# Faithful irreducible representation of quaternion group

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This article describes a particular irreducible linear representation for the following group: quaternion group. The representation is unique up to equivalence of linear representations and is irreducible, at least over its original field of definition in characteristic zero. The representation may also be definable over other characteristics by reducing the matrices modulo that characteristic, though it may behave somewhat differently in these characteristics.
For more on the linear representation theory of the group, see linear representation theory of quaternion group.

## Summary

Item Value
Degree of representation 2
Schur index 2 if the characteristic is 0, 1 if the characteristic is anything other than 0 or 2
Kernel of representation trivial subgroup, i.e., it is a faithful linear representation
Quotient on which it descends to a faithful linear representation quaternion group
Set of character values $\{ 2, -2, 0 \}$ (interpreted/reduced modulo the field or ring)
Characteristic zero: Ring generated: $\mathbb{Z}$: ring of integers, Ideal generated within ring: $2\mathbb{Z}$, Field generated: $\mathbb{Q}$: field of rational numbers
Rings of realization The representation can be realized over any commutative unital ring containing elements $\alpha,\beta$ such that $\alpha^2 + \beta^2 + 1 = 0$. Other rings? In particular, $\mathbb{Z}[\sqrt{-1-m^2}]$ is a ring of realization for any integer $m$, and so is $\mathbb{Z}[t]/(t^2 + t + 1)$ (adjoining a primitive cube root of unity) because if $\omega$ is a primitive cube root of unity, then $\omega^2 + (\omega^2)^2 = -1$.
Fields of realization The representation can be realized over any field containing elements $\alpha,\beta$ such that $\alpha^2 + \beta^2 + 1 = 0$. This includes all finite fields, because every element of a finite field is expressible as a sum of two squares.
Minimal field of realization In characteristic zero: any field of the form $\mathbb{Q}(\sqrt{-1-m^2})$ where $m$ is rational. Examples are $\mathbb{Q}(i), \mathbb{Q}(\sqrt{-2}), \mathbb{Q}(\sqrt{-5})$. Also, any field of the form $\mathbb{Q}(\sqrt{-m^2/2 - 1})$ such as $\mathbb{Q}(\sqrt{-3})$. We can take $\alpha,\beta = (-1 \pm \sqrt{-m^2/2 - 1})/2$
In characteristic $p$: the prime field $\mathbb{F}_p$.
Type of representation Quaternionic, i.e., the character is real-valued but the representation cannot be realized over the reals. See the #Frobenius-Schur indicator section.
Size of equivalence class under automorphisms 1
Size of equivalence class under Galois automorphisms 1 in all characteristics, because the character takes value in the prime subfield. (Note: The question doesn't really make sense in characteristic two, because it's a bad characteristic as explained below)
Bad characteristic 2. In characteristic 2, this generic description fits many different inequivalent representations. None are faithful. None are irreducible.

## Representation table

Suppose $\alpha, \beta$ are elements of a field (or more generally, a commutative unital ring) such that $\alpha^2 + \beta^2 + 1 = 0$, then the representation can be realized in terms of the entries $\{ 0,1,-1,\alpha,-\alpha,\beta,-\beta \}$. The explicit representation involving the Hamiltonian quaternions is the special case of this where $\alpha$ is a square root of $-1$ and $\beta = 0$. Another case is $\alpha^2 = -2$, $\beta = 1$. A third case of interest is $\mathbb{Q}(\sqrt{-3})$, which contains primitive cube roots of unity, where we set $\alpha$ and $\beta$ as distinct primitive cube roots of unity.

Note that the representation makes sense in all characteristics, but there are some problems interpreting it in characteristic two.

