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Near the identity, the matrix logarithm of a unitary matrix is skew-Hermitian. Restrict the logarithm's operator norm below $\pi /n$; determinant one then forces its trace to be zero, since $\det U=\exp (\mathrm{tr}\log U)$ and no nonzero multiple of $2\pi i$ fits this bound. The matrix exponential gives the inverse chart on a ball in the displayed real Lie algebra, of dimension $n^2-1$. Left translation supplies compatible charts on the whole special unitary group.

The algebraic loop algebra of a Lie algebra $\mathfrak{g}$ over a field $k$ consists of finite Laurent polynomials with coefficients in $\mathfrak{g}$. Its Lie bracket is $[x\otimes t^m,y\otimes t^n]=[x,y]\otimes t^{m+n}$, extended bilinearly. Alternation and the Jacobi identity follow from those of $\mathfrak{g}$, since the Laurent polynomial ring is commutative. Appropriate smooth or formal completions give other loop-algebra variants; the affine current algebra adds a central extension to these Fourier modes. For the affine extension, see Kac and Wakimoto, Section 0.2.

In the faithful matrices $g=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$ with $a>0$, the Maurer-Cartan form has entries $da/a$ and $db/a$. Thus the left-invariant differential forms ${\sigma }^D=da/a$, ${\sigma }^T=db/a$ obey $d{\sigma }^D=0$ and $d{\sigma }^T=-{\sigma }^D\wedge {\sigma }^T$. The Lie algebra generators satisfy $[D,T]=T$.

For simply connected $Y$, rational Hurewicz homomorphism identifies ${\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q}$ with the primitive elements of a Hopf algebra of $H_{\ast }(\Omega Y;\mathbb{Q})$ and induces an isomorphism $U({\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q})\cong H_{\ast }(\Omega Y;\mathbb{Q})$ of graded Hopf algebras. The bracket on homotopy is the Samelson product. The underlying algebra theorem says that a connected graded cocommutative Hopf algebra over characteristic zero is the universal enveloping algebra of its primitive Lie algebra. These statements let a free tensor algebra of loop-space homology detect a free graded Lie algebra of homotopy.

A moment map assigns to each element of a Lie algebra a Hamiltonian function for its fundamental vector field, and is equivariant for the coadjoint action. For an abelian Lie group, constant shifts are allowed; the numerical scale also depends on the chosen Lie algebra generators and the normalization of the symplectic form.

A Lie algebra is nilpotent when its lower central series eventually becomes zero.

The abelian Lie algebra $kI\subseteq \mathrm{End}(V)$, for $V\ne 0$, is a nilpotent Lie algebra but contains the nonnilpotent identity map. Thus nilpotence of the abstract Lie algebra does not imply that a particular Lie algebra representation consists of nilpotent endomorphisms. Engel theorem requires nilpotence of every represented endomorphism to obtain strict upper triangularity.

Let $G$ be connected and compact with nonzero irreducible real Adjoint representation. Its Lie algebra has no proper invariant Lie algebra ideal. If it is Abelian, irreducibility forces dimension one, so $G$ is the circle group. Otherwise its centre has zero-dimensional Lie algebra and is finite. A proper closed normal subgroup likewise has zero-dimensional Lie algebra, hence is finite; connectedness of $G$ makes conjugation on that finite subgroup trivial, so the subgroup is central. A Lie group homomorphism onto a nontrivial connected target with surjective differential consequently has finite central group kernel. The qualifications “proper” and “nontrivial” are essential.

A one-parameter subgroup of a Lie group $G$ is a smooth group homomorphism $\gamma :(\mathbb{R},+)\to G$: it satisfies $\gamma (s+t)=\gamma (s)\gamma (t)$, with $\gamma (0)=e$ and $\gamma (-t)=\gamma (t{)}^{-1}$. Every such map has the form $\gamma (t)=\exp (tX)$ for a unique $X={\gamma }^{\prime }(0)$ in the Lie algebra of $G$. Thus its initial tangent vector determines the whole motion, for all real times. The map need not be injective; the word subgroup describes its image. See MIT 18.745, Lecture 7, Proposition 7.1.

The real affine group used here is the orientation-preserving subgroup of the affine group on the real line. It acts by $x\mapsto e^{a}x+b$ and has multiplication $(a,b)(a^{\prime },b^{\prime })=(a+a^{\prime },b+e^a{b}^{\prime })$. The positive dilation distinguishes this connected component from the full group, which also contains negative dilations.

Here is a small Lie group where order matters. Write a point as $(x,1{)}^T$, and letThe one-parameter subgroups $S_a$ and $U_b$ give dilations and translations. With matrix multiplication acting rightmost first, $S_a{U}_b$ sends $x$ to $e^{a}x+e^{a}b$, whereas $U_b{S}_a$ sends $x$ to $e^{a}x+b$. At $a=\log 2$ and $b=1$, translate then double gives $2x+2$, while double then translate gives $2x+1$. Starting at $x=1$ gives endpoints $4$ and $3$.

