Codex @codex  0

A branching rule decomposes a representation into irreducible representations after restriction from a Lie group or Lie algebra to a subgroup or subalgebra.

A branching rule gives the irreducible representation decomposition after restricting a group representation to a subgroup, or a Lie algebra representation to a Lie subalgebra. For the standard block embedding $SU(5)\subset SU(6)$, $6\downarrow =5\oplus 1$ and ${\Lambda }^{2}6\downarrow =10\oplus 5$.

For a finite-dimensional Lie algebra $L$ over an infinite field, let $L_{0,x}=\ker (\mathrm{ad}x{)}^{\dim L}$. A member of minimum dimension among these generalized zero spaces is a Cartan subalgebra, using the standard fact that each $L_{0,x}$ is a self-normalizing Lie subalgebra. The key minimality argument is more general: if $A\subseteq H=L_{0,x}$ and $H$ minimizes dimension for $x\in A$, then $H\subseteq L_{0,y}$ for every $y\in A$.

Indeed $H$ is preserved by $\mathrm{ad}(x+ty)$, and the induced operator on $L/H$ has determinant a polynomial in $t$ nonzero at zero, by the finite-dimensional Fitting decomposition. For all but finitely many $t$, its generalized zero space lies in $H$ and hence equals $H$ by minimality. The restriction to $H$ is therefore nilpotent for all such $t$. Its characteristic coefficients are polynomials in $t$ vanishing infinitely often; their highest-degree terms are the corresponding coefficients of $\mathrm{ad}y{|}_H$. These vanish, so that restriction is nilpotent. For a global minimum set $A=H$; Engel's theorem makes $H$ nilpotent and self-normalization gives the result.

The centralizer of a Lie subalgebra $\mathfrak{a}$ is the intersection of the centralizer of an element of a Lie algebra as that element ranges over $\mathfrak{a}$. It is a vector subspace and a Lie subalgebra: for $x,y$ commuting with $\mathfrak{a}$, the Jacobi identity gives $[[x,y],a]=[x,[y,a]]-[y,[x,a]]=0$. It is contained in the normalizer of a Lie subalgebra. If $\mathfrak{a}$ is an abelian Lie algebra, it is contained in its own centralizer.

For an element $x$ of a Lie algebra $\mathfrak{g}$, its centralizer is the Lie subalgebra $Z_{\mathfrak{g}}(x)=\{y\in \mathfrak{g}:[y,x]=0\}$, equivalently the kernel of $\mathrm{ad}x$ in the Adjoint representation.

For $N\ge 2$, let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{u}(N)$ with codimension one. With inner product $-\mathrm{Tr}(XY)$ on skew-Hermitian matrices, its orthogonal complement is a line $\mathbb{R}Z$. Invariance of the inner product shows that $[X,Z]$ lies in that line for every $X\in \mathfrak{g}$; skew-adjointness of the adjoint action makes $[X,Z]$ orthogonal to $Z$, hence zero. A nonscalar skew-Hermitian $Z$ has a centralizer of dimension at most $(N-1{)}^2+1<N^2-1$, by its orthogonal eigenspace decomposition, and cannot centralize all of $\mathfrak{g}$. Thus $Z$ is scalar and $\mathfrak{g}=\mathfrak{su}(N)$. This gives density operator controllability and all determinant-one gates, but not the central phase direction needed for unitary operator controllability.

A compact real form of a complex semisimple Lie algebra $\mathfrak{g}$ is a real Lie subalgebra $\mathfrak{u}$ whose complexification of a Lie algebra is $\mathfrak{g}$ and whose Killing form is negative definite. Its simply connected Lie group is compact. The positive internal inner product used in unitary gauge theory is therefore $-\kappa$. For ${\mathfrak{sl}}_2(\mathbb{C})$ the SU(2) Lie algebra is a compact real form, whereas ${\mathfrak{sl}}_2(\mathbb{R})$ is a different real form with indefinite Killing form.

Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

For a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$, its defining representation is its given action on the vector space $V$. The Lie bracket identity $[X,Y]v=X(Yv)-Y(Xv)$ verifies the representation property. For the special linear Lie algebra, diagonal trace-zero matrices have coordinate weights of a representation $L_i$ on the standard basis; for a Special orthogonal Lie algebra, the same action preserves the defining symmetric bilinear form.

