Identities of vector calculus

Fundamental correspondence in ${\bf R}^3$

We draw a similar to ${\bf R}^2$ diagram and examine the operators corresponding to the exterior derivative $d$.

$$ \newcommand{\lra}[1]{\!\!\!\!\!\!\!\xleftarrow{}\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ra}[1]{\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{rlllllllllll} & A & \ra{\lambda_0} & \varphi^0 \\ {\rm grad} & \da{} & & \da{d} & \\ & V & \ra{\lambda_1} & \varphi^1 \\ {\rm curl} & \da{} & & \da{d} \\ & W & \ra{\lambda_2} & \varphi^2 \\ {\rm div} & \da{} & & \da{d} \\ & B & \ra{\lambda_3} & \varphi^3 \end{array} $$

Here $A,B$ are functions and $V,W$ are vector fields. Again, the ideas for the arrows come from vector calculus:

• gradient: functions $\mapsto$ vector fields,
• curl: vector fields$\mapsto$ vector fields,
• divergence: vector fields $\mapsto$ functions.

Now, we define the operators $\lambda_0, ...,\lambda_3$ of the fundamental correspondence, in a very simple and natural fashion: $$A \stackrel{\lambda_0}{\mapsto} \varphi^0 = A.$$ $$V = (F,G,H) \stackrel{\lambda_1}{\mapsto} \varphi^1=Fdx+Gdy+Hdz.$$ $$W =(P,R,Q) \stackrel{\lambda_2}{\mapsto} \varphi^2 = Q dx \hspace{1pt} dy + P dy \hspace{1pt} dz + R dz \hspace{1pt} dx.$$ $$B \stackrel{\lambda_3}{\mapsto} \varphi^3 = B dx \hspace{1pt} dy \hspace{1pt} dz.$$

Next, we show that this diagram makes sense, i.e., commutes.

Commutativity of the fundamental correspondence

Next, we take squares of the diagram above and show that they commute.

$$ \newcommand{\lra}[1]{\!\!\!\!\!\!\!\xleftarrow{}\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & A & \ra{\lambda_0} & \varphi^0 \\ {\rm grad} & \da{} & & \da{d} \\ & V & \ra{\lambda_1} & \varphi^1 \end{array} $$

Theorem: The above diagram commutes: $$d \lambda_0 = \lambda_1 {\rm grad}.$$

Proof: Compute $d \lambda_0(A) = dA = {\rm done \hspace{3pt} before} = A_x dx + A_y dy + A_z dz$. Now compute $\lambda_1 {\rm grad} A = \lambda_1(A_x, A_y, A_z) = A_xdx + A_y dy+A_z dz$. $\blacksquare$

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xleftarrow{}\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & V & \ra{\lambda_1} & \varphi^1 \\ {\rm curl} & \da{} & & \da{d} \\ & W & \ra{\lambda_2} & \varphi^2 \end{array} $$

Theorem: The above diagram commutes: $$d \lambda_1 = \lambda_2 {\rm curl}.$$

Proof: Compute $d \lambda_1(F,G,H) = d(F dx + G dy + H dz)$. For $\varphi = Fdx + Gdy + Hdz$ we compute

$$\begin{align*} d\varphi &= dF \hspace{1pt} dx + dG \hspace{1pt} dy + dH \hspace{1pt} dz \\ &= (F_x dx + F_y dy + F_z dz) dx + (G_x dx + G_y dy + G_z dz) dy + (H_x dx + H_y dy + H_z dz)dz) \\ &= F_y dy \hspace{1pt} dx + F_z dz \hspace{1pt} dx + G_x dx \hspace{1pt} dy + G_z dz \hspace{1pt} dy + H_x dx \hspace{1pt} dz + H_y dy \hspace{1pt} dz \\ &=(G_x - F_y) dx \hspace{1pt} dy + (H_y - G_z) dy \hspace{1pt} dz + (F_z - H_x)dz \hspace{1pt} dx \\ &= \lambda_2 {\rm curl}(F,G,H). \end{align*}$$ We used the definition of curl: $${\rm curl}(F,G,H) = (G_x - F_y, H_y - G_z, F_z - H_x).$$ $\blacksquare$

$$ \newcommand{\lra}[1]{\!\!\!\!\!\!\!\xleftarrow{}\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & W & \ra{\lambda_2} & \varphi^2 \\ {\rm div} & \da{} & & \da{d} \\ & C & \ra{\lambda_3} & \varphi^3 \end{array} $$

Theorem: The above diagram commutes: $$d \lambda_2 = \lambda_3 {\rm div}.$$

Exercise: Prove it.

Identities of vector calculus

This is our bird's eye view of the relation between vector calculus and exterior calculus: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} & \Omega ^0 & \ra{FC, \cong} & \{functions\} & \\ & \da{d_0} & & \da{grad} \\ & \Omega ^1 & \ra{FC, \cong} & \{vector fields\} & \\ & \da{d_1} & & \da{curl} \\ & \Omega ^2 & \ra{FC, \cong} & \{vector fields\} & \\ & \da{d_2} & & \da{div} \\ & \Omega ^3 & \ra{FC, \cong} & \{functions\} & \\ \end{array} $$

The operations of vector calculus now become a part of a bigger picture!

For example, what we know about exact forms is that they are all closed. In other words: $$dd=0.$$

This equation has two instances under the fundamental correspondence.

First, $$dd=0 \colon \Omega^0 \rightarrow \Omega^2,$$ which translates, because of the commutativity of the diagram, into $${\rm curl \hspace{3pt} grad} = 0.$$

Second, $$dd=0 \colon \Omega^1 \rightarrow \Omega^3,$$ which translates into $${\rm div \hspace{3pt} curl} = 0.$$

We get the differential identities for free!

Recall also, a vector field $V$ is called conservative if $V = {\rm grad} A$, where $A$ is some function. So, the fundamental correspondence demonstrates the relation:

conservative vector field $\stackrel{{\rm F \hspace{0pt} C}}{\longleftrightarrow}$ exact $1$-form.

In addition, the General Stokes Theorem gives us the integral identities of vector calculus.

Exercises

1. Show that a vector field of the form $V = f(||u||)u$ is conservative.

2. In $R^3 - \{0\}$, construct

• a closed but not exact $2$-form;
• a vector field without sources which is not the curl of any vector field.

3. Are there in $R^n - \{0\}$ closed but not exact forms of degree $n-1$? What about degree $n-1$?