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# Algebra of discrete differential forms

Suppose we have a cubical complex.

## 1 Discrete differential forms as a vector space

If $\varphi, \psi$ are discrete differential forms of the same degree $k$, it is easy to define algebraic operations with them that make the set into a vector space.

First, their addition. The sum $\varphi + \psi$ is a form of degree $k$ too and is computed as: $$(\varphi + \psi)(a) = \varphi(a) + \psi(a).$$

As an example, consider two $1$-forms, $\varphi$, $\psi$. Suppose these are their values defined on the $1$-cells (in green): orange for $\varphi$ and purple for $\psi$:

Then $\varphi + \psi$ is defined on the edges as $-1+1=0, 1+2=3,2+5=7, 2+3=5$. So we compute the values of the new form one cell at a time.

Same for scalar multiplication: $$(\lambda \varphi)(a) = \lambda \varphi(a).$$

Clearly, these operations satisfy the same properties as if they are applied to continuous differential forms: associativity, commutativity, distributivity, etc.

Exercise: What is the dimension of this vector space? Hint: you'll need a basis.

## 2 Cochains

Now we have the continuous differential forms given explicitly as in terms of $dx, dy$, etc, and the discrete cubical forms given as cochains. The latter means that there is a number associated with each cell present in the complex.

Let's have a few examples of how these two can be connected...

Example: What is $dx$? Naturally, the horizontal values are $1$'s and vertical $0$'s:

This is a cochain in the sense that every $1$-cell is assigned a number.

What is $dy$? As a cochain:

Algebraically:

• $dx([m,m+1]\times {n})=1, dx({m} \times [n,n+1] )=0$;
• $dy([m,m+1]\times {n})=0, dy({m} \times [n,n+1] )=1$.

Note that all $1$-forms are "linear combinations" of these two: $$A dx + B dy.$$ Here, we understand $A,B$ as (discrete) functions, not just numbers. They may vary from cell to cell. For example, this could be $A$:

At this point we can integrate this form. For example, suppose $P$ is chain that represents the lower right square on this picture with the $1$-cells oriented, as always, along the axes. Then we can integrate $Adx$, as a line integral, along this curve going clockwise one cell at a time: $$\int _P Adx = 7\cdot 0 + 8\cdot 1 + 0\cdot 0 + 2\cdot (-1) =6.$$ Also $$\int _P Ady = 7\cdot 1 + 8\cdot 0 + 0\cdot (-1) + 2\cdot 0 =7.$$ If $B$ is also provided, the integral $$\int _P Adx+Bdy$$ is a similar sum.

Next we illustrate $2$-forms:

Then the double integral over this square, $S$, is $$\int _S Adxdy = 1+2+0-1=2.$$

Example: What is $dx \hspace{1pt} dy$ as a co-chain?

What is $dydx$?

We get from $dydx$ to $dxdy$ via antisymmetry.

## 3 Wedge product

What about the wedge product of cubical forms?

Just like with continuous forms, the wedge product of two $1$-forms should be a $2$-form.

As a discrete $2$-form, $\varphi \wedge \psi$ is defined on $2$-cells. In this example, there is only one, $\alpha$.

One might try to compute the value of $\varphi \wedge \psi$ at $\alpha$ based on the fact that $\alpha=a \times d$.

First the cochain representation: $$\varphi (a) = 2,$$ $$\psi (d) = 1.$$ What if? \begin{align*} (\varphi \wedge \psi)(\alpha) &= (\varphi \wedge \psi)(a \times d) \\ &= \varphi(a) \psi(d) \\ &= 2 \cdot 1 \\ &= 2. \end{align*} Then the differential form representation is: $$\varphi = 2dx \hspace{6pt} {\rm on \hspace{3pt}} a,$$ $$\psi = 1 dy \hspace{6pt} {\rm on \hspace{3pt}} d.$$ Hence $$\varphi \wedge \psi = 2 dx \hspace{1pt} dy \hspace{6pt} {\rm on \hspace{3pt}} \alpha.$$

Problem? The product doesn't match the wedge product of continuous differential forms. In fact, it's not even antisymmetric!

Recall, that for continuous forms antisymmetry implies some kind of "skew-commutativity": $$\varphi ^m \wedge \psi ^n= (-1)^{mn} \psi ^n \wedge \varphi ^m.$$ We will require the same for cubical forms.

Recall, that for continuous forms antisymmetry implies that $$dxdx=dydy=...=0.$$ What's the analogue of this for discrete forms? $$dx \wedge dx = dy \wedge dy =... =0.$$ This implies that $$\varphi (a) \wedge \psi (a) =0, \varphi (d) \wedge \psi (d) =0.$$

Then, we are left with just two terms: $$\varphi (a) \psi(d) , \varphi (d) \psi(a)$$ We combine these with the minus sign: $$(\varphi \wedge \psi)(\alpha) = (\varphi \wedge \psi)(a \times d)$$ $$=\varphi (a) \psi(d) - \varphi (d) \psi(a).$$ Why minus? Consider: $$(\psi \wedge \varphi)(\alpha) = \psi (a) \varphi(d) - \psi (d) \varphi(a) = -(\varphi \wedge \psi)(\alpha).$$ This way we have the "commutativity" above, for $m=n=1$.

Exercise: Define $\varphi ^0 \wedge \psi ^1$ and $\varphi ^1 \wedge \psi ^2$.

For a cubical complex in the $n$-dimensional space, cochains are defined on the cubes: $$Q=\Pi _{k=1}^{n}A _k,$$ where each $A_k$ is either a vertex or an edge in the $k$th component of the space. If we omit the vertices, a $(p+q)$-cube can be rewritten as $$Q=\Pi _{i=1}^{p+q}I _i,$$ where $I_i$ is its $i$th edge. Now, the cochain of the wedge product is given by $$(\varphi ^p \wedge \psi ^q)(I)=\Sigma _s (-1)^{\pi (s)}\varphi ^p(\Pi _{i=1}^{p}I _{s(i)}) \cdot \psi ^q(\Pi _{i=p+1}^{p+q}I _{s(i)}),$$ with summation taken over all permutations of $\{1,2,...,p+q\}$ and $\pi (s)$ stands for the parity of permutation $s$.

Unfortunately, the discrete wedge product isn't associative!

Exercise. Show: $$\phi ^1 \wedge (\psi ^1 \wedge \theta ^1) \ne (\phi ^1 \wedge \psi ^1) \wedge \theta ^1 .$$

The good news is that it is associative if restricted to closed forms. This leads to the cup product on cohomology.

Summary. The set of all cubical $k$-forms in a cubical complex $|K|$ is denoted by $C^k(K)$. Together these vector spaces (or modules if $R$ is a ring) are called the cochain complex once it's equipped with exterior derivative. It's a graded vector space with an extra operation of wedge product.