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Viewing Homology Geometrically
==============================

Delta sets for the Klein bottle, torus, and projective plane.

Define the delta maps in the natural way.

$$
0  \overset{\partial_3}{\longrightarrow}
C_2 \overset{\partial_2}{\longrightarrow}
C_1 \overset{\partial_1}{\longrightarrow}
C_0
$$

Note that multiplying d1 by d2

$$
0  \overset{\delta_3}{\longleftarrow}
C_2 \overset{\delta_2}{\longleftarrow}
C_1 \overset{\delta_1}{\longleftarrow}
C_0
$$

Fortunately, defining these maps is easy; we can just transpose the matrices by letting $\delta_\bullet = \partial_\bullet {}^T$.

For the projective plane, we end up modding out the base space. But what does this actually look like?

$$
\begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} \left/
\begin{pmatrix} 1 &amp; 1 \\ -1 &amp; 1 \end{pmatrix}
\right.
$$

On the left is the integer lattice and on the right is the sublattice generated by (1, -1) and (1, 1). These can be overlaid to show that there are obviously twice as many points in the former as there are the latter. When we perform the action of "modding out", all of the points in the sublattice go to 0. So in the quotient space, we only have two types of points: ones which were on the sublattice (0) and points which were not (1). But this describes the exact same thing as $Z_2$, which happens to be the homology group under discussion.

The Klein bottle is a little more complicated. In this case, we end up modding a rank 3 space out by a rank 2 space.

$$
\begin{pmatrix} 1 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{pmatrix} \left/
\begin{pmatrix} 1 &amp; -1 \\ 1 &amp; 1 \\ -1 &amp; 1 \end{pmatrix}
\right.
$$

The free component is the line normal to the plane described by the column space. But this is precisely the kernel of the map.

The torsion subgroup turns out to be identical to the one we already discussed for the Klein bottle, only we're in a plane that's slightly skewed with respect to the integer lattice.

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