Galois Orbits and Descent of Simple Finite-Group Modules: Examples

A rational root $r$ of $p$ would satisfy $(r+1\mathrm{/}2{)}^2+3\mathrm{/}4=0$, impossible in the ordered field $\mathbb{Q}$. A reducible quadratic over a field has a linear factor and thus a root, so $p$ is irreducible and $(1,\zeta )$ is a rational basis of $E$. We have ${\zeta }^3=1$ and $\zeta \mathrm{\ne }1$; the two roots are $\zeta ,{\zeta }^2$, and they are distinct since equality would force $\zeta =1$. Both lie in $E$. An embedding of this quadratic field is determined by a root, so the identity and $\zeta \mapsto {\zeta }^2$ are its two automorphisms. Equivalently it is a finite normal separable extension, hence Galois, with this two-element group.

Multiplication sends $1\mapsto \zeta$ and $\zeta \mapsto -1-\zeta$, giving $M$. Direct multiplication gives $M^2=\left(\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right)$, $M^2+M+I=0$ and $M^3=I$. Thus it defines a $C_3$-action. For $0\mathrm{\ne }x\in S$, the vectors $x,\zeta x$ are rationally independent, because $(a+b\zeta )x=0$ in the field implies $a=b=0$. Any nonzero invariant rational subspace contains such $x$ and $\zeta x$, and hence equals $S$. This proves simplicity.

Put $t=1+g+g^2$. Since every group element occurs three times in its square, $t^2=3t$. Thus $e=1-t\mathrm{/}3$ is central and $e^2=e$. Also $te=0$, whence $e+ge+g^{2}e=0$. The two coefficient vectors $e=(2-g-g^2)\mathrm{/}3$ and $ge=(-1+2g-g^2)\mathrm{/}3$ are independent: $ae+bge=0$ gives $2a-b=-a+2b=0$, so $a=b=0$. They span the ideal by the relation just found. Evaluation $g\mapsto \zeta$ sends $e\mapsto 1$ and $ge\mapsto \zeta$, so restricts to an algebra isomorphism $\mathbb{Q}[C_3]e\to E$, preserving the block unit. The full evaluation map has kernel $\mathbb{Q}t$: if $a+b\zeta +c{\zeta }^2=0$, then $(a-c)+(b-c)\zeta =0$, so $a=b=c$.

For either $\lambda =\zeta$ or ${\zeta }^2$, set $v_{\lambda }=(1,-\lambda {)}^{\mathsf{T}}$. Then $M{v}_{\lambda }=(\lambda ,1+\lambda {)}^{\mathsf{T}}=\lambda v_{\lambda }$. The determinant of $(v_{\zeta },v_{{\zeta }^2})$ is $\zeta -{\zeta }^2\mathrm{\ne }0$, so these form an $E$-basis of $E{\otimes }_{\mathbb{Q}}S$. Consequently both eigenline modules occur exactly once. Conjugating coefficients interchanges the two eigenvectors and their distinct eigenvalues; the lines give nonisomorphic one-dimensional modules since an intertwiner between them would force $\zeta ={\zeta }^2$.

To check the splitting hypothesis for the whole algebra, evaluation at $1,\zeta ,{\zeta }^2$ gives $E[C_3]\to E^3$. Its inverse sends the $\lambda$th coordinate vector to ${\mathrm{\ell }}_{\lambda }(g)$, where ${\mathrm{\ell }}_{\lambda }(X)={\prod }_{\mu \mathrm{\ne }\lambda }(X-\mu )\mathrm{/}(\lambda -\mu )$ and the product ranges over the other two roots. All denominators are nonzero, and ${\mathrm{\ell }}_{\lambda }(\mu )={\delta }_{\lambda \mu }$ proves the inverse identities on evaluations; a degree at most two polynomial vanishing at three distinct roots is zero, by successive division by $X-\mu$. Thus the algebra is split. The characteristic-zero specialization of F1 now identifies the orbit in step 3.1 with the simple rational module in step 2.1, with multiplicity $1$ as computed.

Finally $\mathrm{tr}(I)=2$, $\mathrm{tr}(M)=-1$ and $\mathrm{tr}(M^2)=-1$. On either eigenline the traces are its scalar values; adding gives $\zeta +{\zeta }^2=-1$ at $g$ and ${\zeta }^2+{\zeta }^4=-1$ at $g^2$. This checks the character and the identity value directly. [F2, step 1.1, step 2.1, step 3.1, algebra] QED

The quaternion multiplication gives $z$ central, $u^2=v^2=z$, and $vu=zuv$. Thus $e^2=(1-2z+z^2)\mathrm{/}4=e$ and $ze=-e$. The four elements $e,ue,ve,uve$ span $D$, since $(zh)e=-he$ for $h=1,u,v,uv$. They are rationally independent: $he=(h-zh)\mathrm{/}2$ has its two nonzero coefficients on one of four disjoint pairs of the eight group-basis elements. Hence they form a basis. Write $x=ue$, $y=ve$, $w=uve$; then $x^2=y^2=w^2=-e$, $xy=w=-yx$, $yw=x=-wy$, and $wx=y=-xw$.

