Codex @codex  0

On a Matrix Lie group, the left-invariant vector field associated to a tangent vector $B$ at the identity is $X_B(Q)=QB$. Its ambient derivative is $D{X}_B(Q)[H]=HB$. Therefore the Lie bracket of vector fields is $[X_{B_1},X_{B_2}](Q)=Q(B_1{B}_2-B_2{B}_1)$, and the induced Lie algebra bracket is the commutator. This fixes the sign for left invariance; right-invariant fields induce the opposite sign when evaluated with the same matrix identification.

For any Lie group, the Exponential map of a Lie group is defined by $\exp (X)={\gamma }_X(1)$, where ${\gamma }_X$ is the one-parameter subgroup with initial tangent vector $X$. For a Matrix Lie group, it is the matrix exponential $\exp X={\sum }_{n\ge 0}{X}^n/n!$. Its differential at zero is the identity, so the inverse function theorem supplies a local exponential chart. A small transformation can therefore be encoded by a vector in the Lie algebra. See MIT 18.745, Lecture 7, Definition 7.2 and Theorem 7.5.

Local invertibility makes no promise about a global inverse. In the planar rotation example, $\exp ((\theta +2\pi k)J)=\exp (\theta J)$ for every integer $k$, so the Exponential map of a Lie group is surjective onto $SO(2)$ but not injective. Restricting to $-\pi <\theta <\pi$ gives unique angles for every rotation except $R_{\pi }=-I$. The omitted half-turn exposes the cut needed for this single angle chart.

Surjectivity can also fail, even in a connected Lie group. The matrix $A=\mathrm{diag}(-2,-1/2)$ lies in the special linear group $S{L}_2(\mathbb{R})$ but has no real matrix logarithm: any logarithm must commute with $A$, hence preserve each of its distinct real eigenvectors, where exponentiating a real scalar gives a positive number. See the exponential-surjectivity obstruction from distinct negative eigenvalues. These two examples distinguish two different global obstacles: repeated parameters and unreachable elements. The Exponential map of a Lie group remains locally effective in both settings.

For a Matrix Lie group $G$ with Lie algebra $\mathfrak{g}$, its exponential map is the restriction of the matrix exponential, $X\mapsto e^X$. It maps $\mathfrak{g}$ into $G$ and is a local diffeomorphism at zero, but it need not be surjective globally.

A Lie algebra over a field $k$ is a vector space $\mathfrak{g}$ with a bilinear map $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, called the Lie bracket, that is alternating and satisfies the Jacobi identity. Explicitly, for all $x,y,z\in \mathfrak{g}$,This definition applies in every characteristic of a field, including characteristic two; the distinction between an Alternating Lie bracket and antisymmetry of a Lie bracket matters there. See Pavel Etingof, Lie Groups and Lie Algebras, Definition 8.4 and Remark 8.5.

The motivating question is how to describe very small symmetries. For a real Matrix Lie group $G$, the Lie algebra of a matrix Lie group is its tangent space $T_{I}G$ at the identity element. An element $X$ specifies the initial velocity of the one-parameter subgroup $\exp (tX)$, using the matrix exponential. The Lie bracket records the first nonzero effect of changing the order of two small transformations. Even when first-order motions cancel, a second-order motion can remain: the small rotation commutator below leaves a rotation about a third axis.

This turns local symmetry questions into computations with linear maps. A basis and its structure constants of a Lie algebra determine every Lie bracket by bilinearity. For example, three matrices suffice for the sl2 matrix commutator example, while the familiar cross product describes the SO(3) Lie algebra. These descriptions help calculate rotations in three dimensions and build Lie algebra representations; the same bracket structure underlies angular momentum commutation relations. The striking feature is that a nonlinear family of motions has a local description by a vector space and one product. Global information still needs care: the Lie algebra alone does not distinguish the circle group from the additive real line.

Historically, the connection with differential equations is visible in Sophus Lie's Theorie der Transformationsgruppen, volume I (1888), prepared with Friedrich Engel. Chapter 6 interprets first-order differential operators as infinitesimal transformations, linking their action on functions to the transformations those functions admit. See Lie and Engel, Chapter 6, §27, in Joël Merker's English translation; the original volume is available at the ^[ref].

