Codex @codex  0

Use ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. For an actual infinitesimal left Lie group action, define ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$. Its fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ with this convention.

Existence and its global obstruction. Preservation of the symplectic form gives ${\mathcal{L}}_{{\xi }_P}\omega =0$. By Cartan's magic formula, ${\iota }_{{\xi }_P}\omega$ is closed. Locally it equals $d{\mu }_{\xi }$; globally this requires its class in $H_{\mathrm{dR}}^1(P)$ to vanish. If it does for every $\xi$, choose potentials linearly in $\xi$ and defineThis gives the component Hamiltonians of a weakly Hamiltonian action. The assumption in the question does not alone guarantee a global moment map: on $T^2$ with $\omega =dx\wedge dy$, translation in $x$ preserves $\omega$, but its contraction is $dy$, whose integral around the $y$ circle is nonzero. It is not an exact one-form. Thus no global real-valued component Hamiltonian exists for that circle action. The intended construction is valid locally, or globally with this exactness condition, for example if $H_{\mathrm{dR}}^1(P)=0$.

Matching the Lie algebra. On each connected component, the moment-map equivariance obstruction is the constantIndeed, the identities $[X_f,X_g]=-X_{\{f,g\}}$ and $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ show that its differential vanishes. The Jacobi identity for the Poisson bracket saysChanging ${\mu }_{\xi }$ by a linear constant $b(\xi )$ changes $c$ to $c-b([\xi ,\eta ])$. Consequently the necessary and sufficient infinitesimal condition is the cocycle must be a coboundary, so that the constants can be chosen withEquivalently its class in $H^2(\mathfrak{g},\mathbb{R})$ vanishes. For a connected group this infinitesimal equivariance integrates to equivariance under its coadjoint action. A disconnected group can impose additional conditions. Here “coincides” means representing the specified Lie brackets; it does not guarantee injectivity when the group action has ineffective generators.

Proof using the Killing form. A nondegenerate Killing form $B$ makes $\mathfrak{g}$ semisimple. First, it is perfect: the $B$-orthogonal complement of $[\mathfrak{g},\mathfrak{g}]$ is the center by invariance, and the center is zero by nondegeneracy. For symplectic fields,by Cartan's magic formula and the antihomomorphism sign. Every commutator therefore has an exact contraction. Perfectness supplies component Hamiltonians for every generator, even when $P$ is not simply connected: this is Hamiltonian existence for a semisimple symplectic action.

It remains to eliminate the cocycle, not merely to name an existence theorem. Define a linear operator $D$ on $\mathfrak{g}$ by $B(D\xi ,\eta )=c(\xi ,\eta )$. Antisymmetry of $c$ makes $D$ skew for $B$. Combining its cocycle identity with invariance of $B$ givesChoose $Z\in \mathfrak{g}$ by $B(Z,\xi )=\mathrm{tr}(D{\mathrm{ad}}_{\xi })$, possible because $B$ is nondegenerate. Since $D$ is a derivation, $[D,{\mathrm{ad}}_{\xi }]={\mathrm{ad}}_{D\xi }$. Cyclicity of the matrix trace then givesThus $D={\mathrm{ad}}_Z$, and $c(\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes $c$, proving the required condition. This applies to semisimple Lie groups, including $SU(n)$ for $n\ge 2$, $SO(3)$ and $SL(n,\mathbb{R})$ for $n\ge 2$. It is nondegeneracy, not positive definiteness, that matters. Groups with a nonzero central Lie algebra, such as $U(3)$, do not satisfy this Killing-form criterion, although particular actions can still have equivariant moment maps.

