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For strictly decreasing integer tuples $\kappa ,\eta$, let $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ on the diagonal maximal torus of the unitary group. Expanding both determinants, Fourier orthogonality makes the integral zero unless the exponent tuples have the same permutation orbit. Strict decreasing order makes that equivalent to $\kappa =\eta$; the $n!$ matching permutation terms then each contribute one. An alternating Laurent polynomial is a finite linear combination of these alternants. For an irreducible character ${\chi }_{\lambda }$, its highest monomial gives coefficient one for $A_{\lambda +\delta }$ in ${\chi }_{\lambda }{A}_{\delta }$, with $\delta =(n-1,\dots ,0)$. The Weyl integration formula and Schur orthogonality relations force the sum of squared coefficients to be one, so all other coefficients vanish. This proves the unitary Weyl character formula directly.

The free kinetic terms of left- and right-handed Weyl fields are separately invariant under unitary group rotations in flavour space. For $N$ massless Dirac fields, the subgroup preserving common fermion number is $U(N{)}_L\times U(N{)}_R$. A common nonzero Dirac mass term couples the two chirality sectors and leaves its diagonal $U(N)$ subgroup. Gauge interactions can obstruct some of the classical axial currents through a chiral anomaly.

The Peter-Weyl theorem makes finite-dimensional unitary representations separate elements of a compact group. Intersecting their closed kernels gives the identity subgroup. If the descending chain condition holds for closed subgroups, choose finitely many representations whose kernel intersection is already trivial: otherwise repeatedly exclude a surviving nonidentity element to obtain an infinite strict chain. Their direct sum is a faithful representation into a finite-dimensional unitary group. Compactness makes its image closed, and the closed-subgroup theorem gives the original group a compatible Lie group structure.

The complex Grassmannian of $k$-planes in ${\mathbb{C}}^n$ identifies with the displayed Hermitian orthogonal projection matrices. A plane is recovered as the image of its projection. Near a plane $E$, a neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. Its projection is $W(W^{\ast }W{)}^{-1}{W}^{\ast }$ with $W=(I,Z{)}^T$. The unitary group acts transitively by conjugation, and its tangent directions are $[A,P]$ with $A^{\ast }=-A$.

The space of ordered orthonormal complex $k$-frames in ${\mathbb{C}}^n$, for $0\le k\le n$. It is a compact smooth manifold of real dimension $2nk-k^2$. Completing a frame to a unitary basis gives the displayed homogeneous-space description; $V_0$ is a point and $V_n({\mathbb{C}}^n)$ is the unitary group $U(n)$.

The map $(z,U)\mapsto zU$ identifies the unitary group with $(U(1)\times SU(n))/K$, where $K=\{({\zeta }^k,{\zeta }^{-k}{I}_n):\zeta =e^{2\pi i/n}\}$. If an irreducible SU(n) tensor transformation has $b$ boxes in its Young diagram, the central scalar $\zeta I_n$ acts as ${\zeta }^b$. Therefore, for integer $q$, $(z,U)\mapsto z^{q}R(U)$ is trivial on $K$ precisely when ${\zeta }^{q-b}=1$, proving the criterion. Adding a full $n$-box column changes $b$ by $n$ and leaves the condition unchanged. In particular a lift with trivial first-factor U(1) charge descends only if its central action is trivial; this does not prevent other lifts of the same special unitary group representation from descending.

Without fixed gauge charges or interactions distinguishing particles from antiparticles, $N$ free Dirac fields give $2N$ left-handed Weyl fields after charge conjugation of the right-handed fields. Their massless kinetic term has unitary group $U(2N)$. Equal nonzero Dirac masses have equal positive singular values and can be written as $2N$ equal Majorana mass terms; their preserving group is the orthogonal group $O(2N)$. Explicitly, in a basis with symmetric mass matrix $m1$, invariance requires $V^{†}V=V^{T}V=1$, hence real $V$. Restricting to transformations preserving a chosen common fermion number gives the usual $U(N{)}_L\times U(N{)}_R$ at zero mass and $U(N)$ at nonzero mass.