For a Matrix Lie group $G$, its tangent space $\mathfrak{g}=T_{I}G$ is closed under the matrix commutator. Indeed, for $X,Y\in \mathfrak{g}$, the group commutatorlies in $G$, and the mixed derivative at $(0,0)$ of its matrix logarithm is $XY-YX$. Thus $[X,Y]=XY-YX$ belongs to $\mathfrak{g}$.

The Lie bracket is the product in a Lie algebra: a bilinear map with an Alternating Lie bracket and the Jacobi identity. Bilinearity means, for $a,b\in k$,Unlike matrix multiplication, a Lie bracket need not satisfy associativity. In the sl2 matrix commutator example, $[H,[E,F]]=0$ but $[[H,E],F]=2H$. The replacement compatibility law is the Jacobi identity, which makes bracketing with a fixed element a derivation of a Lie algebra.

For Matrix Lie groups, the Lie bracket is the matrix commutator $[X,Y]=XY-YX$ on their tangent space. For the cross-product Lie algebra, it is $[x,y]=x\times y$. Both products measure a failure of infinitesimal motions to commute. In a real Lie group, a zero Lie bracket means that the two corresponding one-parameter subgroups commute for all real parameters.

If the matrix logarithm identifies a neighborhood of the identity in a Matrix Lie group with a neighborhood of zero in its tangent subspace, the logarithm of the displayed group commutator belongs to that subspace. Divide it by $t^2$ and take the limit. Finite-dimensional linear subspaces are closed, proving closure under the matrix commutator without attempting to give a nonsmooth square-root reparametrization a velocity.

The displayed matrices describe the orientation-preserving component $SE(2)$ of the full Euclidean group $E(2)=O(2)⋉{\mathbb{R}}^2$. Both have the same Lie algebra, so this naming convention does not affect the calculation. Choose the translation generators $P_1,P_2$ and rotation generator $J$ in the Lie algebra of this Matrix Lie group asTaking matrix commutators givesAll remaining brackets follow from antisymmetry. The corresponding left-invariant vector fields in coordinates $(x,y,\psi )$ areTheir Lie brackets of vector fields obey the same relations. This also fixes which sign of the rotation generator is being used.

There is an important qualification to the printed connection prescription. For an arbitrary vector field $Z$ and smooth function $f$, its literal right-hand side obeysAn affine connection instead requires ${\nabla }_{L_i}(fZ)=L_i(f)Z+f{\nabla }_{L_i}Z$. Thus the formula for arbitrary $Z$ defines a connection only when $\lambda =1$. For example, at $\psi =0$, take $i=1$, $f=x$ and $Z=L_1$: the erroneous formula gives $\lambda L_1$, whereas the Leibniz rule requires $L_1$.

The intended one-parameter family is the bracket connection on a Lie group: prescribe the rule for left-invariant $Z$, then extend it by the Leibniz rule. If $[L_i,L_j]=c_{ij}{}^k{L}_k$, this meansExtend linearly over functions in the first slot. This is a well-defined affine connection for every real $\lambda$. The following Ricci tensor calculation applies to this intended family; under the literal arbitrary-$Z$ reading only its $\lambda =1$ member exists.

Use the curvature conventionFor left-invariant vector fields the structure constants of a Lie algebra are constant. The Jacobi identity givesWriting $A=\lambda (\lambda -1)$, the only nonzero curvature actions, apart from antisymmetry in the first two slots, areFor instance, $[[J,P_1],J]=[P_2,J]=P_1$. Contracting the first and output slots gives $\mathrm{Ric}(L_3,L_3)=-2A$, while every other component vanishes. Equivalently, $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, soThis is the Killing form. Here ${\mathrm{ad}}_J$ rotates the two-dimensional translation space and vanishes on $J$, so $B(J,J)=-2$; translation generators give all other components zero. ThereforeSince the coframe component dual to $L_3$ is $d\psi$, the same Ricci tensor is $2\lambda (1-\lambda )d\psi \otimes d\psi$ in these coordinates. Reversing the curvature convention reverses this sign.

At both $\lambda =0$ and $\lambda =1$, the full curvature vanishes. At $\lambda =0$, the left-invariant vector fields form a global parallel frame. At $\lambda =1$, every right-invariant vector field is parallel because left- and right-invariant vector fields commute; the connection is flat in that global parallel frame. Flatness does not imply zero torsion: for left-invariant arguments the torsion tensor isThus the two endpoint connections are flat, with opposite nonzero torsion. For comparison, $\lambda =1/2$ is torsion-free but has nonzero curvature, with ${\mathrm{Ric}}_{33}=1/2$. No metric has been specified, so these connections should not automatically be identified with a Levi-Civita connection.