The isotropic oscillator. On $T^{\ast }{\mathbb{R}}^3$, putThe action $z\mapsto Uz$, $U\in U(3)$, preserves the Hermitian norm and the symplectic form, and therefore preserves $H$. A Hermitian matrix $A$ parametrizes the Lie algebra generator $\xi =-iA$. Its unitary moment map of an isotropic oscillator isSince $\{z_j,{\bar{z}}_k\}=-i{\delta }_{jk}$, direct differentiation yieldsThe matrix $-i[A,B]$ is Hermitian and corresponds precisely to $[{\xi }_A,{\xi }_B]$; no central defect remains. The full moment map can be identified, using the pairing with Hermitian matrices, with the rank-one matrix $z{z}^{†}$. The central generator $A=I$ gives ${\mu }_I=H$, whose flow is $z(t)=e^{-it}z(0)$. Globally $U(3)=(SU(3)\times U(1))/{\mathbb{Z}}_3$, so the “$U(1)$ factor” refers to this central circle, rather than a literal direct-product decomposition.

The real orthogonal subgroup acts geometrically by $(x,p)\mapsto (Rx,Rp)$, the cotangent lift of a diffeomorphism of ${\mathbb{R}}^3$. If $K$ is real antisymmetric and $A=iK$, then ${\mu }_A=-x^{T}Kp$. For $Kv=\Omega \times v$, this is $\Omega \cdot (x\times p)$, so its three components are the usual angular momentum $L_i={ϵ}_{ijk}{x}_j{p}_k$.

The remaining six independent Hermitian directions can be taken as real symmetric matrices. Their moment functions spanThe diagonal ones are the individual oscillator energies; their sum is $H$. The off-diagonal ones correlate two oscillator modes. Their flows mix positions and momenta and are not cotangent lifts of the obvious spatial rotation action: for a real symmetric $A$, $\dot{x}=Ap$ and $\dot{p}=-Ax$, rather than $\dot{x}=Kx$, $\dot{p}=Kp$. All nine quadratic functions commute with $H$, since the entire unitary group action preserves the oscillator Hamiltonian.

Use the corrected generating series of the Kostant partition function and the permitted Weyl character formula:For a fixed $w$, the coefficient of $e^{\mu }$ arises exactly whenThere are only finitely many Weyl-group terms, and the partition function vanishes outside the cone of nonnegative root combinations. Coefficient extraction is therefore legitimate in the formal completion used in the preceding part. Summing the signed contributions provesThis proves the Kostant multiplicity formula for every compact connected Lie group, not just the unitary groups. In the presence of a central torus, roots have zero central component, so the partition function automatically gives zero for any incompatible central weight. Also $\rho$ need not itself be an integral weight for the given global group: $w\rho -\rho$ always belongs to the root lattice, so every argument in the formula is nevertheless integral whenever $\lambda$ and $\mu$ are integral group weights.

As a normalization check, for $SU(2)$ write a highest weight as the nonnegative integer $\ell$, with positive root $2$ and $\rho =1$. The two Weyl-group terms giveSince $p(a)=1$ precisely when $a$ is even and nonnegative, this is one for $\mu =\ell ,\ell -2,\dots ,-\ell$ and zero otherwise, including the cancellation below the lowest weight. This recovers the full familiar weight string.

Write the unitary conjugation as $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. The identity component is unchanged, and the generalized Bloch vector transforms byThese coefficients are real, by the same trace argument as in part (a). Unitary conjugation preserves the Hilbert-Schmidt inner product, so the conjugated traceless generators form another orthonormal basis. It follows that $O(U{)}^{T}O(U)=I$. The unitary group is connected, $O(I)=I$, and the determinant of an orthogonal matrix is $\pm 1$; hence $\det O(U)=1$. Thus Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$.

Equivalently, in units $\hslash =1$, the Von Neumann equation givesCyclicity of the trace shows $A_{k\ell }=-A_{\ell k}$, so $A$ is a real skew-symmetric matrix. This is the generator form of Hamiltonian rotations of Bloch vectors, valid also for a time-dependent Quantum Hamiltonian. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator spectrum, not merely its purity of a density operator.

