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For an $H$-representation $V$ and $g\in G$, the conjugate representation ${}^{g}V$ of $gH{g}^{-1}$ is defined by $(gh{g}^{-1})v=hv$. This transports the acting subgroup; it is distinct from conjugating complex matrix entries. It occurs in the Mackey restriction formula.

Let $N_G(D)\le H\le G$ and let $U$ be indecomposable with vertex $D$ over $H$. The Mackey restriction formula writes ${\mathrm{Res}}_H^G{\mathrm{Ind}}_H^{G}U=U\oplus T$, where the vertices in $T$ lie in $H\cap {}^{g}D$ for $g\notin H$. None of these subgroups contains an $H$-conjugate of $D$: equality would put $g$ in $N_G(D)H=H$. Every summand of the induction with a $G$-conjugate of vertex $D$ contributes a vertex-$D$ summand on restriction. Krull-Schmidt decomposition therefore forces exactly one such summand, with multiplicity one. Choosing sources and using transitivity of induction proves the inverse correspondence and preservation of sources.

For rational parabolic subgroups $P=LU$ and $Q=MV$, define $\mathcal{S}(M,L)=\{x:M\cap {}^{x}L\text{ contains a maximal algebraic torus of }G\}$. Then ${}^{\ast }{R}_M^G{R}_L^G$ is the sum over $M^F\backslash \mathcal{S}(M,L{)}^F/L^F$ of $R_{M\cap {}^{x}L}^M{}^{\ast }{R}_{M\cap {}^{x}L}^{{}^{x}L}\mathrm{ad}x$, with the intersection parabolics understood. It follows from ordinary finite-group Mackey restriction formula, taking $V^F$-invariants, and the compatible-Levi intersection lemma for parabolics.

Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral modular representation theory convention. If a completion has several local factors, use the factor supporting the modular block. A block of a group algebra is an indecomposable two-sided ideal $B=RGb$, where $b$ is a primitive central idempotent. The identity decomposes as the sum of orthogonal block idempotents, and $M={⨁}_{b}bM$ for every module. Hence an indecomposable module belongs to $B$ exactly when $bM=M$.

A defect group of a block in characteristic $p$ is a maximal p-subgroup $D$ with ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; the trace criterion for defect groups and conjugacy are proved below. For a complete discrete valuation ring, use the corresponding reduced modular block; modular reduction of block idempotents preserves its defect groups. The equivalent bimodule description says that the module vertices of $B$, with $(x,y)$ acting by $a\mapsto xa{y}^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the bimodule action.

Fix a Sylow subgroup $P$ containing $D$. The regular group algebra is the permutation module $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex: a module source at that module vertex is a summand of the module restriction, and the Mackey restriction formula and minimality prevent all the restricted summands from having smaller module vertices. Thus $B{\downarrow }_{P\times P}$ has an indecomposable summand with module vertex $\Delta D$.

For a finite p-group $Q$, the transitive permutation module $k[Q/L]$ is indecomposable: the group algebra of a p-group in characteristic p is local, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its module vertex is $L$, because it is induced from the trivial $L$-module and its Brauer quotient of a module at $L$ has the nonzero fixed coset $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller module vertex. Reduction gives the same statements for permutation lattices. Therefore Krull-Schmidt decomposition identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.

Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ is a p-subgroup, then $QP$ is a p-subgroup for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

Use finite-dimensional $k$-modules or finite $R$-free lattices. We use the basic relative-projectivity facts that Krull-Schmidt decomposition holds with local endomorphism rings, module vertices are conjugate p-subgroups, and a module source $S$ at a module vertex $D$ satisfies $S\mid M{\downarrow }_D$ and $M\mid S{\uparrow }^G$. Module induction is transitive and Mackey restriction formula gives module vertex containment for restrictions and inductions. In particular, a summand induced from $D$ has module vertex conjugate into $D$, and module restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex. The latter follows from the module source conditions and minimality: restricting further to $D$ must still contain its module source, whose module vertex is $D$.

The module vertex-$D$ form of Green correspondence states that for $N_G(D)\le H\le G$ there is a bijection between indecomposable $RG$-modules with module vertex $D$ and indecomposable $RH$-modules with module vertex $D$. Its maps select the unique module vertex-$D$ summands of module restriction and induction, using $H$-conjugacy for restricted module vertices and $G$-conjugacy for induced module vertices. They occur with multiplicity one and have the same module sources. Other induction summands have module vertices conjugate to proper subgroups of $D$. The module restriction remainder is relatively projective forRelative projectivity for a family means being a summand of a finite sum of inductions from its members. We prove this form; no stronger description of the induction remainder is needed for the application.