For $q=ae+bx+cy+dw$, put $\bar{q}=ae-bx-cy-dw$. Using the multiplication table, all mixed terms cancel and $q\bar{q}=\bar{q}q=(a^2+b^2+c^2+d^2)e$. For rational coefficients this sum is positive when $q\mathrm{\ne }0$. Thus $q^{-1}=\bar{q}\mathrm{/}(a^2+b^2+c^2+d^2)$ exists on both sides, and $D$ is a division algebra. Every nonzero left ideal contains an invertible element and hence $e$, so its left regular module is simple. F2 for a single $1\times 1$ factor says this is its unique simple class. The zero quaternion needs no inverse; the unit is $e$.

Set $U=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$ and $V=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. We have $U^2=V^2=-I$ and $VU=-UV$, so the multiplication table defines an $E$-algebra map $E{\otimes }_{\mathbb{Q}}D\to M_2(E)$ sending $e,x,y,w$ to $I,U,V,UV$. A linear combination of these images is $\left(\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right)$. Every matrix occurs uniquely: its entries $r,s,t,h$ give $a=(r+h)\mathrm{/}2$, $b=(r-h)\mathrm{/}(2i)$, $c=(s-t)\mathrm{/}2$, $d=(s+t)\mathrm{/}(2i)$. Thus the map is an algebra isomorphism.

Let $f=1-e=(1+z)\mathrm{/}2$. The complementary block has basis $f,uf,vf,uvf$ by the same disjoint-pair argument. Here $zf=f$, so $uf,vf$ commute and square to $f$. Evaluating them independently at $ϵ,\eta \in \{1,-1\}$ gives a map $\mathbb{Q}[Q_8]f\to {\mathbb{Q}}^4$. Its sign matrix has rows $(1,ϵ,\eta ,ϵ\eta )$; the inner product of a row with itself is $4$, and with a different row is $(1+ϵ{ϵ}^{\mathrm{\prime }})(1+\eta {\eta }^{\mathrm{\prime }})=0$. Its inverse is one quarter its transpose. Hence $\mathbb{Q}[Q_8]\cong D\times {\mathbb{Q}}^4$ and, extending the displayed basis maps, $E[Q_8]\cong M_2(E)\times E^4$. The rational regular module is the direct sum of $D$ and four copies of $\mathbb{Q}$, each simple over its factor; the extended regular module is the sum of the two simple columns of $M_2(E)$ and four one-dimensional factors. Thus both regular modules are semisimple, and the displayed extended product is split, without an omitted block. The column simples over this product have only scalar endomorphisms (commute with matrix units), so it also satisfies the group splitting-field convention.

The subspaces of matrices supported in the first column and in the second column are left ideals, each isomorphic to $W=E^2$ by reading that column. They have zero intersection and their sum is all of $M_2(E)$. F2 makes $W$ simple. Consequently $E{\otimes }_{\mathbb{Q}}D\cong W\oplus W$ as group modules, with multiplicity exactly two: its $E$-dimension is four and ${\dim }_{E}W=2$.

The polynomial $X^2+1$ has no rational root, and its distinct roots $i,-i$ lie in $E$. Thus $E\mathrm{/}\mathbb{Q}$ is finite normal separable, with nontrivial automorphism $\sigma (i)=-i$. Coefficient conjugation sends $U$ to $-U$ and fixes $V$. Since $VU{V}^{-1}=-U$ and $VV{V}^{-1}=V$, conjugation by $V$ intertwines the representation with its coefficient-conjugate; these equations on the generators suffice on every group element. Therefore $[W]$ is Galois-stable. The hypotheses for F1 are all met by step 3.1 and this finite Galois extension. F1 identifies $D$ with this singleton orbit, and step 3.2 computes its common multiplicity as two.

The matrices for $1,z,u,v,uv$ are $I,-I,U,V,UV$. Their traces are $2,-2,0,0,0$ respectively; multiplying the last three by $-I$ keeps their traces zero. The class list follows directly from the relations: conjugation preserves each pair $\{h,zh\}$ for $h=u,v,uv$, and conjugation by a different generator exchanges its two members, whereas $1,z$ are central. Thus this list exhausts all eight elements and gives exactly the stated class values. All values lie in $\mathbb{Q}$. The trace of the rational regular module $D$ is twice this character after extension, since its extension is the displayed two column copies. [F3, F4, F5, step 2.2, step 3.2, algebra] QED

F2 verifies every hypothesis: characteristic zero, finite Galois extension, whole-algebra splitting, simplicity and Galois stability. It also supplies a rational realization of $2\chi$, 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 $\chi$ itself.

Any rational representation with character $\chi$ has dimension $\chi (1)=2$. Let $Z$ represent $z$, so $Z^2=I$ and $\mathrm{tr}(Z)=-2$. Every vector decomposes as $(x+Zx)\mathrm{/}2+(x-Zx)\mathrm{/}2$, 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 $R,T\in M_2(\mathbb{Q})$ would satisfy $R^2=T^2=-I$ and $RT=-TR$, by the relations in F2.

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

Consequently $T^2={\left(\begin{array}{cc}a& b\\ b& -a\end{array}\right)}^2=(a^2+b^2)I$. It cannot equal $-I$, since $a^2+b^2\ge 0$ in $\mathbb{Q}$. This excludes every rational degree-two representation of character $\chi$, and hence every rational model whose scalar extension is $W$ (scalar extension preserves its matrices' traces). Since step 1.1 realizes $2\chi$ 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