Work with integral cohomology. For a frame $v=(v_1,\dots ,v_k)$ in the complex Stiefel manifold, defineThese fibres form the complement bundle on a complex Stiefel manifold, a complex vector subbundle of the trivial rank-$n$ bundle. Indeedis a smooth orthogonal projector of constant complex rank $n-k$; locally projecting suitable fixed basis vectors and applying Gram-Schmidt process provides bundle frames. The underlying real vector bundle has rank $r=2(n-k)$ and is oriented by its complex structure: a complex change of frame has real determinant equal to the positive number $|{\det }_{\mathbb{C}}A{|}^2$.

A point of its sphere bundle is $(v_1,\dots ,v_k,w)$ with $\Vert w\Vert =1$ and $w$ orthogonal to the frame. Appending $w$ gives a diffeomorphismwhose inverse forgets the last vector while retaining it as the unit vector in the complement. Thus the asserted homotopy equivalence is actually an identification of sphere bundles, for $0\le k<n$.

Start the induction at $V_0({\mathbb{C}}^n)=\{\mathrm{point}\}$, with cohomology the exterior algebra on no generators. SupposeThis is a free abelian group in every degree, and for $k\ge 1$ its smallest positive degree is $2n-2k+1=r+1$. Hence $H^r(V_k;\mathbb{Z})=0$. This vanishing holds at $k=0$ as well, since the base is a point and $r=2n>0$. The Euler class $e(E)$ belongs to this zero group and is therefore zero.

For this oriented rank-$r$ vector bundle, the Gysin sequence of a sphere bundle has segmentBecause $e(E)=0$, this becomesIn particular $H^{\ast }(V_{k+1};\mathbb{Z})$ is torsion-free. At $j=r-1$, choose a class $a\in H^{r-1}(V_{k+1};\mathbb{Z})$ with ${\pi }_{!}a=1$. Its restriction to the sphere fibre is an integral generator. The projection formula for sphere bundle integration, with the conventional fibre-integration sign, isIt can be obtained from the relative cohomology of the disk/sphere pair and the Thom isomorphism theorem. Consequently every cohomology class of $V_{k+1}$ is uniquely expressible as ${\pi }^{\ast }c+{\pi }^{\ast }b⌣a$: subtract the second term using its fibre integral, then use exactness to identify the remaining term; injectivity of ${\pi }^{\ast }$ and the projection formula prove uniqueness. This establishes an additive splitting by the old monomials and the old monomials multiplied by $a$.

To establish the ring, $|a|=r-1$ is odd, so graded commutativity of the cup product gives $a^2=-a^2$, and hence $2a^2=0$. The torsion-freeness just proved now gives $a^2=0$. Graded commutativity also gives the required anticommutation with each old odd generator, and the old relations persist under the injective ring map ${\pi }^{\ast }$. The additive splitting proves that there are no additional relations. This is exactly Gysin ring splitting for an odd-dimensional sphere bundle, and supplies a new exterior generator of degreeInduction therefore proves the full ring calculationOver $\mathbb{Z}$, graded commutativity by itself only kills twice an odd square; torsion-freeness is the essential extra step in this integral cohomology of a complex Stiefel manifold proof.

Finally an ordered orthonormal $n$-frame is a unitary matrix, so $V_n({\mathbb{C}}^n)=U(n)$. Each exterior generator may be included once or omitted from a basis monomial. The rank of degree $j$ is therefore the number of subsets of the generator degrees $1,3,\dots ,2n-1$ summing to $j$. Multiplication of their two-term generating factors givesThe highest exponent is $1+3+\cdots +(2n-1)=n^2$, agreeing with the real dimension of the unitary group, and the total rank is $2^n$.

Polar deformation reduces maps from the circle into $G{L}_N$ to maps into $U(N)$. Define their integer invariant by the winding number of the determinant. It is unchanged under homotopy and stabilization, and determinants multiply under block sum, so it gives a homomorphism $K^1(S^1)\to \mathbb{Z}$. Every integer occurs, via $z\mapsto \mathrm{diag}(z^m,1,\dots ,1)$.

