Differential Forms Generalize R3 Calculus

As a nice brief exercise, we show the following diagram commutes:

The map $β : X (R^{3}) \to Ω^{2} (R^{3})$ is defined by $β (X) = X ┘ d x^{1} \land d x^{2} \land d x^{3}$ ; $β$ is clearly $C^{\infty} (R^{3})$ -linear, and thus its associated pointwise operator, interior multiplication, is a bundle homomorphism from $T R^{3}$ to $Λ^{2} (R^{3})$ . As interior multiplication is injective and both are rank- $3$ bundles, $β$ is a smooth bundle isomorphism between $T R^{3}$ and $Λ^{2} (R^{3})$ . This implies $β$ is a vector space isomorphism between the smooth section spaces. Meanwhile, $* : C^{\infty} (R^{3}) \to Ω^{3} (R^{3})$ is defined by $* (f) = f d x^{1} \land d x^{2} \land d x^{3}$ ; this is clearly a linear isomorphism.

We sequentially show commutativity. As the exterior derivative on $0$ -covectors is simply the differential, the leftmost cycle is trivial (it is well-known that $grad$ and the differential $d$ commute with $♭$ ). Recall that $curl$ is defined as

curl (X) = (\frac{\partial X ^{3}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{3}}) \frac{\partial}{\partial x ^{1}} + (\frac{\partial X ^{1}}{\partial x ^{3}} - \frac{\partial X ^{3}}{\partial x ^{1}}) \frac{\partial}{\partial x ^{2}} + (\frac{\partial X ^{2}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{2}}) \frac{\partial}{\partial x ^{3}}

Thus, $β (curl X)$ satisfies, for any $i$ and $j$ ,

β (curl X) (\frac{\partial}{\partial x ^{i}}, \frac{\partial}{\partial x ^{j}}) = \frac{\partial X ^{3}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{3}} δ_{1}^{i} δ_{1}^{j} \frac{\partial X ^{1}}{\partial x ^{3}} - \frac{\partial X ^{3}}{\partial x ^{1}} δ_{2}^{i} δ_{2}^{j} \frac{\partial X ^{2}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{2}} δ_{3}^{i} δ_{3}^{j}

A simple computation shows

β (curl X) = (\frac{\partial X ^{2}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{2}}) d x^{1} \land d x^{2} + (\frac{\partial X ^{3}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{3}}) d x^{1} \land d x^{3} + (\frac{\partial X ^{3}}{\partial x ^{1}} - \frac{\partial X ^{1}}{\partial x ^{3}}) d x^{1} \land d x^{2}

As the Euclidean metric simply corresponds to $I_{3}$ , the $♭$ operator in canonical coordinates preserves each component. Thus, for any $X \in X (R^{3})$ ,

X^{♭} = X_{i} d x^{i} \Rightarrow d X^{♭} = d X_{i} \land d x^{i} = \frac{\partial X _{i}}{\partial x ^{j}} d x^{j} \land d x^{i}

Therefore, $d X^{♭} = β (curl X)$ as desired.

For any $X \in X (R^{3})$ ,

* (div X) = (\frac{\partial X ^{1}}{\partial x ^{1}} + \frac{\partial X ^{2}}{\partial x ^{2}} + \frac{\partial X ^{3}}{\partial x ^{3}}) d x^{1} \land d x^{2} \land d x^{3}

Meanwhile,

β (X) = X^{1} d x^{2} \land d x^{3} - X^{2} d x^{1} \land d x^{3} + X^{3} d x^{1} \land d x^{2}

d β (X) = d X^{1} \land d x^{2} \land d x^{3} - d X^{2} \land d x^{1} \land d x^{3} + d X^{3} \land d x^{1} \land d x^{2}

Clearly, $d β (X) = * (div X)$ .

As $Id, ♭, β$ , and $*$ are all vector space isomorphisms and $d^{2} \equiv 0$ , this generalization trivializes the identities $div \circ curl \equiv 0$ and $curl \circ grad \equiv 0$ .

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Differential Forms Generalize R3 Calculus

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