Linear representation theory of cyclic group:Z3

This article gives specific information, namely, linear representation theory, about a particular group, namely: cyclic group:Z3.
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This article discusses the linear representation theory of cyclic group:Z3, a group of order three.

Summary

Item	Value
Degrees of irreducible representations over a splitting field	1, 1, 1 maximum: 1, lcm: 1, number: 3, sum of squares: 3
Smallest ring of realization of all representations (characteristic zero)	$\mathbb{Z}[e^{2\pi i/3}]$, also expressible as $\mathbb{Z}[x]/(x^2 + x + 1)$. Degree two integral extension of $\mathbb{Z}$.
Smallest field of realization of all representations (characteristic zero)	$\mathbb{Q}(e^{2\pi i/3})$, also expressible as $\mathbb{Q}[x]/(x^2 + x + 1)$. Degree two cyclotomic extension of $\mathbb{Q}$.
Criterion for a field to be a splitting field	Any field of characteristic not 3 that contains a primitive cube root of unity, i.e., the polynomial $x^2 + x + 1$ splits.
Degrees of irreducible representations over a non-splitting field. Includes the case of field of real numbers $\R$, also field of rational numbers $\mathbb{Q}$	1, 2 maximum: 2, lcm: 2, number: 2
Smallest size splitting field	field:F4, i.e., field with four elements.

Family contexts

Family	Parameter value	General discussion of linear representation theory of family
finite cyclic group	3	linear representation theory of finite cyclic groups
alternating group	3	linear representation theory of alternating groups

Representations

Summary information

Below is summary information on irreducible representations. Note that a particular representation may make sense, and be irreducible, only for certain kinds of fields -- see the "Values not allowed for field characteristic" and "Criterion for field" columns to see the condition the field must satisfy for the representation to be irreducible there.

Name of representation type	Number of representations of this type	Values not allowed for field characteristic	Criterion for field	What happens over a splitting field?	Kernel	Degree	Schur index	What happens by reducing the $\mathbb{Z}$-representation over bad characteristics?
trivial	1	--	any	remains the same	whole group	1	1	--
one-dimensional nontrivial	2	3	contains a primitive cube root of unity	remains the same	trivial subgroup, i.e., it is faithful	1	1	irrelevant question
two-dimensional irreducible	1	--	does not contain a primitive cube root of unity	splits into the two one-dimensional nontrivial representations	trivial subgroup, i.e., it is faithful	2	1	Taking the $\mathbb{Z}$-representation and mapping to field:F3 gives an indecomposable two-dimensional representation that is not irreducible.

Trivial representation

This is a one-dimensional representation sending all the elements to $( 1 )$, and makes sense over any field.

One-dimensional nontrivial representations

The cyclic group of order three has two non-identity elements. Also, in a field with a primitive cube root of unity, there are two such primitive cube roots of unity.

The two one-dimensional nontrivial representations both send the identity element to $(1)$ and the two non-identity elements to the two primitive cube roots of unity. The representations differ in terms of which element of the group is matched with which primitive cube root of unity.

Two-dimensional representation: irreducible in the non-splitting case

If we denote the group as $\{ e,x,x^2\}$ where $e$ is the identity element, we can construct the following representation with integer matrices:

Element	Matrix	Trace	Minimal polynomial
$e$	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$	2	$x - 1$
$x$	$\begin{pmatrix} 0 & -1 \\ 1 & -1 \\\end{pmatrix}$	-1	$x^2 + x + 1$
$x^2$	$\begin{pmatrix} -1 & 1 \\ -1 & 0 \\\end{pmatrix}$	-1	$x^2 + x + 1$

We can interpret this representation over any field $F$, and three cases arise as to the nature of $F$:

Case for the field $F$	What happens in this case
$F$ does not have characteristic $3$ does not contain a primitive cube root of unity	The representation is irreducible. However, it is not absolutely irreducible, it decomposes as a direct sum in a quadratic extension containing a primitive cube root of unity (See next point).
$F$ does not have characteristic $3$ and contains a primitive cube root of unity	The representation decomposes as a direct sum of the two one-dimensional nontrivial representations with kernel V4 in A4
Characteristic of $F$ is $3$	The representation is indecomposable but not irreducible, because there is an invariant one-dimensional subspace.

Our case of current interest is the first one, i.e., where $F$ does not have characteristic $3$ and does not contain a primitive cube root of unity.

In the case $F$ is the field is the field of real numbers, or more generally, any subfield of the reals containing $\mathbb{Q}[\sqrt{3}]$, we can provide an alternative description of this representation as rotations by multiples of $2\pi/3$. Note that this alternative description, though it gives an equivalent representation, does not work over fields that lack a square root of $3$.

Element	Matrix	Trace	Minimal polynomial
$e$	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$	2	$x - 1$
$x$	$\begin{pmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2\\\end{pmatrix}$	-1	$x^2 + x + 1$
$x^2$	$\begin{pmatrix} -1/2 & \sqrt{3}/2 \\ -\sqrt{3}/2 & -1/2 \\\end{pmatrix}$	-1	$x^2 + x + 1$

Character table

Character table over a splitting field

FACTS TO CHECK AGAINST (for characters of irreducible linear representations over a splitting field):
Orthogonality relations: Character orthogonality theorem | Column orthogonality theorem
Separation results (basically says rows independent, columns independent): Splitting implies characters form a basis for space of class functions|Character determines representation in characteristic zero
Numerical facts: Characters are cyclotomic integers | Size-degree-weighted characters are algebraic integers
Character value facts: Irreducible character of degree greater than one takes value zero on some conjugacy class| Conjugacy class of more than average size has character value zero for some irreducible character | Zero-or-scalar lemma

Let $\omega$ be a primitive cube root of unity. The character table over a splitting field is as follows:

Representation/Conjugacy class	$e$ (identity element)	$x$ (generator)	$x^2$ (generator)
trivial representation	1	1	1
one nontrivial representation	1	$\omega$	$\omega^2$
the other (conjugate) nontrivial representation	1	$\omega^2$	$\omega$

Note that this character table is interpreted differently depending on what the splitting field is and which of the primitive cube roots we choose to be $\omega$. Switching the roles of $\omega$ and $\omega^2$ in the above table simply permutes the two nontrivial one-dimensional representations and has no effect on the overall character table.

In characteristic zero, $\omega$ can be taken as $e^{2\pi i/3}$ or $\cos(2\pi/3) + i\sin(2\pi/3)$, which is $(-1 + i\sqrt{3})/2$. $\omega^2$ is the other primitive cube root of unity, and is given as $e^{-2\pi i/3}$ or $\cos(2\pi/3) - i\sin(2\pi/3)$ or $(-1 - i\sqrt{3})/2$.

Character table over a non-splitting field

For a field that is not a splitting field for the group, there are only two equivalence classes of irreducible representations. But also, the number of Galois conjugacy classes is two. The character table looks as follows:

Representation/Galois conjugacy class	Identity element	Non-identity elements
trivial representation	1	1
nontrivial two-dimensional representation	2	-1

Over a finite field, the character values are interpreted as integers modulo the field characteristic; over an infinite field, they are interpreted as rational numbers and hence field elements.

If doing character theory over the real numbers, we know that the number of irreducible representations over reals equals number of real conjugacy classes. The above is the character table both over the rationals and over the reals.