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Old qual problem here, test tomorrow in topology and we barely got to DeRahm Cohomology so I'm not sure how to do this.

Let $G$ be the group of transformations of $\mathbb{R}^3$ generated by \begin{align} A(x,y,z)&=(x+1,y,z)\\ B(x,y,z)&=(x,y+1,z)\\ C(x,y,z)&=(x+y,y,z+1). \end{align} Assume that $M=\mathbb{R}^3/G$ is a three-manifold.

a. Determine $\pi_1(M)$ up to isomorphism by giving a presentation with generators and relations. (note the relations will involve only commutators of generators)

b. Determine $H_{DR}^1(M)$ up to isomorphism and give closed one-forms whose cohomology classes are a basis of $H_{DR}^1(M)$.

Here's all I've got:

We know that $G$ acts properly discontinuously on $\mathbb{R}^3$ so $\pi_1(M)=G$, but I'm not sure how to write $G$.

When we have $G$, we know that $H_{DR}^1(X)=Hom(H_1(X),\mathbb{R})$ where $H_1(X)$ is $\pi_1(X)$ abelianized. I'm also not sure how to find the 1-forms giving a basis.

Thank you.

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By simple computations, one checks $AB=BA$, $AC=CA$, $CB=BCA$ is a presentation. So in the abelianization $A=1$, $BC=CB$ the abelianized group is $\bf Z^2$.

Note that $dy, dz$ are invariant by $G$ and form a base for the de Rham co-homology

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  • $\begingroup$ One last thing- how do I check that dx,dy are invariant by G? $\endgroup$
    – Logan Tatham
    Apr 29, 2016 at 12:48
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    $\begingroup$ $dx$ is not invariant but $dz$ and $dy$ are. For $z\circ A= z= z\circ B, z\circ C=z+1$, so $A^*dz=dz, B^*dz=dz, C^*dz=dz$ ; and the same sort of computations with $dy$ $\endgroup$
    – Thomas
    Apr 29, 2016 at 13:14
  • $\begingroup$ Sorry I meant dy, dz. Thank you. It is all clear now. My final is in a couple hours and we really rushed through that last chapter because we were running behind. Thank you for your help! $\endgroup$
    – Logan Tatham
    Apr 29, 2016 at 14:18

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