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Hodge duality of differential forms

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Recall that the fundamental correspondence establishes this relation:

Forms $\leftrightarrow$ vector fields and functions.

In dimension $2$ we have: $$ \newcommand{\lra}[1]{\!\!\!\!\!\!\!\xleftarrow{}\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} % \begin{array}{llllllllllllll} \Omega^0 & \ra{FC} & \{functions\} & \ra{} & \\ \da{d_0} & & & &\da{} \\ \Omega^1 & \ra{FC} & \{vector fields\} & &\da{same!}\\ \da{d_1} & & & &\da{} \\ \Omega^2 & \ra{FC} & \{functions\} & \la{} & \end{array} $$

The fundamental correspondence, in particular, $$Adx \hspace{1pt} dy \longmapsto A,$$ produces an operator on differential forms only: $$ \Omega^2 \rightarrow \Omega^0 .$$ And back!

For this function we introduce the following new notation: $$(Adx \hspace{1pt} dy)^*=A.$$

More generally, we define the so-called the Hodge duality operator, or Hodge star operator, or simply the $\star$-operator: $$\star \colon \Omega^k \rightarrow \Omega^{2-k},$$ by its action on the basis elements: $$\begin{align*} \Omega^2: & (dx \hspace{1pt} dy)^* = 1,\\ \Omega^0: & 1^* = dx \hspace{1pt} dy,\\ \Omega^1: & dx^*=dy, \\ & dy^*=-dx. \end{align*}$$

It is, in fact, a linear operator on the whole de Rham complex: $$\star \colon \{\Omega^k\} \rightarrow \{\Omega^k\}.$$ So, the Hodge operator is a non-trivial graded automorphism. Explain.

Written with the operator notion, the identities take this form: $$\begin{align*} & \star (dx \hspace{1pt} dy) = 1,\\ & \star 1 = dx \hspace{1pt} dy,\\ & \star dx = dy, \\ & \star dy = -dx. \end{align*}$$

Example: $$\begin{align*} (x^2y dx + xdy)^* &= x^2y(dx)^* + x(dy)^* \\ &=x^2y dy - x dx. \end{align*}$$

The dimensions of any two dual forms are complementary.

Moreover, we notice that the "basic forms" above $\varphi$ and $\varphi^*$ are paired up: $$\varphi \wedge \varphi ^*= dxdy.$$

Let's verify:

  • $dx\wedge (dx)^* \stackrel{?}{=} dx \hspace{1pt} dy$? We know $(dx)^* = dy$, so yes.
  • $dy\wedge (dy)^* \stackrel{?}{=} dx \hspace{1pt} dy$? Use anti-symmetry to get $dy\wedge (-dx) = dx \hspace{1pt} dy$, so yes.
  • $dx \hspace{1pt} dy \wedge (dx \hspace{1pt} dy)^* = dx \hspace{1pt} dy \wedge 1$ and $1 \wedge (1)^* = 1 dx \hspace{1pt} dy$, so yes.

Consider similar rules for other dimensions, $n$.

If $n=1$: $$1 \wedge (1)^* = 1 \cdot dx,$$ $$dx\wedge (dx)^* = dx \cdot 1.$$

If $n=3$: $$\begin{align*} dx\wedge (dx)^* &= dx \hspace{1pt} dy \hspace{1pt} dz = dx \hspace{1pt} dy \hspace{1pt} dz, \\ dy\wedge (dy)^* &= dy \hspace{1pt} dz \hspace{1pt} dx = dx \hspace{1pt} dy \hspace{1pt} dz, \\ dz\wedge (dz)^* &= dz \hspace{1pt} dx \hspace{1pt} dy = dx \hspace{1pt} dy \hspace{1pt} dz ,\\ dx \hspace{1pt} dy\wedge (dx \hspace{1pt} dy)^* &= dx \hspace{1pt} dy \hspace{1pt} dz = dx \hspace{1pt} dy \hspace{1pt} dz. \end{align*}$$

So, we can reduce the whole operator to this list of its values: $$\begin{align*} 1^* &= dx \hspace{1pt} dy \hspace{1pt} dz \\ (dx)^* &= dy \hspace{1pt} dz, \\ (dy)^* &= dz \hspace{1pt} dx, \\ (dz)^* &= dx \hspace{1pt} dy, \\ (dx \hspace{1pt} dy)^* &= dz, \\ (dy \hspace{1pt} dz)^* &= dx, \\ (dz \hspace{1pt} dx)^* &= dy, \\ (dx \hspace{1pt} dy \hspace{1pt} dz)^* &= 1. \end{align*}$$

Note. The resulting forms

  • 1: $\omega=dx$,
  • 2: $\omega=dxdy$,
  • 3: $\omega=dxdydz$,

are called volume forms.