Use the complement bundle on a complex Stiefel manifold inductively. If the ring for $V_k$ is exterior with smallest positive degree $2n-2k+1$, the complement bundle has Euler degree $2(n-k)$, in which base cohomology vanishes. Its Euler class is consequently zero. The Gysin ring splitting for an odd-dimensional sphere bundle adds one exterior generator of degree $2(n-k)-1$. The initial base $V_0$ is a point. This gives the full integral cohomology ring and in particular its torsion-freeness. At $k=n$, the unitary group has Poincare polynomial ${\prod }_{j=0}^{n-1}(1+t^{2j+1})$.

The Lagrangian Grassmannian is the space of all Lagrangian subspaces of a symplectic vector space. The unitary group acts transitively on $\mathrm{LGr}({\mathbb{R}}^{2n})$, and the stabilizer of ${\mathbb{R}}^n$ is the orthogonal group $O(n)$, giving

For a Dirac field, the path integral is a Berezin integral over independent Grassmann fields $\psi$ and $\bar{\psi }$:The vacuum boundary conditions and the Feynman i-epsilon prescription are understood; $\mathcal{N}$ fixes the vacuum normalization. This is the vacuum generating functional without sources, rather than a finite-temperature partition function.

The Dirac adjoint transforms as $\bar{\psi }\mapsto \bar{\psi }{e}^{-i\alpha }$. For constant $\alpha$, ${\partial }_{\mu }(e^{i\alpha }\psi )=e^{i\alpha }{\partial }_{\mu }\psi$. Both the kinetic term and the Dirac mass term therefore retain their values. This proves the global phase symmetry, whose group is the unitary group $U(1)$.

In the Weyl representation of the gamma matrices, write each Dirac field as ${\psi }_i=({\ell }_{i\alpha },r_i^{\dot{\alpha }}{)}^T$, where ${\ell }_i$ and $r_i$ have opposite chirality. With the Weyl sigma matrices ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(1,-\mathbf{\sigma })$,The Dirac mass term couples the two Weyl fields. Under $\ell \mapsto U_L\ell$, $r\mapsto U_{R}r$, its first term acquires $U_L^{†}{U}_R$, so a common nonzero mass requires $U_L=U_R$. At zero mass there is no such constraint:This is the chiral flavour symmetry of massless Dirac fields. Its eight Noether currents are $j_{L,A}^{\mu }={\ell }^{†}{\bar{\sigma }}^{\mu }{T}^A\ell$ and $j_{R,A}^{\mu }=r^{†}{\sigma }^{\mu }{T}^{A}r$. The sum $j_R+j_L$ is the vector current, and $j_R-j_L$ is the axial current, for each flavour generator. This statement concerns the free theory in flat spacetime; an interacting theory can have a chiral anomaly.

For completeness, allowing the extra transformations described in part e gives four left-handed Weyl fields, $q=({\ell }_1,{\ell }_2,r_1^c,r_2^c{)}^T$, where $r_i^c$ is the left-handed charge-conjugate field of $r_i$. After integration by parts their kinetic term is $i{q}^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }q$, whose maximal linear unitary group is $U(4)$. The symmetric mass matrix for equal Dirac masses iswith the antisymmetric spinor contraction implicit. A unitary change of basis makes $M=m{1}_4$; preserving both the kinetic term and $M$ requires $V^{†}V=V^{T}V=1_4$, hence real $V\in O(4)$. Thus the unrestricted free-theory enhancement is $U(4)$ at zero mass and $O(4)$ at equal nonzero mass; the familiar $U(2{)}_L\times U(2{)}_R$ and diagonal $U(2)$ are their common-fermion number-preserving subgroups.

For $z\in U(1)$ and $U\in SU(n)$, the matrix $zU$ is unitary, since $(zU{)}^{†}(zU)=|z{|}^2{U}^{†}U=I_n$. Thus the map has its image in the unitary group $U(n)$. Scalar matrices commute with all matrices, so the direct product of groups multiplication givesAlso $f(1,I_n)=I_n$. HenceThe special unitary group condition $\det U=1$ imposes no requirement that $zU$ itself have unit determinant; its determinant is $z^n$, as expected for a general element of the unitary group.