Element Matrix using $\alpha,\beta$ with $\alpha^2 + \beta^2 = -1$ Matrix using $i = \sqrt{-1}$. Obtained by setting $\alpha = i, \beta = 0$ Matrix using
$\sqrt{2}i = \sqrt{-2}$. Obtained by setting $\alpha = \sqrt{2}i, \beta = 1$
Matrix using primitive cube roots of unity, which we denote by $\omega$ and $\omega^2$, so $\alpha = \omega$, $\beta = \omega^2$. Characteristic polynomial Minimal polynomial Trace, character value Determinant
$\! 1$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ $(t - 1)^2 = t^2 - 2t + 1$ $t - 1$ 2 1
$\! -1$ $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix}$ $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix}$ $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix}$ $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix}$ $(t + 1)^2 = t^2 + 2t + 1$ $t + 1$ -2 1
$\! i$ $\begin{pmatrix} \alpha & \beta \\ \beta & -\alpha \\\end{pmatrix}$ $\begin{pmatrix} i & 0 \\ 0 & -i \\\end{pmatrix}$ $\begin{pmatrix} \sqrt{2}i & 1 \\ 1 & -\sqrt{2}i \\\end{pmatrix}$ $\begin{pmatrix} \omega & \omega^2 \\ \omega^2 & -\omega \\\end{pmatrix}$ $t^2 + 1$ $t^2 + 1$ 0 1
$\! -i$ $\begin{pmatrix} -\alpha & -\beta \\ -\beta & \alpha \\\end{pmatrix}$ $\begin{pmatrix} -i & 0 \\ 0 & i \\\end{pmatrix}$ $\begin{pmatrix} -\sqrt{2}i & -1 \\ -1 & -\sqrt{2}i \\\end{pmatrix}$ $\begin{pmatrix} -\omega & -\omega^2 \\ -\omega^2 & \omega \\\end{pmatrix}$ $t^2 + 1$ $t^2 + 1$ 0 1
$\! j$ $\begin{pmatrix}0 & -1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & -1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & -1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & -1 \\ 1 & 0 \\\end{pmatrix}$ $t^2 + 1$ $t^2 + 1$ 0 1
$\! -j$ $\begin{pmatrix}0 & 1 \\ -1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ -1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ -1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ -1 & 0 \\\end{pmatrix}$ $t^2 + 1$ $t^2 + 1$ 0 1
$\! k$ $\begin{pmatrix} \beta & -\alpha\\ -\alpha & -\beta\\\end{pmatrix}$ $\begin{pmatrix} 0& -i\\ -i & 0\\\end{pmatrix}$ $\begin{pmatrix} 1& -\sqrt{2}i\\ -\sqrt{2}i & -1\\\end{pmatrix}$ $\begin{pmatrix} \omega^2 & -\omega\\ -\omega & -\omega^2\\\end{pmatrix}$ $t^2 + 1$ $t^2 + 1$ 0 1
$\! -k$ $\begin{pmatrix} -\beta & \alpha\\ \alpha & \beta\\\end{pmatrix}$ $\begin{pmatrix} 0 & i\\ i& 0\\\end{pmatrix}$ $\begin{pmatrix} -1 & \sqrt{2}i\\ \sqrt{2}i& 1\\\end{pmatrix}$ $\begin{pmatrix} -\omega^2 & \omega\\ \omega & \omega^2\\\end{pmatrix}$ $t^2 + 1$ $t^2 + 1$ 0 1
Set of values used $\{ 0,1,-1,\alpha,-\alpha,\beta,-\beta \}$ $\{0,1,-1,i,-i\}$ $\{0,1,-1,\sqrt{2}i,-\sqrt{2}i \}$ $\{ 0,1,-1,\omega,-\omega,\omega^2,-\omega^2\}$ -- -- $\{2, -2, 0 \}$ $\{ 1 \}$
Ring generated by values used (characteristic zero) $\mathbb{Z}[\alpha,\beta]$ $\mathbb{Z}[i]$ or $\mathbb{Z}[t]/(t^2 + 1)$ $\mathbb{Z}[\sqrt{2}i]$ or $\mathbb{Z}[t]/(t^2 + 2)$ $\mathbb{Z}[t]/(t^2 + t + 1)$ -- -- $\mathbb{Z}$ $\mathbb{Z}$
Field generated by values used (characteristic zero) $\mathbb{Q}(\alpha,\beta)$ $\mathbb{Q}(i)$ or $\mathbb{Q}[t]/(t^2 + 1)$ $\mathbb{Q}(\sqrt{2}i)$ or $\mathbb{Q}[t]/(t^2 + 2)$ $\mathbb{Q}[t]/(t^2 + t + 1)$, same as $\mathbb{Q}[t]/(t^2 + 3)$ or $\mathbb{Q}(\sqrt{-3})$ -- -- $\mathbb{Q}$ $\mathbb{Q}$

## Frobenius-Schur indicator

For background reference, see indicator theorem, indicator character, and Frobenius-Schur indicator

The Frobenius-Schur indicator of a representation is the inner product of the character of the representation and the indicator character (which is the character that assigns to every element is number of square roots). The Frobenius-Schur indicator can be computed as $\sum \chi(g^2)$ where $\chi$ is the character of the representation.

