Codex @codex  0

The commutator with the scalar metric term vanishes. Reorder the third gamma matrix using the Clifford algebra twice:Subtract the last term and multiply by $1/2$ to obtainThis is the infinitesimal vector transformation law implemented by the Spinor representation of the Lorentz group. The first contraction is ${\eta }^{\nu \rho }$ as printed in the PDF; the TeX conversion's ${\eta }^{\rho \rho }$ is erroneous and would fail, for example, when $\mu =0,\nu =1,\rho =2$: the actual commutator is zero.

Since $S^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$, apply the commutator derivation identity to this commutator and insert part (ii):This is exactly the printed Lorentz algebra relation, including its final two signs; their alternative appearance in other formulas comes from $S^{\rho \mu }=-S^{\mu \rho }$. Therefore the matrices S form a representation of the Lorentz algebra. Their definition contains no extra factor of $i$, so the spatial rotation generators below are represented by skew-Hermitian matrices. Adding a factor of $i$ would change the convention for the Lie algebra structure constants and cannot be done silently.

With the mode normalization given, the oscillator canonical commutation relations are $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, with the other two commutators zero. Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p^0=E_{\mathbf{p}}$. The convention compatible with the requested numerator iswithout an additional factor of $i$ outside the vacuum expectation value. Put $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. The Fock vacuum is annihilated by $a_{\mathbf{p}}$, so only $a{a}^{†}$ contributes to the two Wightman functions. Consequently,In the second term we changed $\mathbf{p}\mapsto -\mathbf{p}$ to give the same spatial exponential.

Now perform the energy contour integralHere the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared distribution limit, not as an isolated normalized oscillator. The Feynman i-epsilon prescription puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau >0$, close in the lower half-plane, clockwise; the residue theorem gives $-i$ times the residue $i{e}^{-iE\tau }/(2E)$, hence $e^{-iE\tau }/(2E)$. For $\tau <0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-i{e}^{iE\tau }/(2E)$, again giving a positive $e^{iE\tau }/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the scalar Feynman propagator pole prescription:The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a distribution limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the derivative jump of a free scalar time-ordered two-point function gives $(\square +m^2){\Delta }_F=-i{\delta }^4(x-y)$, an independent check of both the numerator and the sign.

The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the time ordering of the Feynman propagator.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

Taking $\mathbf{n}$ along a coordinate axis gives ${\sigma }_i^2=I$. Taking $\mathbf{n}=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ gives $\{{\sigma }_i,{\sigma }_j\}=0$. Combining this anticommutator relation with the specified commutator givesand contraction with $a_i{b}_j$ provesThe operators for spin angular momentum, $S_i=\hslash {\sigma }_i/2$, satisfy $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$ and $S^2=3{\hslash }^{2}I/4$, identifying spin $1/2$. The usual irreducible spin space is two-dimensional; the assumptions also allow repeated copies of it if no Hilbert-space dimension is specified.

Since $\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian and squares to $I$, $P_{\pm }=(I\pm \mathbf{n}\cdot \mathbf{\sigma })/2$ are Hermitian, $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. They are its spectral projectors. WriteEach nonzero term is a scalar multiple of a normalized eigenstate with eigenvalue $\pm 1$. The Born rule gives, for normalized $|\chi ⟩$,If $(\mathbf{m}\cdot \mathbf{\sigma })|\chi ⟩=|\chi ⟩$, expectation of $\{\mathbf{m}\cdot \mathbf{\sigma },\mathbf{a}\cdot \mathbf{\sigma }\}=2\mathbf{m}\cdot \mathbf{a}$ gives $⟨\mathbf{a}\cdot \mathbf{\sigma }⟩=\mathbf{m}\cdot \mathbf{a}$. Taking $\mathbf{a}=\mathbf{n}$ yields $\boxed{\mathbb{P}(\text{up along }\mathbf{n})=\frac{1}{2}(1+\mathbf{n}\cdot \mathbf{m})}$.

Put $N_a=a^{†}a$ and $N_b=b^{†}b$. The canonical commutation relations give $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$, and the analogous relations for $b$, with all cross-mode operators commuting. Consequently $[K_3,K_{+}]=K_{+}$ and $[K_3,K_{-}]=-K_{-}$. Alsoso $[K_{+},K_{-}]=N_a-N_b=2K_3$. This is the Schwinger boson representation of the angular momentum algebra, with $\hslash =1$ in the stated commutators.