Let $U$ have module vertex $D$ over $H$. The identity double coset in Mackey restriction formula givesSince $U$ is relatively $D$-projective, a further Mackey decomposition puts every other double coset term in the relatively-$\mathcal{Y}$ family. No member of $\mathcal{Y}$ contains an $H$-conjugate of $D$. Otherwise, conjugation within $H$ would give $D\le H\cap (ha)D(ha{)}^{-1}$ with $h\in H$, $a\notin H$. Equality of orders then makes $ha\in N_G(D)\le H$, a contradiction. Therefore the module restriction above contains exactly one summand with module vertex $H$-conjugate to $D$, namely $U$ once.

Decompose $U{\uparrow }^G$ into indecomposable modules. All their module vertices are conjugate into $D$. Each one with a module vertex $G$-conjugate to $D$ can take $D$ itself as module vertex and contributes a module vertex-$D$ summand on module restriction to $H$. There can thus be at most one such summand and its multiplicity is one. There must be at least one, because smaller module vertices cannot restrict to produce $U$, with module vertex $D$, in the displayed decomposition. Define $g(U)$ to be that summand. Its module restriction contains $U$ once, with remainder a summand of $T$. All other induction summands have strictly smaller module vertices. This is Green correspondence from Mackey multiplicity.

Conversely take $M$ of module vertex $D$ over $G$ with module source $S$. Decompose $S{\uparrow }^H$ into indecomposable modules. Since $M\mid S{\uparrow }^G=(S{\uparrow }^H){\uparrow }^G$, the local endomorphism ring and Krull-Schmidt decomposition give $M\mid U{\uparrow }^G$ for one indecomposable summand $U$ of $S{\uparrow }^H$. Its module vertex is $H$-conjugate to $D$: all module vertices are conjugate into $D$, and a smaller one could not induce $M$. The preceding uniqueness makes $g(U)=M$ and identifies $U$ as the unique module vertex-$D$ summand of $M{\downarrow }_H$. Define $f(M)=U$. This proves $f(g(U))=U$ and $g(f(M))=M$.

If $S$ is a module source of $U$, then $S\mid U{\downarrow }_D\mid M{\downarrow }_D$ and $M\mid U{\uparrow }^G\mid S{\uparrow }^G$. These two conditions and the common module vertex show that $S$ is also a module source of $M$. Thus the correspondence preserves module sources, including trivial source modules, and the proof is complete under the stated coefficient conventions.

For the required expression, a p-local module is a finite sum of inductions from $N_G(Q)$ for nontrivial p-subgroups $Q$. Induct on the order of a module vertex of $M$. Vertex 1 makes $M$ projective, so choose $L_1=L_2=P_1=0$, $P_2=M$.

For module vertex $D\ne 1$, take $U=f(M)$ for $H=N_G(D)$. Its module induction is p-local andwith smaller module vertices for the $X_i$. By induction, writewith $A_i,B_i$ p-local and $P_i,Q_i$ projective. Adding these identities and substituting the induction decomposition givesHence the explicit choiceswork. All sums are finite and module vertex orders strictly decrease, so the induction terminates. If $N_G(D)=G$, induction from that normalizer is still allowed by the definition of p-local.

For the special linear group $G={\mathrm{SL}}_3({\overline{\mathbb{F}}}_q)$, let $N=q-1$, and choose a generator $\chi$ of the character group of ${\mathbb{F}}_q^{\times }$. Every linear character of the diagonal rational maximal torus has a unique expressionThe upper triangular Borel subgroup $B=TU$ is rational. Consequently its Deligne-Lusztig induction is the actual principal series of a finite reductive groupwhere $\tilde{\theta }$ is the inflation of a group representation across $B^F\to T^F$. The geometric reason for this split case is $X(1)=G^F/B^F$: the associated torus covering supplies $\theta$, and the auxiliary unipotent fibres contribute a single even cohomological degree, hence no minus sign.

