Define a relation $\sim$ on $\mathbb{R}^2$ by $$(x_1,x_2)\sim(y_1,y_2) \leftrightarrow (x_1-y_1,x_2-y_2)\in\mathbb{Z}\times\mathbb{Z}$$ Show that this is an equivalence relation on $\mathbb{R}^2$ and determine its quotient space as a subset of $\mathbb{R}^3$

In my understanding to determine the quotient space we first have to know the equivalence classes of $\mathbb{R}^2/\sim$, and then find the inverse of canonical projection map for each classes and their union. The problem is I don't know how to explicitly write out the classes of $\mathbb{R}^2/\sim$.

Is there a trick for determining them?

Thank you

  • $\begingroup$ Observe that $\;(x_1,x_2)\sim(y_1,y_2)\iff x_i=y_i+k\;,\;\;k\in\Bbb Z\;,\;\;i=1,2\;$ , so we're talking of translations by integers here. $\endgroup$ – DonAntonio Nov 14 '16 at 10:53
  • $\begingroup$ A class is a lattice of $\mathbf R^2$, with origin any of its elements., i.e. the set of points translated from a given point $(x,y)$ by a vector with integer coordinates. $\endgroup$ – Bernard Nov 14 '16 at 10:56

HINT: Show that there is one equivalence class for each point in $[0,1)\times[0,1)$.

Further HINT: You get the same quotient space if you start with the closed unit square $I\times I=[0,1]\times[0,1]$ and the equivalence relation $\sim$ defined as follows:

  • $\langle x,y\rangle\sim\langle x,y\rangle$ for each $\langle x,y\rangle\in I\times I$;
  • $\langle 0,y\rangle\sim\langle 1,y\rangle$ for each $y\in[0,1]$; and
  • $\langle x,0\rangle\sim\langle x,1\rangle$ for each $x\in[0,1]$.
  • $\begingroup$ I'm still not able to write out the equivalence classes. Intuitively it seems obvious that if I take an ordered pair in $[0,1) \times [0,1)$ and add any integral multiple to this tuple then I would have an equivalence class. And if we take two distinct ordered pair from $[0,1) \times [0,1)$ their respective classes should be disjoint. The problem is how do I write that out. I'm sorry if this seems dumb but equivalence classes is really not my thing :( $\endgroup$ – Sai Nov 16 '16 at 12:01
  • $\begingroup$ @Sai: You have the right idea. Let $\langle x,y\rangle\in[0,1)\times[0,1)$. Then the equivalence class of $\langle x,y\rangle$ is $$\{\langle x+m,y+n\rangle:m,n\in\Bbb Z\}\;.$$ This is sometimes written $$\langle x,y\rangle+(\Bbb Z\times\Bbb Z)$$ for short. It’s pretty clear that everything in this set is equivalent to $\langle x,y\rangle$, and it’s not hard to check that if you start with different points in $[0,1)\times[0,1)$, you get disjoint sets. Moreover, every point in the plane is in one of these sets: a point $\langle u,v\rangle\in\Bbb R^2$ is in the set generated by the point ... $\endgroup$ – Brian M. Scott Nov 16 '16 at 17:18
  • $\begingroup$ ... $\langle\lfloor u\rfloor,\lfloor v\rfloor\rangle\in[0,1)\times[0,1)$. $\endgroup$ – Brian M. Scott Nov 16 '16 at 17:19

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