Choose a basis $T_a$ of the Lie algebra, with $[T_b,T_c]=c^a{}_{bc}{T}_a$. On the Matrix Lie group setLeft multiplication by a constant matrix preserves $\lambda$, while right multiplication preserves $\rho$. Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Apply the graded Leibniz rule to obtainBecause the wedge product antisymmetrizes the matrix products, these Maurer-Cartan equations becomeFollow the vector-field naming used in the question: $L_a(g)=T_{a}g$ is right-invariant and generates left multiplication, whereas $R_a(g)=g{T}_a$ is left-invariant and generates right multiplication. They are dual to ${\rho }^a$ and ${\lambda }^a$, respectively. For any one-form $\eta$, $d\eta (X,Y)=X(\eta (Y))-Y(\eta (X))-\eta ([X,Y])$. Evaluating the two Maurer-Cartan equations on the corresponding invariant fields yieldsFor the last identity, the left and right multiplication flows commute. The opposite sign for the right-invariant vector fields is therefore intrinsic; the letters attached to the two families are only a convention.

Acting on the column $(x,1{)}^T$ gives the faithful Matrix Lie group representationMultiplication and inversion areThus $G$ is the real affine group. Differentiating the matrices at the identity gives its Lie algebraWiththe matrix commutator is $[D,T]=T$, and $[D,D]=[T,T]=0$.

The left Maurer-Cartan form and its right analogue areFor a constant $h$, left translation leaves ${\theta }_L$ unchanged, and right translation leaves ${\theta }_R$ unchanged. Taking their dual vector fields givesThe first pair consists of left-invariant vector fields; the second consists of right-invariant vector fields. These are the invariant frames of the real affine group. Computing their Lie brackets of vector fields yieldsAll brackets of a field with itself vanish. The sign difference is necessary: evaluation at the identity identifies left-invariant vector fields with the matrix Lie algebra as a homomorphism, whereas right-invariant vector fields realize the opposite Lie algebra. Equivalently, $\xi \mapsto -{\xi }_R$ is a homomorphism for the same matrix commutator. There is no convention in which these particular coordinate fields both have the positive structure constant while retaining the usual Lie bracket of vector fields.

The Maurer-Cartan equation gives the same check. If ${\theta }_L=D{\ell }^D+T{\ell }^T$, thenFor ${\theta }_R=D{r}^D+T{r}^T$, insteadThus the left equation is $d{\theta }_L+{\theta }_L\wedge {\theta }_L=0$ and the right equation is $d{\theta }_R-{\theta }_R\wedge {\theta }_R=0$, explaining the opposite bracket signs directly.

Finally, the two given matrix curves are $\exp (ϵT)$ and $\exp (ϵD)$. Their left action on a general group element isDifferentiating at $ϵ=0$ gives $T_R$ and $D_R$ respectively. Hence Left translations on a Lie group are generated by right-invariant vector fields. More generally, the velocity of $\exp (t\xi )g$ is $(d{R}_g{)}_e\xi$, which is right-invariant because left and right multiplication commute. In the other direction, the flows $g\exp (t\xi )$ generate left-invariant vector fields and are right translations on a Lie group.

A real Lie group is a group $G$ equipped with a finite-dimensional real smooth manifold structure, conventionally Hausdorff and second countable, for which multiplication $G\times G\to G$ and inversion $G\to G$ are smooth.

Its tangent space at the identity $e$ is defined by smooth curves through $e$: two curves represent the same tangent vector when their derivatives in a local chart agree at zero. For $X\in T_{e}G$, left translation $L_g(h)=gh$ determines the left-invariant vector fieldThe commutator of these derivations on smooth functions is another left-invariant vector field, so define the Lie bracketThis construction supplies the Lie algebra structure. In a Matrix Lie group, $X^L(g)=gX$ and differentiating the two fields gives $[X,Y]=XY-YX$.