To prove injectivity, first consider a map with determinant winding zero. Its determinant has a continuous logarithm $ih(z)$ on the circle. Multiplying the first row by $e^{-ith(z)}$ gives a homotopy into $SU(N)$ at $t=1$. If $N=1$ that endpoint is already the identity. For $N\ge 2$, $SU(N)$ is simply connected: $SU(2)\cong S^3$, and the last-column fibrationwith locally trivializations from orthonormal completion, together with its homotopy exact sequence, proves this inductively because the sphere has zero first and second homotopy groups. Thus the remaining circle map is null-homotopic. For a map of winding $m$, multiply by $\mathrm{diag}(z^{-m},1,\dots ,1)$ and apply the zero-winding argument. This provesFor higher spheres, stable unitary homotopy gives $K^1(S^n)={\pi }_n(U)$: a map can be normalized by multiplying by the inverse of its value at the basepoint, a free homotopy because the unitary group is path connected. A free homotopy between based maps can be made based by replacing $H(x,t)$ with $H(x_0,t{)}^{-1}H(x,t)$, without altering its endpoints. The Bott periodicity theorem, in its unitary form ${\pi }_{j+2}(U)\cong {\pi }_j(U)$ with ${\pi }_0(U)=0$ and ${\pi }_1(U)=\mathbb{Z}$, givesThe circle case was proved directly above; the higher-sphere values use the stated general periodicity theorem.

Over $\mathbb{R}$ or $\mathbb{C}$, a Lie algebra is a vector space with a bilinear map $[,]$ which is alternating and obeys the Jacobi identity:Bilinearity and alternation imply $[X,Y]=-[Y,X]$.

For a Matrix Lie group, identify the tangent space $\mathfrak{g}=T_{I}G$ with derivatives ${\gamma }^{\prime }(0)$ of smooth curves through $I$. Product curves show that the sum of two such derivatives is again tangent, and reparametrization supplies scalar multiples. In particular this is a real vector space, even when the matrices have complex entries. For $X,Y\in \mathfrak{g}$, choose a curve $\gamma (s)$ with ${\gamma }^{\prime }(0)=X$. Conjugating a curve with derivative $Y$ shows thatDifferentiate this curve in the finite-dimensional vector space $\mathfrak{g}$. The result isThe matrix commutator is bilinear and alternating, and expanding the six terms proves its Jacobi identity. Thus this construction gives the Lie algebra of a matrix Lie group, with the appropriate bracket, using actual group curves rather than an assumed commutator closure.

For the unitary group, differentiating $\gamma (t{)}^{†}\gamma (t)=I$ gives $X^{†}+X=0$. Conversely, if $X^{†}=-X$, then $e^{tX}$ is unitary and is a curve with derivative $X$. ConsequentlyThis is the unitary Lie algebra. The diagonal entries are purely imaginary, contributing $N$ real parameters, and each upper off-diagonal entry contributes two real parameters. A matrix-unit basis of the unitary Lie algebra isThese $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$ anti-Hermitian matrix units and their combinations are linearly independent over $\mathbb{R}$ and span every allowed entry.

The symplectic stabilizer is a subgroup: the identity preserves $J$, and if $MJ{M}^T=J$ and $NJ{N}^T=J$, thenMultiplying $MJ{M}^T=J$ on the left by $M^{-1}$ and on the right by $M^{-T}$ gives $M^{-1}J{M}^{-T}=J$. Products and inverses remain unitary. This is the compact symplectic group, often denoted $USp(2n)$ or $Sp(n)$.

Differentiating the stabilizer equation gives $XJ+J{X}^T=0$. For $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, this saysTogether with $X^{†}=-X$, these are equivalentlyThese conditions are also sufficient: $e^{tX}$ is unitary, andHence it stays in the subgroup. They characterize its compact symplectic Lie algebra without adding any trace condition. In fact the trace automatically vanishes. The free anti-Hermitian block $A$ has $n^2$ real parameters, and the complex symmetric matrix $B$ has $n(n+1)$ real parameters. Thus

Here is a complete matrix-unit basis of the compact symplectic Lie algebra. The generators supplying $A$ areThe two families supplying the real and imaginary parts of $B$ are, for $1\le i\le j\le n$,The denominator merely avoids double-counting diagonal entries. Each displayed generator obeys both defining tangent conditions. The $H,P,Q$ generators form a real basis of the allowed $A$ blocks, and the $U,V$ generators form a real basis of the complex symmetric $B$ blocks. Thus they are independent and their total number is $n(2n+1)$. They give all generators required by the real compact algebra, including the case $n=1$.

