Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 (Modular Representation Theory)

An $A$-module is a semisimple module when it is a direct sum of simple modules, equivalently when every submodule has a complementary submodule. The finite-dimensional algebra $A$ is a semisimple algebra when its left regular module ${}_{A}A$ is semisimple.

Let $C_{p^n}=⟨g⟩$, put $N=p^n$, and suppose $\mathrm{char}k=p$. The freshman's dream givesso, with $u=g-1$,The indecomposable modules of a cyclic p-group in characteristic p are preciselyIndeed, a module is a vector space with a nilpotent operator $u$, and its decomposition into Nilpotent Jordan blocks gives these modules. Each $M_r$ is a uniserial module, with unique chainTherefore

For a finite-dimensional module over $A$, the radical of a module satisfies $J(M)=J(A)M$. Henceand induction gives $J^r(U\oplus V)=J^r(U)\oplus J^r(V)$ for every $r$. A simple submodule of a direct sum projects into semisimple submodules of each summand, and equivalentlyThus the same argument, or induction through the defining quotients, givesThis is radical and socle series of a direct sum.

Now let $G$ be a finite $p$-group and let $k$ have characteristic $p$. The group algebra of a p-group in characteristic p is local, with unique simple module $k$. The socle of its regular module iswhich is one-dimensional. If $kG=U\oplus V$ with both summands nonzero, finite length gives nonzero socles for $U$ and $V$, and the direct-sum identity would make $\mathrm{Soc}(kG)$ at least two-dimensional. Therefore

The radical of a module is the smallest submodule $N\subseteq M$ for which $M/N$ is a semisimple module. Consequently, ifhas semisimple successive quotients, then $J(M_i)\subseteq M_{i+1}$. Induction givesso the radical series of a module descends at least as fast as every such series.

Dually, the socle is the largest semisimple submodule. Ifhas semisimple successive quotients, induction in $M/N_i$ givesso the socle series of a module ascends at least as fast as every such series.

Both series terminate because $M$ has finite composition length. More precisely,Thus $J^r(M)=0$ exactly when $J(A{)}^r$ annihilates all of $M$, which is exactly when ${\mathrm{Soc}}^r(M)=M$. The two least terminating indices therefore coincide:

For $1\le r\le n$, let $S_r=k$ with a lower triangular matrix $a=(a_{ij})$ acting by the scalar $a_{rr}$. These are pairwise nonisomorphic simple modules, since the diagonal matrix units $e_{ii}$ distinguish them.

Let $N$ be the ideal of strictly lower triangular matrices. It is nilpotent, andis a semisimple algebra. Hence $N$ is the Jacobson radical; explicitly,

For the left regular module, the radical series of a module is obtained by multiplying by powers of the Jacobson radical:Products of matrix units show thatwith $N^0=A$ and $N^n=0$. Thus each step removes one subdiagonal, and the Loewy length of ${}_{A}A$ is $n$.

Order the simple modules as in part (i), so that $S_r$ records the $r$th diagonal character. The quotient $N^i/N^{i+1}$ has basisFor $a\in A$, multiplication givesbecause every term below row $r$ lies on a lower subdiagonal. Hence the line generated by $e_{r,r-i}$ is $S_r$, and the lines are independent. Therefore the radical layers of the lower triangular matrix algebra are

Let $g$ be a p-regular element. Its eigenvalues on $M$ are roots of unity of order prime to $p$. If ${\hat{\lambda }}_1,\dots ,{\hat{\lambda }}_d$ are their Teichmuller lifts to characteristic-zero roots of unity, the Brauer character isIt depends only on the conjugacy class of $g$, is additive in short exact sequences, and equals the restriction of an ordinary character whenever the representation lifts.

For linear independence, choose a splitting p-modular system and let $P_i$ be the projective cover of the simple module $S_i$. A projective lattice lifting $P_i$ has an ordinary character ${\Phi }_i$ that vanishes on p-singular elements. Reduction and ordinary character orthogonality givebecause $S_i$ is the head of $P_i$. Pairing a relation ${\sum }_j{c}_j{\chi }_{S_j}=0$ with every ${\Phi }_i$ yields $c_i=0$ for every $i$. HenceThis is the linear-independence part of the Brauer–Nesbitt theorem.

There is a natural isomorphismobtained by evaluating a homomorphism against a covector. Since $P$ is a direct summand of a free module and tensoring a free $kG$-module with $U^{\ast }$ using the diagonal action again gives a free module, $P\otimes U^{\ast }$ is a projective module.

