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✓ 4 results · all verified · 2 also independently AI-judged

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Schur Indices and Fields of Definition — Examples

1 · Prerequisites

2 · Summary

These examples separate the character field from a field of definition. The cyclic and symmetric-group cases have rational models, while the faithful quaternion character is rational-valued but needs multiplicity two over the rationals.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-09-07Open item page →

The two nontrivial characters of $C_{3}$ form one rational representation

Example

For $C_{3} = ⟨ g ∣ g^{3} = 1 ⟩$ and $ζ = e^{2 π i / 3}$ , the characters $χ (g) = ζ$ and $\overline{χ} (g) = ζ^{2}$ are Galois conjugate. Their sum is afforded over $Q$ by the action of $g$ on $Q^{2}$ with matrix $(\begin{array}{cc} 0 & - 1 \\ 1 & - 1 \end{array}) .$ Thus the rational Galois orbit occurs with multiplicity one; in particular the Schur index of either character over its character field $Q (ζ)$ is one.

Facts & Assumptions

Given: $C_{3}$ , $g$ , and $ζ$ as displayed.

[L1]

Over a nonsplitting field, an irreducible character orbit occurs with its Schur-index multiplicity (Character formula over a nonsplitting field).

Verification

technique · computation

1.1

The displayed matrix has characteristic polynomial $x^{2} + x + 1$ , hence eigenvalues $ζ, ζ^{2}$ , and its cube is the identity. It therefore gives a rational $C_{3}$ -representation with complex character $χ + \overline{χ}$ .

L1algebra

2.1

The orbit appears once, so [L1] identifies its scalar-extension multiplicity over $Q$ as $1$ . Since the one-dimensional character itself is defined over $Q (ζ)$ , its Schur index over its character field is also $1$ .

L1step 1.1algebra∎

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-09-07Open item page →

$S_{3}$ is split over the rationals

Example

The three irreducible complex representations of $S_{3}$ are already defined over $Q$ : the trivial representation, the sign representation, and the two-dimensional standard representation on $W = {(x_{1}, x_{2}, x_{3}) \in Q^{3} : x_{1} + x_{2} + x_{3} = 0}$ by permutation of coordinates. Hence $Q$ is a splitting field for $S_{3}$ .

Facts & Assumptions

Given: The natural coordinate-permutation action of $S_{3}$ on $Q^{3}$ .

[L1]

The number of irreducible complex representations of $S_{3}$ is its number of conjugacy classes, namely three (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , the number of irreducible representations of $G$ equals the number of conjugacy classes).

[L2]

The trivial and sign representations are defined over every field of characteristic zero (The trivial representation, the regular representation, and permutation representations from finite $G$ -sets, The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup).

Verification

technique · computation

1.1

The line $Q (1, 1, 1)$ is trivial and its invariant complement $W$ has dimension two. A transposition has trace $0$ on $W$ , while a $3$ -cycle has trace $- 1$ ; thus $W$ is neither trivial nor sign.

L2algebra

2.1

The three displayed rational models have distinct complex characters, and [L1] says there are no further irreducibles. Therefore every complex irreducible has a rational model, which is exactly that $Q$ is splitting for $S_{3}$ .

L1step 1.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The faithful quaternion character has Schur index two

Example

For $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ , its faithful complex irreducible character has values $χ (1) = 2$ , $χ (- 1) = - 2$ , and $χ (\pm i) = χ (\pm j) = χ (\pm k) = 0$ . It is rational-valued, but $m_{Q} (χ) = 2$ .

Facts & Assumptions

Given: $Q_{8}$ with generators $i, j$ satisfying $i^{2} = j^{2} = - 1$ and $i j = - j i$ .

[L1]

The Schur index is the common scalar-extension multiplicity of the complex constituents of an irreducible representation over the character field (The Schur index of an irreducible character).

Verification

technique · computation

1.1

In the usual complex model, $i$ and $j$ have eigenvalues $i, - i$ , so their traces, and those of their negatives, are $0$ ; $- 1$ acts as $- I_{2}$ . Hence the displayed character is rational-valued.

algebra

1.2

Let $H_{Q} = Q + Q i + Q j + Q k$ and let $Q_{8}$ act on it by left multiplication. The norm $q \overset{ˉ}{q} = a^{2} + b^{2} + c^{2} + d^{2}$ is nonzero for every nonzero rational quaternion, so $H_{Q}$ is a division algebra. A $Q_{8}$ -stable rational subspace is therefore a left ideal (the elements of $Q_{8}$ span $H_{Q}$ ), and this four-dimensional rational representation is irreducible.

algebra

2.1

The trace of left multiplication is $4$ at $1$ , $- 4$ at $- 1$ , and $0$ at the other six elements. Thus the complexification of the irreducible rational representation in step 1.2 has character $2 χ$ (equivalently, $H_{Q} \otimes_{Q} C ≅ M_{2} (C)$ is two copies of the natural module under left multiplication). By [L1], its common scalar-extension multiplicity is $m_{Q} (χ) = 2$ .

L1step 1.1step 1.2algebra∎

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The trivial character has Schur index one

Example

For every finite group $G$ , the trivial character $1_{G}$ has character field $Q$ and Schur index $m_{Q} (1_{G}) = 1$ .

Facts & Assumptions

Given: A finite group $G$ and its one-dimensional trivial complex representation.

[L1]

The Schur index is the least positive multiplicity with a model over the character field (Schur index as minimal realization multiplicity).

Verification

technique · direct

1.1

Every value of $1_{G}$ is $1$ , so its character field is $Q$ . The one-dimensional representation on $Q$ in which every element acts as $1$ is a $Q$ -model.

L1algebra

2.1

Thus multiplicity $1$ is attainable, and [L1] makes the Schur index $1$ .

L1step 1.1∎

Schur Indices and Fields of Definition — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The two nontrivial characters of C3C3​ form one rational representation

Example

Facts & Assumptions

Verification

S3S3​ is split over the rationals

Example

Facts & Assumptions

Verification

The faithful quaternion character has Schur index two

Example

Facts & Assumptions

Verification

The trivial character has Schur index one

Example

Facts & Assumptions

Verification

Sources

The two nontrivial characters of $C_{3}$ form one rational representation

$S_{3}$ is split over the rationals