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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.

For a short Wilson line, a gauge transformation givesUse $V(x+an)=1+i\alpha (x)+ia{n}^{\mu }{\partial }_{\mu }\alpha (x)+O(a^2,{\alpha }^2)$ and $U=1+iga{n}^{\mu }{A}_{\mu }+O(a^2)$. Keeping terms of order $a$, including $a\alpha$, givesSince the Lie bracket is encoded by the Lie algebra structure constants, $i[\alpha ,A_{\mu }{]}^a=f^{abc}{A}_{\mu }^b{\alpha }^c$, and thereforeThis is the infinitesimal form of the Wilson-line gauge transformation.

An almost complex structure on a smooth manifold $M$ is a smooth bundle endomorphism $J:TM\to TM$ satisfying $J^2=-1$. On the complexified tangent bundle, $J$ has the eigenbundle decompositionwith eigenvalues $i$ and $-i$. A differential form of type (p, q) is a section of ${\bigwedge }^p(T^{1,0}M{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}M{)}^{\ast }$. The operators $\partial$ and $\bar{\partial }$ are the type-$(1,0)$ and type-$(0,1)$ components of the exterior derivative.

We prove the three stated conditions are equivalent. If $X,Y$ have type $(1,0)$ and $\alpha$ has type $(0,1)$, the formula for the exterior derivative givesThus closure of $T^{1,0}M$ under the Lie bracket is equivalent to the vanishing of the $(2,0)$ component of $d\alpha$ for every $(0,1)$-form. Complex conjugation of differential-form type gives the corresponding vanishing of the $(0,2)$ component on $(1,0)$-forms. Since every complex one-form is a sum of these two types, this is preciselyThis proves (i)$⟺$(ii).

Under (ii), the $(2,0)$ component of $d^{2}f=0$ is ${\partial }^{2}f$, proving (iii). Conversely, for $X,Y$ of type $(1,0)$,If this vanishes for every smooth complex-valued $f$, then $[X,Y{]}^{0,1}=0$, so (iii) implies (i). These conditions define an integrable almost complex structure.

For a complex manifold, a holomorphic chart $z^j=x^j+i{y}^j$ defines $J_M({\partial }_{x^j})={\partial }_{y^j}$ and $J_M({\partial }_{y^j})=-{\partial }_{x^j}$. The derivative of a holomorphic coordinate change is complex linear, hence commutes with multiplication by $i$; the definitions therefore glue and are independent of coordinates. Locally $T^{1,0}M$ is spanned by the commuting fields $\partial /\partial z^j$, so it is closed under brackets and $J_M$ is integrable. This is the almost complex structure induced by a complex atlas.

We next prove the local Dolbeault-Poincaré lemma. Write a $\bar{\partial }$-closed $(0,q)$-form on a slightly larger polydisc aswhere neither $\alpha$ nor $\beta$ contains $d{\bar{z}}_n$. Apply the supplied one-variable Cauchy-Green operator coefficientwise to $\beta$, obtaining $\gamma$ with $\partial \gamma /\partial {\bar{z}}_n=\beta$. Then $\phi -\bar{\partial }\gamma$ contains no $d{\bar{z}}_n$, and $\bar{\partial }\phi =0$ says that its coefficients are holomorphic in $z_n$ and $\bar{\partial }$-closed in the first $n-1$ variables. Induction on $n$, with the one-variable formula as the base case, makes this remainder $\bar{\partial }$-exact. Hence every $\bar{\partial }$-closed $(0,q)$-form with $q>0$ is $\bar{\partial }$-exact on each bounded polydisc.

Finally, translation by $i$ leaves $d\bar{z}$ unchanged, so this form descends to $\mathbb{C}/(z\sim z+i)$. It is in fact exact on this noncompact complex cylinder, because the invariant function $z+\bar{z}=2\mathrm{Re}z$ descends and satisfies

For an affine algebraic group $G$, its Lie algebra is the tangent space at the identity,Equivalently, it is the space of left-invariant derivations of the coordinate ring $\mathbb{C}[G]$. The Lie bracket is the commutator of derivations,It is antisymmetric because $[Y,X]=-[X,Y]$. Associativity of composition givesafter all six triple products cancel in pairs, proving the Jacobi identity. This construction is the Lie algebra of an affine algebraic group.

A Lie bracket is bilinear, antisymmetric, and satisfies the Jacobi identity. For a basis $T_a$,The structure constant of a Lie algebra coefficients satisfy $f_{ab}{}^c=-f_{ba}{}^c$ and $f_{ab}{}^d{f}_{dc}{}^e+f_{bc}{}^d{f}_{da}{}^e+f_{ca}{}^d{f}_{db}{}^e=0$.

For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

In a coordinate chart centred at the identity, smooth multiplication has a local group lawThe identity gives $F(0,x)=x$ and $F(y,0)=y$, while associativity gives the functional equation $F(u,F(y,x))=F(F(u,y),x)$. Smooth inversion determines coordinates $\bar{x}$ with $F(\bar{x},x)=F(x,\bar{x})=0$. Changes of local coordinates alter the symmetric part of $B^k{}_{ij}$, whereas its antisymmetric part supplies the coordinate-independent Lie bracket on the tangent space at the identity.