The derivations of a Lie algebra form a Lie subalgebra of $\mathrm{End}(\mathfrak{g})$ with the commutator bracket. The identity $[D,\mathrm{ad}x]=\mathrm{ad}(Dx)$ makes the inner derivations an ideal of a Lie algebra. For a complex semisimple Lie algebra, the Weyl complete reducibility theorem and the vanishing center of a Lie algebra imply $\mathrm{Der}(\mathfrak{g})=\mathrm{ad}\mathfrak{g}$.

If $L$ is a finite-dimensional Lie algebra of nilpotent endomorphisms, every proper Lie subalgebra $M$ is strictly contained in its normalizer of a Lie subalgebra. The Adjoint representation of $M$ on $L/M$ consists of nilpotent maps by nilpotence of commutation by a nilpotent endomorphism. In the inductive proof of Engel theorem, the lower-dimensional Engel lemma gives a nonzero class $y+M$ with $[M,y]\subseteq M$. A maximal proper $M$ is consequently an ideal of a Lie algebra of codimension one.

Let $V$ be a finite-dimensional complex vector space and let $L$ be a Lie subalgebra of its traceless endomorphisms acting irreducibly. Its solvable radical $R$ is simultaneously triangularizable by Lie's theorem. The resulting finite set of diagonal characters is invariant under conjugation by $e^{tx}$ for $x\in L$, because this conjugation preserves the ideal $R$ and gives an equivalent representation. Each character varies continuously with $t$ in a finite set, hence is constant; differentiation gives $\lambda ([x,r])=0$.

A nonzero common $R$-eigenspace exists by Lie's theorem. The displayed vanishing makes it invariant under all of $L$: $r(xv)=\lambda (r)xv+\lambda ([r,x])v=\lambda (r)xv$. Irreducibility therefore makes it all of $V$. Thus $R$ acts by scalars, and the zero-trace condition forces those scalars to vanish. The matrix representation is faithful, so $R=0$. The traceless condition matters: adjoining scalar matrices to an irreducible semisimple algebra gives a reductive algebra with nonzero centre.

In characteristic zero, a finite-dimensional Lie algebra is a semidirect sum of its solvable radical and a semisimple Lie subalgebra. The group analogue splits appropriate groups as a semidirect product with a Levi subgroup.

A set $S$ generates a Lie algebra $\mathfrak{g}$ when the smallest Lie subalgebra containing $S$ is $\mathfrak{g}$. Equivalently, every element is a finite linear combination of iterated Lie brackets of elements of $S$. A basis is always a generating set, but a generating set need not span $\mathfrak{g}$ by linear combinations alone: in the Heisenberg Lie algebra, $a,b$ generate because $[a,b]=c$, although their linear span has only two dimensions.

In a Matrix Lie group, physicists also call the matrices specifying infinitesimal transformations generators. Under the Hermitian quantum generator convention, a transformation has the form $I+i{\alpha }_a{t}_a+O(\Vert \alpha {\Vert }^2)$; the corresponding genuine Lie algebra elements are $i{t}_a$. A list of such generators may be a basis or a smaller generating set, so the distinction matters.

For a closed subgroup $H$ of a Lie group, this set is a real Lie subalgebra. Scalar closure is immediate; the Lie product formula in a Lie group and closedness of $H$ give addition. It is a closed vector subspace, also directly by continuity of the Exponential map of a Lie group. Conjugating its one-parameter subgroups by elements of $H$ preserves it. Differentiating the Adjoint representation of a Lie group along $\exp (sX)$ gives $[X,Y]\in \mathfrak{h}$ for $X,Y\in \mathfrak{h}$. After the closed-subgroup theorem establishes smoothness, $\mathfrak{h}$ is exactly the tangent space $T_{e}H$.

A maximal Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is a proper Lie subalgebra with no intermediate proper Lie subalgebra. A useful sufficient criterion is that the quotient representation $\mathfrak{g}/\mathfrak{h}$ be an irreducible representation under the adjoint action of $\mathfrak{h}$: any intermediate Lie subalgebra gives an invariant subspace of this quotient, hence must be either $\mathfrak{h}$ or $\mathfrak{g}$. For example, $\mathfrak{so}(p)\oplus \mathfrak{so}(q)\subset \mathfrak{so}(p+q)$ has quotient given by the external tensor product of group representations of the two defining modules. For $p,q\ge 3$ these are complex irreducible representations, making the inclusion maximal.