The internal direct product theorem states that if $H,K⊴G$, $H\cap K=\{1\}$, and $HK=G$, then $G\cong H\times K$. Indeed, for $h\in H,k\in K$, the commutator $[h,k]$ lies in both $H$ and $K$, so it is trivial. The map $(h,k)\mapsto hk$ is consequently a homomorphism; the intersection condition makes it injective and $HK=G$ makes it surjective. Conversely, the two factors in a direct product satisfy these conditions.

In odd dimension, $-I$ is central and has determinant $-1$. Every $Q\in O(3)$ has a unique expression $Q=(-I{)}^{\epsilon }R$ with $R\in SO(3)$, soBy contrast, $SO(2)\times C_2$ is abelian, while reflections in $O(2)$ conjugate a rotation to its inverse. Thus $O(2)\ncong SO(2)\times C_2$.

The center of the unitary group consists exactly of scalar unitary matrices:The determinant gives a surjective homomorphismwhose kernel is the special unitary group $SU(2)$.

Nevertheless $U(2)$ is not isomorphic to $SU(2)\times \mathbb{T}$. The center of $U(2)$ has exactly one nonidentity element of order two, namely $-I$. The center of the proposed product is $\{\pm I\}\times \mathbb{T}$, which has three nonidentity elements of order two. Since an isomorphism preserves the center and element orders,

A Lie group is a group and smooth manifold whose multiplication and inversion are smooth. A Lie algebra is a vector space with a bilinear alternating bracket satisfying the Jacobi identity. For a Matrix Lie group $G\subset GL(m,\mathbb{C})$, a logarithmic chart of a matrix Lie group near $g$ sends $h$ to $\log (g^{-1}h)\in \mathfrak{g}=T_{I}G$; the matrix exponential is its local inverse, and the Baker--Campbell--Hausdorff formula makes the local group operations smooth.

For $B_1,B_2\in \mathfrak{g}$, consider the group commutatorIts logarithm takes values in the vector space $\mathfrak{g}$, and expansion at $(0,0)$ givesThus $\mathfrak{g}$ is closed under the commutator. Bilinearity, alternation, and the Jacobi identity follow from matrix multiplication, so this proves that the Lie algebra of a matrix Lie group has

A principal bundle $\pi :P\to M$ with structure group $G$ is a smooth fiber bundle with a free right $G$-action, each fiber a single orbit, and equivariant local trivializations ${\pi }^{-1}(U)\cong U\times G$.

For the right action of the unitary group on $GL(n,\mathbb{C})$, defineThe matrix $g{g}^{\ast }$ is a positive-definite matrix and a Hermitian matrix, and $q(gu)=q(g)$ for $u\in U(n)$. The polar decomposition of an invertible complex matrix gives the unique factorizationConsequentlyis a diffeomorphism. If $W\subset H(n)$ is a neighborhood of zero and $N=q^{-1}(W)$, then $N$ is an open neighborhood of $I$, it is a union of complete orbits, andMoreover, two matrices have the same value of $q$ exactly when they differ by right multiplication by a unitary matrix. Thus $q$ induces the smooth identificationunder which $\pi$ is projection $H(n)\times U(n)\to H(n)$. This is the general linear group modulo the unitary group, and $\pi$ is in fact a globally trivial principal $U(n)$-bundle.