The same order dependence survives infinitesimally: the matrix commutator is $[D,T]=DT-TD=T$. Indeed, the group commutator satisfiesAt $a=b=0.1$, the residual translation is $0.01051709$, close to the leading product $ab=0.01$. It measures a second-order failure of the four moves to cancel. This connects finite composition directly to the Lie bracket on the Lie algebra; see Meinrenken, Section 4, Proposition 4.7.

A basis $T_a$ of the Lie algebra of a Lie group gives a global frame of a vector bundle of its tangent bundle by $X_a(g)=(d{L}_g{)}_e{T}_a$. Left translation is a diffeomorphism, so the left-invariant vector fields are independent everywhere. Every Lie group is therefore a parallelizable manifold.

Identify the Lie algebra $\mathfrak{g}$ with the tangent space $T_{e}G$. For $X\in \mathfrak{g}$, its left-invariant vector field is $X^L(g)=(d{L}_g{)}_{e}X$. Let ${\gamma }_X$ be its integral curve with ${\gamma }_X(0)=e$. Uniqueness of integral curves and left invariance give ${\gamma }_X(s+t)={\gamma }_X(s){\gamma }_X(t)$ wherever initially defined. Repetition of a local curve extends it to all real times. Thus ${\gamma }_X$ is the unique one-parameter subgroup with derivative $X$ at zero, and the Exponential map of a Lie group is defined bySmooth dependence of solutions of differential equations on their initial data and parameters makes this map smooth.

The derivative at zero is particularly simple:It is the identity linear map from $\mathfrak{g}$ to $T_{e}G$. The inverse function theorem therefore proves that the Exponential map of a Lie group is a local diffeomorphism at zero, producing a local exponential chart. This proves the requested local assertion.

There is a necessary qualification to the other assertion: the image of every one-parameter subgroup lies in the identity component of a Lie group $G^{\circ }$. The correct statement without a connectedness hypothesis isIndeed, if $\exp$ is a group homomorphism, its image is an abelian subgroup. It contains an identity neighborhood by the local result. An open subgroup of a connected group is the entire group: every coset is open, so its complement is also open. Thus $\exp (\mathfrak{g})=G^{\circ }$, proving that this component is abelian.

Conversely, if $G^{\circ }$ is abelian, then $\exp (tX)\exp (tY)$ is a one-parameter subgroup: its group law follows by commuting the factors. Its derivative at zero is $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$ and, at $t=1$, additivity. This proves the additivity of the Lie exponential map criterion. For connected $G$ it is exactly the stated equivalence with an abelian group. For disconnected $G$ the unrestricted claim is false: $S_3\times \mathbb{R}$ is a nonabelian Lie group, but its exponential is the homomorphism $x\mapsto (e,x)$. The original PDF does not impose connectedness in that sentence, so that qualification cannot be omitted.

Now suppose $G$ is connected and abelian, of dimension $n$. The exponential is a surjective group homomorphism, and its kernel $\Lambda$ is discrete by local injectivity at zero. The induced mapis a bijective Lie group homomorphism and a local diffeomorphism, hence a Lie group isomorphism. To identify the quotient, take linearly independent elements ${\lambda }_1,\dots ,{\lambda }_b\in \Lambda$ spanning the real linear span of $\Lambda$, and let ${\Lambda }_0={\sum }_i\mathbb{Z}{\lambda }_i$. Every coset of ${\Lambda }_0$ in $\Lambda$ has a representative in the bounded fundamental parallelepiped of these vectors. A discrete additive subgroup has finite intersection with a compact set: otherwise differences of arbitrarily close elements would approach zero, contradicting discreteness at zero. Thus $\Lambda /{\Lambda }_0$ is finite.

It follows that $\Lambda$ is a finitely generated torsion-free abelian group of rank $b$, hence has a $\mathbb{Z}$-basis of $b$ elements. These basis elements also span its $b$-dimensional real span and are linearly independent over $\mathbb{R}$. Extending them to a real basis identifies $\Lambda$ with ${\mathbb{Z}}^b\times \{0{\}}^{n-b}$. Consequently the classification of connected abelian Lie groups isThe compact circle factors record the periods of the one-parameter subgroups.

The Lie algebra of SL2R is the space of traceless real two-by-two matrices. For $A$ in this space, the Cayley-Hamilton theorem gives $A^2=cI$, where $c=-\det A$. Expanding the matrix exponential, if $c=s^2>0$ thenIf $c=0$, then $\exp A=I+A$ and its trace is $2$. If $c=-s^2<0$, thenIn every case $\mathrm{tr}(\exp A)\ge -2$. Buthas determinant one and trace $-5/2$. Thus $g\in S{L}_2(\mathbb{R})$ is not an exponential, and the exponential map is not surjective.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.

Choose a basis $x_1,x_2,x_3$ of the Lie algebra and filter its universal enveloping algebra by word length: $F_{0}R=\mathbb{C}$ and $F_{d}R$ is spanned by products of at most $d$ basis elements. The relations $x_j{x}_i-x_i{x}_j=[x_j,x_i]$ replace a degree-two commutator by degree one, so the degree-one symbols commute.

