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Continuous vs discrete differential forms

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We need to establish a relation between continuous differential forms over region $D \subset {\bf R}^n$ which is a realization of a cubical complex $K$ (or cell complex) and discrete differential forms, i.e., cochains, over $K$. So far, we have observed only that they behave in a very similar way.

Recall the diagram of all exterior derivatives of continuous forms and discrete forms: $$ \Omega^0 \stackrel{d_0}{\rightarrow} \Omega^1 \stackrel{d_1}{\rightarrow} \Omega^2 \stackrel{d_2}{\rightarrow} \Omega^3 \stackrel{d_3}{\rightarrow} \ldots ,$$ $$ C^0 \stackrel{d_0}{\rightarrow} C^1 \stackrel{d_1}{\rightarrow} C^2 \stackrel{d_2}{\rightarrow} C^3 \stackrel{d_3}{\rightarrow} \ldots .$$ It's like the second line is a grainy image of the real life in the first line. How is this image made and how much information does it preserve.

Both sequences satisfy the topological property $dd=0$ which makes them cochain complexes.

They main difference is, of course, that the former spaces are infinite dimensional and the latter are finite dimensional. Other than that, continuous and discrete forms are very similar in the sense that they form vector spaces that behave similarly.

As these spaces are already aligned in the diagram above, can we connect them by vertical arrows representing some kind of (linear) functions? To construct such a function we need to be able to match two very different entities -- a continuous form and a discrete one -- one by one. Let's review what they are.

We have thought of continuous differential forms, $\varphi \in \Omega ^k ({\bf R}^n)$, mainly as integrands. Also, the "axiomatic" definition was: they are multilinear antisymmetric functions on real vectors with real values, parametrized by location in ${\bf R}^n$: $$\varphi =\varphi ^k: {\bf R}^n \times ({\bf R}^n)^k \rightarrow {\bf R}.$$

Meanwhile, we have thought of discrete differential forms, $\psi \in C ^k ({\bf R}^n)$, mainly as cochains. Also, the "axiomatic" definition was: they are multilinear antisymmetric functions on integer vectors with real (or possibly integer) values, parametrized by location in ${\bf Z}^n$: $$\psi =\psi ^k: {\bf Z}^n \times ({\bf Z}^n)^k \rightarrow {\bf R}.$$

Based on the axiomatic definitions one might think that discrete forms are restrictions (or samplings) of continuous forms. This seems very simple. We define the "sampling function" $$S:\Omega ^k ({\bf R}^n) \rightarrow C ^k ({\bf R}^n)$$ by $$\psi = S(\varphi ) =\varphi |_{{\bf Z}^n \times ({\bf Z}^n)^k}.$$ It is easy to check the multilinearity and antisymmetry of the resulting function (Exercise).

However, as we have discovered,

sampling a function and its derivative produces two discrete functions that don't satisfy the Fundamental Theorem of Calculus.

These pictures should refresh you memory:

DiscreteForms1.png

DiscreteForms2.png

But the original function and its derivative do satisfy the theorem! The problem is created by the sampling itself. In fact, we can now describe the problem in a more precise way: the diagram below does not commute! $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccccc} \Omega ^0 & \ra{d_0} & \Omega ^{1} \\ \da{S} & \ne & \da{S} \\ C ^0 & \ra{d_0} & C ^{1} \end{array} $$ In other words, $Sd_0f^0\ne d_0Sf^0$. This is important because it shows that the exterior derivative doesn't map to the exterior derivative under the sampling map $S$. It's as if we have two different calculi unrelated to each other...

Why doesn't the diagram commute? Because under sampling, we lose information about the behavior of the function over the interval except for a single point. So, instead we need some kind of averaging. And we know from calc 1 that the average of a continuous function is its integral normalized by dividing by the interval's length.

Integral is the answer! It is given by the de Rham map $R$.

In order to understand where it comes from we, again, consider the simplest case: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccccc} \Omega ^0([0,1]) & \ra{d_0} & \Omega ^{1}([0,1]) \\ \da{R} & & \da{R} \\ C ^0([0,1]) & \ra{d_0} & C ^{1}([0,1]) \end{array} $$ Suppose we start at the left upper corner with a $0$-form $g$: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccccc} g & \ra{d_0} & dg \\ \da{R} & & \da{R=?} \\ f & \ra{d_0} & \phi=df \end{array} $$

