Lie Groups, Invariant Fields, and the Exponential Map — Examples

Addition and negation are smooth on $\mathbb{R}$; multiplication and inversion $a\mapsto 1\mathrm{/}a$ are smooth on each of the open sets ${\mathbb{R}}_{>0}$ and ${\mathbb{R}}^{\times }$. Hence [F1] gives the three one-dimensional Lie groups.

The curve $t\mapsto tx$ is the additive one-parameter subgroup with derivative $x$ at zero. The curve $t\mapsto e^{tx}$ is a multiplicative one-parameter subgroup by [F3], has derivative $x$ at zero, and stays positive. By [F2] their time-one values give the displayed formulas.

All groups are nonempty and one-dimensional; ${\mathbb{R}}^{\times }$ is disconnected but the other two are connected. At $x=0$ all exponentials give the identity. No metric, degeneracy, endpoint issue, or biconditional occurs. The assumed ${\mathrm{A}\mathrm{C}}_{\omega }$ is used by [F2] through its stated supplier chain, with no further choice.

The set $\det \mathrm{\ne }0$ is open in $M_n(\mathbb{R})$, and matrix multiplication and inversion are smooth there, so it is a Lie group with tangent space $M_n(\mathbb{R})$ at $I$. Its left-invariant field generated by $X$ is $A\mapsto AX$; differentiating two such fields gives bracket $XY-YX$.

Expanding the determinant by permutations shows $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$, hence $d(\det {)}_I(X)=\mathrm{tr}X$. At every $A\in {\det }^{-1}(1)$, multiplication by $A$ transports this differential to a nonzero functional, so $1$ is a regular value. By [F3], ${\mathrm{SL}}_n$ is embedded and its tangent space at $I$ is the trace-zero kernel.

Determinant multiplicativity and $\det (A^{-1})=(\det A{)}^{-1}$ make the level set a subgroup, so its induced operations are smooth and it is a Lie group. The commutator bracket preserves trace zero because $\mathrm{tr}(XY)=\mathrm{tr}(YX)$.

For $n=1$, ${\mathfrak{s}\mathfrak{l}}_1=0$. Singular tangent matrices are allowed. No interval, endpoint, metric choice, or biconditional occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used only through the current tangent-bracket interface [F1]; the displayed Euclidean coordinates are finite and add no choice.

Let $F:{\mathrm{GL}}_n(\mathbb{R})\to {\mathrm{Sym}}_n(\mathbb{R})$ be $F(A)=A^{T}A$. Its differential is $d{F}_A(X)=X^{T}A+A^{T}X$. This is surjective: for symmetric $S$, take $X=\frac{1}{2}{A}^{-T}S$. Hence [F3] makes $F^{-1}(I)=O(n)$ an embedded submanifold, and it is a subgroup by [F2].

At $A=I$, the tangent kernel is $X^T+X=0$. It is closed under commutators because $(XY-YX{)}^T=Y^T{X}^T-X^T{Y}^T=-(XY-YX)$ for skew-symmetric $X,Y$.

On $O(n)$, $\det (A{)}^2=\det (A^{T}A)=1$, so determinant takes only the values $1$ and $-1$. Its $1$-fibre is therefore open and closed in $O(n)$ and is an embedded Lie subgroup with the same identity tangent space.

For $n=0$ both groups are the one-point group; for $n=1$, $\mathfrak{s}\mathfrak{o}(1)=0$, $O(1)$ is discrete, and $SO(1)$ is trivial. No interval, endpoint, nondegeneracy beyond invertibility in ${\mathrm{GL}}_n$, metric choice, or biconditional occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is propagated only through [F1].

Regard $M_n(\mathbb{C})$ as ${\mathbb{R}}^{2n^2}$. The complex determinant is a finite polynomial in matrix entries by [F5], hence its real and imaginary parts are real polynomials. By [F2], ${\mathrm{GL}}_n(\mathbb{C})=\{A:\det A\mathrm{\ne }0\}$ is open in this real vector space; multiplication is polynomial and inversion is real smooth there by the adjugate formula. Thus it is a real Lie group by [F1]. Its left-invariant field with identity value $X$ is $A\mapsto AX$; differentiating these linear fields gives the tangent bracket $[X,Y]=XY-YX$ in the convention of [F1]. Now $F(A)=A^{\ast }A$ maps this open group smoothly into the real vector space of Hermitian matrices and has $d{F}_A(X)=X^{\ast }A+A^{\ast }X$. For Hermitian $S$, $X=\frac{1}{2}{A}^{-\ast }S$ maps to $S$, so [F4] makes $U(n)=F^{-1}(I)$ embedded; the adjoint-product identities make it a subgroup. At $I$ its tangent kernel is $X^{\ast }+X=0$.