Lift this projective module to an $\mathcal{O}G$-lattice. Projectivity makes the dimension of the homomorphism space equal to the multiplicity of the trivial representation after extending scalars to $K$. The lifted ordinary character vanishes on p-singular elements, while on p-regular elements it is the productOrdinary character orthogonality and the substitution $g\mapsto g^{-1}$ therefore giveEquivalently, this is the Brauer character inner product between the projective character of $P$ and the Brauer character of $U$.

Apply part (b) with $P=P_i$ and $U=S_j$. A homomorphism $P_i\to S_j$ kills $J(P_i)$ and therefore factors through the head $P_i/J(P_i)\cong S_i$. By Schur lemma over the splitting field,Consequently the two stated bases satisfy the Duality of simple and projective Brauer characters:

The form is nondegenerate directly: if $\varphi \ne 0$, thenEquivalently, in coordinates indexed by the p-regular conjugacy classes it is a positive diagonal Hermitian form with weights $1/|C_G(x)|$.

Writing the Brauer character inner product as a sum over conjugacy-class representatives givesThus the duality from part (c) is exactlyAll three matrices are square. Reversing the two inverse factors gives $D{X}^T\overline{\Pi }=I$, henceTaking complex conjugates yieldsThe $(g,h)$ entry is ${\sum }_S\overline{{\chi }_S(g)}{\chi }_{P_S}(h)$, and $\overline{{\chi }_S(g)}={\chi }_S(g^{-1})$. Therefore Column orthogonality for Brauer characters gives

Give the dual $M^{\ast }={\mathrm{Hom}}_k(M,k)$ the contragredient action $(g\lambda )(m)=\lambda (g^{-1}m)$. If ${\delta }_h$ is the functional dual to the basis element $h\in G$, thenThus the map $h\mapsto {\delta }_h$ extends to a $kG$-isomorphismThis is also the left-module form of the fact that a group algebra is a symmetric algebra.

If $P$ is finitely generated and projective, it is a direct summand of $(kG{)}^r$. Dualizing makes $P^{\ast }$ a direct summand of $((kG{)}^{\ast }{)}^r\cong (kG{)}^r$, so $P^{\ast }$ is projective. The converse follows by dualizing again and using $P\cong P^{\ast \ast }$.

For any finite-dimensional algebra $A$, the module $A^{\ast }$ is injective becauseis exact. Since $kG\cong (kG{)}^{\ast }$, free $kG$-modules are injective, and so are their projective direct summands. Conversely, duality sends injectives to projectives, so every finite-dimensional injective is projective. Hence projective modules over a finite group algebra are injective.

Finally, let $P$ be indecomposable projective. It is also an indecomposable injective. Its nonzero socle contains a simple module $S$, and the injective hull $E(S)$ is a direct summand of $P$. Indecomposability forces $P=E(S)$. Since $S$ is essential in its injective hull, every simple submodule of $P$ equals $S$. Therefore

Write $P\cong kGe$ for a primitive idempotent $e$. The coefficient-of-identity form makes $kG$ a symmetric algebra. Associativity of this nondegenerate form identifies the orthogonal complement of $J(kG)e$ in $kGe$ with the elements annihilated by $J(kG)$, namely $\mathrm{Soc}(P)$. It therefore induces a nondegenerate $kG$-invariant pairing betweenand $\mathrm{Soc}(P)$. Both are simple by projectivity and part (a), and the symmetric form has identity Nakayama permutation. Consequently the head and socle of an indecomposable projective group-algebra module satisfy

Decompose the projective module aswhere $T$ ranges over the simple modules. Since $P_T/J(P_T)\cong T$, the multiplicity of $S$ in the head of a module $P/J(P)$ is $m_S$. Part (b) gives $\mathrm{Soc}(P_T)\cong T$, so the multiplicity in $\mathrm{Soc}(P)$ is the same $m_S$.

The invariant submodule and coinvariant module satisfyThus $\dim P^G$ is the multiplicity of the trivial module in $\mathrm{Soc}(P)$, while $\dim P_G=\dim {\mathrm{Hom}}_{kG}(P,k)$ is its multiplicity in the head. Applying the same argument to the projective module $P^{\ast }$ yields

The dual $P_S^{\ast }$ is indecomposable projective. Its head is dual to $\mathrm{Soc}(P_S)\cong S$, hence is $S^{\ast }$. Uniqueness of projective covers proves the dual of a projective cover over a group algebra:

Put $N_G={\sum }_{g\in G}g$. Its image on every module lies in the invariant submodule, since $h{N}_G=N_G$. In the regular module,and this line is the socle of the projective cover $P_k$ of the trivial module.

