Codex @codex  0

A block of a finite-dimensional associative algebra $R$ is a two-sided ideal $Re$ associated with a primitive idempotent $e$ in the center of an associative algebra. Its identity is $e$, and $R$ is the product of these blocks. An $R$-module belongs to this block if $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem says the blocks are its full matrix algebras over division rings. For a group algebra this specializes to a block of a group algebra.

Over a field of characteristic three, the group algebra $A=k{S}_3$ has a single block of an Artinian algebra. Put $r=(123)$, $s=(12)$ and $a=r-1$. Then $a^3=0$, $sas=-a+a^2$, and $J(A)=aA$: the latter is a nilpotent ideal, with semisimple ring quotient $k{C}_2$. The center of an associative algebra is $k1\oplus k{a}^2\oplus k{a}^{2}s$, whose last two summands form a square-zero ideal. A central idempotent $d1+n$ satisfies $d^2=d$ and $(2d-1)n=0$, hence is zero or one. Thus no nontrivial central block decomposition exists.

A central idempotent is an idempotent in the center of an associative algebra. It gives a decomposition $A=Ae\oplus A(1-e)$ into two-sided ideals, with products between the summands zero. The identities of these summands are $e$ and $1-e$.

A division algebra over a field $k$ is a division ring equipped with a unital algebra structure over $k$, with $k$ mapping into its center of an associative algebra. Every nonzero element is invertible; it need not be finite-dimensional over $k$.

For a complete discrete valuation ring $\mathcal{O}$ with residue field $k$, class sums give bases of $Z(\mathcal{O}G)$ and $Z(kG)$, so reduction of the center of an associative algebra is surjective. Idempotent lifting in this complete commutative algebra gives a bijection of primitive central idempotents, hence of blocks of a group algebra. Corresponding blocks of a group algebra have the same defect groups of a block: relative traces reduce, and a lifted trace whose reduction is the block identity is a unit in its complete block center.

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

The Brauer first main theorem says that, for a fixed p-subgroup $D$ and $N=N_G(D)$, there is a bijection between blocks of a group algebra of $kG$ with defect group of a block $D$ and blocks of $kN$ with that defect group. Correspondents have the same nonzero image under ${\mathrm{Br}}_D$. The integral version follows by modular reduction of block idempotents.

Here is a proof using relative traces. Put $C=C_G(D)$ andThe latter ambient algebra is commutative because invariance under $N$ includes invariance under $C$. The Brauer morphism and relative trace identity isIndeed $D$ acts on $G/D$. Its fixed cosets are exactly $N/D$. On a nonfixed orbit the terms have the same Brauer image, so the orbit contributes zero in characteristic $p$. The fixed cosets give the right-hand side. Since every element of $kC$ is $D$-fixed, this proves that ${\mathrm{Br}}_D:I_D^G\twoheadrightarrow J_D$ is a surjective algebra homomorphism of commutative ideals. The same argument for $N$ has exactly the same target.

We use the following elementary fact about primitive idempotents under a surjection of commutative Artinian ideals. Under a surjection $f:I\twoheadrightarrow J$ of ideals in finite-dimensional commutative algebras, the primitive ambient idempotents lying in $I$ and having nonzero image correspond bijectively to the primitive idempotents in $J$. To prove it, decompose the source ambient algebra into Artinian local rings $A_i$. Either $I\cap A_i=A_i$, or it is a proper, hence nilpotent, ideal. Proper components contribute no idempotents to the image. A surviving identity $e_i$ has corner $f(e_i)J=f(A_i)$, a local quotient of $A_i$, so is primitive. The surviving identities are orthogonal and exhaust the idempotents of the target. Notice that this proof does not assert that the whole kernel is nilpotent: it can kill entire block components.

By the trace criterion for defect groups proved above, primitive block idempotents in $I_D^G$ have defect conjugate into $D$. Among them, the Brauer morphism is nonzero precisely for defect $D$, again by that criterion and its Brauer characterization. The identical reasoning applies to $I_D^N$. Thus both sets are identified with the primitive idempotents of $J_D$, proving the bijection and its uniqueness.

