Clifford Theory over Normal Subgroups

For a simple $N$-submodule $S\subseteq V$, its translate $gS$ is $N$-stable since $n(gs)=g((g^{-1}ng)s)$. The map $s\mapsto gs$ is an isomorphism from ${}^{g}S$ to $gS$, so $gS$ is simple of the conjugate type.

Translating each simple summand in the defining sum gives $g{V}_{\theta }\subseteq V_{{}^g\theta }$. Applying the same argument to $g^{-1}$ gives equality, also when a component is zero.

By complete reducibility applied to $N$ over $\mathbb{C}$, both $V{\mathrm{\mid }}_N$ and $U$ are direct sums of simple modules. Each simple summand of $U$ belongs to the ambient isotypical component of its own type. The ambient directness therefore gives the displayed intersection decomposition; for $U=0$ or $V=0$ it is the zero direct sum.

The sum of the components indexed by the orbit of $\theta$ is nonzero because $V_{\theta }\mathrm{\ne }0$. Translation permutes these components, so their sum is $G$-stable. Irreducibility makes it all of $V$, leaving no other types.

The linear isomorphism $v\mapsto gv$ gives $\dim V_{{}^g\theta }=\dim V_{\theta }$. Conjugate simple modules have the same dimension, since twisting changes only the action. Dividing component dimensions by this common simple dimension proves equality of multiplicities, each positive because its component is nonzero. This also covers a one-element orbit.

Let $V$ afford $\chi$. Its restriction is a direct sum with one common positive multiplicity $e$ for the orbit of $\theta$. The map $gI\mapsto {}^g\theta$ is well defined and bijective: two conjugates agree exactly when the corresponding elements differ on the right by an element of the stabilizer. Additivity of trace now gives the restriction formula.

At the identity every conjugate takes value $\theta (1)$, giving the degree formula. Since $e>0$, precisely $[G:I]$ types occur. Thus an isotypical restriction forces $[G:I]=1$, hence $I=G$; conversely $I=G$ makes the sum a single type. The calculation includes index one and both N=1 and N=G.

If $g\in I_G(\theta )$, then ${}^g\theta =\theta$, so translation gives $g{V}_{\theta }=V_{\theta }$. Thus the inertia group preserves the component.

If $g{V}_{\theta }=V_{\theta }$, translation gives $V_{{}^g\theta }=V_{\theta }\mathrm{\ne }0$. Distinct components have zero intersection, so the two types coincide and $g\in I_G(\theta )$. Restricting the action therefore gives the claimed module. The nonzero hypothesis is essential: the zero subspace has all of $G$ as stabilizer.

The space $W$ is nonzero and $I$-stable. Its inclusion into $V{\mathrm{\mid }}_I$ has a corresponding $G$-map under adjunction. In the stated function model this map is $\mathrm{\Phi }$: replacing $t$ by $ti$ leaves $tif(ti)=tf(t)$ unchanged. For $g\in G$, write $g^{-1}t=t^{\mathrm{\prime }}i$; then $t(gf)(t)=t{i}^{-1}f(t^{\mathrm{\prime }})=g{t}^{\mathrm{\prime }}f(t^{\mathrm{\prime }})$, so reindexing gives $\mathrm{\Phi }(gf)=g\mathrm{\Phi }(f)$.

By transversal evaluation, the functions supported on $tI$ form a copy of $W$, and $\mathrm{\Phi }$ restricts there to the invertible linear map $w\mapsto tw$ onto $tW=V_{{}^t\theta }$. Distinct left cosets give distinct types, and the orbit result says these are all components of $V{\mathrm{\mid }}_N$. Their sum is direct, so $\mathrm{\Phi }$ is bijective before any irreducibility of $W$ is asserted.

Let $U\subseteq W$ be an $I$-submodule. The direct sum ${\sum }_{t\in T}tU$ is $G$-stable: if $gt=t^{\mathrm{\prime }}i$ then $g(tU)=t^{\mathrm{\prime }}U$. For $U\mathrm{\ne }0$ it is nonzero and hence equals $V$. Its dimension is $[G:I]\dim U$, whereas step 2.1 gives $\dim V=[G:I]\dim W$. Therefore $U=W$, proving irreducibility. This argument allows $T=\{1\}$ and $I=N$ without change.

Apply the orbit result inside $I$: every element of $I$ fixes $\theta$, so $W{\mathrm{\mid }}_N$ is a positive number of copies of $\theta$.

