A Bad Attempt At Connecting Differential Forms And Multivariable Calculus

So after taking MVC, we’ve all been through those tedious proofs that \nabla \times \nabla F = 0 and \nabla \cdot \left( \nabla \times F \right) = 0. Here, we give a unified way to view these identities.

We start by giving the notion of a tangent space of a point in \RR^n, which is a vector at that point. Denote this with T_p \RR^n = \lbrace (p, \vec{v}) : \vec{v} \in \RR^n \rbrace, which also has the convenient notation (p, \vec{v}) \equiv v_p. We can then take the union of all these tangent spaces and call it the tangent bundle \displaystyle{T \RR^n = \bigcup_{p \in \RR^n} T_p \RR^n}.

We can view tangent vectors and vector fields as differential operators for a function. The directional derivative of a C^1 function at a point p in direction v is v_p[f] = g'(0), where g(t) = f(p + tv). After some crunching, we arrive at

    \[ v_p[f] = \langle v, \nabla f(p) \rangle. \]

We then naturally extend this to vector fields, namely (Vf)(p) = V(p)[f].

Now we can talk about differential forms. Recall that the total differential of a function f : \RR^3 \to \RR is

    \[ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz. \]

We now define a differential 1-form to be a function \varphi : T \RR^n \to \RR such that when we restrict \varphi to a certain point p and consider \varphi_p : T_p \RR^n \to \RR, \varphi_p is a linear functional. It turns out that we can equip 1-forms with function coefficients. If \varphi is a 1-form, then (f \varphi)(v_p) = f(p) \varphi(v_p). We also let the standard set of differential 1-forms to be the set of functions dx_i((v_1, v_2, \dots, v_n)_p) = v_i, so in \RR^3, dy(v_p) reads out the 2nd coordinate of v. Then the total differential becomes a very natural 1-form.

If we have 1-forms, what about what about 0-forms and 2-forms? We can naively let 0-forms be the set of functions, and higher dimensional forms to be some kind of product of lower dimensional forms, with some sort of derivative operation that moves from one form space to the next. It turns out that when we do this, we get a lot of nice things. We can define a new product of differential forms, called the wedge product, such that df \wedge dg = -dg \wedge df, and takes in more tangent vectors and linear in each component when you fix a point. Then if \varphi = f \, dx + g \, dy + h \, dz is a 1-form, we can write d \varphi = df \wedge dx + dg \wedge dy + dh \wedge dz, and define analogously for higher dimensional spaces (here we took n = 3) and forms. If we crunch out the products, we learn that d^2 = 0, or if \varphi is a differential form, then d(d\varphi) = 0. From this, we can construct an interesting diagram.

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where

    \[ \begin{aligned} f_1(f) &= f \\ f_2(x, y, z) &= x \, dx + y \, dy + z \, dz \\ f_3(x, y, z) &= z \, dx \wedge dy + x \, dy \wedge dz + y \, dz \wedge dx \\ f_4(f) &= f \, dx \wedge dy \wedge dz. \\ \end{aligned} \]

are isomorphisms and \Omega^n denotes the space of n-forms.

Construct a similar chain of morphisms to get good at n-dimensional calculus where n > 3.

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