Suppose $N_{G}M\ne 0$. Choose $m\in M$ with $N_{G}m\ne 0$ and consider the homomorphismIts restriction $f{|}_{P_k}$ is nonzero on $\mathrm{Soc}(P_k)=k{N}_G$. Because $P_k$ is the injective hull of its simple socle, that socle is essential: every nonzero submodule meets it. Hence $\ker (f{|}_{P_k})=0$. The resulting embedding $P_k\hookrightarrow M$ splits because $P_k$ is injective. Since $M$ is indecomposable, $M\cong P_k$.

Conversely, on $P_k$ the image of $N_G$ is its one-dimensional socle. Thus the group norm element detects the trivial projective cover:

An $RG$-module $M$ is relative projective module for $H$ when it has the lifting property for every $H$-split epimorphism: whenever the solid arrows form a commutative diagramwith $\pi :X\twoheadrightarrow Y$ an $RG$-map possessing an $RH$-linear section, there is an $RG$-map $\tilde{f}:M\to X$ such that $\pi \tilde{f}=f$. Equivalently, every $H$-split epimorphism onto $M$ has a $G$-linear section, or

For $\alpha \in {\mathrm{End}}_{RH}(M)$ define the relative traceThe D. Higman criterion is

SetEvery conjugate of $\alpha$ equals $\alpha$, soThe D. Higman criterion therefore gives

If $M$ is projective over $RG$, then its restriction is projective over $RH$ because $RG$ is a free right $RH$-module and a free $RG$-module restricts to a free $RH$-module.

Conversely, suppose ${\mathrm{Res}}_H^{G}M$ is projective. Thenis projective over $RG$. Part (i) says that $M$ is a direct summand of this induced module, so $M$ is projective. Hence projectivity detected on a subgroup of invertible index gives

Let $U\le {\mathrm{SL}}_2(p)$ be the Sylow p-subgroup of upper unitriangular matrices. It is cyclic of order $p$, generated byRealize $S^r(V_2)$ as the homogeneous polynomials of degree $r$ in $X,Y$, with $u$ acting by $X\mapsto X$ and $Y\mapsto X+Y$. For $0\le r<p$, the only vectors fixed by $u$ are the multiples of $X^r$: successive comparison of the coefficients of $Y^r,Y^{r-1},\dots$ proves this. Thus the nilpotent operator $u-1$ has one-dimensional kernel. Its Jordan normal form therefore has a single block, soThis also follows from the indecomposable modules of a cyclic p-group in characteristic p.

For $r=p-1$, the restriction has dimension $p$ and is the regular $kU$-module, hence is projective. Since $[G:U]$ is prime to $p$, part (b)(ii) makes $S^{p-1}(V_2)$ a simple projective $kG$-module. It is therefore a defect-zero representation and lifts to an ordinary irreducible representation of the same dimension. Consequently

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

A block idempotent is a primitive central idempotent $e\in R$. The module $M$ lies in the corresponding block of a group algebra whenequivalently when every other block idempotent annihilates $M$.

If $V$ lies in $e$, then the centrality of $e$ makes both its submodule $U$ and quotient $W$ lie in $e$. Conversely, suppose $eU=U$ and $eW=W$. Then $(1-e)V$ maps to zero in $W$, so $(1-e)V\subseteq U$. But $(1-e)U=0$, and applying the idempotent $1-e$ once more givesThus $eV=V$, proving

Regard the block algebra $RGe$ as an $R[G\times G]$-module through left and right multiplication,A defect group of a block is a p-subgroup $D\le G$ for which $\Delta D=\{(d,d):d\in D\}$ is a vertex of an indecomposable summand determining the block; equivalently, $D$ is maximal withunder the Brauer morphism. The uniqueness of vertices up to conjugacy in $G\times G$, together with the diagonal form of these vertices, shows that any two such $D$ are conjugate in $G$. Hence

Only the classes $1,3,7A,7B$ are 2-regular. The trivial module and the natural three-dimensional module are simple; the dual natural module gives the conjugate three-dimensional character. Restricting the ordinary characters to the odd-order classes and using $\alpha +\overline{\alpha }=-1$ produces the fourth simple character of degree eight. The Brauer character table isThe ordinary degree-eight character restricts exactly to ${\varphi }_8$. Since its degree contains the full 2-part $8$ of $|G|$, this simple module is projective and its singleton block has defect zero. Thus

On the 2-regular classes the six ordinary characters decompose asTherefore the decomposition matrix, with columns ordered $1,3,\overline{3},8$, isThe Cartan matrix of a group algebra is

The rows of the Cartan matrix of a group algebra express the projective characters in the simple Brauer-character basis. ThusEvaluating gives

The eight-dimensional simple module is projective, so its tensor product of group representations with the natural three-dimensional module is projective. Its Brauer character isFrom part (iii),The duality of simple and projective Brauer characters makes the decomposition multiplicities unique. Hence, writing $P_3$ and $P_8$ for the corresponding projective covers,This completes the explicit 2-modular representation theory of GL3 of F2 calculation.

