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Chain operators

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Given a map $f \colon X \rightarrow Y$, we can trace what $f$ does to the homology classes in $X$ as they are mapped to $Y$. This is discussed in Maps and homology.

To study this issue properly, however, we need to limit our attention to $X$ and $Y$ which are realizations of cubical complexes, say $K$ and $L$: $$X = |K|, Y = |L|.$$

Then, we need an analogue of maps in this environment, i.e., given a map $f$ we build such a "combinatorial" construction involving complexes $K$ and $L$ that its "realization" is $f$.

Let's consider maps of the circle to itself $$f \colon X = {\bf S}^1 \rightarrow Y = {\bf S}^1.$$

We represent $X$ and $Y$ as two identical cubical complexes $K$ and $L$ and then find an appropriate representation for $f$.

Map S to S cubical.jpg

We will define function $g$ as follows:

(A) $g$ maps each cell in $X$ to a cell in $Y$ of the same dimension, or $0$.

The latter option is reserved for cells that collapse (just like homology classes collapse as discussed in Maps and homology).

Let's consider a few examples.

Identity: $$\begin{array}{l} g(A) = A, g(B) = B, g(C) = C, g(D) = D, \\ g(a) = a, g(b) = b, g(c) = c, g(d) = d. \end{array}$$

Constant: $$\begin{array}{l} g(A) = A, g(B) = A, g(C) = A, g(D) = A, \\ g(a) = 0, g(b) = 0, g(c) = 0, g(d) = 0. \end{array}$$ All $1$-cells collapse. Unlike with circles, there are only $4$ of such maps for these cubical complexes.

Flip: $$\begin{array}{l} g(A) = D, g(B) = C, g(C) = B, g(D) = A, \\ g(a) = c, g(b) = b, g(c) = a, g(d) = d. \end{array}$$ This is a vertical flip, there are also the horizontal and diagonal flips, total of $4$. Only these four axes allow condition $(A)$ to be satisfied.

Rotation: $$\begin{array}{l} g(A) = ?, g(B) = ?, g(C) = ?, g(D) = ?, \\ g(a) = ?, g(b) = ?, g(c) = ?, g(d) = ?. \end{array}$$

Exercise. How many and what are they?

Let's try this: $$g(A) = A, g(B) = C, ...$$

There is trouble. Even though we can find a cell for $g(a)$, it can't be $[A,C]$. So, $g(a)$ won't be aligned with its end-points. As a result, the "realization" of $g$ won't be continuous. To prevent this from happening we need to require that the end-points of the image of any $1$-cycle in $X$ are the images of the end-point of this cycle. To ensure the cells in all dimensions remain attached after $g$ is applied, we require:

(B) $g$ preserves boundaries.

Algebraically, $$\partial g = g \partial.$$

This condition is the cell analogue of continuity.

Observe now that $g$ is defined on complex $K$ but its values aren't all in $L$. There are also $0$s. They aren't cells, but they are chains! In fact, even though g is defined on cells of $K$ only, it can be extended to all chains, "by linearity": $$g(A + B) = g(A) + g(B), ...$$

Thus, condition $(A)$ simply means that $g$ maps $k$-chains to $k$-chains. More precisely $g$ is a collection of functions: $$g \colon C_k(K) \rightarrow C_k(L), k = 0, 1, 2, ...,$$ where $C_k(K)$ and $C_k(L)$ are the sets of all $k$-chains in $K$ and $L$ respectively. If such a function also preserves boundaries (condition $(B)$), we call it a chain operator or chain map. For brevity we use: $$g \colon C_*(K) \rightarrow C_*(L).$$

Projection: $$\begin{array}{l} g(A) = A, g(B) = A, g(C) = D, g(D) = A, \\ g(a) = 0, g(b) = d, g(c) = 0, g(d) = d. \end{array}$$

Let's verify condition $(B)$: $$\begin{array}{l} \partial g(A) = \partial 0 = 0, \\ g \partial (A) = g(0) = 0. \end{array}$$

Same for the rest of $0$-cells. $$\begin{array}{l} \partial g(a) = \partial (0) = 0, g \partial (a) = g(A + B) = g(A) + g(B) = A + A = 0. \end{array}$$

Same for $c$. $$\begin{array}{l} \partial g(b) = \partial (d) = A + D, \\ g \partial (b) = g(B + C) = g(B) + g(C) = A + D. \end{array}$$ Same for $d$.

