfies three defining rules:

1. Dual pairing: For a 1-form $\alpha \in V^{\ast }$, $i_X\alpha =⟨\alpha |X⟩$.
   1. We needed $1$, and fed $1$, so we get a scalar.
2. Nilpotent: Applying the same vector twice yields zero: $i_X{i}_X\omega =0$.
   1. We fed it the same vector twice.
3. Exterior Leibniz rule:

$$i_X(\alpha \wedge \beta )=(i_X\alpha )\wedge \beta +(-1{)}^{\text{deg}(\alpha )}\alpha \wedge (i_X\beta )$$

The pairing of a $k$-form with $k$ vectors is elegantly equivalent to successively applying the interior product $k$ times:

$$\omega [[X_1,\dots ,X_k]]:=i_{X_k}\dots i_{X_1}\omega$$

Consequently, evaluating a contracted form $i_X\omega$ on $k-1$ vectors is identical to just slotting the contraction vector $X$ into the very first input slot of the original $k$-form:

$$(i_X\omega )[[X_1,\dots ,X_{k-1}]]=\omega [[X,X_1,\dots ,X_{k-1}]]$$

By recursively applying this insertion rule alongside the Exterior Leibniz rule, we mathematically recover the Determinant Formula for evaluating wedge products of 1-forms!

Example (Interior Product in 3D Space)

1. $i_{{\mathbf{u}}_{\text{vec}}}{\mathbf{v}}_{1-\text{form}}=(\mathbf{u}\cdot \mathbf{v}{)}_{0-\text{form}}$
2. $i_{{\mathbf{u}}_{\text{vec}}}{\mathbf{w}}_{2-\text{form}}=(\mathbf{w}\times \mathbf{u}{)}_{1-\text{form}}$
3. $i_{{\mathbf{u}}_{\text{vec}}}{f}_{3-\text{form}}=(f\mathbf{u}{)}_{2-\text{form}}$