The free complex field has the global unitary group $U(1)$ symmetryGauging it permits $\alpha =\alpha (x)$ and introduces a gauge field withThe scalar quantum electrodynamics Lagrangian isIts interaction terms are

With momentum flowing along the scalar arrow, the momentum-space Feynman rules are:

Let $J_0$ be the standard complex structure on ${\mathbb{R}}^{2n}$, let ${\omega }_0$ be the standard symplectic form, and let $g_0$ be the Euclidean inner product. They satisfyThe two-out-of-three property for unitary structures says that a real linear map preserving any two of these structures preserves the third. In terms of the corresponding matrix groups,

For example, if $A$ preserves $J_0$ and ${\omega }_0$, thenso $A$ preserves $g_0$. If it preserves $J_0$ and $g_0$, the second displayed identity shows that it preserves ${\omega }_0$. Finally, $J_0$ is uniquely determined by $g_0(J_{0}u,v)={\omega }_0(u,v)$, so preservation of $g_0$ and ${\omega }_0$ implies $A{J}_0=J_{0}A$. The common intersection is therefore the unitary group.

The unitary group acts on the Lagrangian Grassmannian by $A\cdot L=A(L)$. Every Lagrangian subspace has an orthonormal basis, and adjoining its $J_0$-image gives a unitary basis, so this action is transitive. The stabilizer of the standard real subspace ${\mathbb{R}}^n\subset {\mathbb{C}}^n$ consists exactly of real unitary matrices, namely $O(n)$. Henceis a continuous bijection from a compact space to a Hausdorff space and is therefore a homeomorphism.

For $n=1$, every line in ${\mathbb{R}}^2$ is Lagrangian. ThusConcretely, the line making angle $\theta$ with the real axis corresponds to $e^{2i\theta }$.

Represent the color charges by $r=(1,0,0{)}^T$, $g=(0,1,0{)}^T$ and $b=(0,0,1{)}^T$. The Gell-Mann matrices satisfyThe physical quark-gluon Feynman vertex contains $g_s{T}^a$ with $T^a={\lambda }^a/2$; the off-diagonal combination therefore allows $r\to b$. Color charge conservation assigns the emitted gluon the colour-flow label $r\bar{b}$.

The claimed impossibility of $r\to r$ is false. A direct counterexample from the printed matrices isThus both diagonal gluons have nonzero $r\to r$ vertex elements, $(T^3{)}_{rr}=1/2$ and $(T^8{)}_{rr}=1/(2\sqrt{3})$. They correspond to the traceless combinationsA Gell-Mann matrix is a generator rather than a finite rotation, but exponentiating a diagonal generator also leaves the red basis direction fixed up to a phase. The actual excluded object is the identity generator, proportional to the colour singlet $(r\bar{r}+g\bar{g}+b\bar{b})/\sqrt{3}$. Adding it would extend SU(3) to unitary group and introduce a ninth singlet vector boson. A colour-preserving vertex among the existing eight gluons does not require that extra boson and does not make the ordinary diagonal gluons colour singlets.

The Hamiltonian operator is invariant under the constant global phase symmetry $\psi \mapsto e^{i\theta }\psi$, $\bar{\psi }\mapsto \bar{\psi }{e}^{-i\theta }$. The Noether current of this unitary group $U(1)$ is the Dirac current obtained in b. For fields with vanishing flux at infinity, or periodic boundary conditions, the divergence theorem and that continuity equation implyThe canonical anticommutation relations also give $[Q,\psi ]=-\psi$ and $[Q,{\psi }^{†}]={\psi }^{†}$, hence $e^{i\theta Q}\psi e^{-i\theta Q}=e^{-i\theta }\psi$. Reversing the parameter sign recovers the stated phase convention. Thus $Q$ generates the global phase symmetry. With normal ordering, $Q$ counts fermions minus antiparticles, rather than the sum of both populations. Its conservation law is Dirac fermion number conservation. A position-dependent phase produces derivative terms in this free Hamiltonian operator; the symmetry under discussion is global.

The spectral theorem for a Hermitian matrix gives a unitary diagonalizer $V$ with $V\Phi V^{†}=\mathrm{diag}(x_1,\dots ,x_5)$ and real $x_i$. Similarity preserves the trace, so ${\sum }_i{x}_i=0$.