The rational Bruhat decomposition indexes $B^F\backslash G^F/B^F$ by $W=S_3$. Applying the Mackey restriction formula and Frobenius reciprocity givesEach intersection contains $T^F$, and the two linear characters are trivial on its unipotent part. Its contribution is therefore one exactly when $w$ fixes $\theta$, and zero otherwise. Complete reducibility over $\mathbb{C}$ makes this character inner product the sum of squares of irreducible representation multiplicities. Thus the principal series of a finite reductive group is irreducible exactly when the Weyl group stabilizer is trivial.

To compute that stabilizer, represent $\theta$ by the triple $({\chi }^r,{\chi }^s,1)$. Two triples give the same linear character on determinant-one diagonal matrices exactly when they differ by a common factor $(\nu ,\nu ,\nu )$. For example, triviality of $({\eta }_1,{\eta }_2,{\eta }_3)$ on all such matrices first gives ${\eta }_1={\eta }_3$ by varying $a$, and then ${\eta }_2={\eta }_3$ by varying $b$. The Weyl group permutes the three entries.

A transposition fixes $\theta$ precisely when its two interchanged entries coincide: the third, fixed entry forces the common factor to be one. These three obstructions are $r=0$, $s=0$, and $r=s$. A three-cycle can also fix $\theta$ without any two entries coinciding. Its common factor must satisfy ${\nu }^3=1$; in the nontrivial case the entries, up to order and common factor, are $(1,\nu ,{\nu }^2)$. This happens exactly when $3\mid N$ and $\{r,s\}=\{N/3,2N/3\}$.

The complete condition is thereforeIn particular pairwise distinct coordinate characters alone are insufficient. For $q=4$ the only distinct triples have the order-three symmetry, and no character of this split rational maximal torus gives an irreducible principal series of a finite reductive group.

Let $P=L⋉U$ and $Q=M⋉V$ be rational parabolic subgroups with rational Levi subgroups. Work with complex representations of their finite groups of Lie type. Here Harish-Chandra induction and Harish-Chandra restriction meanThe latter is an $M^F$-module, and it is adjoint to the former by Frobenius reciprocity.

DefineFor $x\in \mathcal{S}(M,L{)}^F$, put $H_x=M\cap {}^{x}L$ and let $\mathrm{ad}x$ transport an $L^F$-module to a $({}^{x}L{)}^F$-module. The Mackey formula for Harish-Chandra induction is the natural functor isomorphismOn characters the direct sum becomes a sum. The intersection parabolics displayed in the subscripts are important: they specify exactly which Harish-Chandra induction and Harish-Chandra restriction occur.

We use the following parabolic-intersection lemmas, stating explicitly the geometric input to the proof. First, representatives of $M^F\backslash \mathcal{S}(M,L{)}^F/L^F$ represent exactly the double cosets $Q^F\backslash G^F/P^F$. Second, for such $x$, the groups $M\cap {}^{x}P$ and $Q\cap {}^{x}L$ are parabolic subgroups of $M$ and ${}^{x}L$, respectively, with common Levi subgroup $H_x$ and respective unipotent radicals $M\cap {}^{x}U$ and $V\cap {}^{x}L$. If $K=Q\cap {}^{x}P$, projection $Q\to M$ maps $K$ onto $M\cap {}^{x}P$ with kernel $K\cap V$. Projection ${}^{x}P\to {}^{x}L$ maps $K\cap V$ onto $V\cap {}^{x}L$ with kernel $V\cap {}^{x}U$. These assertions hold on rational points as well. The surjectivity on rational points uses connectedness of the unipotent kernels and the Lang theorem for algebraic groups. These are the compatible-Levi and rational double-coset lemmas; they follow by applying the root-subgroup decomposition after choosing the common maximal algebraic torus.

For completeness, the representation-theoretic reduction is as follows. The group algebra realization of induced representation splits according to the double cosets:as a $(Q^F,P^F)$-bimodule. Tensoring on the right with an inflated $L^F$-module $E$ identifies each summand with induction from $K^F=(Q\cap {}^{x}P{)}^F$. This proves the finite-group Mackey restriction formula in the formHere ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$.

