Brauer Characters and Decomposition Matrices

Conjugation preserves order: if $y=gx{g}^{-1}$, then $y^n=1$ exactly when $g{x}^n{g}^{-1}=1$, that is, exactly when $x^n=1$. So $\mathrm{\mid }y\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }$.

Let $n=\mathrm{\mid }x\mathrm{\mid }$ and suppose $\gcd (m,n)=1$. Then the cyclic subgroup $⟨x^m⟩$ equals $⟨x⟩$, because some integer $r$ satisfies $mr\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}n)$, hence $x=(x^m{)}^r$. Therefore $\mathrm{\mid }{x}^m\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }$.

Step 1.1 and [F1] prove claim 1, while step 1.2 and [F1] prove claim 2.

The uniqueness clause is already built into [F1]: if $u,v\in {\mathcal{O}}^{\times }$ both have prime-to-$p$ order and reduce to $\lambda$, then both satisfy the defining property of the Teichmuller lift of $\lambda$, so $u=v$.

The product $\tau (\lambda )\tau (\mu )$ reduces to $\lambda \mu$ in $k^{\times }$. It is again a root of unity of order prime to $p$, because it lies in the finite subgroup generated by two prime-to-$p$ roots of unity. By the uniqueness from step 1.1, it must equal the Teichmuller lift of $\lambda \mu$.

Steps 1.1 and 2.1 give the claimed uniqueness and multiplicativity.

Changing basis replaces the matrix of $g$ by a similar matrix. Similar matrices have the same characteristic polynomial, hence the same eigenvalues with the same multiplicities. Therefore the sum in [F1] is basis-independent.

If two splitting realizations identify the same prime-to-$p$ roots of unity in the residue fields, then they identify each eigenvalue of the action of $g$ and, by [L1], identify its unique Teichmuller lift. So the lifted eigenvalue sum computed in [F1] is the same in either realization.

Steps 1.1 and 1.2 prove both independence statements.

In any basis of $V$, the matrices representing $x$ and $y=gx{g}^{-1}$ are similar, because the representation map sends conjugation in $G$ to conjugation in $\mathrm{GL}(V)$. So they have the same eigenvalues with the same multiplicities.

By [F1], the Brauer character is the sum of the Teichmuller lifts of those eigenvalues, and [L1] shows that the value does not depend on which basis was used to see the similarity. Hence ${\phi }_V(x)={\phi }_V(y)$.

Therefore ${\phi }_V$ is a class function on $G^0$.

Fix $g\in G^0$. Choose a basis of $V$ whose first block is a basis of the $g$-stable submodule $U$. In that basis the action matrix of $g$ on $V$ has block upper-triangular form $\left(\begin{array}{cc}A& \ast \\ 0& B\end{array}\right),$ where $A$ is the action on $U$ and $B$ is the induced action on $W$.

A block upper-triangular matrix has eigenvalue multiset equal to the union of the eigenvalue multisets of its diagonal blocks. Therefore the eigenvalues of $g$ on $V$ are exactly those on $U$ together with those on $W$. Using [F1] and [L1], the lifted traces add: ${\phi }_V(g)={\phi }_U(g)+{\phi }_W(g)$.

Since $g\in G^0$ was arbitrary, ${\phi }_V={\phi }_U+{\phi }_W$ on $G^0$. The direct-sum case is the split short exact sequence.

Repeatedly applying [L1] along a composition series of $V$ shows that ${\phi }_V$ is the sum of the irreducible Brauer characters of the composition factors of $V$, counted with multiplicity. The same holds for $W$.

If $V^{\mathrm{s}\mathrm{s}}\cong W^{\mathrm{s}\mathrm{s}}$, then $V$ and $W$ have the same irreducible composition factors with the same multiplicities. Step 1.1 then gives ${\phi }_V={\phi }_W$.

