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In mathematics, the circle group, denoted by ${\displaystyle \mathbb {T} }$ or ${\displaystyle \mathbb {S} ^{1))$, is the multiplicative group of all complex numbers with absolute value 1, that is, the unit circle in the complex plane or simply the unit complex numbers^[1]

$${\displaystyle \mathbb {T} =\{z\in \mathbb {C} :|z|=1\}.}$$

The circle group forms a subgroup of ${\displaystyle \mathbb {C} ^{\times ))$, the multiplicative group of all nonzero complex numbers. Since ${\displaystyle \mathbb {C} ^{\times ))$ is abelian, it follows that ${\displaystyle \mathbb {T} }$ is as well.

A unit complex number in the circle group represents a rotation of the complex plane about the origin and can be parametrized by the angle measure ${\displaystyle \theta }$:

$${\displaystyle \theta \mapsto z=e^{i\theta }=\cos \theta +i\sin \theta .}$$

This is the exponential map for the circle group.

The circle group plays a central role in Pontryagin duality and in the theory of Lie groups.

The notation ${\displaystyle \mathbb {T} }$ for the circle group stems from the fact that, with the standard topology (see below), the circle group is a 1-torus. More generally, ${\displaystyle \mathbb {T} ^{n))$ (the direct product of ${\displaystyle \mathbb {T} }$ with itself ${\displaystyle n}$ times) is geometrically an ${\displaystyle n}$-torus.

The circle group is isomorphic to the special orthogonal group ${\displaystyle \mathrm {SO} (2)}$.

Elementary introduction

One way to think about the circle group is that it describes how to add angles, where only angles between 0° and 360° or ${\displaystyle \in [0,2\pi )}$ or ${\displaystyle \in (-\pi ,+\pi ]}$ are permitted. For example, the diagram illustrates how to add 150° to 270°. The answer is 150° + 270° = 420°, but when thinking in terms of the circle group, we may "forget" the fact that we have wrapped once around the circle. Therefore, we adjust our answer by 360°, which gives 420° ≡ 60° (mod 360°).

Another description is in terms of ordinary (real) addition, where only numbers between 0 and 1 are allowed (with 1 corresponding to a full rotation: 360° or ${\displaystyle 2\pi }$), i.e. the real numbers modulo the integers: ${\displaystyle \mathbb {T} \cong \mathbb {R} /\mathbb {Z} }$. This can be achieved by throwing away the digits occurring before the decimal point. For example, when we work out 0.4166... + 0.75, the answer is 1.1666..., but we may throw away the leading 1, so the answer (in the circle group) is just   ${\displaystyle 0.1{\bar {6))\equiv 1.1{\bar {6))\equiv -0.8{\bar {3))\;({\text{mod))\,\mathbb {Z} )}$   with some preference to 0.166..., because ${\displaystyle 0.1{\bar {6))\in [0,1)}$.

Topological and analytic structure

The circle group is more than just an abstract algebraic object. It has a natural topology when regarded as a subspace of the complex plane. Since multiplication and inversion are continuous functions on ${\displaystyle \mathbb {C} ^{\times ))$, the circle group has the structure of a topological group. Moreover, since the unit circle is a closed subset of the complex plane, the circle group is a closed subgroup of ${\displaystyle \mathbb {C} ^{\times ))$ (itself regarded as a topological group).

One can say even more. The circle is a 1-dimensional real manifold, and multiplication and inversion are real-analytic maps on the circle. This gives the circle group the structure of a one-parameter group, an instance of a Lie group. In fact, up to isomorphism, it is the unique 1-dimensional compact, connected Lie group. Moreover, every ${\displaystyle n}$-dimensional compact, connected, abelian Lie group is isomorphic to ${\displaystyle \mathbb {T} ^{n))$.

Isomorphisms

The circle group shows up in a variety of forms in mathematics. We list some of the more common forms here. Specifically, we show that

$${\displaystyle \mathbb {T} \cong {\mbox{U))(1)\cong \mathbb {R} /\mathbb {Z} \cong \mathrm {SO} (2).}$$

Note that the slash (/) denotes here quotient group.

The set of all 1×1 unitary matrices clearly coincides with the circle group; the unitary condition is equivalent to the condition that its element have absolute value 1. Therefore, the circle group is canonically isomorphic to ${\displaystyle \mathrm {U} (1)}$, the first unitary group.

The exponential function gives rise to a group homomorphism ${\displaystyle \exp :\mathbb {R} \to \mathbb {T} }$ from the additive real numbers ${\displaystyle \mathbb {R} }$ to the circle group ${\displaystyle \mathbb {T} }$ via the map

$${\displaystyle \theta \mapsto e^{i\theta }=\cos \theta +i\sin \theta .}$$

The last equality is Euler's formula or the complex exponential. The real number θ corresponds to the angle (in radians) on the unit circle as measured counterclockwise from the positive x axis. That this map is a homomorphism follows from the fact that the multiplication of unit complex numbers corresponds to addition of angles:

$${\displaystyle e^{i\theta _{1))e^{i\theta _{2))=e^{i(\theta _{1}+\theta _{2})}.}$$

This exponential map is clearly a surjective function from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {T} }$. However, it is not injective. The kernel of this map is the set of all integer multiples of ${\displaystyle 2\pi }$. By the first isomorphism theorem we then have that

$${\displaystyle \mathbb {T} \cong \mathbb {R} /2\pi \mathbb {Z} .}$$

After rescaling we can also say that ${\displaystyle \mathbb {T} }$ is isomorphic to ${\displaystyle \mathbb {R} /\mathbb {Z} }$.

