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Every finite-dimensional Lie algebra over a field of characteristic zero has a finite-dimensional faithful Lie algebra representation. Thus it embeds as a Lie subalgebra of a matrix algebra, using the matrix commutator as bracket. Faithfulness is crucial for detecting central elements: the Adjoint representation alone kills the center of a Lie algebra.

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.

For a nonzero finite-dimensional complex vector space $V$, a unital matrix algebra acting irreducibly on $V$ is the full endomorphism algebra. Its Jacobson radical annihilates $V$: otherwise the radical times $V$ is all of $V$, contradicting nilpotence. Faithfulness kills the radical, and the Artin–Wedderburn theorem then leaves one simple matrix factor acting on its standard column module. This is a matrix-algebra theorem, distinct from the orbit-counting Burnside lemma and the finite-group Burnside's theorem.

Let $k$ be an algebraically closed field and let a block of a group algebra $B$ have a central defect group of a block $D$. Its quotient by $J(kD)B$ is a defect-zero block, hence a full matrix algebra over $k$. The algebra $B$ is free over the central local ring $kD$. Lift a primitive idempotent from this matrix quotient; its corner is free of rank one over $kD$, and is therefore $kD$. Lifting matrix units gives $B\cong {\mathrm{Mat}}_r(kD)$.

Let $N◃G$, let $b$ be a block of a group algebra of $kN$, and let $I$ be its inertia subgroup. A block of $kG$ covering $b$ is a full matrix algebra of degree $[G:I]$ over its corresponding block of $kI$. When $I=N$, the corner is just $kNb$. Let $e_B$ be the global block idempotent and put $c=e_{B}b$. With coset representatives $g_i$, the elements $g_{i}c{g}_j^{-1}$ are matrix units in that global block because different conjugates of $b$ are orthogonal. When $I=N$, one has $c=b$.

A commutative unital subalgebra $D$ of an associative algebra $A$ is maximal commutative if no larger commutative subalgebra contains it. Equivalently $C_A(D)=D$: any element of its centralizer of a subalgebra generates a commutative algebra together with $D$. The full diagonal matrix algebra inside $M_n(\mathbb{C})$ is an example.

Restriction of an irreducible representation to a subgroup is multiplicity-free when its semisimple representation decomposition has no repeated irreducible summands. Over the complex numbers, for finite groups, this is equivalent to the restricted module's endomorphism algebra being commutative. This follows by writing that algebra as a product of matrix algebras of sizes equal to the multiplicities.

Alexandre-Théophile Vandermonde presents a programme in elimination theory: remove unknowns from simultaneous polynomial equations in a way that works uniformly as the number of equations and their polynomial degrees vary. The immediate achievement is narrower than that programme: a general construction for linear equations. The academy year 1772 used in the exam is not the printing date of the memoir, which appeared in 1776. Its compact indices distinguished the equation containing a coefficient from the unknown it multiplied; they were not powers. That notational separation is crucial to interpreting the work.^[ref][ref]

In modern language take a system of linear equations $A\mathbf{x}=\mathbf{b}$ with $n$ unknowns. Its determinant is an alternating expression in the coefficients:Each product selects one coefficient from each equation and each unknown's column. The sign of a permutation determines whether it is added or subtracted. This makes the number $n$ a genuine parameter of a general rule, rather than merely supplying separately memorised formulas for two or three unknowns. The notation here is modern, not a transcription of Alexandre-Théophile Vandermonde's symbols.

For two unknowns, multiply $ax+by=u$ by $d$ and $cx+dy=v$ by $b$, then subtract. Similarly eliminate $x$. This givesThe same principle in arbitrary size is Cramer's rule. If $A_j(\mathbf{b})$ is obtained by replacing column $j$ of $A$ by $\mathbf{b}$, thenAn original verification uses the multilinearity of the determinant: replace $\mathbf{b}$ by the column combination ${\sum }_k{x}_k{A}_k$. Every resulting determinant vanishes through repeated columns except the $k=j$ term, which is $x_j\det A$. This explains how an elimination expression isolates one unknown.

The condition $D\ne 0$ is substantive. If $D=0$, multiplying by an expression which vanishes cannot justify division. For instance, $x+y=1$ and $2x+2y=3$ are inconsistent, whereas replacing $3$ by $2$ gives infinitely many solutions. Thus an elimination formula must distinguish exceptional coefficients and preserve the connection to the original system of linear equations. Nor does the displayed rule solve the much broader problem of arbitrary nonlinear equations.