Write $x=(x_1,x_2,x_3{)}^T$, $v=(x_4,x_5,x_6{)}^T$ and $z=x+iv$. Here products of differentials in a metric tensor are symmetric products, unlike the wedge products used later. ThenIndeed $d{z}_{k}d{\bar{z}}_k=d{x}_k^2+d{x}_{k+3}^2$. A transformation $z^{\prime }=Uz$ preserves this metric whenever $U^{†}U=I$, becauseThus the entire unitary group $U(3)$ preserves it, including its special unitary group $SU(3)$.

To obtain a real matrix embedding, write $U=A+iB$ with $A,B$ real. ThenThe unitarity equations imply $A^{T}A+B^{T}B=I$ and $A^{T}B=B^{T}A$, hence $\rho (U{)}^T\rho (U)=I$. Multiplication of the block matrices gives $\rho (UV)=\rho (U)\rho (V)$, and the blocks recover $U$, proving injectivity. FinallyFor this last identity, over the complex numbers the change of basis from $(x,v)$ to $(z,\bar{z})$ conjugates $\rho (U)$ to $\mathrm{diag}(U,\bar{U})$. This proves that the image lies in the special orthogonal group.

The image is in the unitary group, since $(zU{)}^{†}(zU)=\bar{z}z{U}^{†}U=I$ for $z\in U(1)$ and $U\in SU(n)$. The direct product of groups has multiplication $(z,U)(w,V)=(zw,UV)$. Scalar matrices commute with every matrix, soThus $f$ is a group homomorphism; in particular $f(1,I)=I$.

First prove surjectivity. For any $A\in U(n)$, its determinant has modulus one. Choose $z\in U(1)$ with $z^n=\det A$ and define $U=z^{-1}A$. Then $U^{†}U=I$ and $\det U=z^{-n}\det A=1$, so $A=f(z,U)$.

Next, $f(z,U)=I$ is equivalent to $U=z^{-1}I$. This lies in $SU(n)$ precisely when $z^n=1$. Therefore the kernel of a group homomorphism is the central cyclic groupIts generator is $(\zeta ,{\zeta }^{-1}{I}_n)$, which has order $n$. The inverse phases in the two factors are essential: this is an explicitly embedded subgroup, not a quotient by scalar matrices in just one factor. The first isomorphism theorem now givesThis also holds as a Lie group isomorphism: the induced map is smooth and a continuous bijection from a compact quotient space to the Hausdorff unitary group, and its differential is the Lie algebra isomorphism $(it,X)\mapsto itI+X$. For $n=1$ the kernel of a group homomorphism is trivial, consistently with $SU(1)=\{1\}$.

A universal quantum gate set can approximate any finite-register unitary operator to arbitrary accuracy, disregarding a common global phase. A finite example is the Hadamard gate, T gate and CNOT gate. In the ordered one-qubit basis $(|g⟩,|e⟩)$,In the ordered two-qubit basis $(|gg⟩,|ge⟩,|eg⟩,|ee⟩)$, with the first qubit the control,The one-qubit gates generate a dense set of one-qubit unitary operators, and CNOT gates supply entangling operations. Approximate universality is the appropriate qualification for a finite discrete set: its finite circuits cannot exactly equal every member of a continuous uncountable unitary group.

The conjugate transpose also reverses factor order. If $U$ and $V$ are unitary matrices,Thus their product is unitary. This is closure under multiplication in the unitary group.

Let $V={\mathbb{C}}^n$ and $m\ge 0$. Identify the symmetric power ${\mathrm{Sym}}^{m}V$ with the degree-$m$ part of the symmetric algebra on the basis vectors $e_1,\dots ,e_n$, using formal variables $x_i$ for these vectors. Its monomial basis isThe diagonal maximal torus of the unitary group acts on this monomial by the character $z_1^{a_1}\cdots z_n^{a_n}$. Distinct tuples give distinct characters, so each weight space is one-dimensional.

Suppose $M$ is a nonzero invariant complex subspace. For a nonzero vector $w\in M$, select a monomial with nonzero coefficient and extract it by torus averaging:Fourier orthogonality makes this a nonzero scalar multiple of $x^a$. It belongs to $M$, so $M$ contains a monomial.