Use a left transversal $T$ containing $1$, and write $X={⨁}_{t\in T}{X}_t$, where $X_t$ consists of functions supported on $tI$. For $n\in N$, covariance gives $(nf)(t)=f(n^{-1}t)=f(t(t^{-1}{n}^{-1}t))=(t^{-1}nt)f(t)$. Thus evaluation makes $X_t$ a module of type ${}^t\theta$. These types are distinct for distinct $tI$, so the $X_t$ are exactly the normal isotypical components, and $X_1\cong W$ as an $I$-module.

If $Y\subseteq X$ is a nonzero $G$-submodule, the intersection decomposition gives $Y\cap X_t\mathrm{\ne }0$ for some $t$. Left translation by $t^{-1}$ carries $X_t$ onto $X_1$ and preserves $Y$, so $Y\cap X_1\mathrm{\ne }0$. This intersection is $I$-stable, hence equals $X_1$ by irreducibility of $W$. Translating back fills every $X_t$, giving $Y=X$. This includes $I=G$ (one block) and $I=N$ (induction of the chosen normal type).

Induction sends each element of $\mathrm{Irr}(I\mid \theta )$ to an irreducible lying over $\theta$. Conversely, every irreducible $V$ lying over $\theta$ is induced from its irreducible component $V_{\theta }$. This proves that the displayed map is defined and surjective.

Taking the $\theta$-component of an induced module recovers its inducing module. A $G$-isomorphism preserves these components, since it sends simple $N$-submodules to simple submodules of the same type. Hence isomorphic induced modules have isomorphic inducing modules. This proves injectivity and the specified inverse.

The restriction formula shows that occurrence of one type is equivalent to occurrence of each conjugate, and excludes every type outside that orbit. Every irreducible $G$-module has a nonzero finite-dimensional restriction: an $N$-stable nonzero subspace of least possible dimension is simple, so a constituent exists. Consequently the orbit-indexed sets cover $\mathrm{Irr}(G)$ and are pairwise disjoint. No step requires I to be proper in G or to properly contain N.

Let $S$ afford $\theta$ and let $T$ meet the left cosets of $N$. On the functions supported on $tN$, evaluation satisfies $(nf)(t)=(t^{-1}nt)f(t)$. Therefore restriction of induction has character ${\sum }_{t\in T}{}^t\theta$, by taking traces on this finite direct sum.

Applying the multiplicity formula to a simple module shows that two irreducible characters have inner product one when equal and zero otherwise. Conjugate characters are irreducible. Thus reciprocity gives $⟨\mathrm{Ind}\theta ,\mathrm{Ind}\theta {⟩}_G={\sum }_{t\in T}⟨\theta ,{}^t\theta {⟩}_N$. Exactly the cosets $tN$ in $I_G(\theta )\mathrm{/}N$ contribute one, giving the asserted index.

The induced character is an actual nonzero character. If it is irreducible its norm is one, hence $[I_G(\theta ):N]=1$ and $I_G(\theta )=N$. Conversely that equality makes its norm one, hence it is irreducible. This includes $N=G$, when induction is identity, and $N=1$, when the norm is $\mathrm{\mid }G\mathrm{\mid }$.

Decompose the nonzero induced module into simple $G$-modules by complete reducibility. The coefficient of an irreducible character $\psi$ is $⟨{\mathrm{Ind}}_N^G\theta ,\psi {⟩}_G=⟨\theta ,{\mathrm{Res}}_N^G\psi {⟩}_N$. The latter is the nonnegative integer multiplicity of $\theta$: conjugate symmetry of the inner product does not change that real integer. It is zero exactly outside the lying-over set and equals $e_j$ for $\psi ={\chi }_j$. This proves the first identity and also that the finite list is nonempty.

Applying the multiplicity formula to each simple module itself gives $⟨{\chi }_j,{\chi }_k{⟩}_G={\delta }_{jk}$. Taking the norm of the finite sum in step 1.1 therefore gives ${\sum }_j{e}_j^2$. The normal-induction norm formula identifies this with $[I_G(\theta ):N]$. If the index is one there is exactly one term with e=1; the formula also includes N=1 and N=G.

The map $(s,f)\mapsto f(s)$ is complex bilinear, so the tensor universal property defines evaluation. It is complex linear since this is true on elementary tensors, and it is $N$-linear because $f(ns)=nf(s)$ and $N$ acts trivially on the second factor.

Complete reducibility and the isotypical hypothesis give $U\cong S^{\oplus m}$ for some integer $m\ge 0$. Fix inclusions $j_1,\dots ,j_m:S\to U$ from such a decomposition. Each component of an $N$-map $S\to U$ is a scalar endomorphism of $S$, so the $j_a$ form a basis of $M$. Evaluation sends $s\otimes j_a$ to $j_a(s)$ and is therefore an isomorphism. When $m=0$, both spaces and this map are zero.