The passage's importance is the move from particular elimination calculations towards a theory of alternating expressions. It would be misleading to turn this into the claim that the familiar Vandermonde determinant, formed from powers of variables, is what this extract displays. A positional upper index and an exponent have different meanings; the history of that eponym requires separate evidence. Likewise, modern matrix algebra is a useful explanatory language here, not the institutional or conceptual setting of an eighteenth-century memoir.^[ref]

A central simple algebra over $K$ is a finite-dimensional associative $K$-algebra with no nonzero proper two-sided ideals and with center $K$. The Artin–Wedderburn theorem writes it uniquely as $M_r(D)$ with $D$ a central division algebra. After a finite separable splitting field extension it becomes $M_d$, so its dimension is $d^2$, and $d$ is its degree of a central simple algebra. Passing from $D$ to a matrix algebra over $D$ does not change its Brauer class.

The Brauer group consists of these classes, with product induced by tensor product. Extending to a common splitting field shows that a tensor product of central simple algebras is again central simple; its two-sided ideals and center descend from the corresponding full matrix algebra. The neutral class is $[K]$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$, becauseThe displayed map becomes the standard isomorphism for matrices over a splitting field, so it is already an isomorphism over $K$. Thus Brauer classes measure precisely the obstruction to being a split matrix algebra.

An explicit family is supplied by cyclic algebras. Let $L/K$ be cyclic of degree $n$, with generator $\sigma$, and let $a\in K^{\times }$. DefineThis is an associative algebra: form the skew polynomial ring with relation $ux=\sigma (x)u$ and quotient by the central relation $u^n-a$. Reduction of powers gives the displayed $n^2$-dimensional $K$-basis. An element commuting with every $x\in L$ has no $u^i$ coefficient for $i\ne 0$, since ${\sigma }^i$ is not the identity. Commuting also with $u$ forces the remaining coefficient into $K$. Hence the center is $K$.

Here is an explicit splitting construction. On $L^n$, with basis $e_0,\dots ,e_{n-1}$, putThen $UD(x)=D(\sigma x)U$ and $U^n=aI$. After extending scalars to $L$, the separable decomposition $L{\otimes }_{K}L\cong {\prod }_{j=0}^{n-1}L$ supplies all diagonal matrices, and their products with powers of $U$ supply all matrix units. ThereforeAny nonzero proper two-sided ideal of $A$ would extend to one of $M_n(L)$, which is impossible. Thus this construction really gives a central simple algebra, with splitting field $L$.

Its central arithmetic feature is the splitting criterion for a cyclic algebra:If $a=Nc$, let $L$ act by multiplication on the $K$-vector space $L$, and let $u$ act as $c\sigma$. Its nth power is multiplication by $Nc=a$, so this gives a representation $A\to {\mathrm{End}}_K(L)$. It is injective by simplicity and an isomorphism by equal dimension. Conversely, if $A\cong {\mathrm{End}}_K(V)$ with ${\dim }_{K}V=n$, its embedded field $L$ makes $V$ one-dimensional over $L$. The relation for $u$ says that $u$ is $\sigma$-semilinear. In an $L$-basis it must have the form $c\sigma$, whose nth power is $Nc$. The relation $u^n=a$ forces $a=Nc$.

More generally the crossed-product algebra of a Galois extension attaches an algebra to a multiplicative two-cocycle. Changing the cocycle by a coboundary rescales its basis and preserves the Brauer class. Descent of a split matrix algebra identifiesFor cyclic $L/K$, cyclic Tate cohomology identifies this second cohomology group with $K^{\times }/N{L}^{\times }$. In a crossed product, rescale successive cyclic basis elements to powers of a single $u$; the remaining datum is $u^n=a$, changed by a field norm when $u$ is rescaled. This explains why cyclic algebras give every class split by a cyclic extension, and why multiplication of their parameters is addition of their Brauer classes.

For a finite field $K={\mathbb{F}}_q$, every finite splitting field $L={\mathbb{F}}_{q^n}$ is cyclic over $K$. The field norm is the power map $x\mapsto x^{(q^n-1)/(q-1)}$ on its cyclic multiplicative group, and is onto ${\mathbb{F}}_q^{\times }$. Thus every relative Brauer class is zero, and every central simple algebra has zero class:

For a non-Archimedean local field $K$, the valuation classification of central division algebras shows that every Brauer class splits over a finite unramified extension. For the degree-$n$ unramified extension $L/K$ with arithmetic Frobenius $\sigma$, units are norms and norm valuations are multiples of $n$. Hence the cyclic relative Brauer group is $\mathbb{Z}/n\mathbb{Z}$, represented by $a={\pi }^r$. The compatible local Brauer invariant isThe union over all $n$ givesA class with invariant $r/n$ in lowest terms has a central division algebra representative of degree of a central simple algebra $n$. This is the local valuation classification used above; the explicit cyclic algebras exhibit all of its invariants. At the Archimedean places, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$: its nontrivial class is the quaternion algebra $(\mathbb{C}/\mathbb{R},\text{conjugation},-1)$, since complex norms are positive.