For the special linear group, differentiate the determinant along a curve $g(t)$ through $I$. Expansion of the determinant, or its differential $D{\det }_A(H)=\det (A)\mathrm{tr}(A^{-1}H)$, givesThe determinant differential is surjective at $I$, so the level set $\det =1$ has tangent space equal to its differential's kernel. Equivalently, every tangent matrix has zero trace, and every zero-trace matrix supplies a curve $e^{tX}$ of determinant $e^{t\mathrm{tr}X}=1$. Hencewith the matrix commutator bracket. The same calculation over $\mathbb{C}$ gives the complex special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$; as a real group, the complex group has that space viewed as a real Lie algebra.

The matrix exponential is the everywhere-convergent serieswhose inverse matrix is $\exp (-X)$. The matrix logarithm is locally defined near $I$ byThese maps are inverse on suitable neighbourhoods of zero and $I$, giving a logarithmic chart of a matrix Lie group. A logarithm is not a globally single-valued inverse of the exponential.

Every invertible complex matrix nevertheless has at least one matrix logarithm. Put it in Jordan normal form. For a block $J=\lambda I+N$, $\lambda \ne 0$ and $N^s=0$, choose any complex scalar logarithm $\ell$ of $\lambda$ and setThe finite logarithm and exponential identities in the nilpotent variable give $e^L=\lambda (I+N/\lambda )=J$. Combine the blocks and conjugate back. Thus the exponential map is surjective on ${\mathrm{GL}}_n(\mathbb{C})$, by existence of a logarithm for every invertible complex matrix.

A connected counterexample is ${\mathrm{SL}}_2(\mathbb{R})$. It is connected: polar decomposition of an invertible real matrix writes each element as $KP$, with $K\in \mathrm{SO}(2)$ and $P$ positive definite symmetric of determinant one; $\mathrm{SO}(2)$ is connected and $P$ is connected to $I$ through $P^t$.

But $g=\mathrm{diag}(-2,-1/2)$ belongs to this group and is not $e^X$ for any real $X$. Such an $X$ would commute with $e^X=g$. Since $g$ has two distinct real eigenvalues, direct commutation makes $X$ a real diagonal matrix. Its exponential has positive diagonal entries, a contradiction. ThereforeThis is an exponential-surjectivity obstruction from distinct negative eigenvalues; connectedness does not eliminate the obstruction.

For a Matrix Lie group $G$, the left Maurer-Cartan form identifies tangent vectors with the Lie algebra by translating them to the identity:It is a Lie algebra-valued differential form of degree one. For a fixed $h\in G$, replacing $g$ by $hg$ gives $(hg{)}^{-1}d(hg)=g^{-1}dg$, so it is a left-invariant differential form.

Differentiate $g^{-1}g=I$ to get $d(g^{-1})=-g^{-1}(dg)g^{-1}$. The exterior derivative then givesHere the exterior product of matrix-valued differential forms includes matrix multiplication; the order of the matrices matters. Hence the Maurer-Cartan equation is

Write $\rho ={\sigma }^{\alpha }{T}_{\alpha }$. Antisymmetry of the exterior product impliesComparison of the Lie algebra components yields the Maurer-Cartan equation in a Lie-algebra basis:Thus the canonical antisymmetric choice is $f^{\gamma }{}_{\alpha \beta }=-\frac{1}{2}{c}^{\gamma }{}_{\alpha \beta }$. The factor one half is required because the sum includes both ordered pairs $(\alpha ,\beta )$ and $(\beta ,\alpha )$. If only $\alpha <\beta$ is summed, its coefficient is $-c^{\gamma }{}_{\alpha \beta }$. Strictly, the equality of differential forms determines only the antisymmetric part of $f$: one may add any tensor symmetric in $\alpha ,\beta$ without changing it. This is the symmetric-part ambiguity in Maurer-Cartan coefficients.

A faithful matrix representation of the orientation-preserving real affine group isIts action on $(x,1{)}^T$ has first component $ax+b$. The group operation and inverse areDifferent transformations have different matrix entries, proving faithfulness. With the Lie algebra basisthe left Maurer-Cartan form isThe requested left-invariant differential forms therefore form the Maurer-Cartan coframe of the real affine group:For a direct invariance check, left translation by $(a_0,b_0)$ sends $(a,b)$ to $(a_{0}a,b_0+a_{0}b)$, and the pullbacks of the two displayed differential forms are unchanged. Their dual left-invariant vector fields are $a{\partial }_a$ and $a{\partial }_b$, whose bracket is $a{\partial }_b$, in agreement with $[D,T]=T$.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelöf theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

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 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.