We next record the elementary normal-subgroup identity underlying Harish-Chandra restriction. If $N◃D$ are finite groups, with $N$ a normal subgroup, and $C\subset D$, then for a complex $C$-module $A$,To see this, replace invariants by coinvariants using the averaging idempotent $|N{|}^{-1}{\sum }_{n\in N}n$. Taking coinvariants in $\mathbb{C}[D]{\otimes }_{\mathbb{C}[C]}A$ first replaces $\mathbb{C}[D]$ by $\mathbb{C}[D/N]$, and forces $C\cap N$ to act trivially on $A$. Thus it yields $\mathbb{C}[D/N]{\otimes }_{\mathbb{C}[CN/N]}{A}_{C\cap N}$; averaging over $C\cap N$ identifies this last module with the one displayed. This also shows why characteristic zero makes the argument exact.

Apply this identity to $D=Q^F$, $N=V^F$, $C=K^F$. By the intersection lemmas, $CN/N=(M\cap {}^{x}P{)}^F$. Since ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$, its $(K\cap V{)}^F$-invariants are exactlyMoreover $(M\cap {}^{x}U{)}^F$ acts trivially on this module, so induction from $(M\cap {}^{x}P{)}^F$ to $M^F$ is precisely Harish-Chandra induction from $H_x$. Substituting into each double-coset summand gives the boxed formula. This proves the formula for rational parabolics; it does not assume a general Mackey formula for nonrational Deligne-Lusztig induction.

For a finite group $G$, subgroups $H,K$, and a complex group representation $V$ of $H$, the Mackey restriction formula isChoose one representative $g$ of each double coset $KgH$. The conjugate subgroup representation ${}^{g}V$ is a representation of $gH{g}^{-1}$ on the same vector space, with $(gh{g}^{-1})v=hv$. Restriction of a representation keeps the vector space and restricts the acting subgroup. induced representation is $\mathbb{C}G{\otimes }_{\mathbb{C}H}V$, and analogously for each subgroup of $K$. The displayed isomorphism is an isomorphism of $K$-representations. These definitions specify every summand and its subgroup.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.

Let $P$ be a Sylow p-subgroup. Since $[G:P]$ is invertible in $k$, every $kG$-module is relatively $P$-projective.

Suppose first that $kP$ has only finitely many indecomposable modules $U_1,\dots ,U_t$. For every indecomposable $kG$-module $M$, decompose ${\mathrm{Res}}_P^{G}M$ into the $U_i$. Relative projectivity makes $M$ a summand of the corresponding finite direct sum of the ${\mathrm{Ind}}_P^G{U}_i$. The Krull–Schmidt theorem leaves only finitely many possible indecomposable summands, so $kG$ has finite representation type.

Conversely, suppose $kG$ has finitely many indecomposables. For an indecomposable $kP$-module $U$, the identity double coset in the Mackey restriction formula shows that $U$ is a direct summand ofDecomposing the induced module into the finitely many $kG$-indecomposables and restricting them shows, again by Krull–Schmidt, that only finitely many $U$ can occur. Thus

If $P$ is cyclic, the indecomposable modules of a cyclic p-group in characteristic p form a finite list. If $P$ is noncyclic, its Frattini quotient has rank at least two and therefore has a quotient $C_p\times C_p$. Inflation preserves indecomposability and nonisomorphism, while $k(C_p\times C_p)$ has infinitely many indecomposable modules. The Higman criterion for finite representation type of a group algebra now gives

Existence of a minimal subgroup follows because $G$ has only finitely many subgroups and every module is relatively $G$-projective. Suppose an indecomposable module $M$ is relatively projective for both $H$ and $K$. Then $M$ is a summand of ${\mathrm{Ind}}_H^{G}U$ and of ${\mathrm{Ind}}_K^{G}V$ for suitable modules $U,V$. Applying the Mackey restriction formula and the Krull–Schmidt theorem shows that $M$ is relatively projective for some subgroupIf $H$ and $K$ are minimal, this forces $H\subseteq {}^{g}K$. Reversing their roles gives the reverse containment after conjugacy; since the groups are finite, $H$ and $K$ are conjugate. Thus vertices form a unique conjugacy class.

Let $Q$ be a vertex and let $P$ be a Sylow p-subgroup of $Q$. Since $[Q:P]$ is invertible in $k$, every $kQ$-module is relatively $P$-projective. Transitivity of relative projectivity makes $M$ relatively $P$-projective, so minimality forces $Q=P$. Hence every vertex is a p-group.

