Schur Indices and Fields of Definition

Choose $F$-bases of $V$ and $W$, and put $H={\mathrm{Hom}}_F(V,W)$. For $g\in G$, let $T_g:H\to H$ send $f$ to $f{\rho }_V(g)-{\rho }_W(g)f$. By [L2], ${\mathrm{Hom}}_G(V,W)={\bigcap }_{g\in G}\ker T_g$. Since $H$ is finite-dimensional, so is ${\mathrm{End}}_F(H)$. Therefore finitely many $T_{g_1},\dots ,T_{g_r}$ span the subspace generated by all the $T_g$, and ${\mathrm{Hom}}_G(V,W)={\bigcap }_{i=1}^r\ker T_{g_i}$.

The chosen bases identify $E{\otimes }_{F}H$ with ${\mathrm{Hom}}_E(E{\otimes }_{F}V,E{\otimes }_{F}W)$ and identify each $E{\otimes }_F{T}_g$ with the corresponding intertwining equation over $E$. Gaussian elimination over $F$ gives a basis of the finite simultaneous kernel in step 1.1 and extends it to a basis of $H$; the same row operations remain valid over $E$. Hence scalar extension preserves this finite kernel. Every $T_g$ is an $F$-linear combination of the selected $T_{g_i}$, so the simultaneous kernel over $E$ for all $g$ equals that finite kernel and, by [L2], is ${\mathrm{Hom}}_G(E{\otimes }_{F}V,E{\otimes }_{F}W)$.

The resulting identification sends $a\otimes f$ to the displayed map, which agrees with [L1]. Therefore it is the asserted $E$-linear isomorphism.

By [L1], write $E{\otimes }_{F}V\cong {⨁}_X{m}_{X}X$ as a direct sum over its irreducible constituents.

Apply $\sigma$ 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 ${⨁}_X{m}_X{}^{\sigma }X$.

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

By [L4], decompose $E{\otimes }_{F}V$ into irreducibles and choose a constituent $U$. The Galois action permutes its isomorphism classes.

Let $X$ be the sum of the isotypic components in the Galois orbit of $U$. The canonical semilinear $\mathrm{\Gamma }=\mathrm{Gal}(E\mathrm{/}F)$-action on $E{\otimes }_{F}V$ preserves $X$. To make descent explicit, choose trace-dual $F$-bases $(x_i)$ and $(y_i)$ of $E$. For $w\in X$, every $T(y_{i}w)={\sum }_{\sigma \in \mathrm{\Gamma }}\sigma (y_{i}w)$ lies in $X^{\mathrm{\Gamma }}$, and the separability identity ${\sum }_i{x}_i\sigma (y_i)={\delta }_{\sigma ,1}$ gives $w={\sum }_i{x}_{i}T(y_{i}w)$. Hence $X=E{\otimes }_F{X}^{\mathrm{\Gamma }}$. Under $(E{\otimes }_{F}V{)}^{\mathrm{\Gamma }}=V$, the $G$-stable space $X^{\mathrm{\Gamma }}$ is an $F$-subrepresentation of $V$; irreducibility therefore forces $X=E{\otimes }_{F}V$.

By [L3], every member of this orbit has one common multiplicity $m$. The orbit is indexed without repetition by $\mathrm{Gal}(E\mathrm{/}F)\mathrm{/}\mathrm{Stab}(U)$, which gives the formula.

Let $\bar{E}$ be an algebraic closure of $E$. For any irreducible constituent $U$, [L1] and [L2] give ${\mathrm{End}}_G(\bar{E}{\otimes }_{E}U)\cong \bar{E}{\otimes }_E{\mathrm{End}}_G(U)=\bar{E}\mathrm{.}$ The scalar extension is semisimple by [L4]; if it were reducible, projection onto a proper summand would be a nonscalar endomorphism. Hence $U$ is absolutely irreducible. Distinct orbit points are inequivalent by the definition of the stabilizer.

For $\sigma \in \mathrm{Gal}(E\mathrm{/}F)$, equivalent $E$-representations have equal characters. Conversely, if $\sigma (\chi (g))=\chi (g)$ for every $g\in G$, [L1] and [L2] make the complex scalar extensions of ${}^{\sigma }U$ and $U$ isomorphic. By [L3] with the extension $\mathbb{C}\mathrm{/}E$, a nonzero complex intertwiner comes from a nonzero $E$-intertwiner. Both $E$-representations are irreducible, so this intertwiner is an isomorphism. Thus $\sigma \in \mathrm{Stab}(U)⟺\sigma (\chi (g))=\chi (g)$ for every $g\in G$.

The right side says exactly that $\sigma$ fixes the field generated by $F$ and all character values, namely $F(\chi )$.