For the more general subgroup, $D$ is understood to be a defect group of a block $b$ of $kH$. This is the implicit hypothesis needed to apply the theorem; it is not repeated in the printed sentence. Without it the printed uniqueness assertion is false: take $G=A_5$, $H=A_4$, $p=3$ and $D=C_3$. The centralizer of $D$ in $A_5$ is $D$, but the unique three-dimensional defect-zero block of $k{A}_4$ is the bimodule restriction of each of the two three-dimensional defect-zero blocks of $k{A}_5$. The two ordinary degree-three characters both restrict to the degree-three character of $A_4$, with values $3,-1,0$ on its identity, involutions and elements of order three. Defect-zero reduction gives the same simple projective $k{A}_4$-module, so both restricted block bimodules are this same matrix block. This illustrates why block induction requires the block's defect centralizer. Write $e_b$ for its block idempotent and $K=N_H(D)$. Since $D{C}_G(D)\le H$, we have $D{C}_G(D)\le K\le N_G(D)$. Apply the theorem inside $H$ to obtain the unique corresponding block $\beta$ of $kK$. Its center of an associative algebra has a residue field $F_{\beta }$, and the quotient map is its central character of a block ${\omega }_{\beta }:Z(kK)\to F_{\beta }$. No assumption that $k$ is algebraically closed is needed here. DefineThe Brauer image is in $(k{C}_G(D){)}^{N_G(D)}\subseteq Z(kK)$, so this is a unital algebra homomorphism. Among the orthogonal primitive central idempotents of $kG$, exactly one has image $1$. The block it selects is $b^G$. If $H$ contains $N_G(D)$ this is the inverse of the bijection just proved. For arbitrary $H$, the resulting block can have a strictly larger defect group. Block induction is transitive on admissible intermediate subgroups. For the normalizer chain used in the proof below, the required transitivity will be verified directly with Brauer images.

To prove the bimodule characterization, we explicitly state the standard permutation-module fact used here. For a permutation module, the Brauer quotient of a module at $P$ has as basis the permutation-basis points fixed by $P$. An indecomposable summand has nonzero quotient at $P$ exactly when its vertex of an indecomposable module contains a conjugate of $P$. For summands with vertex exactly $P$, taking this quotient, with its $N(P)/P$-action, detects the summand and detects which orthogonal idempotent projection acts as its identity. These facts hold for trivial source modules; the fixed-basis description follows by deleting proper transfers, and the detection assertion is the permutation form of the first-main-theorem argument, applied to their endomorphism algebra.

As an $H\times H$ permutation module, with $(h_1,h_2)g=h_{1}g{h}_2^{-1}$, the group algebra decomposes asThe block bimodule $b$ is an indecomposable module with vertex of an indecomposable module $\Delta D=\{(d,d):d\in D\}$. None of the other double-coset modules can contain $b$: a basis element fixed by $\Delta D$ centralizes $D$, hence lies in $C_G(D)\subseteq H$, whereas those double cosets lie outside $H$. Thus $b$ occurs exactly once as an indecomposable summand of this restriction, in the $kH$ term. Decomposing $kG$ into its global block bimodules therefore puts this occurrence in precisely one of them, say $B$.

Under the fixed-basis description, the quotient of $kG$ at $\Delta D$ is $k{C}_G(D)$, and multiplication by $e_B$ induces multiplication by ${\mathrm{Br}}_D(e_B)$. The quotient of $b$ is ${\mathrm{Br}}_D^H(e_b)k{C}_G(D)=e_{\beta }k{C}_G(D)$. Consequently the detecting projection for this occurrence of $b$ satisfieswhich is equivalent to ${\omega }_b^G(e_B)=1$. Thus $B=b^G$, proving the requested unique-block direct-summand characterization. Here the restriction is to $H\times H$; the extra subscript in the converted TeX is a transcription error, not a different subgroup.