Differentiating the unitary group action makes $M$ invariant under $\mathfrak{u}(n)$ and hence under its complexification ${\mathfrak{gl}}_n(\mathbb{C})$. The matrix unit $E_{ij}$ replaces an occurrence of $e_j$ by $e_i$, so its action on the symmetric algebra isWhenever $a_j>0$, this transfers one unit of degree from coordinate $j$ to coordinate $i$ with a nonzero coefficient. Starting with any monomial, such transfers can produce every other tuple of nonnegative integers summing to $m$. Thus $M$ contains every monomial and equals ${\mathrm{Sym}}^{m}V$.

ConsequentlyFor $m=0$ this is the one-dimensional trivial representation, and for $n=1$ every symmetric power is also one-dimensional. This weight-transfer proof of irreducibility of symmetric powers uses neither the Weyl character formula nor the Weyl dimension formula, so no unproved dimension formula enters the argument.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

For a group action of $G$ on $X$, the orbit of a group action through $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $H_x=\{g:gx=x\}$. A transitive group action has one orbit. A simply transitive group action has exactly one element carrying any given point to any other; equivalently it is transitive and every stabilizer subgroup is trivial.

In the modern terminology, a multiply transitive group action is transitive on ordered $k$-tuples of distinct points for some $k\ge 2$. In older geometric usage, the phrase can instead mean transitive but not simply transitive, so that there are several elements carrying one point to another. Under either interpretation the action is transitive, and that alone is what the following argument needs. These meanings should not be confused: a transitive action with nontrivial stabilizers need not be two-transitive.

Fix $x_0\in X$ and put $H=H_{x_0}$. DefineThis is well-defined because every element of $H$ fixes $x_0$. It is surjective by transitivity, and $g{x}_0=g^{\prime }{x}_0$ is equivalent to $g^{\prime -1}g\in H$, proving injectivity. For a smooth Lie group action, $H$ is closed, since it is the inverse image of $x_0$ under the orbit map. The orbit map has constant rank: translations in $G$ and the action diffeomorphisms of $X$ intertwine its differentials. Transitivity makes this rank $\dim X$; otherwise every point would be a critical value, contradicting Sard theorem. Its local submersion charts give smooth local sections and hence a smooth inverse for $F$. ThereforeChoosing $x_1=a{x}_0$ replaces $H$ by $aH{a}^{-1}$. Thus the stabilizer subgroup is determined by the base point, and intrinsically only up to conjugacy. A simply transitive group action gives $H=\{e\}$.

For a finite quantum system with Hilbert space $\mathcal{H}={\mathbb{C}}^{n+1}$, a pure state is a nonzero vector modulo multiplication by a nonzero complex number. Equivalently, it is a unit vector modulo its physically irrelevant phase. Consequently the space of pure states is the space of complex lines,The qualification “pure” is essential: arbitrary quantum states include mixed density matrices and do not form Complex projective space.

The unitary group $U(n+1)$ acts transitively on these lines: choose unit vectors in two given lines, complete each to an orthonormal basis, and map one basis to the other by a unitary transformation. The stabilizer of the line $\mathbb{C}{e}_0$ consists exactly of block matrices $\mathrm{diag}(z,V)$ with $z\in U(1)$ and $V\in U(n)$. Indeed a unitary transformation preserving the line also preserves its orthogonal complement. The homogeneous space description is thereforeIts real dimension is $(n+1{)}^2-1-n^2=2n$. This action is generally not two-transitive: the absolute inner product between unit representatives of two lines is invariant.

The Hopf fibration is the projection $\pi :S^{2n+1}\to {\mathbb{CP}}^n$, $z\mapsto [z]$. The right action $z\cdot \lambda =z\lambda$, $\lambda \in S^1$, is free, and its orbits are exactly the fibres. On the chart $U_i$ where $z_i\ne 0$, write $w_j=z_j/z_i$ for $j\ne i$. A smooth local section isEvery unit vector over this chart has the unique expression $z={\sigma }_i([z])\lambda$, where $\lambda =z_i/|z_i|$. These formulas provide local trivializations ${\pi }^{-1}(U_i)\cong U_i\times S^1$, proving that this is a principal bundle with fibre $S^1$. For $n\ge 1$ it has no global section: a global section would give $S^{2n+1}\cong {\mathbb{CP}}^n\times S^1$, contradicting their fundamental groups. For $n=0$ the base is a point and the bundle is trivial.