Thus $\mathrm{Stab}(U)=\mathrm{Gal}(E\mathrm{/}F(\chi ))$; [L4] gives $E^{\mathrm{Stab}(U)}=F(\chi )$.

If $V$ is absolutely irreducible, then $\bar{F}{\otimes }_{F}V$ is irreducible for an algebraic closure supplied by [L2] under [A1]. By [L3] its endomorphism algebra is $\bar{F}$, and [L1] gives $\bar{F}{\otimes }_F{D}_V\cong \bar{F}$. Since $D_V\subseteq {\mathrm{End}}_F(V)$ is finite-dimensional, comparison of $\bar{F}$-dimensions gives ${\dim }_F{D}_V=1$. The scalar copy of $F$ in $D_V$ is therefore all of $D_V$.

Conversely assume $D_V=F$, and let $E\mathrm{/}F$ be any field extension. By [L1], ${\mathrm{End}}_G(E{\otimes }_{F}V)\cong E{\otimes }_F{D}_V\cong E$, so every $G$-endomorphism is scalar. By [L4], $E{\otimes }_{F}V$ is completely reducible.

If $E{\otimes }_{F}V$ were reducible, complete reducibility would split it into two nonzero $G$-subrepresentations. Projection onto one would be a nonzero, nonidentity idempotent $G$-endomorphism and therefore could not be scalar over the field $E$, contradicting step 1.2. Thus $E{\otimes }_{F}V$ is irreducible for every extension $E\mathrm{/}F$, which is absolute irreducibility.

Under AC, the attached $V$ exists by The Schur index of an irreducible character. Write $E{\otimes }_{K}V$ as $m$ times its orbit of pairwise inequivalent absolutely irreducible constituents, using [L2].

Each constituent remains irreducible after extension from $E$ to the splitting field $L$, and distinct constituents remain distinct; therefore [L1] writes $L{\otimes }_{K}V$ with the same coefficient $m$.

Applying [L2] directly over $L$ identifies its common multiplicity with that coefficient. Hence both choices give $m$, proving independence.

By [L1] and [L2], taking characters of the scalar-extension decomposition gives the displayed character identity.

Evaluating that identity at $1_G$ gives ${\dim }_{F}V=m[\mathrm{Gal}(E\mathrm{/}F):H]\chi (1)$.

By [L3] and the finite Galois correspondence, the index $[\mathrm{Gal}(E\mathrm{/}F):H]$ equals $[F(\chi ):F]$. Substitute this into step 2.1.

Under AC the attached $V$ and $m_K(\chi )$ exist by the definition. Since $K(\chi )=K$, [L1] gives ${\dim }_{K}V=m_K(\chi )\chi (1)$. The integer $\chi (1)$ is positive, so this expresses ${\dim }_{K}V$ as an integer multiple of $m_K(\chi )$.

Hence $m_K(\chi )\mid {\dim }_{K}V$.

Under the stated AC premise, the attached $V$ and $m_K(\chi )$ exist. By [L2] and [L4], ${\mathrm{End}}_G(E{\otimes }_{K}V)\cong {\mathrm{End}}_G(U^{\oplus m_K(\chi )})\cong M_{m_K(\chi )}(E)$. Thus [L1] gives $E{\otimes }_K{D}_V\cong M_{m_K(\chi )}(E)$.

If $z\in Z(D_V)$, its image in the matrix algebra of step 1.1 commutes with every matrix, so it is $\lambda I$ for some $\lambda \in E$. Because $1\otimes z$ is fixed by $\mathrm{Gal}(E\mathrm{/}K)$, so is $\lambda$; hence $\lambda \in K$. Since $K$ already acts by scalar endomorphisms, $Z(D_V)=K$.

Taking $E$-dimensions in step 1.1 gives ${\dim }_K{D}_V=m_K(\chi {)}^2$. With step 2.1, [L3] therefore gives $\mathrm{ind}(D_V)=m_K(\chi )$.

Under the stated AC premise, let $V$ be the irreducible $K$-representation from the Schur-index definition. By [L1], its scalar extension has character $m_K(\chi )\chi$, so $m_K(\chi )\chi$ is afforded over $K$.

Conversely, if $r\chi$ is afforded by a $K$-representation $W$, [L4] decomposes $W$ into irreducible $K$-summands, while [L5] identifies its complex scalar extension with $U^{\oplus r}$. By [L3], every summand that contributes $U$ is isomorphic to $V$, and by [L1] each copy contributes $U$ with multiplicity $m_K(\chi )$. Hence $m_K(\chi )\mid r$.

Step 1.1 attains $r=m_K(\chi )$ and step 1.2 excludes every smaller positive $r$, so this is the least such multiplicity. With $r=1$, [L2] gives exactly the stated realizability criterion.