For the conjugation action, fixed elements have coefficients constant on $G$-conjugacy classes. Therefore $(\mathcal{O}G{)}^G=Z(\mathcal{O}G)$. More generally an $\mathcal{O}$-basis of $(\mathcal{O}G{)}^H$ consists of sums of the $H$-conjugation orbits on $G$.

For $a\in (\mathcal{O}G{)}^H$ define the relative trace, or transfer, byThe expression is independent of the coset representatives and is $G$-fixed. If $z\in Z(\mathcal{O}G)$, then $z{\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_H^G(za)$; hence its image is a transfer ideal of conjugation-fixed elements in the center of an associative algebra.

Let $P$ be a Sylow $p$-subgroup of $H$. Trace transitivity gives ${\mathrm{Tr}}_P^G={\mathrm{Tr}}_H^G{\mathrm{Tr}}_P^H$, proving inclusion of the $P$-image in the $H$-image. Conversely $[H:P]$ is a unit of $\mathcal{O}$, and for an $H$-fixed element $a$,Thus ${\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_P^G(a/[H:P])$, proving the reverse inclusion and

Write $N=N_G(D)$, $C=C_G(D)$, and use the commutative transfer idealsThe target fixed algebra is commutative, because $C\le N$ makes every $N$-fixed element of $kC$ central in $kC$.

The Brauer morphism and relative trace identity isIndeed let $D$ act on $G/D$ by left multiplication in the trace sum. Its fixed cosets are exactly $gD$ with $g\in N$; every other orbit has size divisible by $p$. After Brauer projection the contributions on each such nonfixed orbit are equal, so they cancel. Fixed cosets give precisely the right-hand side. Since the Brauer projection of $(kG{)}^D$ is all of $kC$, this proves a surjection $I_D\twoheadrightarrow J_D$.

We use the trace criterion for defect groups: a block idempotent $e$ belongs to $I_D$ exactly when its defect group is conjugate into $D$; among these, ${\mathrm{Br}}_D(e)\ne 0$ exactly when its defect group is $D$ up to conjugacy. The criterion can be read from the class-sum basis: if $E=C_D(x)$ thenThis coefficient is nonzero exactly when $E$ is a Sylow $p$-subgroup of $C_G(x)$. Thus $I_D$ is spanned by classes whose class defect is conjugate into $D$, and the kernel of its Brauer projection is the sum of the corresponding proper-subgroup transfer ideals. The block version follows by refinement of the central idempotents in these ideals.

For clarity, the refinement fact needed here is primitive idempotents under a surjection of commutative Artinian ideals: a surjection between ideals of commutative Artinian algebras bijects the primitive idempotents in the module source whose images are nonzero with the primitive idempotents in the target. Decompose the algebras into local factors. An ideal in a local Artinian factor either contains the identity and is the whole factor, or lies in its nilpotent maximal ideal and has no nonzero idempotents. A surviving local factor has a local quotient, so its identity remains primitive. This proves the stated refinement fact without imposing nilpotence on the entire kernel.

The primitive idempotents in $I_D$ are the block idempotents with defect conjugate into $D$, and the surviving ones have defect $D$. Applying refinement to the surjection above givesThis is the Brauer morphism on a defect transfer ideal; the target is the transfer ideal $J_D$, not the entire fixed center of an associative algebra.

The Jacobson radical is $J(A)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals of $A$; equivalently, it is the intersection of the annihilators of all simple modules. It is a two-sided ideal. A projective module has the lifting property against every surjective R-module homomorphism. A finitely generated module is a projective module precisely when it is a direct summand of a finitely generated free module. An indecomposable module is nonzero and admits no direct sum decomposition into two nonzero submodules.

To calculate the top of an indecomposable projective module, use the right Artinian ring hypothesis, namely the descending chain condition on right ideals. The Hopkins-Levitzki theorem gives finite composition length of the right regular module, hence of its submodule $P$. Also $J=J(A)$ is a nilpotent ideal, and $A/J$ is a semisimple ring. Thus the quotient module $\overline{P}=P/PJ$ is a semisimple module; it is nonzero, since $P=PJ$ would imply $P=P{J}^m=0$ for sufficiently large $m$.