For a number field, the Albert-Brauer-Hasse-Noether theorem describes the Brauer group by the exact sequenceThus its classes are exactly finite-support families of local invariants with sum zero; real invariants are $0$ or $1/2$ and complex invariants are zero. Every global Brauer class has a cyclic splitting extension, so cyclic algebras also realize all global classes. If $\chi (\sigma )=1/n$, their local invariants are ${\chi }_v({\mathrm{rec}}_{K_v}(a))$, with ${\chi }_v$ restricted to the local decomposition subgroup. The sum-zero relation for a principal parameter is the invariant form of Artin reciprocity. Injectivity supplies the local-to-global splitting test, which yields the Hasse norm theorem for cyclic extensions as explained in the preceding solution. Here the local valuation classification and the global invariant theorem are the structural theorems used to describe the Brauer groups; the construction and norm-splitting calculation have been proved explicitly.

A central simple algebra over $K$ is a finite-dimensional associative unital $K$-algebra $A$ with no nonzero proper two-sided ideals and center exactly $K$. Its dimension is a square, ${\dim }_{K}A=n^2$. The Artin–Wedderburn theorem writes it as $M_r(D)$ for a uniquely determined central division algebra $D$ up to isomorphism. One obtains this structure by taking a minimal left ideal and its commuting division endomorphism ring; the simple algebra acts as all endomorphisms over that ring. Over a suitable finite splitting field it becomes a full matrix algebra.

The Brauer group identifies two central simple algebras if their division-algebra parts agree, equivalently if tensoring each with suitable matrix algebras makes them isomorphic. The product is tensor product, the identity is the class of $K$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$. Indeedis an isomorphism for a central simple algebra, so the product class is split. This also explains why matrix size is irrelevant to the Brauer class. The order of the class is its period, while the degree of its division-algebra representative is its index; these are different notions in general.

For the cyclic algebra construction, take a cyclic extension $L/K$ of degree $m$, a generator $\sigma$ and $a\in K^{\ast }$. SetThe multiplication is obtained by moving coefficients past $u$ and replacing every $u^m$ by $a$. It is associative: reducing a product of three monomials gives the same total exponent, the same power of $a$ and the same iterated automorphisms in either order. The scalar $a$ is fixed by $\sigma$ and ${\sigma }^m=1$, so reduction is consistent.

An element commuting with every $\ell \in L$ has only its $u^0$ coefficient nonzero, because ${\sigma }^i\ne 1$ for $0<i<m$. Commuting with $u$ then forces that coefficient into $K$, proving that the center is $K$. To prove simplicity, choose a nonzero element of a nonzero ideal with the fewest nonzero monomial coefficients. Since $u$ is invertible and each nonzero coefficient in $L$ is invertible, normalize one term to be the constant one. If another term remains, its automorphism differs from the identity; commuting with a suitable element of $L$ eliminates the constant term but not that other term. This gives a shorter nonzero element, a contradiction. The minimal element is therefore a unit, and the ideal is all of $A$. Thus $A$ is central simple of dimension $m^2$.

It splits over $L$. For example, over $L$ represent $\ell$ by the diagonal matrix with entries $\ell ,\sigma (\ell ),\dots ,{\sigma }^{m-1}(\ell )$, and let $u$ shift the standard basis down one place, sending the first vector to $a$ times the last. These matrices satisfy the defining relations. A primitive element of the separable extension $L/K$ has distinct conjugates; the Vandermonde matrix of their powers is invertible. Thus the diagonal images span all diagonal matrices after scalar extension to $L$. Multiplying the diagonal matrix units by powers of the shift gives every matrix unit, proving $A{\otimes }_{K}L\cong M_m(L)$.

The particularly useful splitting criterion isIf $a=N(b)$, let the algebra act on $L$ by multiplication for $\ell$ and by the semilinear operator $b\sigma$ for $u$. Its $m$th power is multiplication by $N(b)=a$, so this gives an isomorphism with ${\mathrm{End}}_K(L)$, by simplicity and dimension. Conversely, a split algebra acts on an $m$-dimensional $K$-space $V$. Restriction to the subfield $L$ makes $V$ one-dimensional over $L$, and $u$ is a nonzero $\sigma$-semilinear operator. In an $L$-basis it is $b\sigma$, so $u^m=a$ forces $a=N(b)$. More generally the relative Brauer group split by a cyclic $L/K$ is $K^{\ast }/N_{L/K}{L}^{\ast }$, with multiplication of parameters corresponding to addition of classes.

For a finite field, a finite division algebra is commutative by Wedderburn's little theorem. Therefore every central simple algebra is a matrix algebra and $\boxed{\mathrm{Br}({\mathbb{F}}_q)=0}$. The cyclic-algebra viewpoint gives the same conclusion for cyclic classes because the norm between finite fields is surjective on their cyclic multiplicative groups.