Write $g(x)=I+x^i{T}_i+O(x^2)$ in a matrix representation and $[T_i,T_j]=c^k{}_{ij}{T}_k$. Direct multiplication givessoThe cancellation of the constant and linear terms explains why the group commutator first detects the Lie bracket at quadratic order.

For the sl2 Lie algebra, let $v_{\lambda }$ be the highest-weight vector. The Poincaré-Birkhoff-Witt theorem gives the basis $(f^r{v}_{\lambda }{)}_{r\ge 0}$, and the defining Lie brackets implyA positive-degree basis vector is singular exactly when $r=\lambda +1$ is a positive integer. Therefore $M_{\lambda }$ is irreducible when $\lambda \notin {\mathbb{Z}}_{\ge 0}$.

If $\lambda =m\in {\mathbb{Z}}_{\ge 0}$, the vector $f^{m+1}{v}_{\lambda }$ has weight $-m-2$ and generates a submodule isomorphic to $M_{-m-2}$. The latter is irreducible because $-m-2\notin {\mathbb{Z}}_{\ge 0}$. Every nonzero submodule contains a singular vector by the preceding part, and the displayed coefficient shows that this is the only possible proper singular vector. Henceis the unique proper nonzero submodule, as summarized by the Reducibility of an sl2 Verma module.

For the trace trilinear form of a Lie algebra representation, the trace of a commutator vanishes:Taking $X=T_b$, $Y=T_a$, $Z=T_c$, $W=T_d$ and expanding each Lie bracket givesSince $f_a{}^{bc}$ is antisymmetric in $b,c$,Raising indices with the inverse Killing form gives $f_a{}^{bc}{f}_{bc}{}^e={\delta }_a{}^e$ in the stated normalization. Using $H(T_e,T_d)=-\mu {\delta }_{ed}$ proves

A Poisson algebra is a commutative associative algebra with a Lie bracket that is a derivation in each argument. Smooth functions on a symplectic manifold are its standard example. The linear span of moment-map components may form a finite-dimensional Lie subalgebra, while closing those components under multiplication normally produces a much larger Poisson algebra.

A Lie algebra over the p-adic integers whose underlying module is free of finite rank and whose Lie bracket satisfies $[L,L]\subseteq p^{ϵ}L$, where $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. Finite freeness is part of this convention: merely requiring a small Lie bracket on a torsion module would not correspond to uniform pro-p groups.

For a Lie algebra ideal $I$, the quotient vector space $\mathfrak{g}/I$ has Lie bracket $[x+I,y+I]=[x,y]+I$. The ideal condition makes this independent of representatives. Its Lie algebra ideals correspond to the ideals of $\mathfrak{g}$ containing $I$, by inverse image under the quotient Lie algebra homomorphism.

Use the cross-product Lie algebra convention $X=\widehat{e_1}$, $Y=\widehat{e_2}$, $Z=\widehat{e_3}$, so $[X,Y]=Z$. Put $R_x(t)=\exp (tX)$ and $R_y(t)=\exp (tY)$, using the matrix exponential. The finite group commutator isHere $\epsilon \to 0$ and big O notation refers to any fixed matrix norm. Active rotation matrices act on column vectors from right to left: first rotate about $y$ by $-\epsilon$, then about $x$ by $-\epsilon$, then about $y$ by $+\epsilon$, and finally about $x$ by $+\epsilon$. Swapping $x$ and $y$ replaces the product by its inverse and changes the leading Lie bracket to $-Z$.

Writing $c=\cos \epsilon$ and $s=\sin \epsilon$, direct matrix multiplication givesFor $\epsilon =0.1$ radians, $C_{\epsilon }{e}_1\approx (0.999950208,0.009966462,0.000501230)$, whereas $e_1+{\epsilon }^{2}Z{e}_1=(1,0.01,0)$. Thus the residual is a positive rotation about $z$ to leading order; at finite $\epsilon$ it also has smaller components. The Lie bracket captures the leading effect of the four moves rather than their entire finite motion. See Etingof, Section 7.2, Proposition 7.11.

The left panel projects the actual four arcs onto the $yz$-plane for $\epsilon =0.25$; the start and endpoint are close but distinct. The right panel divides the residual coordinates by ${\epsilon }^2$, making the limit $(\Delta y/{\epsilon }^2,\Delta z/{\epsilon }^2)\to (1,0)$ visible. This is a numerical picture of $[X,Y]=Z$ in the SO(3) Lie algebra.

This Lie group acts on the plane by $v\mapsto x+\rho v$ with $\rho >0$. Its multiplication is $(\rho ,x)({\rho }^{\prime },x^{\prime })=(\rho {\rho }^{\prime },x+\rho x^{\prime })$, and its inverse is $({\rho }^{-1},-{\rho }^{-1}x)$. It is a semidirect product in which positive dilation acts on the translation subgroup. A dilation generator $D$ and translation generators $T_1,T_2$ have Lie brackets $[D,T_a]=T_a$ and $[T_1,T_2]=0$.