If complex numbers are realized as 2×2 real matrices (see complex number), the unit complex numbers correspond to 2×2 orthogonal matrices with unit determinant. Specifically, we have

$${\displaystyle e^{i\theta }\leftrightarrow {\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix))=f\left(e^{i\theta }\right).}$$

This function shows that the circle group is isomorphic to the special orthogonal group ${\displaystyle \mathrm {SO} (2)}$ since

$${\displaystyle f\left(e^{i\theta _{1))e^{i\theta _{2))\right)={\begin{bmatrix}\cos(\theta _{1}+\theta _{2})&-\sin(\theta _{1}+\theta _{2})\\\sin(\theta _{1}+\theta _{2})&\cos(\theta _{1}+\theta _{2})\end{bmatrix))=f\left(e^{i\theta _{1))\right)\times f\left(e^{i\theta _{2))\right),}$$

${\displaystyle \times }$

This isomorphism has the geometric interpretation that multiplication by a unit complex number is a proper rotation in the complex (and real) plane, and every such rotation is of this form.

Properties

Every compact Lie group ${\displaystyle \mathrm {G} }$ of dimension > 0 has a subgroup isomorphic to the circle group. This means that, thinking in terms of symmetry, a compact symmetry group acting continuously can be expected to have one-parameter circle subgroups acting; the consequences in physical systems are seen, for example, at rotational invariance and spontaneous symmetry breaking.

The circle group has many subgroups, but its only proper closed subgroups consist of roots of unity: For each integer ${\displaystyle n>0}$, the ${\displaystyle n}$-th roots of unity form a cyclic group of order ${\displaystyle n}$, which is unique up to isomorphism.

In the same way that the real numbers are a completion of the b-adic rationals ${\displaystyle \mathbb {Z} [{\tfrac {1}{b))]}$ for every natural number ${\displaystyle b>1}$, the circle group is the completion of the Prüfer group ${\displaystyle \mathbb {Z} [{\tfrac {1}{b))]/\mathbb {Z} }$ for ${\displaystyle b}$, given by the direct limit ${\displaystyle \varinjlim \mathbb {Z} /b^{n}\mathbb {Z} }$.

Representations

The representations of the circle group are easy to describe. It follows from Schur's lemma that the irreducible complex representations of an abelian group are all 1-dimensional. Since the circle group is compact, any representation

$${\displaystyle \rho :\mathbb {T} \to \mathrm {GL} (1,\mathbb {C} )\cong \mathbb {C} ^{\times ))$$

${\displaystyle {\mbox{U))(1)\cong \mathbb {T} }$

For each integer ${\displaystyle n}$ we can define a representation ${\displaystyle \phi _{n))$ of the circle group by ${\displaystyle \phi _{n}(z)=z^{n))$. These representations are all inequivalent. The representation ${\displaystyle \phi _{-n))$ is conjugate to ${\displaystyle \phi _{n))$:

$${\displaystyle \phi _{-n}={\overline {\phi _{n))}.}$$

These representations are just the characters of the circle group. The character group of ${\displaystyle \mathbb {T} }$ is clearly an infinite cyclic group generated by ${\displaystyle \phi _{1))$:

$${\displaystyle \operatorname {Hom} (\mathbb {T} ,\mathbb {T} )\cong \mathbb {Z} .}$$

The irreducible real representations of the circle group are the trivial representation (which is 1-dimensional) and the representations

$${\displaystyle \rho _{n}(e^{i\theta })={\begin{bmatrix}\cos n\theta &-\sin n\theta \\\sin n\theta &\cos n\theta \end{bmatrix)),\quad n\in \mathbb {Z} ^{+},}$$

${\displaystyle \mathrm {SO} (2)}$

${\displaystyle n}$

${\displaystyle \rho _{-n))$

${\displaystyle \rho _{n))$

Group structure

The circle group ${\displaystyle \mathbb {T} }$ is a divisible group. Its torsion subgroup is given by the set of all ${\displaystyle n}$-th roots of unity for all ${\displaystyle n}$ and is isomorphic to ${\displaystyle \mathbb {Q} /\mathbb {Z} }$. The structure theorem for divisible groups and the axiom of choice together tell us that ${\displaystyle \mathbb {T} }$ is isomorphic to the direct sum of ${\displaystyle \mathbb {Q} /\mathbb {Z} }$ with a number of copies of ${\displaystyle \mathbb {Q} }$.^{[citation needed]}

The number of copies of ${\displaystyle \mathbb {Q} }$ must be ${\displaystyle {\mathfrak {c))}$ (the cardinality of the continuum) in order for the cardinality of the direct sum to be correct. But the direct sum of ${\displaystyle {\mathfrak {c))}$ copies of ${\displaystyle \mathbb {Q} }$ is isomorphic to ${\displaystyle \mathbb {R} }$, as ${\displaystyle \mathbb {R} }$ is a vector space of dimension ${\displaystyle {\mathfrak {c))}$ over ${\displaystyle \mathbb {Q} }$. Thus

$${\displaystyle \mathbb {T} \cong \mathbb {R} \oplus (\mathbb {Q} /\mathbb {Z} ).}$$

The isomorphism

$${\displaystyle \mathbb {C} ^{\times }\cong \mathbb {R} \oplus (\mathbb {Q} /\mathbb {Z} )}$$

can be proved in the same way, since ${\displaystyle \mathbb {C} ^{\times ))$ is also a divisible abelian group whose torsion subgroup is the same as the torsion subgroup of ${\displaystyle \mathbb {T} }$.