For a nonarchimedean local field, the local Brauer invariant givesAn unramified cyclic extension of degree $m$, with arithmetic Frobenius generator, gives invariant $v_K(a)/m$ to its cyclic algebra. Question 2 shows that all units are norms; norm valuations are multiples of $m$. Thus the relative classes are $\mathbb{Z}/m\mathbb{Z}$, represented by powers of a uniformizer. Allowing all $m$ realizes all invariants. At the infinite local fields, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})\cong \mathbb{Z}/2\mathbb{Z}$, with the nonzero class represented by Hamilton's quaternions and invariant $1/2$.

For a number field, the Albert-Brauer-Hasse-Noether theorem supplies the exact sequenceHence a global algebra is determined by its finitely many nonzero local invariants, whose sum is zero; it is split precisely when all its local classes vanish. Every division algebra over a number field is cyclic, so cyclic algebras provide representatives for the global classes as well. Their norm parameters translate splitting into local norm conditions and the reciprocity constraint on the sum of invariants. The local invariant theorem, Wedderburn's little theorem and the global invariant theorem are the standard structural results used here; their role in the requested computation is explicit.

Start with the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$. For the dimension count, complexify and rescale the timelike matrix so that the generators obey the Euclidean complex relation $\{{\gamma }_a,{\gamma }_b\}=2{\delta }_{ab}$. This changes neither the complex representation dimension nor its decomposition under the complexified even algebra. In $D=2k$, pair the generators intoDirect substitution gives $\{a_i,a_j\}=\{b_i,b_j\}=0$ and $\{a_i,b_j\}={\delta }_{ij}$. The commuting operators $n_j=b_j{a}_j$ satisfy $n_j^2=n_j$, so their occupation eigenvalues are zero or one. An irreducible module contains an all-zero occupation state $|0⟩$, obtained by lowering the occupied entries of a simultaneous eigenvector. The statesare nonzero and independent by their distinct joint occupation eigenvalues. They span an invariant module; irreducibility makes this the full module. Conversely this Fock construction supplies a representation on exactly these $2^k$ states. Products of occupation projectors and raising/lowering operators give its matrix units, so the generated algebra is the full matrix algebra. Therefore the minimal complex Dirac spinor has

This module is irreducible under the full Clifford algebra, but the Spinor representation of the Lorentz group is generated by the even bilinears ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\gamma }^{\mu },{\gamma }^{\nu }]$. The chirality matrix in the Euclideanized basis isIts two eigenspaces therefore remain invariant under Lorentz transformations. Each has half the Fock states, since any gamma matrix bijects them. The even algebra connects every pair of occupation states of the same parity and has no further invariant subspaces within either one. HenceThese are the irreducible Weyl spinors. Globally they are ordinary representations of $\mathrm{Spin}(D-1,1)$, the double cover; calling them representations of $SO(D-1,1)$ uses the conventional projective spinor terminology.

For $D=2k+1$, adjoin ${\gamma }_{2k+1}=\pm {\gamma }_{\ast }$. It anticommutes with the old matrices and gives a Clifford module of the same dimension $2^k$. The new even generators $[{\gamma }_a,{\gamma }_{2k+1}]$ interchange the two old chirality eigenspaces, so the odd-dimensional Lorentz spinor is irreducible of complex dimension $2^k$, with no Weyl splitting. The product of all odd-dimensional gamma matrices is central and acts as a scalar; it cannot be a new anticommuting chirality operator. Its two signs distinguish the two complex odd-Clifford modules, whose restrictions to Spin group are equivalent. Majorana spinor reality conditions, when the signature permits them, are additional conditions and should not be confused with complex chirality.

For the reduction, take the massless zero modes on a flat seven-torus with no flux or projections. The eleven-dimensional fields are the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. Split $M=(\mu ,i)$ with $\mu =0,\dots ,3$ and $i=1,\dots ,7$. The metric supplies $G_{\mu \nu }$, seven vectors $G_{\mu i}$ and 28 scalar fields $G_{ij}=G_{ji}$. The three-form supplies a four-dimensional three-form $A_{\mu \nu \rho }$, seven two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars $A_{ijk}$.

A massless two-form in four dimensions has one physical particle polarization and can be dualized to a scalar by $H_3=\ast d\sigma$; the seven two-forms thus supply seven more scalars. The four-dimensional three-form has no local particle polarization, since its four-form field strength is locally a constant times the volume form. The total bosonic content is consequentlyThe eleven-dimensional Majorana spinor has 32 real components, decomposing into eight four-dimensional Majorana spinors. Accordingly ${\Psi }_{\mu }$ supplies eight gravitini, while ${\Psi }_i$ supplies $7\times 8=56$ spin-$1/2$ fields after diagonalizing the gravitino/matter kinetic terms. The resulting theory has 32 real supercharges, hence is four-dimensional N=8 supergravity, withThe on-shell counts check the reduction: the metric has $11(11-3)/2=44$ polarizations and the three-form has $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, totaling 128. In four dimensions $2+28\cdot 2+70=128$ bosonic polarizations match $8\cdot 2+56\cdot 2=128$ fermionic ones. Higher Kaluza–Klein levels, fluxes or projections would be additional choices beyond this massless toroidal reduction.

