# Exterior Forms

$$\def\Real{\mathbb{R}}$$

1. To generalize the notion of 1-forms, we need to develop some algebraic apparatus. It is called Exterior Calculus.
2. Consider a vector space $$V\cong \mathbb{R}^n$$.
We say that the functional
$$\omega:\underbrace{V\times \cdots \times V}_{k \text{ times}}\to \mathbb{R}$$
is an exterior $$k$$-form if it is:

1. Linear with respect to each of its entries:
$$\omega(\ldots,\alpha v + \beta v’,\ldots) = \alpha \omega(\ldots,v ,\ldots) +\beta \omega(\ldots,\beta v’,\ldots)$$
2. Skew symmetric: if the same input is plugged twice or more then:
$$\omega(\ldots, v,\ldots ,v ,\ldots) = 0$$
(or equivalently, if we interchange two entries $$v_i \leftrightarrow v_j$$ the sign of $$\omega(v_1,\ldots,v_k)$$ flips).
3. Examples:
1. Exterior $$1$$-forms (which we already have seen) are merely linear functionals on $$V$$: the condition b) is vacuous.
2. Exterior $$n$$ forms are proportional to determinants of the operators sending the basis vectors to the collection $$v_1,\ldots,v_n$$..
3. Exterior $$2$$-forms in $$V=\mathbb{R}^3$$: the signed area of the projection of the parallelogram
spanned by $$A$$ and $$B$$ to any of the 3 coordinate planes.
4. The wedge product. We create forms by combining
forms of lower degrees – an operation known as a wedge product (or exterior product).
We begin with combining two 1-forms into a 2-form: let $$\omega_1,\omega_2$$ be two 1-forms, and define the 2-form:
$$(\omega_1\wedge \omega_2)(v_1,v_2) = \left|\begin{matrix} \omega_1(v_1) & \omega_1(v_2) \\ \omega_2(v_1) & \omega_2(v_2) \end{matrix} \right|$$
This is indeed a two form, as the determinant is linear with respect to both entries, and
$$(\omega_1\wedge \omega_2)(v,v) = 0$$ as required.
5. More generally, if $$\omega_1$$ is a $$k$$-form, and $$\omega_2$$ is an $$l$$-form, then $$\omega_1\wedge\omega_2$$
is a $$(k+l)$$-form, and its value on a collection of $$(k+l)$$ vectors is given by
$$(\omega_1\wedge \omega_2)(v_1,v_2,\ldots,v_{k+l})= \sum_{I\amalg J} (-i)^{s(I,J)}\omega_1(v_{i_1},\ldots,v_{i_k})\omega_2(v_{j_1},\ldots,v_{j_l}),$$
where the summation is over all partitions of the set $$\{1.\ldots,m\}$$ into disjoint subsets $$I,J$$ of sizes $$k$$ and
$$l$$, respectively, and $$s(I,J)$$ is the parity of the permutation $$(i_1,i_2,\ldots,i_k, j_1,\ldots,j_l)$$.

The wedge product is clearly linear in each argument. Further, it is associative and graded commutative: if $$\omega_1$$ is a $$k$$-form, and $$\omega_2$$ is an $$l$$-form,
$$\omega_1\wedge\omega_2=(-1)^{kl}\omega_2\wedge\omega_1$$.

6. Exercise: let $$\omega=dx\wedge dy+ds\wedge dt+du\wedge dv$$. Find $$\omega^3$$.

7. In $$V\cong \mathbb{R}^3$$, the algebra of exterior forms is closely related to the classical vector analysis. (Unfortunately, not generalizable to other dimensions.)

Let $$A=(A_1,A_2,A_3)$$ be a vector in $$V$$. We set 1-form $$\alpha_A=A_1dx_1+A_2dx_2+A_3dx_3$$, and 2-form $$\beta_A=A_1dx_2\wedge dx_3+A_2dx_3\wedge dx_1+A_3dx_1\wedge dx_2$$. Then
$\alpha_A\wedge\alpha_B=\beta_{A\times B},$
$\alpha_A\wedge\beta_B=(A\cdot B)dx_1\wedge dx_2\wedge dx_3.$

8. Let $$Q^k$$ be the collection of all $$k$$-forms over $$V$$. It form, clearly, a linear space.
There is a natural map
$$Q^1\times \cdots \times Q^1 \to Q^k$$
given by by taking wedge product of $$k$$ 1-forms.

Expanding each of the 1-form in the wedge product as a sum of the basis covectors $$dx_k,k=1,\ldots,n$$, we see that each such form is a linear combination of coordinate monomials $$\xi_I$$ defined as
$\xi_I=dx_{i_1}\wedge\ldots\wedge dx_{i_k},$
where $$i_1<i_2<\ldots<i_k$$. These monomials are linearly independent, as we'll see next.

9. Proposition: $$\eta_I$$ form a basis.
Proof: plugging into a linear combination of monomials different collections of the coordinate vectors, we see that if they all are vanishing, then the coefficients should be zeros. On the other hand, any collection of k vectors, whenplugged into a form, can be expanded into a sum of the values of the form on collections of coordinate vectors – this are exactly the monomials.
10. Corollary: the dimension is given by all possible
ways to choose $$k$$ coordinates out of $$n$$:
$$\text{dim}(Q^k(\mathbb{R}^n)) = {n \choose k}$$
11. For example, for $$\mathbb{R}^3$$ with three dual basis vectors $$\eta_1,\eta_2,\eta_3$$ we have
$$Q^0 = \text{span}\{1\} =\mathbb{R}$$
(this is a convention that will be later explained), and
\begin{align} Q^1 &= \text{span}\{\eta_1,\eta_2,\eta_3\} \\ Q^2 &= \text{span}\{\eta_1\wedge \eta_2 ,\eta_2 \wedge \eta_3 ,\eta_1\wedge \eta_3\} \\ Q^3 &= \text{span}\{\eta_1\wedge \eta_2\wedge \eta_3\} \end{align}

If $$\omega_1 = A_1\eta_1 + A_2 \eta_2 + A_3 \eta_3$$
and $$\omega_2 = B_1\eta_{23} + B_2 \eta_{31} + B_3 \eta_{12}$$
Then
$$\omega_1\wedge \omega_2 = (A_1\eta_1 + A_2 \eta_2 + A_3 \eta_3)(B_1\eta_1 + B_2 \eta_2 + B_3 \eta_3) = (A_1B_2 – A_2B_1) \eta_1 + (A_2B_3 – A_3B_2) \eta_2 + (A_3B_1 – A_1B_3) \eta_3$$
which is the cross product of the coordinates of the 1-forms.

Now let $$\omega^1$$ be a 1-form and $$\omega^2$$ be a 2-form:

$$\omega^1 \wedge \omega^2 = (A_1B_1+A_2B_2+A_3B_3) \eta_1\wedge \eta_2 \wedge \eta_3$$