Defining Geometric Shapes using Group Theory

Today we will go through a nice application of group theory - defining geometric shapes through the language of group isomorphisms. This article is based off this video by Michael Penn.

Preliminary Definitions

If you do not have a basic understanding of group theory, I’d recommend you learn the basics before embarking on this article. A great introductory video can be found here and a fantastic series with a focus on intuition can be found here.

Groups

A group is simply a collection, or set, of objects equipped with some operation between each object. This operation must:

  • Take in two objects in the set and return a third object in the set (closure)
  • If you perform the operation on three (or more) objects, the order in which you do so does not matter (associativity)
  • There is a special object in the set, say \(\mathit{e}\), such that when you perform the operation with this object and another, say \(\mathit{g}\), you get \(\mathit{g}\) back. \(\mathit{e}\) is called the identity of \(\mathit{G}\).
  • For each object in the set \(\mathit{g}\), there exists another object in the set \(\mathit{g^{-1}}\) such that when you perform the operation on \(\mathit{g}\) and \(\mathit{g^{-1}}\), you get that special element \(\mathit{e}\) back. \(\mathit{g^{-1}}\) is called the inverse of \(\mathit{g}\).

More formally, a set \(\mathit{G}\) and an operation \(\cdot\) form a group \((\mathit{G}, \cdot)\) if it satisfies the following:

For all \(g, h \in G, g \cdot h \in G\) (closure)
For all \(g, h, k \in G, (g \cdot h) \cdot k = g \cdot (h \cdot k)\) (associativity)
There exists \(e \in G\) such that for all \(g \in G, g \cdot e = e \cdot g = g\) (identity)
For all \(g \in G\), there exists \(g^{-1} \in G\) such that \(g \cdot g^{-1} = e\) (inverse)

A subgroup \(H\) of a group \(G\) is, as the name suggests, a subset of the group \(G\) that follows the group definition in its own right.

A handy criterion we have for determining subgroups is as follows:

\(H\) is a subgroup of \(G\) (written \(H \le G\)) if and only if for all \(x,y \in G\), if \(x \cdot y \in G\), then \(x \cdot y^{-1} \in G\).

This criterion will come in handy later.

Another incredibly useful (and beautiful) theorem in group theory is the first isomorphism theorem. Unfortnately, from this point onwards you’ll probably want a basic understanding of group homomorphims, quotient groups, images and kernels. If I try to explain everything, this post will be of infinite length.

The first isomorphism theorem states:

If \(\phi : G \to H\) is a group homomorphism between groups \(G\) and \(H\), then \(G/Ker(\phi) \cong Im(\phi)\)

Isomorphisms To Circles

Let us define the surface of a sphere, \(S^1\), as the collection of complex numbers \(\Bbb{C}\) at a distance of 1 from the origin - our good friend, the unit sphere.

\[S^1 := \{z \in \Bbb{C} : |z| = 1\}\]

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Note that even though this is 2-dimensional shape - the circle - the boundary of the circle is what we are interested in. In this case, it is the 1-dimensional line that borders the circle. Hence the name \(S^1\) for our surface.

We consider the multiplicative group of \(S^1\), that is the group operation of multiplication of complex numbers.

Next, we claim that \(S^1\) is a subgroup of \(\Bbb{C}^+\), the multiplicative group of the complex numbers, excluding \(0\). This may seem a trivial fact, but can easily be proven.

To prove this, we will use the criterion listed above.

Suppose that \(z,w \in S^1\), so \(z \cdot w \in S^1\). By the defining property of \(S^1\), \(|z| = |w| = 1\). We have that \(|z \cdot w^{-1}| = \frac {|z|}{|w|} = \frac 11 = 1\), Hence we have that \(z \cdot w \in S^1\) implies \(z \cdot w^{-1} \in S^1\), so \(S^1 \le \Bbb{C^+}\), as stated.

We finally get to our first big claim - that the quotient group of the additive real numbers by the integers, is isomorphic to the multiplicative unit circle. How cool is that! More precisely, we will show that

\[\Bbb{R}/\Bbb{Z} \cong S^1\]

More intuitively, one can think of taking this quotient group as setting each integer within the reals to be ‘equivalent’. So 0 gets ‘identified’ to 1, 1 gets ‘identified’ to 2, and so on. When you identify each of these elements, you ‘wrap’ them around to each other - 1 gets ‘connected’ to 0, 2 gets ‘connected’ to 1, etc. This forms a circle.

