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A group representation is a group homomorphism from a group to the general linear group of a vector space. It realizes abstract group elements as invertible linear transformations.

The defining identities are $\rho (e)=I$ and $\rho (gh)=\rho (g)\rho (h)$, so an inverse element is represented by a matrix inverse. A choice of basis gives a matrix representation, but the underlying action does not depend on that choice. The action need not distinguish every group element: the trivial representation sends all of them to $I$.

The defining SU(3) representation restricts to the given matrix representation. All its restricted matrices are real, so the conjugate representation restricts identically. Hence the first two branchings areFor the adjoint representation of SU(3), use $3\otimes \overline{3}=1\oplus 8$. This follows by splitting endomorphisms into scalar and traceless matrices. Subtracting one trivial constituent from c givesAs an independent check, its restricted character is $|{\chi }_R{|}^2-1=(8,0,0,0,0)$, the regular character of $G$. Thus this is precisely the regular representation, whose representation multiplicities equal the corresponding dimensions.

A group representation on a vector space $V$ over a field $K$ is a group homomorphism $R:G\to \mathrm{GL}(V)$. In particular, $R(gh)=R(g)R(h)$, $R(e)=I$ and $R(g^{-1})=R(g{)}^{-1}$. Choosing a basis identifies it with a matrix representation.

Two group representations $R$ on $V$ and $S$ on $W$ are equivalent representations when an invertible linear map $T:V\to W$ satisfies $TR(g)=S(g)T$ for every $g$. Thus $S(g)=TR(g)T^{-1}$; they differ only by a choice of basis.

A nonzero group representation is a reducible representation if there is a vector subspace $U$ with $0\ne U\ne V$ and $R(g)U\subseteq U$ for every $g$. It is an irreducible representation if it has no such invariant subspace. Reducibility alone does not assert the existence of an invariant complementary vector subspace; that stronger conclusion says that the representation is a semisimple representation.

The Schur lemma states that an intertwiner $T$ between irreducible representations is either zero or an isomorphism. Consequently an intertwiner between inequivalent irreducible representations is zero. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:For the first assertion, the kernel and image of $T$ are invariant subspaces, so a nonzero $T$ is injective and surjective. For the scalar conclusion, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has a nonzero kernel, so it must vanish. The finite-dimensional complex hypothesis matters: the scalar conclusion need not hold over the real numbers, as a real planar rotation through $\pi /2$ is an irreducible representation of the cyclic group of order four, while the rotation itself is a commuting nonscalar endomorphism.

Now $m_1=m_2=n$ and the first two matrix representations agree in the chosen bases. Each of $P_{11},P_{12},P_{21},P_{22}$ is therefore a commuting endomorphism of that common irreducible representation, so each is scalar by the Schur lemma. All blocks connecting either of these copies with the third vanish, since the third irreducible representation is inequivalent. HenceEvery such matrix commutes with $R(g)$. In particular the off-diagonal scalars $b,c$ can mix the two identical copies. The commutant of a semisimple representation is $M_2(\mathbb{C})\oplus \mathbb{C}$, of complex dimension five. If the first two representations were merely equivalent, rather than literally equal, the mixed blocks would instead be scalar multiples of a fixed invertible intertwiner and its inverse.

There are three complex irreducible representations, because there are three conjugacy classes. Two are the trivial representation $1$ and the sign representation $\epsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the one-dimensional representations, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.

The defining two-dimensional matrix representation $D$ has characterIts character inner product with itself isso it is an irreducible representation. Directly, the two coordinate lines are the distinct eigenspaces of $r$, and $s$ exchanges them, leaving no common invariant line. The sum of squares of irreducible representation dimensions is $1^2+1^2+2^2=6$, proving completeness. In the conjugacy class order of solution, the character table isRepresentatives $r,s$ in this table stand for their full conjugacy classes. This lists all irreducible representations up to equivalence of group representations.

The two invariant subspaces areCoordinate permutations preserve both equal coordinates and the coordinate sum. Their dimensions add to three and their intersection is zero, so ${\mathbb{C}}^3=U_0\oplus U_2$. On $U_0$, the basis $u_0=(1,1,1{)}^T$ gives $R_{U_0}(\tau )=(1)$ for every transposition.

On $U_2$ choose the basis $u=e_1-e_3$, $v=e_2-e_3$. Directly permuting their entries givesThus, taking images as columns, the matrices of the matrix representation areEach has trace zero and squares to $I$. Their three-cycle products have trace $-1$, so this is exactly the two-dimensional irreducible representation of solution. These are all nonzero invariant subspaces affording individual irreducible representations, since each occurs once in solution.