For $A\in U(n)$, $\mathrm{\mid }\det A{\mathrm{\mid }}^2=\det (A^{\ast }A)=1$, so determinant maps $U(n)$ into the unit circle. Near $1$ that circle has the real coordinate $z\mapsto \mathrm{Im}z$ on the arc $\mathrm{Re}z>0$. Differentiating the finite determinant formula at $I$ gives $d(\det {)}_I(X)=\mathrm{tr}X$; on skew-Hermitian $X$ this is imaginary and every imaginary scalar occurs from a diagonal $X$. Left multiplication by any $A\in SU(n)$ transports this surjectivity to $A$. Hence $1$ is a regular value of the circle-valued determinant map and [F4] makes $SU(n)$ embedded in $U(n)$, with tangent kernel $\mathrm{tr}X=0$ at $I$. Its subgroup operations are smooth by restriction.

Both tangent spaces are closed under commutator: adjoint reverses products, and trace of a commutator vanishes by finite reindexing. Hence they are the asserted Lie algebras.

At $n=1$, $\mathfrak{u}(1)=i\mathbb{R}$ and $\mathfrak{s}\mathfrak{u}(1)=0$; $n=0$ is excluded by the Statement. No interval, endpoint, arbitrary metric choice, or biconditional occurs. The ambient determinant is nonzero exactly on ${\mathrm{GL}}_n(\mathbb{C})$ by [F2]. ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1]; finite coordinates add no choice.

Let ${\mathrm{Skew}}_{2n}(\mathbb{R})$ be the vector space of skew-symmetric matrices and define $F:{\mathrm{GL}}_{2n}(\mathbb{R})\to {\mathrm{Skew}}_{2n}(\mathbb{R})$ by $F(A)=A^{T}JA$; the codomain is correct because $J^T=-J$. Then $d{F}_A(X)=X^{T}JA+A^{T}JX$. At a point of $F^{-1}(J)$ write $X=AZ$; then the differential is $Z^{T}J+JZ$. Every skew-symmetric $S$ occurs by taking $Z=\frac{1}{2}{J}^{-1}S$. Hence the differential is surjective onto its stated codomain along the level, and [F3] makes it embedded.

The equations $(AB{)}^{T}J(AB)=J$ and $(A^{-1}{)}^{T}J{A}^{-1}=J$ show that the level is a subgroup, so [F1] makes it a Lie group. At $I$, the tangent kernel from step 1.1 is exactly $X^{T}J+JX=0$.

If $X$ and $Y$ satisfy that equation, direct expansion gives $(XY-YX{)}^{T}J+J(XY-YX)=0$, so the tangent space is closed under the commutator bracket.

For $n=0$ the group is trivial. Singular tangent matrices are allowed, while group matrices are invertible because the defining equation gives an explicit inverse. No interval, endpoint, metric choice, or biconditional occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is propagated only through [F1].

Matrix multiplication gives $h(x,y,z)h(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }})=h(x+x^{\mathrm{\prime }},y+y^{\mathrm{\prime }},z+z^{\mathrm{\prime }}+x{y}^{\mathrm{\prime }})$ and $h(x,y,z{)}^{-1}=h(-x,-y,-z+xy)$. Thus ${\mathbb{R}}^3$ with these polynomial formulas is a Lie group embedded in ${\mathrm{GL}}_3$.

Differentiation at the identity gives the span of $E_{01},E_{12},E_{02}$. From the product law in step 1.1, the corresponding left-invariant fields are $X^L={\mathrm{\partial }}_x$, $Y^L={\mathrm{\partial }}_y+x{\mathrm{\partial }}_z$, and $Z^L={\mathrm{\partial }}_z$. Their commutators are $[X^L,Y^L]=Z^L$ and $[X^L,Z^L]=[Y^L,Z^L]=0$. Hence [F1] gives $[X,Y]=Z$ and $Z$ central.

The group is nonempty and three-dimensional; its Lie algebra is two-step nilpotent but the bracket is degenerate because $Z$ is central. No metric, interval, endpoint, or biconditional occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is propagated only through the current tangent-bracket supplier, and finite coordinates add no choice.

Multiplication and inversion are $(a,b)(c,d)=(ac,ad+b)$ and $(a,b{)}^{-1}=(a^{-1},-a^{-1}b)$. These are smooth for $a,c>0$. The chart $(a,b)\mapsto (\log a,b)$ identifies the underlying manifold with ${\mathbb{R}}^2$, hence it is connected.

