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Galois conjugates have equal scalar-extension multiplicity

Statement

Let $E / F$ be finite Galois of characteristic $0$ , let $G$ be finite, and let $V$ be a finite-dimensional $F$ -representation. If $U$ is an irreducible constituent of $E \otimes_{F} V$ , then every $^{σ} U$ is a constituent with the same multiplicity as $U$ .

Facts & Assumptions

Given: $E / F$ , $G$ , $V$ , and $U$ as in the statement.

[L1]

In characteristic not dividing $∣ G ∣$ , finite-dimensional representations of $G$ are completely reducible (If $char k ∤ ∣ G ∣$ , every finite-dimensional representation of $G$ is completely reducible).

[L2]

Galois conjugation applies an automorphism entrywise and preserves equivalence (Galois conjugates of a representation).

Proof

technique · direct

1.1

By [L1], write $E \otimes_{F} V ≅ ⨁_{X} m_{X} X$ as a direct sum over its irreducible constituents.

L1given

2.1

Apply $σ$ entrywise to this decomposition. Since the matrices of $V$ have entries in $F$ , [L2] identifies the conjugate of the left side with itself, while the right side becomes $⨁_{X} m_{X}^{σ} X$ .

L2step 1.1

3.1

Uniqueness of multiplicities in a completely reducible decomposition now gives $m_{^{σ} U} = m_{U}$ , as required.

L1step 2.1∎

Depends on

Used by

Scalar extension of an irreducible finite-group representation Theorem

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gabor Wiese, Galois Representations, Lemma 2.2.9 (standard reference, not scraped)
Weizhe Zheng, Lectures on Algebra, Proposition 4.3.2 (standard reference, not scraped)