For this faithful irreducible representation of the quaternion group, the Frobenius-Schur indicator can be computed as follows:

Element $g$ Element $g^2$ $\chi(g^2)$
1 1 2
-1 1 2
$i$ -1 -2
$-i$ -1 -2
$j$ -1 -2
$-j$ -1 -2
$k$ -1 -2
$-k$ -1 -2
Total -- -8

The inner product is $-8/8$ which is $-1$. Thus, by the indicator theorem, the representation is a quaternionic representation, i.e., it has a real-valued character but cannot be realized over the real numbers. In this case, the representation in fact has rational character value and cannot be realized over the real numbers.

## Realizability information

We see from the above that the representation can be realized over any field where $-1$ is a sum of two squares. Are there other fields over which it can be realized?

### Finite fields

Since every element of a finite field is expressible as a sum of two squares, we can always, for a finite field, find $\alpha, \beta$ such that $\alpha^2 + \beta^2 = -1$, so this representation can be realized over any finite field. See below:

Field size $q$ mod 8 Is -1 a square (so representation realized using square root of -1)? Is -2 a square (so representation realized using square root of -2)? Is -1 a sum of two squares?
1 Yes Yes Yes
3 No Yes Yes
5 Yes No Yes
7 No No Yes

### Fields over which it cannot be realized

The representation cannot be realized over any formally real field, and in particular over any subfield of the field of real numbers, even though the character of the representation can be realized over such a field. The Schur index of the representation in characteristic zero is 2, which means we need to take a suitable quadratic extension of such a field (for instance, by adjoining the square root of $-1$ or $-2$ or $-1-m^2$ for any $m$ in the field), in order to realize the representation.

Over a formally real field, there is a four-dimensional irreducible representation that is not absolutely irreducible, and that, over a quadratic extension of the sort described above, splits into two copies of the two-dimensional irreducible representation discussed on the current page. For more on the four-dimensional irreducible (but not absolutely irreducible) representation, see four-dimensional irreducible representation of quaternion group.

## Interpretations

### Interpretation as Hamiltonian quaternions

The two-dimensional representation of the quaternion group can be described in a number of explicit ways. One such way is by viewing the Hamiltonian quaternions as a two-dimensional right vector space over the complex numbers, and viewing the actions of the elements of the quaternion group on this vector space by left multiplication.

The specific matrices for the representation depend on how we think of the Hamiltonians as a right vector space over the complex numbers. The typical way is to identify $\mathbb{C}$ as the subspace spanned by 1 and $i$, and take a basis as $1$ and $j$ for the vector space. Then, we have:

$1 = (1,0); i = (i,0); j = (0,1); k = (0,-i)$

We can now compute the action of the elements $\pm 1, \pm i, \pm j, \pm k$ by left multiplication on this vector space, and write the matrices.

## In characteristic two

The general description of the representation can be applied to give a representation in characteristic two, by finding elements $\alpha,\beta$ in a field of characteristic two such that $\alpha^2 + \beta^2 + 1 = 0$ and constructing the matrices. However, there are some crucial differences:

Field of characteristic not two case Explanation Field of characteristic two case Explanation You were wondering ...
All representations are equivalent as linear representations, regardless of the choice of $\alpha$ and $\beta$ We see from the table that the representations all have the same character, and character determines representation when the characteristic of the field does not divide the order of the group. So, any two such representations are equivalent. The representations need not be equivalent. For instance, consider the representation obtained by setting $\alpha = 1, \beta = 0$. This has kernel $\langle i \rangle$. On the other hand, the representation obtained by setting $\alpha = 0, \beta = 1$ has kernel $\langle k \rangle$. Why doesn't the conjugating matrix that establishes the equivalence of representations in characteristic zero reduce to a matrix in characteristic two that establishes the equivalence of representations? The key is that the matrix (and its inverse) establishing equivalence of representations have entries in $\mathbb{Q}(\alpha,\beta)$ and not all in $\mathbb{Z}[\alpha,\beta]$, and it cannot be reduced modulo 2.
All representations are faithful. None of the elements other than the identity maps to the identity matrix, regardless of the choice of $\alpha, \beta$. None of the representations are faithful. Since $1 = -1$ in characteristic two, the elements $\{ 1, -1 \}$ of the quaternion group (notation confusing here) both map to the identity matrix. Hence, the kernel contains the center of quaternion group and the representation is not faithful. As the examples $\alpha = 1, \beta = 0$ and $\alpha = 0, \beta = 1$ illustrate, the kernel may in fact be strictly bigger and equal one of the cyclic maximal subgroups of quaternion group. There are examples (realized over field:F4) where the kernel is precisely the center of quaternion group.
All representations are irreducible. Can be verified by computing the character norm, or by noting that there is no invariant one-dimensional subspace. All representations are indecomposable but not irreducible. Follows from general facts about action on $p$-groups: the fixed-point subspace must be nonzero and is hence an invariant one-dimensional subspace.

## Character

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

### Character values and interpretations

The character can be computed using any of the interpretations provided. See below:

Conjugacy class Size of conjugacy class Character value Size-degree-weighted character value Philosophical justification
$\{ 1 \}$ 1 2 1 Must equal the degree.
$\{ -1 \}$ 1 -2 -1 This is the only irreducible representation that separates the two central elements $\{ 1, -1 \}$ in the quaternion group. Thus, $-1$ is forced to map to a central element not equal to 1, and which squares to 1, so the only choice is the scalar $2 \times 2$ matrix with $-1$s on the diagonal.
$\{ i,-i \}$ 2 0 0 Galois interpretation, automorphism interpretation, induction interpretation
$\{ j,-j \}$ 2 0 0 Galois interpretation, automorphism interpretation, induction interpretation
$\{ k,-k\}$ 2 0 0 Galois interpretation, automorphism interpretation, induction interpretation
Total 8 -- 0 (as expected, since the character is orthogonal to the trivial character)

## Embeddings

### Embeddings in general linear groups and special linear groups

As discussed above, this representation can always be realized over a finite field of characteristic not equal to two. Moreover, for any finite field of characteristic not equal to two, the representation is faithful and gives an embedding of the quaternion group in the general linear group of degree two over the field. In fact, the embedding goes inside the special linear group of degree two, because all matrices have determinant 1. Further, since the representation is unique up to equivalence of representations, and it is the only faithful representation of degree two, this forces the subgroup to be an isomorph-conjugate subgroup in the general linear group (though it's unclear whether the conjugation can be done within the special linear group as well).

We look at some particular finite fields:

Field size Field information General linear group of degree two Special linear group of degree two Embedding in general linear group Embedding in special linear group
3 field:F3 general linear group:GL(2,3) special linear group:SL(2,3) Q8 in GL(2,3) Q8 in SL(2,3)
5 field:F5 general linear group:GL(2,5) special linear group:SL(2,5) Q8 in GL(2,5) Q8 in SL(2,5)

## Relation with supergroups

### Representations of supergroups that restrict to this representation on the quaternion group

Note that by Frobenius reciprocity, it also turns out that these representations are contained in the induced representation from the quaternion group.

Group Order Quaternion group as subgroup of this group Quaternionic representation(s) (if any) that restricts to the representation on the quaternion group Other representations that restrict to the representation on quaternion group
semidihedral group:SD16 16 Q8 in SD16 faithful irreducible representation of semidihedral group:SD16 (two representations)
generalized quaternion group:Q16 16 Q8 in Q16 faithful irreducible representation of generalized quaternion group:Q16 (two representations)
special linear group:SL(2,3) 24 Q8 in SL(2,3) quaternionic representation of special linear group:SL(2,3) two other representations