Suppose $\overline{P}$ were not a simple module. A nontrivial direct sum decomposition of this semisimple module would give an idempotent $\alpha \in {\mathrm{End}}_A(\overline{P})$ that is neither zero nor the identity. Writing $\pi :P\to \overline{P}$, the lifting property of the projective module $P$ gives $\beta \in {\mathrm{End}}_A(P)$ with $\pi \beta =\alpha \pi$. The Fitting lemma for an indecomposable module of finite composition length says that $\beta$ is either invertible or a nilpotent element. Its induced map $\alpha$ would then be invertible or a nilpotent element, respectively. Neither is possible for a nontrivial idempotent. Therefore $P/PJ(A)$ is simple. This argument does not assume that an embedded projective module automatically splits off from the ambient module.

A block of an Artinian algebra is a nonzero two-sided direct summand $Ae$ determined by a primitive central idempotent $e$: $e$ cannot be written as a sum of two nonzero orthogonal central idempotents. Its identity is $e$. For a finite-dimensional associative algebra, the blocks of an Artinian algebra give its unique decomposition as a finite product of indecomposable algebras, or equivalently as a direct sum of two-sided ideals.

For the block of S3 in characteristic three, put $A=k{S}_3$, $r=(123)$, $s=(12)$ and $a=r-1$. The group algebra has basis $r^i,r^{i}s$ for $0\le i<3$. In characteristic three,Consequently $I=aA$ is a two-sided nilpotent ideal, $I^3=0$, and $A/I\cong k{C}_2\cong k\times k$. A nilpotent ideal lies in the Jacobson radical, and a quotient that is a semisimple ring forces the reverse inclusion. HenceDefine orthogonal idempotents $e_{+}=(1+s)/2$ and $e_{-}=(1-s)/2$. They sum to one, so the right regular module decomposes asThe two summands are the indecomposable projectives of S3 in characteristic three. Each direct summand is a projective module, with basis $e_{\pm },e_{\pm }r,e_{\pm }{r}^2$. Its quotient module modulo multiplication by $J(A)$ is one-dimensional: the trivial representation for $e_{+}A$, and the sign representation for $e_{-}A$. Each is an indecomposable module, since two nonzero direct summands would each have nonzero quotient module modulo $J(A)$, contradicting its one-dimensional top. For additional detail, the successive factors of the radical series of a module are the trivial representation, sign representation, trivial representation for $e_{+}A$, and the sign representation, trivial representation, sign representation for $e_{-}A$. Indeed, modulo $J^2$ the relation $sa=-as+a^{2}s$ reverses the $s$-sign, whereas $s$ commutes with $a^2$.

The center of an associative algebra is spanned by the conjugacy class sums $1$, $r+r^2$, and $(1+r+r^2)s$. Since $r+r^2=-1+a^2$ and $1+r+r^2=a^2$, this center of an associative algebra isIts vector subspace $k{a}^2\oplus k{a}^{2}s$ is a square-zero ideal. If $d1+n$ is a central idempotent, then $d^2=d$ and $(2d-1)n=0$. Thus $d=0$ or $1$ and $n=0$. There is no nontrivial central idempotent, so the whole group algebra is its single block. The two three-dimensional projective modules above are a decomposition of the regular module, not two blocks of an Artinian algebra.

Write $A=kG$, $N=N_G(D)$ and $C=C_G(D)$. The Brauer morphism is the coefficient projectionwhere the superscript $D$ denotes invariance under conjugation. This is an algebra homomorphism: for $c\in C$, the pairs $(x,y)$ with $xy=c$ are permuted by $D$; their coefficient products are constant on each orbit. Every nonfixed orbit has size divisible by $p$ and contributes zero. The fixed pairs are exactly $x,y\in C$, giving the coefficient of $c$ in the product of the projections. It is surjective since $kC\subset A^D$.