Conversely, suppose ${\phi }_V={\phi }_W$. Subtracting the two expansions from step 1.1 gives a linear relation among irreducible Brauer characters. By [L2], those characters are linearly independent, so every coefficient is $0$. Hence the composition multiplicities in $V$ and $W$ agree, and therefore $V^{\mathrm{s}\mathrm{s}}\cong W^{\mathrm{s}\mathrm{s}}$.

Steps 2.1 and 2.2 prove the equivalence.

By [L1], every irreducible Brauer character is a class function on $G^0$, so the family lies in the target vector space. By [F5], it is linearly independent.

Let $f:G^0\to \mathbb{C}$ be a class function. Extend it to a class function $\tilde{f}:G\to \mathbb{C}$ by setting $\tilde{f}(g)=0$ on the $p$-singular elements. Then [L3] gives $\tilde{f}={\sum }_{\chi }{a}_{\chi }\chi$ as a linear combination of ordinary irreducible characters. Restricting to $G^0$ yields $f={\sum }_{\chi }{a}_{\chi }{\chi }^0\mathrm{.}$

By [F4], each ${\chi }^0$ is the Brauer character of some finite-dimensional $kG$-module. A composition series for that module and [L2] write ${\chi }^0$ as a linear combination of irreducible Brauer characters. Step 1.2 therefore expresses $f$ as a linear combination of irreducible Brauer characters.

Steps 1.1 and 2.1 show that the irreducible Brauer characters are linearly independent and span the class functions on $G^0$, hence form a basis.

By [A1], the vector space of class functions on $G^0$ has dimension equal to the number of $p$-regular conjugacy classes.

By [L1], the irreducible Brauer characters form a basis of that same space. The number of basis vectors is therefore the number of irreducible Brauer characters, equivalently the number of simple $kG$-modules.

Hence the number of simple $kG$-modules equals the number of $p$-regular conjugacy classes.

By [A1], after replacing $(L,L^{\mathrm{\prime }})$ by $({\pi }^{r}L,L^{\mathrm{\prime }})$ if necessary, we may assume $L\subseteq L^{\mathrm{\prime }}$. Multiplication by the scalar ${\pi }^r$ does not change the class of the reduction in the Grothendieck group, because ${\pi }^{r}L\mathrm{/}\mathfrak{m}{\pi }^{r}L\cong L\mathrm{/}\mathfrak{m}L$ as $kG$-modules.

Put $A:=L^{\mathrm{\prime }}\mathrm{/}L$. Reduction of the inclusion $L\hookrightarrow L^{\mathrm{\prime }}$ has kernel $(L\cap \pi L^{\mathrm{\prime }})\mathrm{/}\pi L$ and cokernel $L^{\mathrm{\prime }}\mathrm{/}(L+\pi L^{\mathrm{\prime }})$. Multiplication by $\pi$ identifies the kernel with $A[\pi ]:=\{a\in A:\pi a=0\}$, while the cokernel is $A\mathrm{/}\pi A$. Thus $[L\mathrm{/}\pi L]-[L^{\mathrm{\prime }}\mathrm{/}\pi L^{\mathrm{\prime }}]=[A[\pi ]]-[A\mathrm{/}\pi A]$ in $R_k(G)$.

Multiplication by $\pi$ on the finite-length $\mathcal{O}G$-module $A$ gives exact sequences $0\to A[\pi ]\to A\to \pi A\to 0,0\to \pi A\to A\to A\mathrm{/}\pi A\to 0.$ Their Grothendieck-group identities give $[A[\pi ]]=[A\mathrm{/}\pi A]$. Both end terms are annihilated by $\pi$, so this equality is an equality in $R_k(G)$. Step 2.1 therefore gives $[L\mathrm{/}\mathfrak{m}L]=[L^{\mathrm{\prime }}\mathrm{/}\mathfrak{m}{L}^{\mathrm{\prime }}]$.

Hence the Grothendieck-group class in [F1] is independent of the stable lattice.

By [F1], $d_{\chi \phi }$ is the multiplicity of the simple module $S_{\phi }$ among the composition factors of the reduced lattice representing $\chi$.