The Brauer group of a field consists of equivalence classes of central simple algebras, with addition induced by tensor product, zero represented by matrix algebras, and inverse represented by the opposite algebra. Localization sends a global algebra to its scalar extensions at the completions. For a non-Archimedean completion of a number field, the local Brauer invariant is an isomorphismNormalize it so that an unramified cyclic algebra of degree $m$, with arithmetic Frobenius generator and parameter $a$, has invariant $v(a)/m$. For a real completion, $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2$ with the Hamilton quaternion algebra mapped to $1/2$; for a complex completion the Brauer group is zero.

The global computation is the exact sequence in the Albert-Brauer-Hasse-Noether theorem:Thus a global Brauer class is determined uniquely by a finite-support family of local invariants. Such a family occurs if and only if its sum is zero; at real places the only allowed entries are $0,1/2$, and at complex places the entry is zero. Injectivity is a local-global splitting principle; the middle exactness is the reciprocity constraint; surjectivity of the sum follows already from any one finite local summand. These are separate assertions, not merely different descriptions of the sum formula.

For an explicit group description, choose one finite place $v_0$. Assign arbitrary finite-support invariants at all other finite places and arbitrary allowed invariants at the real places, then set the invariant at $v_0$ equal to minus their sum. This givesThe description by the entire invariant family is canonical; this displayed elimination of one coordinate depends on the choice of $v_0$.

The connection with Artin reciprocity is especially transparent for cyclic algebras. Let $\chi :\mathrm{Gal}(L/K)\hookrightarrow \mathbb{Q}/\mathbb{Z}$ describe a cyclic extension, with $\chi (\sigma )=1/n$. Write $(\chi ,a)$ for the class of $(L/K,\sigma ,a)$. The local reciprocity pairing isIt follows from the local fundamental class construction of the Local Artin map; on an unramified character it is precisely the normalization $v(a_v)/n$. The group kernel in the parameter is the local norm group, and restriction/corestriction give the norm compatibility of this pairing.

For a principal parameter $a\in K^{\times }$, the zero-sum identity for its global cyclic algebra therefore readsOn the other hand the idelic Artin reciprocity law isand evaluating this product by $\chi$ gives exactly that zero-sum identity. Conversely, testing every cyclic character of a finite abelian Galois group separates its elements, so the zero-sum identities for cyclic algebras imply this principal-idele reciprocity law for every finite abelian extension. At almost all places a unit and an unramified character pair trivially, so the sums and products are finite.

The full idelic statement identifies the group kernel as $K^{\times }{N}_{L/K}{J}_L$ and induces $J_K/(K^{\times }{N}_{L/K}{J}_L)\cong \mathrm{Gal}(L/K)$. It explains why the local norm-residue pairings must glue globally. The global Brauer theorem supplies, in addition, the injectivity and realization statements needed to compute all algebra classes; the principal-product identity alone would not prove those statements. For cyclic algebras its injectivity yields the Hasse norm theorem, as in Question 3. In degree two, an invariant is either zero or $1/2$, so the zero-sum law becomes the Hilbert reciprocity law ${\prod }_v(a,b{)}_v=1$. For example, a quaternion algebra over $\mathbb{Q}$ can be ramified at precisely a prescribed finite set of places, including possibly the real place, if and only if that set has even cardinality; the invariants uniquely determine its Brauer class. This illustrates how arithmetic reciprocity becomes a concrete constraint on global algebra structure.

The map $T\mapsto T^{\mathsf{T}}A+AT$ is linear, so its kernel $L$ is a vector subspace of the matrix algebra. If $T_1,T_2\in L$, use $T_i^{\mathsf{T}}A=-A{T}_i$ to computeThus $L$ is closed under the commutator. This bracket is bilinear and satisfies $[T,T]=0$. Associativity of matrix multiplication gives the Jacobi identity: expanding $[T_1,[T_2,T_3]]+[T_2,[T_3,T_1]]+[T_3,[T_1,T_2]]$ makes each of the six ordered triple products occur once with each sign. Therefore $L$ is a Lie algebra. No nonsingularity or symmetry assumption on $A$ was needed.