For the trivial module $k$, every $kH$-endomorphism is scalar, and its relative trace to $G$ is multiplication by $[G:H]$. The D. Higman criterion says that $k$ is relatively $H$-projective exactly when $p\nmid [G:H]$. The minimal such subgroups are precisely the Sylow p-subgroups. Therefore the vertex of an indecomposable module gives

For subgroups $H,K\le G$ and an $H$-representation $W$, Mackey restriction formula isFrobenius reciprocity statesApplying these with $K=H$ givesThe identity double coset contributes one exactly when $W$ is irreducible, and every summand is a nonnegative integer. This proves Mackey irreducibility criterion: the induced representation is irreducible exactly when $W$ is irreducible and every nonidentity double-coset summand vanishes.

Now take $G=S_n$, $H=S_{n-1}$, and $x=(n-1n)\notin H$. Thenthe subgroup fixing both $n-1$ and $n$. Conjugation by $x$ fixes every element of this subgroup, so the two representations appearing in the corresponding Mackey summand are both ${\mathrm{Res}}_{S_{n-2}}^{S_{n-1}}W$. Their inner product is positive because this restriction is nonzero. Hence the Mackey irreducibility criterion fails whenever $W$ is irreducible. If $W$ is reducible, induction preserves its direct-sum decomposition, so the induced representation is again reducible. Therefore ${\mathrm{Ind}}_{S_{n-1}}^{S_n}W$ is never irreducible for $n\ge 2$.

The Mackey restriction formula says, for an $H$-representation $V$,Frobenius reciprocity saysApplying both with an irreducible character $\chi$ givesThe identity double coset contributes one, and all terms are nonnegative integers. Therefore Mackey irreducibility criterion says that ${\mathrm{Ind}}_H^G\chi$ is irreducible exactly when every term belonging to a nonidentity double coset is zero.

Now take $G=S{L}_2(k)$. Writeso $B=T⋉U$. For $q\ge 4$, commutators of $T$ with $U$ generate $U$, because conjugation sends $u(b)$ to $u(a^{2}b)$ and one can choose $a^2\ne 1$. Every degree-one character of $B$ is therefore trivial on $U$ and has the formwhere $\theta :k^{\times }\to {\mathbb{C}}^{\times }$ is a multiplicative character.

The Bruhat decomposition of SL2 over a finite field has two double cosets,and $B\cap wB{w}^{-1}=T$. Conjugation by $w$ inverts $T$, soThe only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. ConsequentlyThus the trivial character is excluded, as is the unique quadratic character when $q$ is odd. All other degree-one characters of $B$ induce irreducibly; the induced representations have degree $[G:B]=q+1$. This is the irreducible principal series of finite SL2.

The Mackey restriction formula says that for subgroups $H,K\le G$ and a $K$-representation $W$,where ${}^{x}W(xk{x}^{-1})=W(k)$.

For $H=K=B$, the Bruhat decomposition of SL2 over a finite field givesThus $B\backslash G/B$ has representatives $1,w$. The identity double coset contributes $\theta$, whilecontributes an induced representation from the diagonal subgroup. ThereforewhereFor $t(a)=\mathrm{diag}(a,a^{-1})$, one has $w^{-1}t(a)w=t(a^{-1})$.

For coset permutation modules, Mackey restriction formula and Frobenius reciprocity identify the homomorphism space with a free module having one orbit-sum basis vector per double coset. The same basis works over the residue field, so reduction of homomorphisms is surjective. Applied to endomorphisms of a permutation module, idempotent refinement theorem lifts each modular direct summand to a trivial source lattice. Lifting inverse isomorphisms between reductions shows uniqueness among trivial source lattices.

For a rational Borel subgroup $B=TU$, the principal series is ${\mathrm{Ind}}_{B^F}^{G^F}\tilde{\theta }$, with $\tilde{\theta }$ inflated from a linear character of $T^F$. The Mackey restriction formula and rational Bruhat decomposition give its self-inner-product as the size of the Weyl stabilizer of $\theta$. It is irreducible precisely when that stabilizer is trivial.

Over the complex numbers, the signed Young permutation module isThe isomorphism is the tensor identity for induced representations. The partition index is unchanged by this twist; conjugation instead changes the labels of its irreducible constituents. For the conjugate partition ${\lambda }^{\prime }$, the character inner product of $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ is one: Mackey restriction formula reduces it to the unique zero-one matrix with row margins $\lambda$ and column margins ${\lambda }^{\prime }$.