A composition multiplicity counts how many times a simple factor occurs, so it is an integer and cannot be negative. Hence $d_{\chi \phi }\in {\mathbb{Z}}_{\ge 0}$.

Therefore all decomposition numbers are nonnegative integers.

Write $P_{\phi }$ as the image of an idempotent matrix over $kG$. Because $\mathcal{O}$ is complete by [F3], idempotents lift through the ideal $\mathfrak{m}{M}_n(\mathcal{O}G)$. A lifted idempotent has image a projective $\mathcal{O}G$-lattice ${\widehat{P}}_{\phi }$ whose reduction is $P_{\phi }$.

Let $L_{\chi }$ be a stable lattice affording $\chi$. Since ${\widehat{P}}_{\phi }$ is projective, formation of $\mathrm{Hom}$ from it commutes with extension to $K$ and reduction to $k$. Hence ${\dim }_K{\mathrm{Hom}}_{KG}(K\otimes {\widehat{P}}_{\phi },V_{\chi })={\dim }_k{\mathrm{Hom}}_{kG}(P_{\phi },L_{\chi }\mathrm{/}\mathfrak{m}{L}_{\chi })\mathrm{.}$ The left side is the multiplicity of $V_{\chi }$ in the split semisimple module $K\otimes {\widehat{P}}_{\phi }$.

The functor ${\mathrm{Hom}}_{kG}(P_{\phi },-)$ is exact, and on a simple module $S_{\psi }$ it has dimension $1$ for $\psi =\phi$ and $0$ otherwise, because every map from $P_{\phi }$ to a simple module factors through its head $S_{\phi }$. The right side of step 2.1 is therefore the composition multiplicity of $S_{\phi }$ in $L_{\chi }\mathrm{/}\mathfrak{m}{L}_{\chi }$, namely $d_{\chi \phi }$ by [F1].

Thus the multiplicity of every $\chi$ in the ordinary character of $K\otimes {\widehat{P}}_{\phi }$ is $d_{\chi \phi }$. These coefficients depend only on $P_{\phi }$, so the character is independent of the chosen lift and equals ${\mathrm{\Phi }}_{\phi }={\sum }_{\chi }{d}_{\chi \phi }\chi$.

Fix $\phi$. Decompose the projective indecomposable character as ${\mathrm{\Phi }}_{\phi }={\sum }_{\chi }{d}_{\chi \phi }\chi$ by [L1].

The Cartan entry $c_{\phi \psi }$ is the multiplicity of $S_{\psi }$ in the projective cover $P_{\phi }$ by [F1]. Apply the decomposition map [F2] to the ordinary-character expansion in step 1.1. Its left side becomes $[P_{\phi }]$, while the right side becomes ${\sum }_{\chi }{d}_{\chi \phi }{\sum }_{\psi }{d}_{\chi \psi }[S_{\psi }]$. Comparing coefficients of $[S_{\psi }]$ gives $c_{\phi \psi }={\sum }_{\chi }{d}_{\chi \phi }{d}_{\chi \psi }\mathrm{.}$

This is exactly the $(\phi ,\psi )$ entry of $D^{\mathsf{T}}D$, so $C=D^{\mathsf{T}}D$.

Put $A=\mathcal{O}H$. It is finite free over $\mathcal{O}$, hence complete and separated for the $\mathfrak{m}$-adic topology by F1. Every element of $\mathfrak{m}A$ lies in $J(A)$: if $x\in \mathfrak{m}A$ and $y\in A$, then $(yx{)}^n\to 0$, so $1-yx$ has inverse ${\sum }_{n\ge 0}(yx{)}^n$. This is the Jacobson-radical test.