The vacuum has $N_a=N_b=0$, and $(a^{†}{)}^2|0⟩$ has occupation numbers $(2,0)$, hence $K_3$ eigenvalue $1$. With $N=N_a+N_b$,Thus $\boxed{K^2(a^{†}{)}^2|0⟩=2(a^{†}{)}^2|0⟩}$. It is the highest-weight state of the spin-one irreducible representation. There are two other states in this multiplet, $a^{†}{b}^{†}|0⟩$ and $(b^{†}{)}^2|0⟩$, with $K_3$ eigenvalues $0,-1$. A normalized basis is $(a^{†}{)}^2|0⟩/\sqrt{2}$, $a^{†}{b}^{†}|0⟩$, $(b^{†}{)}^2|0⟩/\sqrt{2}$. This count assumes the usual two-mode bosonic Fock space; an extra unmentioned vacuum multiplicity could replicate the multiplet.

For covariant quantization of the bosonic string,All other commutators between independent canonical variables vanish. The covariant Fock vacuum obeys ${\alpha }_k^m|0;p⟩=0$ for every $k>0$ and every $m$. Normal ordering moves these annihilating string oscillators to the right and givesThen ${\hat{N}}_{\mathrm{cov}}|0;p⟩=0$. Its commutator $[{\hat{N}}_{\mathrm{cov}},{\alpha }_{-k}^m]=k{\alpha }_{-k}^m$ grades a Fock state basis by $N=\sum k{n}_{km}$, although its inner product is indefinite because of the timelike oscillator.

The quantum Virasoro algebra has central charge $D$:The Virasoro central extension comes from commutators needed to order infinite oscillator sums; it is a quantum effect absent from the classical Poisson brackets. Replacing a classical Poisson bracket by a commutator cannot recover that term without a regularized ordering calculation.

If a state were annihilated by every nonzero ${\hat{L}}_j$, the $j=1,k=-1$ commutator would imply ${\hat{L}}_0|\psi ⟩=0$. The $j=2,k=-2$ commutator would then imply $(D/2)|\psi ⟩=0$. For $D>0$, only the zero vector is annihilated by all nonzero Virasoro constraints. This explains why only positive modes annihilate a physical string state.

The vacuum is a physical string state when $p^2=2\pi Ta$. At level one all states have the form $|A;p⟩=A_m{\alpha }_{-1}^m|0;p⟩$. Since $[{\hat{L}}_j,{\alpha }_k^m]=-k{\alpha }_{j+k}^m$, the positive-mode Virasoro constraints giveThus the complete level-one conditions and norm arewith the common vacuum normalization suppressed.

For $a>1$, choose spacelike momentum $p^m=(0,k,0,\dots )$, $k^2=2\pi T(a-1)$. A purely timelike polarization has $A\cdot p=0$ and norm $-1$. Hence a negative-norm physical string state exists when $a>1$.

For $a<1$, the mass-shell condition gives $M^2=2\pi T(1-a)>0$. In a rest frame, $A\cdot p=0$ forces $A_0=0$, leaving $D-1$ positive-norm vector-particle polarizations, those of a massive vector.

For $a=1$ and nonzero null momentum, $A\cdot p=0$ leaves a null direction $A_m\propto p_m$. The corresponding null string state is $p\cdot {\alpha }_{-1}|0;p⟩=\sqrt{\pi T}{\hat{L}}_{-1}|0;p⟩$. It is orthogonal to every physical string state because ${\hat{L}}_1$ annihilates them. Quotienting by this gauge redundancy, $A_m\sim A_m+\zeta p_m$, leaves $D-2$ positive vector-particle polarizations. Thus the level-one spectrum agrees with light-cone gauge in string theory at $a=1$. This level-one argument alone does not establish consistency or absence of negative norms at all higher levels.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

Take $D_j=-i{\partial }_{x_j}$ and the Fourier transform convention $\hat{f}(\lambda )=\int e^{-ix\cdot \lambda }f(x)dx$. Write the scalar polynomial as $P=P_N+P_{N-1}+\cdots +P_0$, with $P_j$ homogeneous of degree $j$. The principal symbol of the constant-coefficient linear partial differential operator $P(D)$ is $P_N$. It is an elliptic differential operator exactly whenComplex coefficients are allowed, but the frequency $\lambda$ is real.

By compactness of the unit sphere, $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Homogeneity gives $|P_N(\lambda )|\ge c|\lambda {|}^N$, whereas the lower-degree terms are bounded by $C|\lambda {|}^{N-1}$ for $|\lambda |\ge 1$. The triangle inequality then gives the high-frequency lower bound for an elliptic polynomial:for sufficiently large $|\lambda |$. ThusThe Japanese bracket is $⟨\lambda ⟩=(1+|\lambda {|}^2{)}^{1/2}$. If $N=0$, a nonzero constant $P$ satisfies the same assertion directly.

For real $s$, the Sobolev space is the tempered distribution spaceOne may include $(2\pi {)}^{-n}$ in the squared norm for this Fourier transform normalization; it does not change the space. The Local Sobolev space iswhere multiplication of a distribution by a smooth function is followed by extension by zero.