To ensure that the diagonalizer belongs to SU(5) rather than only the unitary group $U(5)$, write $\det V=e^{i\alpha }$ and put $U=e^{-i\alpha /5}V$. Its determinant is one, and the scalar phase cancels from its conjugation action. ThusThe same argument shows why no additional reality condition beyond Hermiticity is needed. Permuting the real eigenvalues merely changes their order in the diagonal form.

For the unitary group $U(n)$ with $n\ge 3$, the third tensor power of its defining representation decomposes as displayed. The first two summands have highest weights $(3,0,\dots )$ and $(1,1,1,0,\dots )$ and dimensions $\left(\genfrac{}{}{0ex}{}{n+2}{3}\right)$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The mixed summand has highest weight $(2,1,0,\dots )$ and dimension $n(n^2-1)/3$. In the three-dimensional weight space spanned by permutations of $e_1\otimes e_1\otimes e_2$, the vectors killed by all positive root vectors have coefficient sum zero, giving multiplicity two. For $n=2$, the exterior cube vanishes and $V_{(2,1)}=\det \otimes {\mathbb{C}}^2$.

The real Lie algebra of the unitary group is $\mathfrak{u}(N)=\{X:X^{†}=-X\}$, of real dimension $N^2$. Differentiating unitarity gives necessity; exponentiating any skew-Hermitian matrix gives a curve in the group, proving sufficiency.

On the canonical cotangent bundle $T^{\ast }{\mathbb{R}}^n$, $\omega =i\sum d{z}_j\wedge d{\bar{z}}_j$ and $H=z^{†}z$. The unitary group acts by $z\mapsto Uz$, preserving both. For a Hermitian $A$, the generator $-iA$ has Hamiltonian ${\mu }_A=z^{†}Az$. The Poisson bracket satisfies $\{{\mu }_A,{\mu }_B\}={\mu }_{-i[A,B]}$, with no central term. The central generator has moment map $H$; imaginary antisymmetric matrices give ordinary angular momentum, while real symmetric matrices give quadratic observables mixing position and momentum.

In the symmetric power ${\mathrm{Sym}}^m{\mathbb{C}}^n$ of the defining unitary group representation, diagonal torus weights are indexed by nonnegative integer tuples summing to $m$, with one monomial per weight space. Torus averaging extracts a monomial from any nonzero invariant complex subspace. The complexified Lie algebra acts by $E_{ij}=x_i{\partial }_{x_j}$ on the symmetric algebra. Whenever $a_j>0$, the displayed action moves one unit of degree from $j$ to $i$ with nonzero coefficient. Finite transfers connect every degree-$m$ monomial, so the subspace is the entire representation. The case $m=0$ is the one-dimensional trivial representation. This proof needs neither the Weyl character formula nor the Weyl dimension formula.

The Weyl denominator is the alternating expression in the Weyl denominator formula. Its product form uses the Weyl vector $\rho$, half the positive-root sum. On a compact maximal torus, the factor $e^{\rho }$ has unit modulus; if it is only defined on a covering torus, the squared modulus still descends. Thus $|\Delta (t){|}^2={\prod }_{\alpha >0}|1-\alpha (t){|}^2$ is a globally defined density in the Weyl integration formula. For the unitary group an integer central shift gives the Vandermonde determinant $\det (z_i^{n-j})$, whose modulus is the same.

For the unitary group, the diagonal maximal torus has coordinates $z_i=e^{i{\theta }_i}$ and Weyl group $S_n$. The Weyl integration formula weights independent uniform angles by $n{!}^{-1}{\prod }_{i<j}|z_i-z_j{|}^2$. The Vandermonde determinant $\det (z_i^{n-j})$ has squared torus integral $n!$, by Fourier orthogonality. Multiplying an irreducible character of highest weight $\lambda$ by this determinant gives an alternating Laurent polynomial. Its leading alternant has coefficient one; character orthogonality for compact groups forces every other alternant coefficient to vanish. This yields the quotient $\det (z_i^{{\lambda }_j+n-j})/\det (z_i^{n-j})$.