The divisibility in (i) is the standard dimension test for a projective modular representation. If $P$ is a projective $kG$-module and $S$ is a Sylow p-subgroup of order $p^d$, then ${\mathrm{Res}}_S^{G}P$ is projective over $kS$. The group algebra of a p-group in characteristic p is local, so every finitely generated projective $kS$-module is free. ConsequentlyThis will apply to $P=\overline{W}$ in the implication (v)$\Rightarrow$(i).

Assume (ii), and write the ring decomposition asThe natural column module $W\cong {\mathcal{O}}^n$ is projective over $M_n(\mathcal{O})$ by Morita equivalence; extending it by zero across $B$ makes it projective over $\mathcal{O}G$. After extending scalars to $K$, the action factors through $M_n(K)$ on its natural module $K^n$, which is simple. Therefore

Assume (iii). Reduction of the projective lattice $W$ is projective, because a direct-summand decomposition of a free $\mathcal{O}G$-module remains one after tensoring with $k$. Completeness of $\mathcal{O}$ and idempotent lifting show that $\overline{W}$ is indecomposable: otherwise a nontrivial idempotent of ${\mathrm{End}}_{kG}(\overline{W})$ would lift and split $W$, hence split the simple $KG$-module $M$.

Let $\chi$ be the ordinary irreducible character of $M$. Since $W$ is projective, $\chi$ vanishes on p-singular elements and restricts to the projective character $\Phi$ of $\overline{W}$. ThusAn indecomposable projective module that is not simple has, besides its identity, a nonzero noninvertible endomorphism obtained by projecting onto its simple head and embedding the isomorphic simple socle. Hence its endomorphism algebra has dimension at least two. Therefore $\overline{W}$ is simple as well as projective, proving

Under (ii), tensor the direct-product decomposition with the residue field $k$. Since $W$ is free over $\mathcal{O}$,Henceand the reduced action map is the projection onto the first factor. This proves

Assume (v). If the generic fibre $M=K{\otimes }_{\mathcal{O}}W$ had a nonzero proper $KG$-submodule, intersecting it with $W$ and rescaling to obtain a saturated lattice would give a nonzero proper $kG$-submodule of $\overline{W}$. Thus $M$ is simple.

Because $\overline{W}$ is projective, its restriction to a Sylow p-subgroup $S$ is projective. The local algebra $kS$ has only free finitely generated projectives, so $|S|=p^d$ dividesConsequently

Let $D$ act on the group algebra $kG$ by conjugation. Its fixed-point algebra isand the centralizer $C_G(D)$ consists of the elements of $G$ commuting with every element of $D$. The Brauer morphism isThus $\beta$ deletes the coefficients of basis elements outside $C_G(D)$. The nonfixed $D$-orbits have cardinality divisible by $p$, so the usual orbit argument shows that this projection is a unital ring homomorphism. Hence

Put $N=N_G(D)$. For an algebra on which a group acts by conjugation, writefor its transfer ideal of conjugation-fixed elements. In the diagram, $\tau ={\mathrm{Tr}}_D^G$, ${\tau }^{\prime }={\mathrm{Tr}}_D^N$, the upper map is $\beta ={\mathrm{Br}}_D$, and ${\beta }^{\prime }$ is the restriction of ${\mathrm{Br}}_D$ from $(kG{)}^G$ to $(kG{)}_D^G$. Since $N$ normalizes both $D$ and $C_G(D)$, the lower map lands in $(k{C}_G(D){)}_D^N$.

For $a\in (kG{)}^D$, let $D$ act on the left cosets $G/D$. A coset $gD$ is fixed precisely when $g^{-1}Dg\le D$, hence, because the two groups have the same order, precisely when $g\in N$. Every nonfixed orbit has size divisible by $p$. Moreover, after applying ${\mathrm{Br}}_D$, all summands indexed by one $D$-orbit are equal: conjugation by an element of $D$ acts trivially on $k{C}_G(D)$. Those orbits therefore contribute zero in characteristic $p$, while the fixed cosets contribute the trace over $N/D$. ConsequentlyThis is the Brauer morphism and relative trace identity, so the diagram commutes.

The decomposition matrix separates into two connected components. The first contains ${\varphi }_1,{\varphi }_2$ and ${\chi }_1,{\chi }_{3A},{\chi }_{3B},{\chi }_4$; it is the principal block $B_0$. The second contains only ${\varphi }_3$ and ${\chi }_5$; since ${\varphi }_3(1)=5$ contains the full 5-part of $|A_5|=60$, this is a defect-zero representation and its block $B_1$ has defect group $1$.