We need two trace facts. The kernel isIndeed, a basis of $A^D$ consists of conjugation-orbit sums. Nonfixed orbit sums are traces from their proper stabilizers, while fixed basis elements are exactly the elements of $C$. Traces from proper subgroups have zero coefficient at every fixed element, proving the reverse inclusion. Second, Brauer morphism and relative trace givesOne can see this by letting $D$ act on the cosets $G/D$: fixed cosets are exactly $N/D$, and all other orbit contributions vanish after projection. More generally, ${\mathrm{Br}}_E^G({\mathrm{Tr}}_D^G(A^D))=0$ unless $E$ is conjugate into $D$.

Set $I_D={\mathrm{Tr}}_D^G(A^D)$ and $J_D={\mathrm{Tr}}_D^N(kC)=(kC{)}_D^N$. The first is an ideal of the center of an associative algebra $Z(A)$; the second is an ideal of $(kC{)}^N$, which is commutative because $C\subset N$. The displayed trace identity and surjectivity of the Brauer projection give a surjective algebra homomorphismThe ideals can lack identities, so we justify the required idempotent argument. A finite-dimensional commutative algebra is a product of Artinian local rings. Its ideal intersects each local factor either in the whole factor, containing its identity, or in a proper nilpotent ideal. Under a surjection of such ideals, the proper parts have nilpotent image. Each nonzero image of a local-factor identity has a local corner algebra, a quotient of that factor, and is therefore primitive. If $u$ is the sum of these nonzero images, then $j-uj$ lies in the nilpotent image of the proper-factor ideals for every $j$ in the target; an idempotent has $j-uj=0$. Its components in the local corners are consequently zero or the corner identities. These orthogonal images account for every primitive idempotent in the target, because what remains is nilpotent. Thus primitive idempotents under a surjection of commutative Artinian ideals apply: the primitive idempotents of $J_D$ correspond exactly to primitive central idempotents $b\in I_D$ with ${\mathrm{Br}}_D^G(b)\ne 0$.

To identify these with the required blocks, we prove the trace criterion for defect groups. Choose a subgroup $D_0$ of smallest order with $b\in I_{D_0}$. Such a subgroup exists: for a Sylow $p$-subgroup $P$, $b={\mathrm{Tr}}_P^G(b/[G:P])$. If ${\mathrm{Br}}_{D_0}^G(b)=0$, write $b={\mathrm{Tr}}_{D_0}^G(a)$ and replace $a$ by $ba$. The kernel formula and transitivity of trace would giveMultiplying by $b$ gives the identity of the local algebra $bZ(A)$ in a sum of the ideals $b{I}_Q$. If all were proper, they would lie in its unique maximal ideal; hence some $b{I}_Q$ contains $b$, contradicting minimality. Therefore ${\mathrm{Br}}_{D_0}^G(b)\ne 0$.

The general trace-vanishing assertion shows that any $E$ with nonzero Brauer image is conjugate into $D_0$. Conversely, every conjugate of $D_0$ has nonzero image. Hence the maximal such subgroups, each called a defect group of a block, are precisely the conjugates of $D_0$. It also follows, using trace transitivity, that $b\in I_D$ if and only if a defect group is conjugate into $D$. Combining this with ${\mathrm{Br}}_D^G(b)\ne 0$ forces equality of the subgroup orders, so the idempotents singled out above are exactly those with defect group $D$.

We have proved the desired bijectionFinally, Brauer first main theorem states that these blocks correspond bijectively to the blocks of $k{N}_G(D)$ with defect group $D$. Apply the same argument to $N$: its normalizer of $D$ is itself and its centralizer of $D$ is still $C$. The target ideal is therefore the same $J_D$. Matching the two bijections proves the theorem and gives the Brauer correspondence, characterized by equal nonzero Brauer images. Notice that we never assume the entire kernel on $I_D$ is nilpotent; it may kill whole blocks of smaller defect.

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

Conjugation by a permutation merely permutes the transpositions. Their sum $C$ is therefore a conjugacy class sum and belongs to the center of an associative algebra $\mathbb{C}{S}_n$. Since the complex Specht module is an irreducible representation, Schur lemma says that $C$ acts on it as a scalar, say $c(\lambda )$.