13.6 Boundary
We've already defined the boundary operator, \( \boundary \), for blips and intervals.
\[
\begin{align}
&\boundary [x_0]_{x} = 0
\\
&\boundary [x_1, x_2]_{x} = [x_2]_{x} - [x_1]_{x}
\end{align}
\]
Let's extend the boundary operator to all boxes by using laws similar to the differential laws for polyrulers.
Law.
The boundary law for addition states
\[
\begin{align}
\boundary (q + r) &= \boundary q + \boundary r
\end{align}
\]
where \( q : \BoxType_{}^{} \) and \( r : \BoxType_{}^{} \) are boxes.
We must be careful with our multiplication laws!
Laws.
We have two boundary laws for multiplication
\[
\begin{align}
\boundary (b \mult q) &= \boundary b \mult q + b \mult \boundary q
\\
\boundary (i \mult q) &= \boundary i \mult q - i \mult \boundary q
\end{align}
\]
where
\[
\begin{align}
b &: \Blip_{},
&
i &: \Interval_{},
&
q &: \BoxType_{}^{}
\end{align}
\]
are a blip, an interval, and a box.
We can make sense of these laws with an example.
Example.
Let's take the boundary of the degree two box
\[
\begin{align}
[1, 4]_{x} \mult [2, 3]_{y} &: \BoxType_{(x, y)}^{2}
\end{align}
\]
pictured as follows.
We calculate.
\[
\begin{align}
\boundary \bigl([1, 4]_{x} \mult [2, 3]_{y}\bigr) &= \boundary [1, 4]_{x} \mult [2, 3]_{y} - [1, 4]_{x} \mult \boundary [2, 3]_{y}
\\
&= \bigl([4]_{x} - [1]_{x}\bigr) \mult [2, 3]_{y} - [1, 4]_{x} \mult \bigl([3]_{y} - [2]_{y}\bigr)
\\
&= [4]_{x} \mult [2, 3]_{y} - [1]_{x} \mult [2, 3]_{y} - [1, 4]_{x} \mult [3]_{y} + [1, 4]_{x} \mult [2]_{y}
\\
&= [4]_{x} \mult [2, 3]_{y} + [1]_{x} \mult [3, 2]_{y} + [4, 1]_{x} \mult [3]_{y} + [1, 4]_{x} \mult [2]_{y}
\end{align}
\]
The boundary is the sum of four boxes, each with degree one.
Notice how the boundary cycles, matching the box's orientation.
If we take the boundary of a boundary, we'll find many cancellations!
Theorem.
Suppose \( q : \BoxType_{(x, y)}^{2} \) is a box with degree two. The second boundary of \( q \) is zero.
\[
\begin{align}
\boundary \boundary q &= 0
\end{align}
\]
Why?
When we take the second boundary, we'll see each corner twice: once with a positive orientation and once with a negative orientation. These each cancel. If you wish to see the details, write \( q \) as
\[
\begin{align}
q &= [x_1, x_2]_{x} \mult [y_1, y_2]_{y}
\end{align}
\]
and compute \( \boundary \boundary q \).