For a unital star-subalgebra $A\subseteq B(H)$, the Von Neumann double commutant theorem statesHere $A^{\prime }$ is the commutant of an operator algebra, consisting of operators commuting with every element of $A$. The double commutant is closed in either topology: multiplication by a fixed bounded operator is continuous in both. Therefore both closures of $A$ lie in $A^{\prime \prime }$.

For the converse, let $T\in A^{\prime \prime }$ and choose finitely many vectors ${\xi }_1,\dots ,{\xi }_n$. In $H^n$ putIt is invariant under the diagonal action of $A$ and its adjoints, so its orthogonal projection $Q$ commutes with that diagonal action. Consequently each matrix entry $Q_{ij}$ belongs to $A^{\prime }$. The diagonal operator with entry $T$ therefore commutes with $Q$. Since $I\in A$, the tuple $\xi =({\xi }_1,\dots ,{\xi }_n)$ belongs to $K$, and hence so does $(T{\xi }_1,\dots ,T{\xi }_n)$. By the definition of $K$, one element $a\in A$ approximates this entire tuple arbitrarily closely. This is exactly approximation in every strong-operator neighborhood of $T$. It proves $A^{\prime \prime }\subseteq {\overline{A}}^{\mathrm{SOT}}$ and thus both asserted equalities. In particular, $M=A^{\prime \prime }$ is a Von Neumann algebra.

The Kaplansky density theorem strengthens the result by preserving norm bounds:It also holds for the self-adjoint and positive unit balls; the full ball can in fact be approximated in the strong-star topology, meaning strong convergence of both operators and adjoints. We prove the bound rather than merely scaling the unbounded approximants furnished by the double commutant theorem.

First put $C={\overline{A}}^{\Vert \cdot \Vert }$, a unital C-star algebra. Its self-adjoint part is weak-operator dense in $M_{\mathrm{sa}}$: symmetrize any weak approximating net. For a convex subset, strong and weak operator closures agree. To see this explicitly, separation from a strong closure supplies a continuous real functional on finitely many image vectors, hence one of the form $\mathrm{Re}{\sum }_j⟨a{\xi }_j,{\eta }_j⟩$, which is also weak-operator continuous. The Hahn-Banach separation theorem would then separate the point from the weak closure as well. Thus every self-adjoint $x\in M$ is a strong limit of self-adjoint $a_i\in C$, initially without a common norm bound.

The needed functional-calculus continuity follows directly from resolvents. The resolvent identity givesThe first factor has norm at most one, so the right side tends strongly to zero. The same holds at $-i$. Both resolvent nets are uniformly bounded, so their products converge strongly. Their scalar functions generate a self-adjoint algebra separating points of $\mathbb{R}$ and vanishing at infinity. The Stone-Weierstrass theorem, applied to the one-point compactification, makes this algebra uniformly dense in $C_0(\mathbb{R})$. Uniform approximation and the functional-calculus norm bound therefore showThis proves strong continuity of functional calculus through resolvents without an unjustified uniform bound on the original $a_i$.

If $x=x^{\ast }$ and $\Vert x\Vert \le 1$, choose a real compactly supported continuous $f$ equal to the identity on $[-1,1]$ and bounded in absolute value by one. Then $f(a_i)\in C$ are self-adjoint contractions and converge strongly to $f(x)=x$. For a positive contraction $x$, choose instead $0\le f\le 1$ equal to the identity on $[0,1]$; this gives positive contractions. These prove the two restricted versions for $C$.

For an arbitrary contraction $x\in M$, apply the self-adjoint result in the matrix algebra $M_2(C)$ toThe strong closure of $M_2(C)$ is $M_2(M)$, since finitely many entries may be approximated simultaneously. The upper-right entries $c_i$ of the approximating self-adjoint contraction matrices belong to $C$, satisfy $\Vert c_i\Vert \le 1$, and converge strongly to $x$. The lower-left entries are $c_i^{\ast }$ and converge strongly to $x^{\ast }$. This proves the full strong-star assertion for $C$.

Finally transfer the bounds from $C$ to the possibly nonclosed algebra $A$. A contraction $c\in C$ has an approximant $b\in A$ with $\Vert b-c\Vert <\epsilon$, so $b/(1+\epsilon )$ is an $A$-contraction within $2\epsilon$ of $c$. Self-adjoint approximants are obtained by symmetrizing first. For a positive contraction, approximate $c^{1/2}$ by $b\in A$ and use the positive contraction $b^{\ast }b/(1+\epsilon {)}^2$, which tends in norm to $c$ as $\epsilon \downarrow 0$. Combining these norm approximations with each finite-vector strong approximation proves all versions for $A$. The reverse inclusions follow because the relevant norm bounds and the inequalities defining a positive operator are preserved by strong limits.

