I'm a beginner in Algebraic topology and reading Armstrong as it has a nice algebraic flavor. The shaded areas are the fundamental regions in the groups of isometries of the plane. I understand how we get Torus and Sphere but unable to visualize or to draw Klein bottle's fundamental polygon. Kindly give some hints, Thank you!
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It seems the two half arrows in (c) symbolise barely the two generators $g_1, g_2$ of the plane symmetry group $\mathrm{pg}$. These are two different (anti-parallel) glide reflection planes. They generate the translation $t$ by $g_1 g_2 t = 1$ . Thats all what you can see in this diagram. Another question is how they can be found to be consistent with the fundamental region of the Klein bottle (Thats kind of obvious I'd say).
The meaning of Figure 4.5 (c) in Armstrong's text is as follows.
Let's identify the vertices of the figure with the set $S = \{(m-n, m+n) \mid m,n \in \mathbb{Z}\}$. In other words, if we let $v = (-1,1)$ and $w = (1,1)$ then the vertices are all integer linear combinations of $v$ and $w$. You should think of the lower corner of the shaded diamond as the origin and $v$ and $w$ as vectors representing the bottom left and bottom right sides, respectively.
The dark shaded segments with the half arrowheads are meant to suggest glide reflections. The one on the left is given by $a(x,y) = (-1-x,y+1)$ and the one on the right by $b(x,y) = (1-x,y+1)$.
It's easy to see that $a$ and $b$ preserve the set $S$. With some further thought, you can deduce that the group $G$ generated by $a$ and $b$ (meaning just allow all composites of $a$ and $b$ and their inverses) acts transitively on $S$, meaning that the orbit of the origin under $G$ is all of $S$. More subtle is that this is a simply transitive action, meaning there are no fixed points.
To see the more familiar fundamental domain of the Klein bottle, take the diamond and draw a horizontal segment from the left corner to the right corner. Orient this segment from left to right and label it by $E$. Orient the lower left diagonal edge of the diamond upward and label it by $F$; do the same for the lower right edge and label it $G$. Since the action of $G$ carries the lower left edge to the upper left edge with a flip, this upper left edge is also labeled by $F$ and is still oriented upwards. The same happens for the upper right edge, except it is labeled by $G$.
At this point you have a diamond with boundary (clockwise starting at the bottom vertex) label $GGF^{-1}F^{-1}$. You also have a horizontal segment labeled $E$ and oriented from left to right.
Cut the diamond along $E$ and remember the edge label and orientations. This makes two oriented labeled triangles. Flip one of them over. Then paste along the oriented edges labeled by $F$. You then have the familiar opposite edge pairing where the $G$ labels are identified by a translation and the $E$ labels are identified by a glide reflection.