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An abelian Lie group has commutative multiplication. Its Lie algebra is an abelian Lie algebra, and its Lie exponential map is a group homomorphism from the additive Lie algebra. Conversely, an additive Lie exponential map forces the identity component of a Lie group to be abelian, but does not force all disconnected components to commute.

The Exponential map of a Lie group is a group homomorphism from the additive Lie algebra exactly when its identity component of a Lie group is abelian. If it is additive, its image is an abelian subgroup containing an identity neighborhood by the local exponential chart; connectedness makes that image the entire identity component. Conversely, in an abelian identity component the product $\exp (tX)\exp (tY)$ is a one-parameter subgroup with derivative $X+Y$, giving the identity by uniqueness. The disconnected Lie group $S_3\times \mathbb{R}$ shows that the full group need not be abelian.

The action $(A,B):M\mapsto AM{B}^{-1}$ preserves the quadratic form $-\det M$ on real two-by-two matrices. Its kernel consists of the simultaneous scalar pairs $(I,I)$ and $(-I,-I)$. The differential $(a,b):M\mapsto aM-Mb$ is injective on the two traceless Lie algebras, and both source and target have dimension six. Its image is therefore open and, since the source is connected, equals the identity component of a Lie group $S{O}_0(2,2)$. This gives a two-sheeted covering map onto the identity component, not onto the disconnected full special orthogonal group $SO(2,2)$.

Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.

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.

For $X\in \mathfrak{g}=T_{e}G$, let $X^L$ be the left-invariant vector field with value $X$ at the identity. Its integral curve through the identity is the unique one-parameter subgroup ${\gamma }_X$ with ${\gamma }_X^{\prime }(0)=X$. It exists for all real time: local existence followed by translation and the identity ${\gamma }_X(t+s)={\gamma }_X(t){\gamma }_X(s)$ extends it indefinitely. The Exponential map of a Lie group isIt is smooth and $d_0\exp ={\mathrm{id}}_{\mathfrak{g}}$, because differentiating $t\mapsto \exp (tX)$ at zero gives $X$. The inverse function theorem therefore makes it a local diffeomorphism at zero.

If $G$ is abelian, the product $t\mapsto \exp (tX)\exp (tY)$ is a one-parameter subgroup with initial tangent $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$, henceConversely, if this additive homomorphism identity holds, all $\exp (tX)$ and $\exp (sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the identity component of a Lie group, so the identity component is abelian. Conversely, if the Lie algebra is abelian, its left-invariant vector fields commute, so their flows commute. Hence all one-parameter subgroups commute; their product with initial tangent $X+Y$ is the one-parameter subgroup for $X+Y$. The Lie exponential map is therefore additive, and its image generates an abelian identity component. Thus the stated equivalence holds for connected $G$. Without connectedness the exact assertion is that the Lie exponential map is a homomorphism precisely when $G^{\circ }$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the Lie algebra consists of $tJ$, whereRotations satisfy $R_{t+s}=R_t{R}_s$, but a reflection $S$ satisfies $S{R}_t{S}^{-1}=R_{-t}$, so the full orthogonal group is not abelian.

Now take a connected abelian Lie group of dimension $n$. Its Lie exponential map is a homomorphism whose image contains an identity neighbourhood, hence an open subgroup. A connected group has no proper open subgroup, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective local diffeomorphism and therefore a Lie group isomorphismHere is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose ${\lambda }_1,\dots ,{\lambda }_r\in \Lambda$ that form a real basis of that span. Their integer span ${\Lambda }_0$ has finite index in $\Lambda$: subtract integer multiples of the ${\lambda }_i$ to put each coset representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the structure theorem for finitely generated abelian groups makes it free abelian of rank $r$. Its integer basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to ${\mathbb{Z}}^r\times \{0\}$ in ${\mathbb{R}}^n$.

This proves the classification of connected abelian Lie groups:The isomorphism depends on the choice of lattice basis and complementary vector space. Compactness is equivalent to $s=0$.

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 $T_e{G}_Q=M_n(\mathbb{R})$.