Interpret the involution as the adjoint for the Hilbert-space structure on $V$. The hypotheses make $\Omega$ a cyclic vector for an operator algebra for both $A$ and $A^{\prime }$. They also make it a separating vector for each: if $a\Omega =0$, then $a$ vanishes on $A^{\prime }\Omega =V$, and similarly for $A^{\prime }$. Thus the Tomita operatoris well defined, antilinear, invertible and satisfies $S^2=I$. Define the modular operator and modular conjugation byWe calculate them explicitly to prove the commutant assertion.

The finite-dimensional structure of a unital star-algebra gives an orthogonal decompositionThis is the usual matrix-block structure: minimal central projections separate the simple summands, and each simple matrix algebra acts as its defining representation with a multiplicity space. Identify each vector block with an $n_j$-by-$m_j$ matrix $D_j$ with its Hilbert-Schmidt inner product. The two algebras act by left and right multiplication. If $D_j$ has rank $r_j$, the left orbit has dimension $n_j{r}_j$ and the right orbit dimension $m_j{r}_j$. Cyclicity on both sides therefore forces $r_j=m_j=n_j$. Every $D_j$ is square and invertible.

Right multiplication by a suitable unitary is a unitary change of this identification commuting with the left algebra. The polar decomposition of each $D_j$ therefore lets us assume $D_j>0$. In one such block, for an arbitrary matrix $X=aD$, the Tomita operator isIts polar factors areIndeed, $J$ is an antiunitary involution and $J{\Delta }^{1/2}(X)=D^{-1}{X}^{\ast }D$. The fact that ${\Delta }^{1/2}$ is a positive operator is transparent in a basis diagonalizing $D$: it multiplies the matrix unit $E_{kl}$ by the positive ratio $d_k/d_l$. These are therefore the unique polar factors of $S$.

Finally, if $L_a$ denotes left multiplication,As $a$ ranges over the full matrix block, these are exactly its right multiplications. Applying this on every block and transporting back by the unitary identifications provesThis also establishes the required modular definitions intrinsically, independent of the chosen matrix coordinates.

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.

We first prove the simple Artinian case of the Artin–Wedderburn theorem. This part does not need finite dimension over $k$. Choose a minimal nonzero left ideal $L$, using the descending chain condition. Simplicity gives $LA=A$. Also $L^2\ne 0$: otherwise $LAL=(LA)L=L$ would be zero, since each $la{l}^{\prime }$ lies in $L^2$. Choose $x\in L$ for which right multiplication $r_x:L\to L$ is nonzero. By simplicity of $L$, it is an automorphism. Choose $e\in L$ with $ex=x$. Then $(e^2-e)x=0$, and injectivity of $r_x$ gives $e^2=e$. Thus $L=Ae$.

Right multiplication by $e$ splits the left regular module as $A=Ae\oplus A(1-e)$. Repeat within the remaining left ideal: if it contains a minimal left ideal $Af$ generated by an idempotent, the map $v\mapsto vf$ splits off $Af$. Each nonzero remaining complement is a strictly smaller left ideal, so the Artinian condition terminates the process. Therefore the regular module is a finite direct sum of minimal left ideals.

All these simple summands are isomorphic. For primitive idempotents $e,f$, simplicity gives $AeA=A$, so $fAe\ne 0$; otherwise $fAeA=fA$ would be zero. A nonzero element of $fAe$ defines a nonzero homomorphism $Af\to Ae$, which is an isomorphism between simple modules. Thus $A\cong L^{\oplus n}$ as a left module. Put $\Delta ={\mathrm{End}}_A(L)$, a division algebra by the Schur lemma. Right multiplication givesTaking opposites, with matrix transposition reversing products, provesThe construction is $k$-linear and embeds $k$ into the center of $D$.

A central simple algebra is a finite-dimensional simple $k$-algebra whose center is exactly $k$. The Brauer group consists of such algebras modulo matrix stabilization: $A,B$ represent the same class if $M_r(A)\cong M_s(B)$ for some positive integers $r,s$. Its operation and inverse areThe tensor product is central simple, and $A\otimes A^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ by $(a\otimes b)(x)=axb$, explaining the inverse. Both facts can be checked after a splitting field of a central simple algebra extension, where they are matrix-algebra identities, and descended by faithful scalar extension. Every class has a unique central division-algebra representative, by the regular-module description above.

To prove infinitude over $\mathbb{Q}$ explicitly, put $E=\mathbb{Q}(i)$ and, for $a\ne 0$, letThis is a quaternion algebra. It is central simple: the center is $\mathbb{Q}$, and after adjoining $i$ its generators have the faithful matrix modelThese generate $M_2(E)$; consequently a nonzero proper two-sided ideal over $\mathbb{Q}$ is impossible. The quaternion conjugation givesA nonzero element has an inverse exactly when this norm is nonzero. Thus $A_a$ is a division algebra when $a$ is not an $E/\mathbb{Q}$ norm; if $a$ is a norm it has a nonzero zero divisor and is $M_2(\mathbb{Q})$ by the classification just proved.

