User:Fly by Night/sandbox



Consider the chain complex given by
 * $$ \{0\} \longrightarrow \Delta_2(X;\Z) \longrightarrow \Delta_1(X;\Z) \longrightarrow \Delta_0(X;\Z) \longrightarrow \{0\} \, . $$

Using the decomposition given in the diagram, we have
 * $$ \Delta_2(X;\Z) = \Z\langle \alpha, \beta \rangle \,, $$
 * $$ \Delta_1(X;\Z) = \Z\langle A, B, C, D, E, F \rangle \,, $$
 * $$ \Delta_0(X;\Z) = \Z\langle a, b, c, d \rangle \, . $$

The boundary operators are given as follows:
 * $$ \partial_3 : \{0\} \longrightarrow \Delta_2(X;\Z) \,, $$
 * $$ \partial_2 : \Delta_2(X;\Z) \longrightarrow \Delta_1(X;\Z) \,, $$
 * $$ \partial_1 : \Delta_1(X;\Z) \longrightarrow \Delta_0(X;\Z) \,, $$
 * $$ \partial_0 : \Delta_0(X;\Z) \longrightarrow \{0\} \, . $$

The homology groups are, by definition, given by $$H_i(X;\Z) \cong \ker(\partial_i)/\text{im}(\partial_{i+1})$$ for all $$0 \le i \le 2$$. The homology groups are well defined since $$\partial^2 \equiv 0$$ and so $$\text{im}(\partial_{i+1}) \subseteq \ker(\partial_i)$$ for all $$0 \le i \le 2$$.

Let us first consider $$\partial_3$$. Trivially, $$\text{im}(\partial_3) = \ker(\partial_3) = \{0\}.$$ Next consider $$\partial_2$$:
 * $$ \partial_2\alpha = A+C+D+B \,, \ \ \

\partial_2\beta = -(B + E + D + F) \, .$$ As elements of $$\Z\langle A, B, C, D, E, F \rangle$$ both $$\partial_2\alpha$$ and $$\partial_2\beta$$ are linearly independent, and so $$\text{im}(\partial_2) \cong \Z^2$$, while $$\ker(\partial_2) = \{0\}$$.

Next we consider $$\partial_1$$. We have the following:
 * $$\partial_1A = d-a\,, \ \ \ \partial_1B = a-b \, , \ \ \ \partial_1C = b-c\, , $$
 * $$\partial_1D = c-d\,, \ \ \ \partial_1E = b-c \, , \ \ \ \partial_1F = d-a\, . $$

Clearly $$\partial_1E = \partial_1C$$ and $$\partial_1F = \partial_1A$$. Thus neither $$\partial_1E$$ nor $$\partial_1F$$ contribute towards the rank of $$\partial_1$$, and may be discounted from further consideration. Consider the system of equation $$\partial_1A = d-a$$, $$\partial_1B = a-b$$, $$\partial_1C = b-c$$ and $$\partial_1D = c-d$$. These may be re-written in matrix notation as follows:
 * $$\partial_1\left( \left[\begin{array}{c} A \\ B \\ C \\ D \end{array}\right]\right) =

\left[\begin{array}{cccc} -1 & 0 & 0 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \end{array}\right] \left[\begin{array}{c} a \\ b \\ c \\d \end{array}\right]. $$ The 4-by-4 matrix on the right hand side has rank four, meaning that $$\partial_1$$ must also have rank three. It follows that $$\text{im}(\partial_1) \cong \Z^3$$ while $$\ker(\partial_1) \cong \Z^3$$.

Finally, we consider $$\partial_0$$. Since $$\partial_0 : \Delta_0(X;\Z) \longrightarrow \{0\}$$, it follows that $$\text{im}(\partial_0) \cong \{0\}$$ while $$\ker(\partial_0) = \Delta_0(X;\Z) \cong \Z^4$$.

Using the aforementioned definitions of the homology groups, we have
 * $$ H_2(X;\Z) := \{0\}/\{0\} \cong \{0\} \,, \ \ \

H_1(X;\Z) := \Z^3/\Z^2 \cong \Z \,, \ \ \ H_0(X;\Z) := \Z^4/\Z^3 \cong \Z \,. $$

As a trivial consequence, the Euler characteristic:
 * $$ \chi(X) = \dim(H_0) - \dim(H_1) + \dim(H_2) = 1 - 1 + 0 = 0 \, . $$