Check Your Understanding

This discussion primarily involves Topology, so minimal prerequisites are necessary for comprehension. However, there is substantial context provided in the preceding articles.

Boundary Analysis of k-Cells

Utilize the boundary of a k-chain formula to demonstrate that the boundary of a 2-cell and a 3-cell equals zero.

k-Cells and Calculus III Pullbacks

Employ concepts from this article to clarify how an integral over a specific filled region can be transformed into an integral over the Cartesian product of two closed intervals.

Coordinate Invariance

Prove the invariance of fundamental Vector Calculus operators under coordinate transformations. [Hint: Leverage properties of the pullback and revisit the article on the exterior derivative.]

2D Boundary Morphing

You can create all the regions below by morphing a two-chain (think of a rectangle):

• A ring (Refer to article examples)
• A circle (Refer to article examples)
• The side of a cylinder (like a soup can label)
• The surface of a torus (akin to a doughnut)
• The surface of a sphere.

As you manipulate these shapes, consider the changes to the oriented boundary and visualize the transformation step by step.

3D Boundary Morphing

Similar transformations can be applied to a three-chain (consider a rectangular prism/cuboid):

• A sphere
• A cylinder
• A torus

Again, reflect on how the oriented boundary shifts as you morph these shapes.

Möbius Strip Boundary Analysis

Without delving into two-chains, what would you predict the boundary of a Möbius strip to be?

Challenges with Möbius Strips

While one can create a Möbius strip from a two-chain, akin to twisting a strip of paper and connecting the ends, complications arise regarding the boundary. Discuss what occurs with the boundary and why it diverges from the cases outlined in the Boundary Test sections.

Prerequisites for This Discussion

A fundamental understanding of the boundary of a region is necessary, which can be visualized as follows:

In simple terms, you are on a region's boundary if you can consistently identify nearby points both inside and outside the region, regardless of your selected neighborhood. If you find this concept confusing, I recommend reviewing my article on An Intro to Topology.

Additionally, it may be beneficial to revisit the previous three articles to address some discussion questions and to grasp the context surrounding k-cells and k-chains.

Defining k-Cells and k-Chains

In this section, we will establish a formal definition for a box and its boundary, followed by proving the Generalized Stokes' Theorem for boxes across arbitrary dimensions and subsequently for any collection of boxes.

k-Cells

A k-cell is defined as a k-dimensional cuboid or rectangular prism, created by taking the Cartesian product of closed intervals. These shapes have no assigned orientation, which will be significant later. Essentially, it's just a box, denoted by σ.

k-Chains

A k-chain is a linear combination of k-cells, represented by Ξ (the capital Greek letter Xi) because it has a unique appearance and is rarely used.

Since adding and scaling boxes isn't inherently logical, we can approach this issue in two ways.

k-Chains as Inputs

First, consider the integral as having two inputs:

1. The object being integrated (e.g., a scalar or vector field).
2. The region of integration.

As previously noted, an integral is linear concerning what is being integrated. Formally, we can express this relationship.

We can also assert that an integral is linear with respect to the region of integration. For instance, the total mass of two regions with a given density function can be represented as either the sum of integrals over each region or as a single integral over both regions combined. Similarly, one can visualize a region with a hole as an integral of the entire region minus the integral over the hole. To manage both scenarios, we define the addition, subtraction, and scaling of regions in terms of the corresponding operations on the integrals.

By employing addition and scaling, we can create a vector space that allows for linear combinations.

Visualizing k-Chains

If two k-cells are not adjacent, envision their sum as merely placing them in the same space. If σ₁ is one box and σ₂ is another, then σ₁ + σ₂ represents both in the same spatial area. The white lines on the box edges help with clarity. Additionally, these are three-dimensional boxes that encompass their interiors, thus categorizing them as three-chains.

If two k-cells are adjacent, think of their addition as gluing them together.

To remove a k-cell from a larger k-cell that contains it, one must create an opening in the larger k-cell. Here, the darker green area illustrates σ₁ — σ₂.

While there are further operations possible with k-cells, these are the essentials.

Understanding Boundaries of k-Cells

The boundary of a k-cell is a (k–1)-chain composed of (k–1)-cells, akin to taking the endpoints of an interval, the sides of a rectangle, or the faces of a cuboid.

Addressing Boundary Issues

If we define the boundary of a k-cell as the sum of all (k–1)-cells, complications arise when forming k-chains. For instance, consider the two-cell displayed here. Its boundary should consist of its four sides, which might lead one to think we can achieve this by summing the one-cells.

This method presents a challenge. To illustrate, we can also express it as the sum of multiple two-cells.

If we add the boundaries of these two-cells, we should obtain the boundary of our original two-cell; however, we end up with additional one-cells due to the dissection of the two-cell (specifically yielding an extra 2σ₁₉ term).

To resolve this, we must redefine the boundary operator to eliminate those additional one-cells. We achieve this by assigning opposite signs to the boundaries of k-cells.

Formal Definition of the k-Cell Boundary Operator

Given that our k-cell is represented as [a₁, b₁] × [a₂, b₂] × … × [aₙ, bₙ], the boundary can be expressed as a (k–1)-chain defined by the following.

This can be likened to the product rule, but with alternating signs for each term. For example, if our two-cell is defined by [2, 5] × [-3, 1], then our boundary is as follows.

Let's break this down:

• The first term consists of all pairs (x, y) where x = 5 and y lies between -3 and 1. As the cell is positive, we move in the positive y direction.
• The second term contains pairs (x, y) where x = 2 and y ranges from -3 to 1. Here, the cell is negative, leading us in the negative y direction.
• The third term consists of pairs (x, y) where x is between 2 and 5 and y = 1. Being negative, we proceed in the negative x direction.
• The fourth term involves pairs (x, y) with x between 2 and 5 and y = -3. This time, the cell is positive, directing us positively along the x-axis.

