Schur Indices and Fields of Definition — Examples

The displayed matrix has characteristic polynomial $x^2+x+1$, hence eigenvalues $\zeta ,{\zeta }^2$, and its cube is the identity. It therefore gives a rational $C_3$-representation with complex character $\chi +\bar{\chi }$. Since $x^2+x+1$ has no rational root, this representation has no invariant rational line and is irreducible over $\mathbb{Q}$.

The orbit appears once, so [L1] identifies its scalar-extension multiplicity over $\mathbb{Q}$ as $1$. Over $K=\mathbb{Q}(\zeta )$, the maps $g\mapsto \zeta$ and $g\mapsto {\zeta }^2$ each define an irreducible one-dimensional $K$-representation whose complex scalar extension has multiplicity one. These explicit computations use no choice principle. Under AC, [L2] identifies each attached representation and multiplicity with its Schur index over $K$, giving the two asserted values.

The line $\mathbb{Q}(1,1,1)$ is trivial and its invariant complement $W$ has dimension two. With $u=(1,-1,0)$ and $v=(0,1,-1)$ as a basis of $W$, the transposition $s=(12)$ and cycle $r=(123)$ act by $S=\left(\begin{array}{cc}-1& 1\\ 0& 1\end{array}\right)$ and $R=\left(\begin{array}{cc}0& -1\\ 1& -1\end{array}\right)$. In particular $R$ has no eigenvalue $1$ over $\mathbb{C}$.

Every one-dimensional complex representation of $S_3$ is trivial or sign: conjugate transpositions have one common image $\epsilon$ with ${\epsilon }^2=1$, and every $3$-cycle is a product of two transpositions, so acts as $1$. A proper invariant line in $W_{\mathbb{C}}$ would therefore be fixed by $r$, contrary to step 1.1. Thus $W_{\mathbb{C}}$ and $W$ are irreducible.

The two central averages $e_{+}=\frac{1}{6}{\sum }_{g\in S_3}g$ and $e_{-}=\frac{1}{6}{\sum }_{g\in S_3}\mathrm{sgn}(g)g$ act respectively as $(1,0,0)$ and $(0,1,0)$ under the algebra map $\mathrm{\Phi }:\mathbb{Q}[S_3]\to \mathbb{Q}\oplus \mathbb{Q}\oplus {\mathrm{End}}_{\mathbb{Q}}(W)$ defined by the three displayed representations; they kill $W$ because $W$ has neither a trivial nor a sign line. The four matrices $I,S,R,SR$ are linearly independent over $\mathbb{Q}$ (their coordinate determinant is $-3$), so the $W$-projection of $\mathrm{\Phi }$ is all $M_2(\mathbb{Q})$. The central averages separate the scalar factors, making $\mathrm{\Phi }$ surjective; both sides have dimension $6$, hence $\mathrm{\Phi }$ is an isomorphism.

Tensoring step 3.1 with $\mathbb{C}$ gives $\mathbb{C}[S_3]\cong \mathbb{C}\oplus \mathbb{C}\oplus M_2(\mathbb{C})$. By [L1] over both fields, the three displayed models are all irreducibles over $\mathbb{Q}$ and $\mathbb{C}$. Each simple module of $\mathbb{Q}\oplus \mathbb{Q}\oplus M_2(\mathbb{Q})$ has endomorphism ring $\mathbb{Q}$: this is immediate for the scalar factors, while an endomorphism of the column module commuting with every matrix unit is scalar. The splitting-field definition therefore makes $\mathbb{Q}$ a splitting field for $S_3$.

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.

Let ${\mathbb{H}}_{\mathbb{Q}}=\mathbb{Q}+\mathbb{Q}i+\mathbb{Q}j+\mathbb{Q}k$ and let $Q_8$ act on it by left multiplication. The norm $q\bar{q}=a^2+b^2+c^2+d^2$ is nonzero for every nonzero rational quaternion, so ${\mathbb{H}}_{\mathbb{Q}}$ is a division algebra. A $Q_8$-stable rational subspace is therefore a left ideal (the elements of $Q_8$ span ${\mathbb{H}}_{\mathbb{Q}}$), and this four-dimensional rational representation is irreducible.

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 $2\chi$ (equivalently, ${\mathbb{H}}_{\mathbb{Q}}{\otimes }_{\mathbb{Q}}\mathbb{C}\cong M_2(\mathbb{C})$ is two copies of the natural module under left multiplication). This explicit multiplicity computation uses no choice principle. Under AC, [L1] identifies this irreducible rational module with the attached one, so its multiplicity is $m_{\mathbb{Q}}(\chi )=2$.

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

The rational one-dimensional model has scalar-extension multiplicity one by direct inspection, without any choice principle. Under AC, [L1] makes the Schur index $1$.