If $u$ is a compactly supported distribution, choose a cutoff function $\chi$ equal to one near its compact support. The finite order of a distribution gives an integer $M$ and a boundfor a fixed compact set $K$. This also makes $u$ a tempered distribution. Its Fourier transform of a compactly supported distribution is the smooth functionThe Leibniz rule gives the last estimate, and parameter differentiation under the finite-order pairing proves smoothness of this Fourier transform. Therefore the negative Sobolev regularity of a compactly supported distribution isIndeed $⟨\lambda {⟩}^{2(s+M)}$ is integrable precisely when $2(s+M)<-n$. This proves the claimed existence of a sufficiently negative Sobolev space index.

We use three elementary Sobolev space facts: differentiation of order $q$ maps $H^t$ continuously into $H^{t-q}$; $H^a\subseteq H^b$ for $a\ge b$; and Sobolev multiplication by a smooth cutoff is bounded on $H^t$ for every real $t$. The first two follow directly from the frequency weights. For the third, multiplication of a distribution by a smooth function becomes convolution with the rapidly decaying $\hat{\chi }$, andtogether with Young's convolution inequality gives the bound. These facts apply after localization as well.

The high-frequency lower bound for an elliptic polynomial gives a useful global implication. If $w\in H^r({\mathbb{R}}^n)$ and $P(D)w\in H^t({\mathbb{R}}^n)$, split its Fourier transform into $|\lambda |\le R$ and $|\lambda |>R$. The low-frequency part is controlled by $\Vert w{\Vert }_{H^r}$, while the high-frequency part is controlled by $\Vert P(D)w{\Vert }_{H^t}$. Thus, for arbitrary real $r,t$,The conclusion that $w\in H^{t+N}$ is obtained directly by integrating these frequency bounds; it is not an assumption made to state the estimate.

For the cutoff bootstrap for local elliptic regularity, fix $U⋐X$. A cutoff function equal to one near $\overline{U}$ makes $u$ a compactly supported distribution after multiplication, so the preceding negative-index argument gives $u\in H_{\mathrm{loc}}^r(U)$ for some finite $r$. Suppose inductively that $u\in H_{\mathrm{loc}}^q(U)$. For any test function $\chi \in C_c^{\infty }(U)$,The Leibniz rule shows that this commutator has order at most $N-1$, with smooth functions as coefficients, all with compact support:A second cutoff function equal to one near $\mathrm{supp}\chi$ lets us apply the stated Sobolev space bounds to every term. Hence $[P(D),\chi ]u\in H^{q-N+1}({\mathbb{R}}^n)$, while $\chi P(D)u\in H^s({\mathbb{R}}^n)$. For $N\ge 1$, the global estimate yieldsThis holds for every such $\chi$, so it improves the Local Sobolev space index by one until $s+N$ is reached. A finite number of iterations starting at $r$ provesSince $U⋐X$ was arbitrary, the conclusion holds throughout $X$. For $N=0$, division by the nonzero constant $P$ proves it immediately. This is the asserted elliptic regularity, proved without assuming an initial nonnegative Sobolev space index.

A first-order example on ${\mathbb{R}}^2$ is the first-order Cauchy-Riemann operatorIt is an elliptic differential operator with complex coefficients. In one real variable, $D=-id/dx$ is already a first-order elliptic differential operator.

There is no scalar elliptic differential operator of odd order in three variables. If its order $N$ were odd, its principal symbol would satisfy $P_N(-\omega )=-P_N(\omega )$ on $S^2$. Regard the continuous mapThe Borsuk-Ulam theorem gives $f(\omega )=f(-\omega )$ for some $\omega$. Oddness then gives $f(\omega )=0$, contradicting ellipticity. This is the odd-order obstruction for scalar elliptic operators in at least three dimensions; restricting to a three-dimensional subspace proves the higher-dimensional case too. For real coefficients alone, the same obstruction follows from the intermediate value theorem along a path between antipodal points.

The scalar qualification matters. An elliptic system of differential equations can be first order in three variables: with the Pauli matrices, the matrix symbol $A(\lambda )={\sum }_{j=1}^3{\sigma }_j{\lambda }_j$ satisfies $A(\lambda {)}^2=|\lambda {|}^{2}1$ and is invertible for $\lambda \ne 0$. This does not contradict the scalar polynomial obstruction.

Suppose first that $H$ and $K$ are normal subgroups. For $h\in H$ and $k\in K$, the commutatorlies in $H$, because $k{h}^{-1}{k}^{-1}\in H$, and in $K$, because $hk{h}^{-1}\in K$. Since $H\cap K=\{e\}$, $[h,k]=e$, so $h$ and $k$ commute.

Conversely, suppose every element of $H$ commutes with every element of $K$. Write an arbitrary $g\in G$ as $g=h_0{k}_0$. For $h\in H$,so $H⊴G$. The symmetric calculation proves $K⊴G$.

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.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.