Exercise. Try the "diagonal fold": $A$ goes to $C$, while $C, B$ and $D$ stay.

In each of these examples, an idea of a map $f$ of the circle/square was present first, then $f$ was realized as a chain map $g$ which can be denoted by $$g = f_* \colon C_*(K) \rightarrow C_*(L).$$

Next, suppose we already have a chain operator $$g \colon C_*(K) \rightarrow C_*(L),$$ what is a possible realization of $g$: $$|g| \colon |K| \rightarrow |L|?$$

The idea is simple: if we know where each vertex goes under $f$, we can construct the rest of $f$ by linearity, i.e., interpolation.

A simple example first. Suppose $$K = \{A, B, a | \partial a = A+B \}, L = \{C, D, b| \partial b = C+D \}$$ are two complexes representing the closed intervals. Define a chain operator: $$g(A) = C, g(B) = D, g(a) = b.$$

If the first two identities is all we know, we can still build a continuous function $f \colon |K| \rightarrow |L|$ such that $f_* = g$. The third identity will be taken care of because of condition $(B)$.

Map interval to interval.png

If we include $|K|$ and $|L|$ as subsets of the $x$-axis and the $y$-axis respectively, the solution becomes obvious: $$f(x) = C + \frac{D-C}{B-A} \cdot (x-A).$$

This approach allows us to give a single formula for realizations of all chain operators: $$f(x) = g(A) + \frac{g(B)-g(A)}{B-A} \cdot (x-A).$$

For example, suppose we have a constant map: $$g(A) = C, g(B) = C, g(a) = 0 {\rm \hspace{3pt} (collapse)}.$$ Then $$f(x) = C + \frac{C-C}{B-A} \cdot (x-A) = C.$$

Example. Let's consider a chain map $g$ of the complex $K$ representing the "filled" square/$2$-cell.

Closed 2cell.png

Knowing the values of $g$ on the $0$-cells of $K$ gives us the values of $f=|g|$ at those points. How do we extend it to the rest of $|K|$?

An arbitrary point $u$ in $|K|$ is represented as a convex combination of $A, B, C, D$: $$u = sA + tB + pC + qD, {\rm \hspace{3pt} with \hspace{3pt}} s + t + p + q = 1.$$

Then we define $f(u)$ as $$f(u) = sf(A) + tf(B) + pf(C) + qf(D).$$

Thus, $f(u)$ is a convex combination of $f(A), f(B), f(C), f(D)$. But all of these are vertices of $|L|$, so $f(u) \in |L|$.

For example, suppose we are interested in the projection: $$\begin{array}{l} g(A) = A, g(B) = A, g(C) = D, g(D) = A, \\ g(a) = 0, g(b) = d, g(c) = 0, g(d) = d. \\ g(τ) = 0. \end{array}$$ Then $$\begin{array}{} f(u) &= sf(A) + tf(B) + pf(C) + qf(D) \\ &= sA + tA + pD + qD \\ &= (s+t)A + (p+q)D. \end{array}$$

Since $(s+t) + (p+q) = 1$, $f(u)$ belongs to the interval $[A,D]$. (See also Extensions.)

To summarize the algebra of chain maps consider the commutative diagram:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} \da{\partial_3} & & \da{\partial_3} \\ C_2(K) & \ra{g_2} & C_2(L) \\ \da{\partial_2} & & \da{\partial_2} \\ C_{1}(K) & \ra{g_{1}} & C_{1}(L)\\ \da{\partial_1} & & \da{\partial_1} \\ C_{0}(K) & \ra{g_{0}} & C_{0}(L)\\ \da{\partial_0} & & \da{\partial_0} \\ 0 & & 0\\ \end{array} $$

This is an expanded version of condition $(B)$: $$\partial g = g \partial.$$

The condition indicates that each square in the diagram commutes: $$\partial_k g_{k-1} = g_k \partial_k.$$

Then, $\{C_*(K), \partial \}$ is called the chain complex of $K$. It is used to compute the homology $\{H_*(K) \}$ of $K$. Meanwhile, a chain operator is a map between chain complexes $$g \colon \{ C_*(K), \partial \} \rightarrow \{C_*(L), \partial \}$$ that preserves boundaries. It is used to compute the homology operator $$g_* \colon H(K) \rightarrow H(L).$$