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Use Euclidean coordinates $(x^1,x^2,x^3,x^4)$ and orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. For $D_{\mu }={\partial }_{\mu }+A_{\mu }$, the gauge curvature is $F_{\mu \nu }=[D_{\mu },D_{\nu }]$. The ASDYM equations are $F=-\ast F$, equivalentlyPut $w=x^1+i{x}^2$, $z=x^3+i{x}^4$, with ${\partial }_w=({\partial }_1-i{\partial }_2)/2$ and similarly for $z$. The bars label the complex conjugate coordinates on the Euclidean real slice; after complexification they can be treated as independent coordinates. A Lax pair for the anti-self-dual Yang-Mills equations isAt infinity rescale the operators and use the other affine coordinate. Their commutator isCompatibility for every $\lambda$ is therefore precisely $F_{wz}=F_{\bar{w}\bar{z}}=0$ and $F_{w\bar{w}}+F_{z\bar{z}}=0$. To verify the real component signs,The conjugate equation supplies the remaining real components. This is compatibility of an overdetermined system, not an additional field equation on $\Psi$ for only one value of the spectral parameter.

Impose invariance under translations in $x^1,x^2,x^3$, writing $t=x^4$. In a symmetry-adapted gauge the potentials depend only on $t$. A further $t$-dependent Yang-Mills gauge transformation sets $A_4=0$ locally, by solving the usual parallel transport ordinary differential equation; it does not reintroduce spatial dependence. ThenThe three real equations above become ${\dot{A}}_1=[A_2,A_3]$, ${\dot{A}}_2=[A_3,A_1]$ and ${\dot{A}}_3=[A_1,A_2]$, namelyThese are the Nahm equations. The orientation chosen above gives the requested plus sign; reversing orientation exchanges self-duality of gauge curvature and anti-self-duality of gauge curvature. Translation invariance here is a local dimensional reduction and is not a claim of finite four-dimensional action over all three translation directions.

To prove conservation, abbreviate $P=A_1+i{A}_2$, $Q=A_1-i{A}_2$ and $R=A_3$, so $A(\lambda )=P+2R\lambda -Q{\lambda }^2$ and $B(\lambda )=-iR+iQ\lambda$. Expanding the commutator, including the cancellation of the cubic term, givesThe Nahm equations giveThus the polynomial Lax representation of the Nahm equations isTake a finite-dimensional matrix representation of the Lie algebra and its complexification. For every positive integer $p$, differentiating the product and using cyclicity of the matrix trace givesThis proves the trace invariants of a Lax equation directly. Since $\mathrm{Tr}(A(\lambda {)}^p)$ is a polynomial of degree at most $2p$ in the affine coordinate $\lambda$, its derivative vanishes identically only if the derivative of every coefficient vanishes. Therefore every trace-polynomial coefficient is independent of $t$. At projective infinity these polynomials are sections of $\mathcal{O}(2p)$; they need not be constant functions of $\lambda$ on the whole projective sphere.

A smooth fiber bundle $\pi :E\to M$ with typical fiber $F$ is locally a product: every base point has a neighborhood $U$ and a local trivialization ${\pi }^{-1}(U)\cong U\times F$ commuting with projection to $U$. Its overlap maps act smoothly on $F$. A principal bundle with structure Lie group $G$ has a free right $G$-action, each fiber is one orbit, and the local trivializations are equivariant for right multiplication on $G$. A smooth section of a fiber bundle is a map $s:M\to E$ with $\pi \circ s={\mathrm{id}}_M$.

For a principal bundle, a global section of a fiber bundle gives the mapThe free transitive action on each fiber makes $\Phi$ bijective. In a local trivialization, write $s(x)=(x,a(x))$; then $\Phi (x,g)=(x,a(x)g)$ and its inverse is $(x,k)\mapsto (x,a(x{)}^{-1}k)$. Both are smooth and $\Phi$ is equivariant. Conversely, a product principal bundle has the section $x\mapsto (x,e)$. This proves a principal bundle is trivial exactly when it has a global section. The argument depends on the free transitive action: an arbitrary bundle section need not trivialize its bundle.

For an $n$-dimensional smooth manifold, the frame bundle has fiber over $x$ consisting of ordered bases of $T_{x}M$, equivalently linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. Right composition by $GL(n,\mathbb{R})$ makes it a principal bundle. If a metric has signature $(p,q)$, its pseudo-orthonormal frame bundle consists of the isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$, with structure group $O(p,q)$. In the positive-definite case this is the usual orthonormal frame bundle. Choosing space and time orientations reduces the Lorentzian structure group to its corresponding connected subgroup.