Let $\bar{e}\in A\mathrm{/}\mathfrak{m}A=kH$ be idempotent and choose any lift $e_1\in A$. Inductively, if $e_n^2-e_n\in {\mathfrak{m}}^{n}A$, apply the Newton correction $e_{n+1}=e_n-(2e_n-1)(e_n^2-e_n)\mathrm{.}$ Because $e_n$ commutes with $e_n^2-e_n$, direct expansion shows that $e_{n+1}^2-e_{n+1}$ is a multiple of $(e_n^2-e_n{)}^2$, while $e_{n+1}-e_n\in {\mathfrak{m}}^{n}A$. Thus the errors tend to zero and $(e_n)$ is Cauchy. Completeness gives a limit $e$ with $e^2=e$ and reduction $\bar{e}$.

Suppose now that $\bar{e}$ is central. For $X=eA(1-e)$ its reduction is $\bar{e}(kH)(1-\bar{e})=0$, hence $X=\mathfrak{m}X$: the nontrivial inclusion uses that $\mathfrak{m}=(\pi )$ in the DVR, since $x=\pi a=e(\pi a)(1-e)=\pi (ea(1-e))$. The finite $\mathcal{O}$-module $X$ has generators $x_1,\dots ,x_t$ with $x_i=\pi {\sum }_j{a}_{ij}{x}_j$. Multiplying $(I-\pi (a_{ij}))(x_j)=0$ by its adjugate shows that $\det (I-\pi (a_{ij}))x_j=0$ for every $j$. The determinant is congruent to $1$ modulo $\mathfrak{m}$, hence is a unit, so $X=0$. Applying the same argument to $(1-e)Ae$ gives that space zero too. Therefore $ea=eae=ae$ for every $a\in A$, and $e$ is central.

If central idempotents $e,f\in A$ have the same reduction, then $e-f\in \mathfrak{m}A\subseteq J(A)$. The commuting products $e(1-f)$ and $f(1-e)$ are idempotents in $J(A)$, and an idempotent in the Jacobson radical is zero. Hence $e=ef=f$. Thus every central idempotent of $kH$ has exactly one central lift.

Central primitivity is preserved. If a central lift $e$ decomposed into two nonzero orthogonal central idempotents, neither summand could reduce to zero, since step 1.1 puts its kernel inside $J(A)$; their reductions would decompose $\bar{e}$. Conversely, a central decomposition of $\bar{e}$ lifts termwise by steps 2.1–3.1, and uniqueness in step 4.1 makes the lifted sum equal to $e$. Hence $e$ is primitive exactly when $\bar{e}$ is primitive. Together with F2 this proves the bijection and the asserted unique block lift, including the trivial-group case. The constructions are finite or sequential limits fixed by explicit formulas, so no choice principle is used.

Because the primitive central idempotents are central, pairwise orthogonal, and sum to $1$, every module $M$ decomposes as the direct sum of the eigenspaces $eM$.

If $M$ is simple, only one summand in step 1.1 can be nonzero; otherwise $M$ would split as a nontrivial direct sum of submodules. So each simple $kG$-module, hence each irreducible Brauer character by [L1], lies in exactly one block.

By [F2], each modular block idempotent has a unique lift in $\mathcal{O}G$. These lifted idempotents remain central, orthogonal, and sum to $1$, and they act on $KG$-modules through $\mathcal{O}G\subseteq KG$. Thus step 1.1 applies to a simple $KG$-module using the lifted block idempotents: exactly one summand is nonzero. Its ordinary irreducible character consequently belongs to that unique block.

If $d_{\chi \phi }\mathrm{\ne }0$, then by [F1] the simple module $S_{\phi }$ occurs in the reduction of a stable lattice affording $\chi$. A composition factor of a module lies in the same block as the module itself, so $\chi$ and $\phi$ must belong to the same block.

Therefore entries connecting two different blocks are zero. After reordering rows and columns by blocks, all nonzero entries lie inside the matching block sectors, so the matrix is block diagonal.

The element $u$ has order $p$, so it is $p$-singular. The displayed matrix still has an ordinary trace, namely $2$ in $k$.

But [F1] defines the Brauer character only on $G^0$, and $u\mathrm{\notin }{G}^0$. So the trace value at $u$ is not a Brauer-character value.

Therefore the statement is false.