For a prime $p\equiv 3(\mathrm{mod}4)$, every nonzero rational sum $x^2+y^2$ has even $p$-adic valuation. Divide out the minimum valuation of $x,y$; cancellation modulo $p$ is impossible because $-1$ is not a square modulo $p$. Therefore $p$ is not an $E/\mathbb{Q}$ norm, and $A_p$ is a nontrivial division-algebra Brauer class. This is the quaternion norm obstruction at primes congruent to three modulo four.

We also prove the product rule needed to distinguish classes. In $A_a\otimes A_b$, write the generators as $i_1,j_1,i_2,j_2$. The commuting subalgebras generated respectively byare $A_{ab}$ and the split quaternion algebra $(1,b{)}_{\mathbb{Q}}\cong M_2(\mathbb{Q})$. The latter has the matrix model $\mathrm{diag}(1,-1)$ and $\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$. Together the two subalgebras generate the whole tensor product, since they recover $i_2$ and $j_1$. Dimensions are sixteen on both sides, givingQuaternion conjugation identifies $A_a$ with its opposite, so these classes have order at most two. For distinct primes $p,q\equiv 3(\mathrm{mod}4)$, the parameter $pq$ has odd valuation at $p$, so $A_{pq}$ is a division algebra. Thus $[A_p]\ne [A_q]$. There are infinitely many such primes: if their list were $p_1,\dots ,p_m$, the integer $4p_1\cdots p_m-1$ would have a prime factor congruent to three modulo four outside that list. We concludeIndeed the same valuation argument for any nonempty finite product gives independent quaternion classes over rational numbers.

For $\mathbb{R}$, we prove the required division-algebra classification. Let $D$ be finite-dimensional and central over $\mathbb{R}$. If $D\ne \mathbb{R}$, choose a nonreal element; its irreducible real minimal polynomial is quadratic, so translating and rescaling it gives $i\in D$ with $i^2=-1$. The centralizer $D_{+}$ of $i$ is a finite-dimensional division algebra over $\mathbb{C}=\mathbb{R}(i)$, with that field central. Every element has a split minimal polynomial over $\mathbb{C}$, hence belongs to $\mathbb{C}$, so $D_{+}=\mathbb{C}$.

Conjugation by $i$ is an involution, giving $D=D_{+}\oplus D_{-}$, where $D_{-}=\{x:ix=-xi\}$. If $D_{-}=0$, then $D=\mathbb{C}$, contrary to its center being $\mathbb{R}$. Choose $j\ne 0$ in $D_{-}$. For any $x\in D_{-}$, $x{j}^{-1}$ commutes with $i$, so $D_{-}=\mathbb{C}j$. Also $j^2\in \mathbb{C}$ and commutes with $j$, forcing $j^2\in \mathbb{R}$. It cannot be positive, since $(j-\sqrt{j^2})(j+\sqrt{j^2})=0$, or zero, since $D$ is a division algebra. Rescale to $j^2=-1$. ConsequentlyConversely the positive definite quaternion norm makes $\mathbb{H}$ a division algebra. The two possible central division representatives are therefore $\mathbb{R}$ and $\mathbb{H}$. They are not equivalent, and conjugation identifies $\mathbb{H}$ with its opposite, soThis uses the finite-dimensional real division-algebra classification directly, not as an unexplained theorem citation.

Finally let $k_s$ be a separable closure and $\Gamma =\mathrm{Gal}(k_s/k)$. The cohomological description isHere $k_s^{\times }$ has the discrete topology and its natural Galois action. For a finite Galois splitting field $L$, identify $A{\otimes }_{k}L$ with a matrix algebra. Its descent action is implemented by matrices $u_{\sigma }$, unique up to scalars, soThe inverse convention here agrees with the crossed-product convention below. The semilinear operators $u_{\sigma }\sigma$ commute with the descended algebra $A$; their crossed product has class $[A{]}^{-1}$ by the commuting-factor decomposition of the full endomorphism algebra. Inverting its scalar cocycle therefore gives the class $[A]$. Associativity gives the two-cocycle identity, and changing the representatives changes $c$ by a coboundary. Conversely a two-cocycle defines the crossed-product algebra of a Galois extension, with $u_{\sigma }a=\sigma (a)u_{\sigma }$ and $u_{\sigma }{u}_{\tau }=c(\sigma ,\tau )u_{\sigma \tau }$. It is central simple and split by $L$. Galois descent identifies coboundaries with split matrix-algebra classes and gives $H^2(\mathrm{Gal}(L/k),L^{\times })\cong \ker (\mathrm{Br}(k)\to \mathrm{Br}(L))$. Every central simple algebra has a finite separable, hence finite Galois, splitting field, yielding the displayed limit.