For Minkowski spacetime, use a fixed origin and a standard Lorentzian frame. Every pair consisting of a point and a pseudo-orthonormal frame is the image of this reference pair under a unique affine isometry. Thus, for the full groups,Here the first expression denotes the total space of the orthonormal frame bundle, not merely the Lorentz group. Projection sends an affine isometry to the image of the reference point, and the right action of the stabilizer changes its frame.

Realize de Sitter spacetime of radius $\ell$ as ${\eta }_{AB}{X}^A{X}^B={\ell }^2$ in ${\mathbb{R}}^{1,4}$. The ambient orthogonal group is transitive on this hyperboloid. Its point stabilizer acts on the Lorentzian tangent space as $O(1,3)$; choosing a tangent orthonormal frame removes that stabilizer completely. ConsequentlyFor oriented, time-oriented frames, replace these groups by $S{O}_0(1,4)$ and $S{O}_0(1,3)$, and similarly use the proper orthochronous Poincaré group for Minkowski spacetime. The isometries are uniquely determined by their action on a point and its frame, which explains these bundle identifications.

Finally, choose a basis $v_1,\dots ,v_n$ of the Lie algebra $T_{e}G$. The left-invariant vector fields $X_i(g)=(d{L}_g{)}_e{v}_i$ form a basis at every $g$, because $L_g$ is a diffeomorphism. They give a global section of a fiber bundle of the frame bundle, so that bundle is trivial. More explicitly,Thus every Lie group is a parallelizable manifold; connectedness is not needed for this construction.

Use the degree-one convention implicit in this enveloping-algebra assertion: a degree-one almost commutative algebra has $F_{0}R=k$, a finite-dimensional $F_{1}R$ containing $1$ which generates $R$, and commutative associated graded ring. Taking the leading symbol of a commutator givesin particular $[F_{1}R,F_{1}R]\subseteq F_{1}R$. Thus $\mathfrak{g}=F_{1}R$, equipped with $[u,v]=uv-vu$, is a finite-dimensional Lie algebra. Bilinearity and antisymmetry are immediate, and expanding the three nested commutators proves the Jacobi identity.

Its inclusion into the associative algebra $R$ is a Lie algebra homomorphism. The universal enveloping algebra is $T(\mathfrak{g})$ modulo the relations $u\otimes v-v\otimes u-[u,v]$. Sending the tensor generators to their elements of $R$ kills these relations and gives an algebra homomorphism $U(\mathfrak{g})\to R$. It is surjective because $F_{1}R$ generates $R$. ThereforeThe vector $1\in F_{1}R$ is retained as a central Lie algebra generator and maps to the associative identity; its corresponding relation in the kernel is that generator minus $1_{U(\mathfrak{g})}$. One should not first replace $F_{1}R$ by $F_{1}R/k1$, since commutators of generators can be nonzero scalars.

Terminology is important here. Merely requiring a commutative finitely generated associated graded ring for an arbitrary weighted filtration is a broader definition of almost commutative algebra; it does not supply the finite-dimensional bracket-closed first step used in this argument. The degree-one assumptions make the requested enveloping presentation precise. The Noetherianity in part (b) also follows directly under the broader finite-generation convention.

Work in finite dimension over $\mathbb{C}$; the characteristic-zero real version follows by complexification. The Cartan solvability criterion isIts semisimplicity consequence, also often called Cartan's criterion, is $\mathfrak{g}$ is semisimple if and only if its Killing form is nondegenerate. We prove both forms.

The auxiliary results used are as follows. The Lie theorem simultaneously triangularizes every finite-dimensional representation of a complex solvable Lie algebra. The Engel theorem says that a matrix Lie algebra consisting entirely of nilpotent endomorphisms has a common annihilated nonzero vector and a basis making all its matrices strictly upper triangular; in particular it is a nilpotent Lie algebra. Nilpotent means that the lower central series terminates, which implies that the derived series terminates as well. Finally, the Additive Jordan decomposition writes any complex endomorphism uniquely as $x=x_s+x_n$, where $x_s$ is diagonalizable, $x_n$ is nilpotent and they commute. On a generalized eigenspace $V_{\lambda }$, the two parts are $\lambda I$ and $x-\lambda I$, and both are polynomials in $x$. On endomorphisms, $\mathrm{ad}{x}_s$ is diagonalizable, $\mathrm{ad}{x}_n$ is nilpotent, and these are the Jordan parts of $\mathrm{ad}x$: the first assertion follows by splitting into maps between eigenspaces, and the second by expanding repeated commutators with the nilpotent $x_n$.