Let $V$ be the $2$-dimensional $k{C}_p$-module with generator acting by $I+N=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, and let $W=k\oplus k$ be the direct sum of two trivial modules. Since $N^2=0$ and $\mathrm{char}k=p$, $(I+N{)}^p=I$, so this is a representation of $C_p$. The modules are not isomorphic: the generator acts nontrivially on $V$ and trivially on $W$.

For each group element $g^a$, the matrices are $(I+N{)}^a=I+aN$ on $V$ and $I$ on $W$, so their ordinary-style traces are both $2$ in $k$. The invariant line $k{e}_1\subset V$ and its quotient are both trivial, so $V$ and $W$ also have the same semisimplification. The only $p$-regular element of $C_p$ is the identity, where both Brauer characters, when defined through a splitting system, take the value $2$. These conclusions follow directly from the matrices and require no splitting-system hypothesis on $k$.

Thus ordinary-style character data do not recover the full modular representation; even Brauer characters cannot distinguish these two modules. The statement is false.

The space $K{\otimes }_{\mathcal{O}}L$ is the two-dimensional sum-zero subspace of $K^3$. The cycle $(123)$ has distinct eigenlines there with eigenvalues ${\zeta }_3$ and ${\zeta }_3^2$ (for example, they are spanned by $(1,{\zeta }_3^2,{\zeta }_3)$ and $(1,{\zeta }_3,{\zeta }_3^2)$). A transposition conjugates the cycle to its inverse and swaps these eigenlines. Thus neither line is stable under $S_3$; a proper nonzero submodule of a two-dimensional representation would be a stable line, so this ordinary representation is irreducible.

By [F1], reducing modulo the maximal ideal gives the $k{S}_3$-module $\bar{L}=L\mathrm{/}(1-{\zeta }_3)L\mathrm{.}$ Coordinate reduction identifies $\bar{L}$ with the two-dimensional sum-zero subspace of $k^3$: if all coordinates of a vector in $L$ are divisible by $1-{\zeta }_3$, dividing them by this element still gives a vector in $L$. The image of $(1,1,-2)\in L$ is the nonzero vector $(1,1,1)$ because $k={\mathbb{F}}_3$. It is fixed by every permutation matrix, so its line is a nonzero proper invariant subspace of $\bar{L}$. Thus the reduction is reducible.

Thus an ordinary irreducible representation can have reducible reduction modulo $p$. In the decomposition process recorded by [F2], this reduction is therefore not irreducible. The statement is false.

The group $C_2$ has exactly two ordinary irreducible characters over $K$: in any representation the operator representing $g$ satisfies $(T-1)(T+1)=0$, and these factors are coprime because $2$ is invertible in $K$, so the representation splits into its $+1$ and $-1$ eigenspaces. An irreducible representation is therefore one-dimensional, with $g$ acting by $+1$ or $-1$; both occur and are distinct.

The only $2$-regular element of $C_2$ is the identity, so [F2] gives exactly one irreducible Brauer character. By [F1] and step 1.1, $D$ has two rows and one column; by [F3], $C=D^{\mathsf{T}}D$ has one row and one column.

A $2\times 1$ matrix cannot equal a $1\times 1$ matrix. Thus this finite group and prime refute the universal statement.

The group algebra $k{C}_p$ has only one irreducible Brauer character, namely the trivial one: in characteristic $p$, a generator acts on every simple module as $1$, since $(g-1{)}^p=0$. Over a splitting field of characteristic $0$, $C_p$ has $p$ distinct ordinary irreducible characters, all linear. Each has a stable rank-one lattice, whose reduction is a one-dimensional $k{C}_p$-module and hence the trivial Brauer character. Thus every ordinary character has decomposition number $1$ in the unique Brauer column.

By [L1] and [L2], a nonzero decomposition number joins an ordinary character to a Brauer character in the same block. Step 1.1 therefore places all $p$ ordinary irreducibles in the single block containing the trivial Brauer character. For any prime $p\ge 2$, that block contains more than one ordinary irreducible.

Therefore the statement is false.