For an arbitrary pair of forms $\varphi,\psi$ of the same degree the above rule takes a more complex form: $$\varphi \wedge \psi ^*= <\varphi,\psi^*>\omega,$$ when an inner product is provided.

If now we think of functions $A,B,C$ as coefficients of these "basic" forms we can recover the full correspondence: $$Adx+Bdy+Cdz \stackrel{\star}{\longleftrightarrow} Adydz+Bdxdz+Cdxdy,$$ $$Adxdydz \stackrel{\star}{\longleftrightarrow} A.$$ It is clear now that Hodge duality is independent from the fundamental correspondence.

Another observation: $$\star\star = (-1)^?Id.$$ The sign depends on the sign convention.

The definition of Hodge duality for forms of all degrees as an operator $$\star : \Omega ^k \rightarrow \Omega ^{n-k}$$ relies on Hodge duality in linear algebra.

What's Hodge duality for?

First, it reveals a certain symmetry of the de Rham complex. Indeed, for $n$ the dimension of the space, consider this (non-commutative) Hodge duality diagram: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} & \Omega ^{0}& \ra{d} & ... & \ra{d}& \Omega ^k & \ra{d} & \Omega ^{k+1} & \ra{d} &...& \ra{d} & \Omega ^{n}\\ & \da{\star} & & & & \da{\star} & \ne & \da{\star} & & & & \da{\star} \\ & \Omega ^{n}& \la{d}& ... & \la{d}& \Omega ^{n-k}& \la{d} & \Omega ^{n-(k+1)}& \la{d} &...& \la{d} & \Omega ^{0} \\ \end{array} $$ Both the top and bottom rows represent the same complex!

Keeping in mind that the star operator is an isomorphism, the result is a "mirror" symmetry, illustrated for dimensions $2$ and $3$: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % n=2: \begin{array}{ccccccccccc} & \Omega ^{0},& \Omega ^1, & \Omega ^2 & &\\ & \da{} & & \ua{} & &\\ & & \ra{\cong}& & &\\ \end{array} n=3: \begin{array}{ccccccccccc} & \Omega ^{0},& \Omega ^1 & \ra{\cong} &\Omega ^2, & \Omega ^3 \\ & \da{} & & & & \ua{} \\ & & \ra{} & \ra{\cong} & \ra{} & \\ \end{array} $$ The second diagram explains why vector fields appear in two very different integral theorems of vector calculus (as $1$-forms and as $2$-forms): the duality operator matches the algebra but, since the exterior derivative is reversed, not the calculus. Compare also to Poincare duality of homology vs cohomology.

Exercise. Explain the mirror symmetry of the exterior derivative.

Here's another place for Hodge duality. Given a function $f$, we know that $ddf=0$. So the "second exterior derivative" is always trivial. But how do I study, for example, the concavity of $f$ without some kind of second derivative?!

The answer is to

  • "dualize" the exterior derivative $df$,
  • take the exterior derivative of the result, and then
  • "dualize" it back:

$$f \stackrel{d}{\longmapsto} df = f' dx \stackrel{\star}{\longmapsto} f' \stackrel{d}{\longmapsto} f' ' dx \stackrel{\star}{\longmapsto} f' '.$$

So, for a $0$-form this is, again, a $0$-form. This procedure has this effect for all forms: the "dualized second derivative" of a $k$-form $$\varphi ' ' =\star d \star d \varphi$$ is a $k$-form. We just make a full circle around one of the squares in the duality diagram above. If first take Hodge dual and then exterior derivative and continue around the adjacent square, the result is another "dualized second derivative" $$\varphi ' ' =d \star d \star \varphi,$$ which is also a $k$-form. Combining these two is used to define the analogue of the Laplacian for differential forms.