If $U_0\subseteq U$ is an $N$-submodule, it is completely reducible. Every simple summand $R$ of $U_0$ is isomorphic to $S$: some coordinate projection $R\to S$ is nonzero, and its kernel and image are submodules, forcing an isomorphism. Apply step 2.1 to $U_0$. Inclusion of its Hom space into $M$ intertwines the two evaluation maps on every elementary tensor, hence gives equality of actual subspaces $U_0=E_U(S\otimes M_0)$, with $M_0={\mathrm{Hom}}_N(S,U_0)$.

For any subspace $M_0\subseteq M$, the space $E_U(S\otimes M_0)$ is $N$-stable. The natural map $M_0\to {\mathrm{Hom}}_N(S,E_U(S\otimes M_0))$, sending $m$ to $(s\mapsto E_U(s\otimes m))$, is an isomorphism by the scalar-coordinate calculation of step 2.1, and agrees with inclusion into $M$. Thus recovery of $M_0$ is exact and unique, including $M_0=0$ and $M_0=M$.

Choose simple decompositions of $U$ and $U^{\mathrm{\prime }}$. An $N$-map between them is a matrix whose entries are endomorphisms of $S$, hence scalars. Those scalar matrices are exactly the linear maps $a:M\to M^{\mathrm{\prime }}$. This proves the map assertion and uniqueness, also if either multiplicity space is zero. Composition satisfies $(1_S\otimes b)(1_S\otimes a)=1_S\otimes (ba)$ on elementary tensors, proving compatibility.

Write $\rho (i)$ for the fixed extension on $S$. For any $\theta$-isotypical $I$-module $U$, put $M={\mathrm{Hom}}_N(S,U)$ and define $(i\cdot f)(s)=i\cdot f(\rho (i{)}^{-1}s)$. For $n\in N$, one has $\rho (i{)}^{-1}\rho (n)=\rho (i^{-1}ni)\rho (i{)}^{-1}$, so $(i\cdot f)(\rho (n)s)=n\cdot (i\cdot f)(s)$. Thus the formula stays in $M$.

The action law holds because $i\cdot (j\cdot f)(s)=ij\cdot f(\rho (j{)}^{-1}\rho (i{)}^{-1}s)=((ij)\cdot f)(s)$, and the identity acts identically. For $n\in N$, $n\cdot f(\rho (n{)}^{-1}s)=f(s)$ by $N$-linearity. Hence $N$ acts trivially on $M$ and this is a representation of $I\mathrm{/}N$.

The evaluation isomorphism is $I$-equivariant for the diagonal action on $S\otimes M$: $E_U(\rho (i)s\otimes (i\cdot f))=i\cdot f(s)$. Every $N$-submodule is $E_U(S\otimes M_0)$ for a unique $M_0\subseteq M$. Since $\rho (i)$ is invertible, its translate is $E_U(S\otimes i{M}_0)$. Uniqueness shows that this submodule is $I$-stable exactly when $M_0$ is stable under $I\mathrm{/}N$. For nonzero $U$, the multiplicity space is nonzero, so $U$ is irreducible if and only if $M$ is irreducible.

Conversely start with a quotient module $M$ and form $\tilde{S}\otimes \mathrm{Inf}M$. The map $m\mapsto f_m$, where $f_m(s)=s\otimes m$, is an isomorphism $M\to {\mathrm{Hom}}_N(S,\tilde{S}\otimes M)$ by the evaluation lemma and scalar-coordinate identification. The action constructed above satisfies $i\cdot f_m=f_{im}$. Therefore this recovers the quotient module, and step 3.1 proves irreducibility for every irreducible parameter. An isomorphism of $I$-modules induces an isomorphism of their Hom spaces by composition, respecting the quotient action; thus distinct quotient parameters cannot give isomorphic $I$-modules.

Every irreducible $I$-module above $\theta$ is $\theta$-isotypical, so steps 1.1–3.1 apply and its evaluation isomorphism supplies the required tensor form. This proves exhaustivity as well as injectivity. Taking tensor-product characters yields the stated character map.

Clifford correspondence now supplies the bijection after induction. Its inverse identifies the $\theta$-component with the inducing tensor module, whose restriction to $N$ is $\dim M=\eta (1)$ copies of $S$. Thus its ramification index is $\eta (1)$. If $I=N$, the quotient is trivial and only $S$ occurs; if $I=G$, induction is identity. An extension was assumed throughout, not obtained merely from invariance.