Homology in dimension 1

Redirect to:

• Oriented chains

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,

the boundary of a oriented (directed) $1$-cells is the $0$-chain: endpoint - initial point.

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):

if $a = \displaystyle\sum_i s_i \sigma_i, s_i \in {\bf Z}, $ then $\partial(a) = \displaystyle\sum_i s_i \partial(\sigma_i)$.

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.