When we graph these one-cells, we obtain a clear representation.

Why Choose These Boundary Orientations?

While we must choose the orientations for k-cells so that opposing sides negate each other, we could alternatively decide to make the larger endpoint consistently point in the positive direction.

This approach resolves the boundary issue but complicates our work with the exterior derivative, which is defined in relation to the integral over the green box.

Visualizing k-Cell Boundaries and Chains

As illustrated in the graph, one-cells can be represented with arrows. To depict two-cells, we draw arrows that signify the boundary. If the arrows follow a clockwise pattern, we label the two-cell as clockwise; otherwise, we mark it as counterclockwise.

To represent the orientation of a three-cell, we draw all one-cells with their respective orientations. Care must be taken as the boundary of a boundary is void, meaning we need to arrange one-cells to ensure they cancel out. If executed correctly, all two-cells will exhibit a consistent orientation observable through the right-hand rule, Hodge duals, or normal vectors.

A three-cell with an "outward" orientation is illustrated here. A notable aspect of 3D graphics is their reliance on the principles discussed in previous articles. For example, Blender automatically assigns the correct orientation to each texture's side due to its employment of the same internal logic for k-cell orientation that we have examined here.

While I presume that four-cells could be portrayed through a similar methodology, most individuals struggle with three dimensions, so I will refrain from further discussion. We can graphically combine k-cells by assembling them and negating any overlapping arrows that face opposite directions.

The boundary remains unchanged even if all sides possess red and green arrows, as they counterbalance each other. A similar phenomenon occurs with a three-chain, as the one-cells on the edges and two-cells on the surfaces will also negate.

Boundaries of Mapped k-Chains

An advantageous feature of k-chains is that if we can smoothly map a k-chain to a spatial region, the k-chain's boundary will correspond to the boundary of that spatial region. For instance, consider a ring.

We could construct a two-chain from two-cells entirely contained within the ring. All edges of the two-cells within the boundary negate each other since they point in opposing directions.

We can then smoothly transform the two-chain into the ring, resulting in equivalent boundaries.

This is my initial attempt at hand-animated content, and I take pride in how much I've improved my ability to draw smooth lines. I first sketched the two circles and the blocky k-chain, then added other elements afterward. The outer boundary of the k-chain possesses a counterclockwise (i.e., positive) orientation, while the inner boundary is clockwise (i.e., negative). This reflects the notion that the region represented by the inner boundary is subtracted from the outer boundary's region.

Alternatively, we can utilize a polar mapping starting with the rectangle [R₁, R₂] × [0, 2π] and employing the mapping (x, y) = (ρ cos φ, ρ sin φ). The outer boundary remains counterclockwise, and the inner boundary continues to be clockwise.

Note that when the two sides of the two-chain are glued together, they cancel out due to their opposing orientations.

Closing Gaps

As R₁ approaches zero, the entire bottom of the rectangle converges to a single point. However, we need not be concerned since every point on this boundary connects to a point on the opposite side (termed its antipode), alongside every other boundary point. Each point has an orientation that directly opposes its antipode, leading to a complete cancellation. Essentially, we disregard the point from the boundary. This cancellation occurs whenever a hole is closed.

Understanding the Pullback

Imagine we have a region R in standard Euclidean space, which we can smoothly translate into a k-chain (this "smooth method…" is defined as a diffeomorphism). We denote this "smooth method" by Φ* and refer to it as the pullback. Generally speaking, the pullback represents a change of coordinates.

We’ve previously engaged with this concept implicitly when deriving the volume element, as it was the transformation used to convert a differential into a total differential. Formally, we can state that the pullback of a wedge product equals the wedge product of the pullback applied to the forms within it.

Moreover, we can demonstrate that the pullback of the exterior derivative of a form corresponds to the exterior derivative of the pullback of that form.

Discussions surrounding pullbacks essentially communicate that coordinate transformations should not alter outcomes.

The Significance of the Pullback

Assuming we have a k-form field, we can transport the portion of it within region R during the transition to a k-chain. Note that the arrows (representing some k-form) in the k-chain differ in length compared to their representation in the ring. To rectify this, we must multiply by the volume element.

When undertaking this transformation, we must consider two factors: any new boundaries created by the incision in the region and the volume elements. For example, if R is a disk, we slice along the line φ = 0, ρ > 0, introducing two new one-cells that correspond to the new boundary.

It's crucial to account for any stretching or compressing of our k-form during this transition, which is what the volume element achieves. Applying the pullback to the one-forms that constitute the k-form addresses the volume element (as previously derived), but I emphasize this step because neglecting it can lead to complications. By multiplying our original k-form by the volume element, we generate a new k-form adjusted to yield the same results when integrating the new k-form over the k-chain as if we had integrated the original k-form over the region R.

What Lies Ahead?

I acknowledge that this is not the first time I've postponed the article on the Generalized Stokes' Theorem (as it was previously anticipated three articles ago), but I believe I have a solid rationale for doing so. The content is nearly complete, with most text and all but a few images and animations prepared for the article, as the topics covered here were originally part of the GST article. The only missing element is an extension of the proof for manifolds. I initially opted to delay discussing manifolds until their relevance in Hamiltonian Mechanics was evident, as symplectic manifolds are vital in that context. However, I realized that I can address it in the forthcoming article, leading to a comprehensive proof of the GST.

In essence, the next article will be An Intro to Manifolds. While it may take some time to compose and publish, I have several articles ready to follow the release of An Intro to Manifolds.