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A Galois-stable character need not descend with multiplicity one

Statement refuted

If $E / F$ is finite Galois and splits a finite group's algebra in characteristic zero, then every Galois-stable simple $E$ -representation descends to an $F$ -representation with multiplicity one. In particular, the asserted conclusion would make every such $F$ -valued irreducible character realizable over $F$ .

Facts & Assumptions

[F1]

The orbit theorem allows a common positive multiplicity, not necessarily one: Galois orbits classify simple modules after splitting base change.

[F2]

For $F = Q$ , $E = Q (i)$ and $Q_{8}$ , the simple $E$ -module $W$ has rational-valued character $χ = (2, - 2, 0, 0, 0)$ on the five classes, is Galois-stable, and the rational module $D$ satisfies $E \otimes_{Q} D ≅ W \oplus W$ . The extension is finite Galois and splits the whole group algebra: The rational simple quaternion block.

Counterexample

Given: The module $W$ in F2, with matrices $U = diag (i, - i)$ and $V = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ for the generators $u, v$ .

1.1

F2 verifies every hypothesis: characteristic zero, finite Galois extension, whole-algebra splitting, simplicity and Galois stability. It also supplies a rational realization of $2 χ$ , namely the four-dimensional simple rational module $D$ . F1 identifies its descent multiplicity with two, in agreement with the explicit column decomposition. It remains to rule out a rational representation of character $χ$ itself.

F1F2given

1.2

Any rational representation with character $χ$ has dimension $χ (1) = 2$ . Let $Z$ represent $z$ , so $Z^{2} = I$ and $tr (Z) = - 2$ . Every vector decomposes as $(x + Z x) / 2 + (x - Z x) / 2$ , in the $+ 1$ and $- 1$ eigenspaces respectively, whose intersection is zero. If their dimensions are $r, s$ , then $r + s = 2$ and $r - s = - 2$ , giving $r = 0$ , $s = 2$ . Thus $Z = - I$ . Its generator matrices $R, T \in M_{2} (Q)$ would satisfy $R^{2} = T^{2} = - I$ and $R T = - T R$ , by the relations in F2.

F2algebra

2.1

For any nonzero rational vector $x$ , the vectors $x, R x$ are independent. Otherwise $R x = r x$ with $r \in Q$ , and $R^{2} x = r^{2} x = - x$ would force $r^{2} = - 1$ , impossible. In the basis $(x, R x)$ , $R$ is therefore $J = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ . Write $T = (\begin{array}{cc} a & b \\ c & d \end{array})$ in this same rational basis. The equation $J T + T J = 0$ is $(\begin{array}{cc} b - c & - d - a \\ a + d & b - c \end{array}) = 0$ , so $c = b$ and $d = - a$ .

step 1.2algebra

3.1

Consequently $T^{2} = {(\begin{array}{cc} a & b \\ b & - a \end{array})}^{2} = (a^{2} + b^{2}) I$ . It cannot equal $- I$ , since $a^{2} + b^{2} \geq 0$ in $Q$ . This excludes every rational degree-two representation of character $χ$ , and hence every rational model whose scalar extension is $W$ (scalar extension preserves its matrices' traces). Since step 1.1 realizes $2 χ$ and no positive integer lies strictly between $1$ and $2$ , the least positive rational realization multiplicity is exactly two. The excluded conclusion is multiplicity one, not the common-positive-multiplicity conclusion of F1. [step 1.1, step 1.2, step 2.1, algebra] QED

Remarks

Wiese, Corollary 2.5.10, p.40, discusses the general realizability obstruction; the proof above establishes this witness directly and does not consume a later Schur-index theorem. If “Schur index” is expressed as the minimum degree of a realization field over the character field, the same example also has index two: its character field is $Q$ , degree one is excluded above, and $Q (i)$ is an explicit degree-two realization field.

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Sources

Weizhe Zheng, Lectures on Algebra (10 January 2025) (standard reference, not scraped)
Gábor Wiese, Galois Representations (standard reference, not scraped)