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

Using the Levi-Civita symbol and the contraction of two Levi-Civita symbols, the $i$th component of the vector triple product isThus the triple-product identity is $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Apply this with the unit vector $n$ to obtain $\Pi (x)=x-(n\cdot x)n$. Dot products and scalar multiplication are linear in $x$, so $\Pi (sx+ty)=s\Pi (x)+t\Pi (y)$. Its matrix representation isThe fixed-vector equation is equivalent to $(n\cdot x)n=0$. Since $n\ne 0$, all its solutions are $x\in n^{\perp }$. Also $\boxed{\Pi (n)=0}$. Every vector decomposes as $x=\Pi (x)+(n\cdot x)n$, with $\Pi (x)$ orthogonal to $n$. Hence $\Pi$ is the orthogonal projection onto the plane through the origin normal to $n$: it fixes that plane and removes the perpendicular component.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

For a Matrix Lie group $G$, the left Maurer-Cartan form identifies tangent vectors with the Lie algebra by translating them to the identity:It is a Lie algebra-valued differential form of degree one. For a fixed $h\in G$, replacing $g$ by $hg$ gives $(hg{)}^{-1}d(hg)=g^{-1}dg$, so it is a left-invariant differential form.

Differentiate $g^{-1}g=I$ to get $d(g^{-1})=-g^{-1}(dg)g^{-1}$. The exterior derivative then givesHere the exterior product of matrix-valued differential forms includes matrix multiplication; the order of the matrices matters. Hence the Maurer-Cartan equation is

Write $\rho ={\sigma }^{\alpha }{T}_{\alpha }$. Antisymmetry of the exterior product impliesComparison of the Lie algebra components yields the Maurer-Cartan equation in a Lie-algebra basis:Thus the canonical antisymmetric choice is $f^{\gamma }{}_{\alpha \beta }=-\frac{1}{2}{c}^{\gamma }{}_{\alpha \beta }$. The factor one half is required because the sum includes both ordered pairs $(\alpha ,\beta )$ and $(\beta ,\alpha )$. If only $\alpha <\beta$ is summed, its coefficient is $-c^{\gamma }{}_{\alpha \beta }$. Strictly, the equality of differential forms determines only the antisymmetric part of $f$: one may add any tensor symmetric in $\alpha ,\beta$ without changing it. This is the symmetric-part ambiguity in Maurer-Cartan coefficients.

A faithful matrix representation of the orientation-preserving real affine group isIts action on $(x,1{)}^T$ has first component $ax+b$. The group operation and inverse areDifferent transformations have different matrix entries, proving faithfulness. With the Lie algebra basisthe left Maurer-Cartan form isThe requested left-invariant differential forms therefore form the Maurer-Cartan coframe of the real affine group:For a direct invariance check, left translation by $(a_0,b_0)$ sends $(a,b)$ to $(a_{0}a,b_0+a_{0}b)$, and the pullbacks of the two displayed differential forms are unchanged. Their dual left-invariant vector fields are $a{\partial }_a$ and $a{\partial }_b$, whose bracket is $a{\partial }_b$, in agreement with $[D,T]=T$.

The columns of $P$ are the coordinates of the vectors ${\mathbf{f}}_j$, so they form a basis exactly whenFor a linear transformation $\alpha$, define its matrix representation in the standard basis byThe change-of-basis matrix satisfies $[\mathbf{v}{]}_{\mathbf{e}}=P[\mathbf{v}{]}_{\mathbf{f}}$. Thereforeand multiplication by $P^{-1}$ gives the similar matrix relation

The characteristic polynomial of the displayed $A$ isEigenvectors $(2,-1,2{)}^T$, $(1,0,0{)}^T$, and $(0,-1,3{)}^T$ giveIndeed, $\det P=1$ andso $P^{-1}AP=B$. Finally, the diagonalization of a matrix gives

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

Write $g(x)=I+x^i{T}_i+O(x^2)$ in a matrix representation and $[T_i,T_j]=c^k{}_{ij}{T}_k$. Direct multiplication givessoThe cancellation of the constant and linear terms explains why the group commutator first detects the Lie bracket at quadratic order.

For $q=2^{2a+1}\ge 8$, the Suzuki group $Sz(q)={}^2{B}_2(q)$ is a twisted finite group of Lie type. In characteristic two, the algebraic symplectic group over a field $S{p}_4$ has an exceptional endomorphism $\sigma$ satisfying ${\sigma }^2=F_2$. Fixed points of ${\sigma }^{2a+1}$ give $Sz(q)$; its natural matrix representation has degree four.^[ref]

It is simple, of order $q^2(q^2+1)(q-1)$, and acts two-transitively on the $q^2+1$ points of a Suzuki ovoid. A point stabilizer has order $q^2(q-1)$ and contains a subgroup of order $q^2$ acting regularly on the remaining points. At $q=8$ the degree is $65$ and the order is $29120$. The parameter $q=2$ instead gives a nonsimple group of order $20$.