The Poincaré-Birkhoff-Witt theorem says the ordered monomials $x_1^{a_1}{x}_2^{a_2}{x}_3^{a_3}$ are a basis. Its relevant proof can be seen by reordering an inverted neighboring pair using the displayed relation. Reordering terminates because same-length terms have fewer inversions and bracket terms have smaller word length. Disjoint swaps commute; the only overlapping three-letter reorderings differ by $[x_i,[x_j,x_k]]+[x_j,[x_k,x_i]]+[x_k,[x_i,x_j]]=0$. The Jacobi identity therefore makes the reduced ordered expression independent of the reordering path. Ordered monomials have no remaining relations and form the basis. Their degree-$d$ symbols giveFor generators $m_1,\dots ,m_q$ of the finitely generated module $M$, put $M_d={\sum }_j{F}_{d}R{m}_j$. This is a good filtration, and $\mathrm{gr}M$ is finite over the three-variable polynomial ring. Define the dimension of a filtered module as the Krull dimension of its associated graded module, equivalently the degree of the eventual cumulative polynomial ${\dim }_{\mathbb{C}}{M}_d$, or the Gelfand–Kirillov dimension of a module. Two good filtrations have bounded shifts, so the growth degree is independent of the generators. Sinceits polynomial degree is at most three. Therefore every nonzero finitely generated $R$-module has dimension at most $3$. The zero module may be assigned the usual separate empty-support convention. This bound also follows immediately from its support of a module inside affine three-space.

Choose Hermitian generators $T^a$ of the gauge Lie algebra, with $[T^a,T^b]=i{f}^{abc}{T}^c$, and write $A_{\mu }=A_{\mu }^a{T}^a$. For the usual compact gauge algebra choose $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$. More generally use a nondegenerate invariant symmetric bilinear form on the algebra in place of this trace contraction; a completely arbitrary component metric would not give gauge invariance.

A matter vector is useful to fix conventions: with ${\psi }^{\prime }=U(x)\psi$ and $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Acting on an arbitrary vector gives the Yang-Mills gauge transformationFor $U=1+ig\theta +O({\theta }^2)$ this becomesHere the second $D_{\mu }$ denotes the induced derivative in the Adjoint representation.

Define the curvature by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$. Expanding the commutator derivesSince commutators of the transformed derivatives conjugate, $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$. Cyclicity of the trace proves the invariant bilinear-form construction of a Yang-Mills action:For a general invariant form $h$, the corresponding proof is $h([\theta ,X],Y)+h(X,[\theta ,Y])=0$, applied to the two curvature factors. No matter field is required in this pure-gauge Lagrangian.

Introduce the adjoint Faddeev-Popov ghost $c^a$ and antighost ${\bar{c}}^a$, independent Grassmann-odd scalar fields of ghost numbers $+1$ and $-1$, together with the Grassmann-even Nakanishi-Lautrup field $b^a$ of ghost number zero. The bar labels the independent antighost, not an additional propagating complex-conjugate matter field. Choose a left BRST symmetry differential:It is an odd derivation: $s(XY)=(sX)Y+(-1{)}^{|X|}X(sY)$ for homogeneous Grassmann parity $|X|$. The two relevant properties are $s{\mathcal{L}}_{\mathrm{YM}}=0$ and off-shell nilpotence $s^2=0$. The first is gauge invariance with parameter replaced by the ghost, and the second follows from the Lie algebra's graded Jacobi identities. ThusAlgebraically this holds for any $\Psi$; a physical gauge-fixing addition uses an odd gauge-fixing fermion of ghost number $-1$ so that $s\Psi$ is even and has ghost number zero.

For BRST-exact covariant gauge fixing, chooseThe graded Leibniz rule supplies the essential minus sign:Use Gaussian gauge fixing with an auxiliary field to integrate $b$. Pointwise,Translation of the regulated oscillatory Gaussian leaves only a field-independent normalization. The remaining Lagrangian is thereforeThis is covariant Feynman gauge, with gauge parameter one. Eliminating $b$ need not retain manifest off-shell nilpotence in the reduced field variables; the original auxiliary-field action gives the off-shell BRST symmetry formulation.

Finally, the ungauge-fixed quadratic momentum kernel isGauge directions are null vectors, so there is no inverse defining a vector propagator, and the functional integral also counts an infinite gauge orbit volume. The added gauge-fixing term cancels the longitudinal part, leaving $Q_{\mu \nu }^{ab}=-{\delta }^{ab}{p}^2{\eta }_{\mu \nu }$. With the vacuum prescription, the propagators areThe ghost action represents the Faddeev-Popov determinant and supplies ghost-vector vertices; closed ghost loops have the fermionic minus sign. Expanding the cubic and quartic Yang-Mills terms and this ghost interaction now defines consistent perturbative Feynman rules. Physical mass-shell poles remain, whereas the gauge degeneracy has been removed. Possible global gauge-fixing obstructions and boundary zero modes are separate from the perturbative inverse around the trivial vacuum.