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Proof of Isomorphism

Now for the proof. Our strategy will be to construct a group homomorphim having image \(S^1 \le \Bbb{C^+}\) and kernel \(\Bbb{Z}\). Then, by the first isomorphism theorem, we will have the desired result.

Let \(\phi : \Bbb{R} \to \Bbb{C^+}\) be a group homomorphim, with \(\phi(x) = e^{2 i \pi x}\). Notice the jump of operation - the operation of the domain is addition, and the operation of the codomain is multiplication.
We check that this is indeed a group homomorphism.
Recall, to have a homomorphism, we must have that \(\phi(x \cdot y) = \phi(x) \cdot \phi(y)\) for all \(x,y\). For any \(x,y \in \Bbb{R}\), we have that \(\phi(x y) = e^{2 i \pi (x+y)} = e^{2 i \pi x} e^{2 i \pi y} = \phi(x)\phi(y)\) as required.
Now we calculate the image of \(\phi\), \(Im(\phi)\). This will be the subgroup of \(\Bbb{C^+}\) such that for all \(y \in Im(\phi)\), there exists some \(x \in \Bbb{R}\) with \(\phi(x) = y\)
Any complex number \(z\) on the unit circle can be expressed as \(e^{2 i \pi x}\) for some \(x \in \Bbb{R}\). This is true by Euler’s formula, which states that \(e^{2 i \pi x} = \cos(2 i \pi x) + i \sin(2 i \pi x)\) for \(x \in \Bbb{R}\).
Hence, we see that \(Im(\phi) = S^1\), the unit circle.
Next, we calculate the kernel of \(\phi\), \(Ker(\phi)\).
Note that the group \(S^1\) has the multiplicative identity \(1\). Recall that \(Ker(\phi) = \{x \in \Bbb{R} : e^{2 i \pi x} = 1\}\).
Since \(e^{2 i \pi x} = \cos(2 i \pi x) + i \sin(2 i \pi x)\), we see that we must have \(x \in \Bbb{Z}\) for \(e^{2 i \pi x}\) to be \(1\). In other words, the kernel of \(\phi\) is precisely the integers.
We now have all the components we need to apply the first isomorphism theorem. Since \(\phi\) is a homomorphsm from \(\Bbb{R}\) to \(\Bbb{C^+}\), with kernel \(\Bbb{Z}\) and image \(S^1\), we have that \(\Bbb{R}/\Bbb{Z} \cong S^1\) - the desired result!

We have now shown that the quotient group of the reals by the integers is none other than the unit circle!

Higher Dimensions

Let’s consider a higher dimensional example. Recall that the Gaussian integers \(\Bbb{Z}[i]\) is the set of ‘integer’ complex numbers. That is:

\[\Bbb{Z}[i] = \{a + bi : a,b \in \Bbb{Z}\}\]

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This fascinating set forms a lattice of points within the complex plane (as shown above), and the set \(\Bbb{Z}[i]\) clearly forms a subgroup of \(\Bbb{C}\). So what happens when we take the quotient group of \(\Bbb{C}\) by \(\Bbb{Z}[i]\)?

More precisely, what group \(G\) satisfies the following:

\[\Bbb{C}/\Bbb{Z}[i] \cong G\]

I won’t answer this question as rigourously as the previous example, but I will provide some geometric intuition.

When we take the quotient group \(\Bbb{C}/\Bbb{Z}[i]\), we ‘identify’ all elements that differ by an element of \(\Bbb{Z}[i]\). This can be seen more clearly if we zoom into the lattice of the Gaussian integers.

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The red line segments between \(0\) and \(1\), and \(i\) and \(1+i\), are ‘identified’ with each other, and the blue line segments between \(0\) and \(i\), and \(1\) and \(1+i\) are ‘identified’ with each other. When we ‘glue’ the red identfied line segments, we get a cylinder. When we ‘glue’ the blue identified segments, we join the ends of the cylinder to form none other than a torus! This process repeats infinitely across all lattice points.

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In other words, this quotient group is isomorphic to a torus! The surface of a torus can be described as \(S^1 \times S^1\), so more precisely we have:

\[\Bbb{C}/\Bbb{Z}[i] \cong S^1 \times S^1\]

Conclusion

In this article, we saw how we can describe the unit circle as the quotient group of the reals by the integers - an interesting result in its own right. We then saw how we can construct the torus using the Gaussian integers. These results help to build a bridge between the fields of geometry and number theory - cool stuff!

Micahel Penn video

Group theory introduction

Group theory intuition