First prove the linear trace version: if $L\subseteq \mathfrak{gl}(V)$ andthen $L$ is solvable. Fix $x\in [L,L]$ and let $V=⨁V_{\lambda }$ be its generalized eigenspace decomposition. Define $y$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. We do not assume that $y$, $x_s$ or $x_n$ belongs to $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}x$ has the form $(\lambda -\mu )I+N$ with $N$ nilpotent, whereas $\mathrm{ad}y$ is multiplication by $\overline{\lambda -\mu }$. Choose a polynomial $q$ with $q(d)=\bar{d}$ for every occurring eigenvalue difference $d$, and with all its derivatives of positive order up to the relevant nilpotency index zero at $d$. Such a polynomial exists by Hermite interpolation, or the Chinese remainder theorem for the pairwise coprime powers of $(t-d)$. At $d=0$ its value is zero, so $q(0)=0$. Expanding $q(dI+N)$ now givesBecause $x\in L$ and $[L,L]$ is an ideal, every positive power of $\mathrm{ad}x$ sends $L$ into $[L,L]$. The zero constant term therefore gives $[y,L]\subseteq [L,L]$.

Express $x$ as a sum of commutators ${\sum }_j[a_j,b_j]$ with $a_j,b_j\in L$. Cyclicity of trace and the hypothesis implyBut on $V_{\lambda }$, $xy$ has trace $\dim (V_{\lambda })|\lambda {|}^2$, since the nilpotent part has trace zero. Henceforcing every eigenvalue of $x$ to vanish. Thus every member of the derived algebra $[L,L]$ is a nilpotent endomorphism. By the Engel theorem, $[L,L]$ is nilpotent and therefore solvable; its derived series terminates, and adjoining the initial term $L$ shows that $L$ is solvable.

Conversely, if $L$ is solvable, the Lie theorem makes it upper triangular. Commutators of upper-triangular matrices have zero diagonal, and multiplying such a commutator by an upper-triangular matrix still has zero diagonal. Therefore the trace condition holds. Apply this equivalence to $L=\mathrm{ad}\mathfrak{g}$. Its trace pairing is the Killing form, and solvability of its image is equivalent to solvability of $\mathfrak{g}$: the kernel is its abelian centre, so a terminating derived series in the quotient terminates after at most one more step in $\mathfrak{g}$. This proves the solvability criterion.

For the semisimplicity criterion, let $R$ be the radical of the Killing form. Invariance makes $R$ a Lie algebra ideal. For $x,y\in R$, their adjoint actions on $\mathfrak{g}/R$ vanish, so computing trace in a basis adapted to $R$ gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$. The solvability criterion applied to $R$ shows it is a solvable ideal. Thus a semisimple Lie algebra has $R=0$ and a nondegenerate Killing form.

Conversely, any nonzero solvable ideal has a last nonzero term $A$ in its derived series; this is an abelian ideal of $\mathfrak{g}$. For $x\in A$, $y\in \mathfrak{g}$, the map $\mathrm{ad}x\mathrm{ad}y$ sends $\mathfrak{g}$ into $A$ and vanishes on $A$, so has trace zero. Hence $A\subseteq R$. A nondegenerate Killing form therefore rules out nonzero solvable ideals, exactly the definition of semisimplicity.

For classification, nondegeneracy of the Killing form provides the duality and root-space pairings used to extract a reduced crystallographic root system from a semisimple Lie algebra. Classification then reduces to the finite Dynkin diagrams and reconstruction of the Lie algebra from its root data.

For a Lie group $G$ with identity $e$, its Lie algebra as a set is the tangent space $T_{e}G$:The ambient embedding identifies these tangent vectors with vectors in ${\mathbb{R}}^N$. For a matrix Lie group, the identity is $I$ and the tangent vectors are matrices.

If $\gamma (t)\in {\mathrm{SL}}_n$ with $\gamma (0)=I$, put $A={\gamma }^{\prime }(0)$. The determinant expansion $\det (I+tA+o(t))=1+t\mathrm{tr}A+o(t)$ follows directly from its permutation formula: to first order only the diagonal entries contribute. Since $\det \gamma (t)=1$, differentiation gives $\mathrm{tr}A=0$. ThusIn fact equality holds: if $\mathrm{tr}A=0$, then $\exp (tA)$ has determinant $\exp (t\mathrm{tr}A)=1$ and derivative $A$ at zero.