De Rham map.png

The image $f$ of $g$ under $R$ will be still sampling at the end points: $$f(0)=g(0), f(1)=g(1).$$ We also understand the exterior derivatives at the top and the bottom: $$df[0,1]=f(1)-f(0)=g(1)-g(1).$$ Altogether: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccccc} g & \ra{d_0} & dg \\ \da{sampling} & & \da{R=?} \\ f:f(0)=g(0), f(1)=g(1) & \ra{d_0} & \phi:\phi [0,1]=g(1)-g(1) \end{array} $$ It is $R$ on the right that needs to be figured out: $Rdg=\phi$. For the diagram to commute we need: $$(Rdg)[0,1]=g(1)-g(0).$$ The answer is obvious: $R$ should be given by the integral over $[0,1]$ and the match comes from the Fundamental Theorem of Calculus: $$\int_{[0,1]}dg=g(1)-g(0).$$ So we can define $$R(\psi)[0,1]=\int_{[0,1]} \psi$$ for any $1$-form $\psi$.

To complete this analysis we just need to observe that, even though the de Rham in degree $0$ is defined via sampling, it is also an integral: $$\int_{\{m\}} g^0=g(m).$$

So, the link from continuous to discrete forms is given by the function $R:\Omega ^k \rightarrow C^k,k=0,1,...$ by integrating the form: $$R (\omega )(c) =\int _c \omega,$$ where $\omega$ is a differential form of degree $k$ and $c$ is a $k$-chain. It is called the de Rham map.

The de Rham map is linear (Exercise).

Theorem. The de Rham map commutes with the exterior derivative: $$dR=Rd.$$ Proof. The proof of this statement for degrees $0$ and $1$ is above. Now we rely on Stokes' theorem instead. Let's unwrap the simple equation above. Both sides are functions, so they are equal when their values are equal, for all inputs: $$(d_kR_{k})(\psi ^k)=(R_{k+1}d_k)(\psi ^k),$$ for all $k$-forms $\psi ^k$. But this is still a function equation because a form is also a function. So we have $$(d_kR_{k})(\psi ^k)(c_{k+1})=(R_{k+1}d_k)(\psi ^k)(c_{k+1}),$$ for all $k$-forms $\psi ^k$ and all $(k+1)$-chains $c_{k+1}$. Consider the left-hand side and apply the fact $d=\partial ^*$ to the form $R_{k}(\psi ^k)$: $$(d_kR_{k})(\psi ^k)(c_{k+1})=d_k(R_{k}(\psi ^k))(c_{k+1})=\partial^*_k(R_{k}(\psi ^k))(c_{k+1})=(R_{k}(\psi ^k))(\partial_{k+1}c_{k+1}).$$ By definition of the de Rham map this is $$(R_{k}(\psi ^k))(\partial_{k+1}c_{k+1})=\int_{\partial_{k+1}c_{k+1}}\psi ^k=\int_{c_{k+1}}d_k\psi ^k,$$ if we apply the Stokes Theorem. Finally, we apply the definition of the de Rham map again: $$\int_{c_{k+1}}d_k\psi ^k = (R_{k+1}d_k)(\psi ^k)(c_{k+1}).$$ $\blacksquare$

As a result, the two cochain complexes form one commutative diagram: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \ra{d} & \Omega ^k & \ra{d} & \Omega ^{k+1} &\ra{d} \\ & \da{R} & & \da{R} \\ \ra{d} & C ^k & \ra{d} & C ^{k+1} & \ra{d} \end{array} $$ In other words the de Rham map is a cochain map.

In addition, we will see that the de Rham map establishes an isomorphism between the de Rham cohomology $H^*_{dR}$ and cubical cohomology $H^*$.

Theorem. The de Rham map is onto.

The main interpretation:

  • physical quantities are represented by differential forms, while
  • the de Rham map represents how humans measure these quantities.

Now the inverse link. Since the de Rham map is many-to-one, it has no inverse. But, since it's onto, it has right inverses: $$RR^{-1}=Id.$$ One of them is called the Whitney map: $$W:C^k(K) \rightarrow \Omega ^k (D),k=0,1,...$$ It is defined via "interpolation". Specifically, one interpolates cochains by interpolating the generators, piecewise linearly.

In dimension $1$, $0$-cochains are defined on ${...,-2,-1,0,1,2,3...}$ and the generators are: $$g_i(n)=\delta _{in}.$$ The corresponding $0$-form is a "tooth": $$ W(g_i)(x) = \begin{cases} 0, & \text{for }x<n-1 \\ x-n+1, & \text{for }n-1 \le x \le n \\ -x+n+1, & \text{for }n \le x \le n+1 \\ 0, & \text{for }x \le n+1. \\ \end{cases} $$

In the simplest setting:

Whitney map.png


To summarize:

Note: Any differentiable manifold can be triangulated.