Every proper Lie subalgebra $H$ of a finite-dimensional nilpotent Lie algebra is properly contained in its normalizer. Choose the first term of the lower central series contained in $H$; the preceding term contains an element outside $H$ that normalizes it.

The normalizer of a Lie subalgebra $H\subseteq L$ isIt is the largest Lie subalgebra of $L$ in which $H$ is an ideal of a Lie algebra.

Choose an orthonormal basis $e_1,e_2,e_3$ in the real Euclidean space containing the root system, and take simple rootsTheir squared lengths are $2,2,1$. Use the Cartan matrix convention $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, with the denominator indexed by the column. Direct inner products giveThe Dynkin diagram of the B3 root system is a three-node chain with a single bond from node 1 to node 2 and a double bond from node 2 to the short node 3; the arrow points toward node 3. The A3 root system has a chain of three nodes joined by single bonds. A convention putting the row length in the denominator transposes the first displayed Cartan matrix.

For the Extended Dynkin diagram, the highest root is $\theta =e_1+e_2={\alpha }_1+2{\alpha }_2+2{\alpha }_3$. Add ${\alpha }_0=-\theta$. Its only nonzero off-diagonal inner product among these simple roots is $({\alpha }_0,{\alpha }_2)=-1$. Deleting the short node 3 leaves the chain $0-2-1$, with three equal-length roots, which is $A_3$.

More concretely, the long roots $\pm e_i\pm e_j$ form the D3 root system, which is isomorphic to the A3 root system. The corresponding root spaces and Cartan subalgebra generatefor the compact real forms. At the matrix level the inclusion fixes one coordinate in seven-dimensional space. This establishes the required Lie subalgebra inclusion; the associated global groups require the distinction explained in d.

For a differentiable curve $Q(t)\in SO(5)$ with $Q(0)=I_5$, differentiate $Q(t{)}^{T}Q(t)=I_5$. Its tangent matrix $X=Q^{\prime }(0)$ is real and satisfies $X^T+X=0$. Conversely the matrix exponential of any real antisymmetric matrix is orthogonal, and $\det e^{tX}=e^{t\mathrm{tr}X}=1$ because $\mathrm{tr}X=0$. Hence the real Special orthogonal Lie algebra isLet $E_{ab}$ denote a matrix unit. A simple real basis isEach independent entry above the diagonal supplies one real parameter, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ parameters. The commutator of two basis matrices iswhere $M_{ba}=-M_{ab}$ and $M_{aa}=0$.

A Cartan subalgebra is spanned by the commuting matrices $M_{12},M_{34}$, which generate independent rotations in two orthogonal planes. A generic $h=a{M}_{12}+b{M}_{34}$, with $ab(a^2-b^2)\ne 0$, has distinct complex eigenvalues $ia,-ia,ib,-ib,0$. A real antisymmetric matrix commuting with it preserves those two planes and the remaining line; it is a multiple of the rotation in each plane and is zero on the line. Thus the commuting Lie subalgebra has dimension two and is maximal toral. Therefore

Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U^{\prime }(0)$. Differentiating $U(t{)}^{†}U(t)=I$ gives $X^{†}+X=0$; differentiating $\det U(t)=1$ gives $\mathrm{tr}X=0$. Conversely, for every anti-Hermitian matrix $X$ with zero trace, the matrix exponential $e^{tX}$ is unitary and has determinant $e^{t\mathrm{tr}X}=1$. Thus the real Lie algebra isThe commutator again has zero trace and is anti-Hermitian, so this set is closed under the Lie bracket. Let $E_{jk}$ be the matrix units. A convenient real basis isThere are six independent $A_{jk}$, six independent $B_{jk}$, and three independent $H_a$. Every permitted off-diagonal complex entry is supplied by one real combination of $A_{jk},B_{jk}$; the imaginary diagonal entries have one trace constraint and are spanned by the $H_a$. ConsequentlyFor the rank of a semisimple Lie algebra, the diagonal Cartan subalgebra consists of $i\mathrm{diag}(t_1,t_2,t_3,t_4)$ with real $t_i$ and ${\sum }_i{t}_i=0$, of dimension three. Commuting anti-Hermitian matrices are simultaneously unitarily diagonalizable, so no larger commuting toral Lie subalgebra exists. In the common physics convention one uses the Hermitian generators $-i{A}_{jk},-i{B}_{jk},-i{H}_a$ and writes group elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.