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# Homology in dimension 1

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Recall that given a cell complex $K$, a $k$-chain is a "formal" linear combination of finitely many oriented $k$-cells, such as $3a + 5b - 17c$ with, typically, integer coefficients. Then the set of all $k$-chains $C_k(K)$ is an abelian group with respect to chain addition generated by the $k$-cells of $K$: $$C_k(K) = \left\{ \displaystyle\sum_i s_i \sigma_i \colon s_i \in {\bf Z}, \sigma_i {\rm \hspace{3pt} is \hspace{3pt} a \hspace{3pt}} k{\rm -cell \hspace{3pt} in \hspace{3pt}} K \right\}.$$

Now a relation will be established between cells/chains of different dimensions in order to capture the topology of the cell complex. This relation is given by the *boundary operator*.

The boundary of a vertex empty, so the boundary operator of a $0$-chain is $0$: $$\partial (A) = 0 {\rm \hspace{3pt} for \hspace{3pt} any \hspace{3pt}} A \in C_0(K).$$

The boundary of a $1$-cell consists of its two end-points, so in the *binary* setting this was simple:
$$\partial (a) = \partial (AB) = A + B.$$

In the *integral* setting, the direction of the edge matters ($AB \neq BA$), so we define:
$$\partial (a) = \partial (AB) = B - A {\rm \hspace{3pt} for \hspace{3pt} any \hspace{3pt}} a = AB \in C_1(K).$$
On the other hand,
$$\partial (BA) = A - B,$$
which makes sense since $BA = -AB$.

Let's define the boundary *operator*. First,

Or $$\partial (AB) = B - A$$

Now extend this definition to the whole chain complex, by assuming that $\partial$ is a homomorphism (or a linear operator in the vector space case):

In other words we define the boundary operator on the generators (cells) of the group and then extend it to the whole group *by linearity*.

In the above example, the boundary operator evaluated on the generators: $$\partial (a) = \partial (AB) = B - A,$$ $$\partial (b) = \partial (CB) = B - C,$$ $$\partial (c) = \partial (AC) = C - A;$$

on other chains: $$\partial (a + b) = \partial (a) + \partial (b) = B - A + B - C = 2B - A - C,$$ $$\partial (a + b + c) = \partial (a) + \partial (b) + \partial (c) = B - A + B - C + C - A = 2B - 2A.$$

It may appear that in the above picture, that $a, b$, and $c$ form a circle. They do unless take into account their directions. To confirm this algebraically we need to see what's the boundary of the chain $a + b + c$: $$\partial (a + b + c) = B - A + B - C + C - A = -2A + 2B.$$

This isn't zero so $a + b + c$ isn't a cycle.

Which chain is? It's $a - b - c$: $$\partial (a - b - c) = 0.$$

But so is $2a - 2b - 2c, 3a - 3b - 3c$, and all other of its multiples. So, the cycles is a subgroup generated by $a - b - c$: $$Z_1(K) = <a - b - c>.$$

**Definition.** A chain is called a *cycle* if its boundary is $0$.

**Theorem.** The cycles form a subgroup $Z_k(K)$ of the chain group $C_k(K)$ which is the kernel of the boundary operator:
$$Z_k(K) = {\rm ker}( \partial ).$$

This subgroup is called the cycle group.

In the above example, we have: $$Z_1(K) = <a - b - c> = {\bf Z} < C_1(K) = {\bf Z}^3.$$

We showed that all of these chains are cycles, but how do we know that we've found all of them? Let's solve the problem properly.

**Topology problem.** Evaluate $Z_1(K)$ for $K$ given above.

For simplicity we assume that these are *real* chains:
$$\displaystyle\sum_i s_i \sigma_i, s_i \in {\bf R}.$$

Then this is what we know: $$\partial \colon C_1(K) \rightarrow C_0(K) = {\bf R}^3$$ is a linear operator between two copies of ${\bf R}^3$ with bases $\{a, b, c \}$ and $\{A, B, C \}$ respectively. The values of $\partial$ on the basis elements are given above. Now to solve our problem, we observe that: $$x \in C_1(K) {\rm \hspace{3pt} is \hspace{3pt} a \hspace{3pt} cycle \hspace{3pt} iff \hspace{3pt}} \partial x=0.$$ But $x$ can be expressed as a linear combination of the elements of the basis: $$x = ua + vb + wc.$$ Then the condition above can be rewritten as new $$\partial (ua + vb + wc) = 0,$$ or $$\partial (ua) + \partial (vb) + \partial (wc) = 0$$ or $$u(B - A) + v(B - C) + w(C - A) = 0.$$ After collecting similar terms, we can restate or problem.

**Linear algebra problem.** Find real numbers (turns out integers) $u, v$, and $w$ such that
$$(-u-w)A + (u+v)B + (-v+w)C = 0.$$

Since $\{A, B, C \}$ is a basis, these elements are linearly independent. Hence each of their coefficients is $0$: $$-u - w = 0,$$ $$u + v = 0,$$ $$-v + w = 0.$$

This is a (homogeneous) system of linear equations. It can be solved in a number of ways. For example, take $w = v$ from the last equation and substitute it into the first, then we have $$-u - v = 0,$$ $$u + v = 0.$$ Substitute again and we have $u = -v$. Then, the solutions are $$u = -t, v = t, w = t {\rm \hspace{3pt} for \hspace{3pt}} t \in {\bf R}.$$ Or, the solution set is span$\{(-1, 1, 1) \}$.

Now, back to topology: $$Z_1(K) = {\rm span}\{(-a+b+c)\}.$$ Since there are no $2$-cells, no two $1$-chains are homologous. So $$H_1(K) = Z_1(K) = {\rm span}\{(-a + b + c) \},$$ or, in the group case, $$H_1(K) = <(-a + b + c)>.$$

Adding however a face to this triangle will make $-a + b + c$ homologous to $0$. See Homology in dimension 2.

**Exercise.** Compute the homology (by solving a system of linear equations) of the figure eight, represented by a cell complex with (a) $5$ edges, (b) $2$ edges.

For a more concrete way of capturing $0$- and $1$-cycles, see Homology of images.