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Quotient of ℂ by the Eisenstein integers
The quotient of the complex plane $ℂ$ by the lattice containing all Eisenstein integers is a complex torus of real dimension 2. This is one of two tori with maximal symmetry among all such complex tori. This torus can be obtained by identifying each of the three pairs of opposite edges of a regular hexagon.
unitsize(2cm);
pair[] p;
for(int k:sequence(6)){
pair pt=dir(60k);
p.push(pt);
label((string)(k+1),pt,pt);
}
draw(p[1]--p[0],blue,MidArrow);
draw(p[3]--p[4],blue,MidArrow);
draw(p[5]--p[0],red,MidArrow);
draw(p[3]--p[2],red,MidArrow);
draw(p[2]--p[1],green,MidArrow);
draw(p[4]--p[5],green,MidArrow); | unitsize(2cm);
pair[] p;
picture pic1;
unitsize(pic1,2cm);
for(int k:sequence(6)){
pair pt=dir(60k);
p.push(pt);
label(pic1,(string)(k+1),pt,pt);
}
path tri=p[2]--p[3]--p[4]--cycle;
fill(tri,gray(.8));
fill(shift(p[0]-p[4])*tri,gray(.8));
draw(p[1]--p[0],blue,MidArrow);
draw(p[3]--p[4],blue,MidArrow);
draw(p[5]--p[0],red,MidArrow);
draw(p[3]--p[2],red,MidArrow);
draw(p[2]--p[1],green,MidArrow);
draw(p[4]--p[5],green,MidArrow);
add(pic1.fit(),(0,0)); | unitsize(2cm);
pair[] p;
for(int k:sequence(6)){
pair pt=dir(60k);
p.push(pt);
if(k==3){pt=2p[1];}
label((string)(k+1),pt,unit(pt));
}
path tri=p[2]--p[3]--p[4]--cycle;
draw(p[5]--p[0],red,MidArrow);
draw(shift(p[0]-p[4])*(p[3]--p[2]),red,MidArrow);
draw(p[0]-p[4]+p[2]--p[0],MidArrow);
draw(p[2]--p[4],MidArrow);
draw(p[2]--p[1],green,MidArrow);
draw(p[4]--p[5],green,MidArrow); |
What surface do we get by joining the opposite edges of a hexagon?
You can show it is a torus by cut and paste methods. These are the same methods used in the proof of the classification of surfaces in the comment of @Bacon, however you do not need to cite the classification of surfaces. Instead you can show by hand that this is a torus, which will have the side benefit of teaching you a little about how the classification of surfaces is proved. Roughly speaking that proof is an algorithmic process which let's you take any polygon gluing diagram and cut and paste it to put it into a normal form which let's you recognize the surface.
To do the cut and paste, first number the vertices in your hexagon picture in clockwise order as $1,2,3,4,5,6$ where 1 is the rightmost vertex. Cut the hexagon along the line segment from vertex 3 to vertex 5, thus cutting off a triangle (the left portion) and a pentagon (the right portion). Label the two cut edges with a quadruple arrow so as to remember how to repaste them at a later time. But do not repaste them yet! Instead, paste the triple arrow edge of the triangle to the triple arrow edge of the pentagon. You now have a different gluing pattern on a hexagon. You should be able to see how this different hexagon gluing pattern is the same as an ordinary square gluing pattern, hence you get the torus.
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