For a finite-dimensional Lie algebra, define its Killing form byThe trace identity $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ proves symmetry, and linearity of trace proves bilinearity. The Jacobi identity gives $\mathrm{ad}[x,y]=[\mathrm{ad}x,\mathrm{ad}y]$. Consequently, by cyclic invariance of trace,This is the invariance identity. Equivalently $B([h,x],y)+B(x,[h,y])=0$ for all three elements.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

In the truncated tensor algebra $T^{(N)}({\mathbb{R}}^d)$, multiplication isLet ${\mathfrak{g}}^N({\mathbb{R}}^d)$ be the Lie algebra generated by the degree-one space, with commutator $[u,v]=u\otimes v-v\otimes u$ and every term of degree greater than $N$ set to zero. The free step-N nilpotent Lie group isIts operation is the truncated tensor product, its identity is $e=(1,0,\dots ,0)$, and ${\exp }_N(\ell {)}^{-1}={\exp }_N(-\ell )$. Truncated multiplication is associative; the finite Lie exponential and logarithm identify this group with the free nilpotent Lie algebra as a manifold.

The Chow theorem says that on a connected manifold a smooth bracket-generating family of vector fields joins every pair of points by a piecewise smooth horizontal path. Applied to the degree-one left-invariant fields on this group, it says that every $g\in G^N({\mathbb{R}}^d)$ is $S_N(x{)}_{0,1}$ for a piecewise smooth, and hence suitably parametrized Lipschitz continuous, control path $x$ starting at zero.

For $N=d=2$, every Lie element is $v+a[e_1,e_2]$, with $v\in {\mathbb{R}}^2$ and $a\in \mathbb{R}$. The step-2 planar group coordinates giveWe realize this element explicitly. First traverse the rectangleIf $a=0$, omit the loop. Its displacement is zero, and its second-level entries areThus the loop signature is $\exp (a[e_1,e_2])$. Follow it by the straight segment from zero to $v$, whose signature is $\exp (v)=1+v+\frac{1}{2}v\otimes v$. By the Chen identity, concatenation has signaturebecause the commutator lies in the central second layer. Every group element has therefore been realized, proving this case of Chow's theorem.

The Carnot-Carathéodory distance is the infimum of lengths of horizontal controls realizing the relative group element. The construction gives the upper bound $\Vert \exp (v+a[e_1,e_2]){\Vert }_{CC}\le |v|+4\sqrt{|a|}$. Conversely, a control of length $L$ has displacement at most $L$, and its antisymmetric area satisfiesHere the control starts at zero, so its maximum Euclidean distance from zero is at most $L$. These estimates show that the homogeneous group norm is comparable to $|v|+|a{|}^{1/2}$.

The given path is the pure-area rough path ${\mathbf{y}}_t=\exp (t[e_1,e_2])$. Centrality givesThe graded dilations ${\delta }_r(v,a)=(rv,r^{2}a)$ scale horizontal lengths by $r$. Inversion preserves the norm. Hence, with $c=d_{CC}(e,\exp ([e_1,e_2]))>0$,Thus the path is exactly $1/2$-Hölder continuous: on a compact time interval it has every exponent at most $1/2$, and none greater.

Its p-variation is finite precisely when $p\ge 2$. Indeed, the $p$th power of the variation sum is $c^p{\sum }_j(t_{j+1}-t_j{)}^{p/2}$. For $p\ge 2$ it is bounded by $c^p{T}^{p/2}$; for $p<2$, the partition into $m$ equal intervals gives $c^p{T}^{p/2}{m}^{1-p/2}\to \infty$. In the usual step-two rough-path range it is weak geometric for $2\le p<3$, including $p=2$. For $p\ge 3$, the same formula $\exp (t[e_1,e_2])$ in $G^{\lfloor p\rfloor }$ gives the additional levels and a weak geometric $p$-rough path there. The original $G^2$ data alone are a truncated object when those additional levels are required.

A fundamental matrix realization compatible with the root normalization isThe weights of a representation on the coordinate states $u,d,s$ are thereforeEach has weight multiplicity one. The weight diagram of the defining SU(3) representation is the corresponding equilateral triangle. These are $(H_1,H_2)$ coordinates; the familiar vertical flavor hypercharge coordinate is $Y=2H_2/\sqrt{3}$.

For the adjoint representation of SU(3), the states are elements of the Lie algebra. The two Cartan generators commute with both $H_i$, giving representation weight $(0,0)$ with multiplicity two. Each root generator $E_{\pm }^r$ is an eigenvector of the Cartan adjoint action with representation weight $\pm {\alpha }_r$. ConsequentlyThe six nonzero representation weights form a regular hexagon, with a doubly occupied centre. Its highest weight is ${\alpha }_3={\omega }_1+{\omega }_2$, so its Dynkin labels are $(1,1)$ and its dimension is eight. The triplet and octet panels in the preceding figure give these two weight diagrams explicitly.

Use $\omega ={\sum }_{i}d{q}_i\wedge d{p}_i$, ${\iota }_{X_f}\omega =df$ andThen $X_f(g)=\{g,f\}$ and $[X_f,X_g]=-X_{\{f,g\}}$. For a left group action, let ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$; its fundamental fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$.