Tangent matrices at the identity are $u{E}_{00}+v{E}_{01}$. In the $(a,b)$ coordinates, the left-invariant fields generated by $H$ and $X$ are $H^L=a{\mathrm{\partial }}_a$ and $X^L=a{\mathrm{\partial }}_b$. Their commutator is $[H^L,X^L]=a{\mathrm{\partial }}_b=X^L$, so [F1] gives $[H,X]=X$.

The group is nonempty and two-dimensional; its nonabelian bracket has the one-dimensional ideal spanned by $X$. No metric, nondegeneracy, interval, endpoint, or biconditional occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited only through [F1], with no further choice.

Integer translations preserve the standard smooth charts, so addition and negation descend to smooth operations on the quotient; hence [F1] gives an $n$-dimensional Lie group with tangent space ${\mathbb{R}}^n$ at the identity.

For $X\in {\mathbb{R}}^n$, the curve $t\mapsto [tX]$ is a one-parameter subgroup with initial velocity $X$. By [F2], its time-one point is ${\exp }_{{\mathbb{T}}^n}(X)=[X]$. This equals the identity exactly when $X\in {\mathbb{Z}}^n$.

At $n=0$ the torus and kernel are trivial; at $n=1$ this is the circle quotient. The lattice is discrete but no nondegeneracy is asserted. There is no metric, endpoint issue, or biconditional beyond the direct kernel calculation. The assumed ${\mathrm{A}\mathrm{C}}_{\omega }$ is used by [F2] through its stated supplier chain, and the fixed integer lattice adds no choice.

Suppose first that $n\ge 1$ and put $\parallel A{\parallel }_{\mathrm{\infty }}={\max }_{i,j}\mathrm{\mid }{a}_{ij}\mathrm{\mid }$. The row-by-column formula [F4] gives $\parallel AB{\parallel }_{\mathrm{\infty }}\le n\parallel A{\parallel }_{\mathrm{\infty }}\parallel B{\parallel }_{\mathrm{\infty }}$, and hence $\parallel X^k{\parallel }_{\mathrm{\infty }}\le n^{k-1}\parallel X{\parallel }_{\mathrm{\infty }}^k$ for $k\ge 1$. Thus [F2] dominates the matrix series and its termwise derivative uniformly on every compact $t$-interval by scalar exponential series. Consequently, for $E(t)={\sum }_{k\ge 0}{t}^k{X}^k\mathrm{/}k!$, termwise differentiation is valid for every real $t$, and [F4] gives $E^{\mathrm{\prime }}(t)=XE(t)=E(t)X$ and $E(0)=I$. When $n=0$ this is the constant identity curve directly.

Let ${\gamma }_X:\mathbb{R}\to G$ be the subgroup from [F1], viewed as a matrix curve. Its velocity at $t$ is the left translate of $X$, so ${\gamma }_X^{\mathrm{\prime }}(t)={\gamma }_X(t)X$ and ${\gamma }_X(0)=I$. Transposing gives $({\gamma }_X^T{)}^{\mathrm{\prime }}=X^T{\gamma }_X^T$. Step 1.1 likewise gives $(E^T{)}^{\mathrm{\prime }}=X^T{E}^T$ with $E(0{)}^T=I$. On every compact interval containing zero, [F3] makes these transposed solutions equal. Thus ${\gamma }_X(t)=E(t)$ for all $t\in \mathbb{R}$.

Evaluating step 2.1 at $t=1$ proves the displayed formula and, because ${\gamma }_X(1)\in G$, also proves $e^X\in G$. For $n=0$ both sides are the identity of the one-point group, and for $X=0$ both equal $I$; no invertibility or spectral hypothesis on $X$ is used. The parameter is global, so $t=1$ is not an endpoint issue. This is an equality, not an iff claim. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used exactly through [F1]; the series and ODE comparison add no choice.

The tangent curve $c(t)=I+tX+o(t)$ gives $C_g(c(t))=I+t(gX{g}^{-1})+o(t)$. By [F1], differentiating at zero proves ${\mathrm{Ad}}_{g}X=gX{g}^{-1}$.

By [F3], a curve through the identity with velocity $X$ is $e^{tX}=I+tX+o(t)$, whose inverse is $e^{-tX}=I-tX+o(t)$. Step 1.1 therefore gives ${\mathrm{Ad}}_{e^{tX}}Y=e^{tX}Y{e}^{-tX}=Y+t(XY-YX)+o(t)$.

Differentiating step 2.1 at zero yields $d(\mathrm{Ad}{)}_I(X)(Y)=XY-YX$; [F2] identifies the left side with ${\mathrm{ad}}_X(Y)$ and proves the second formula. For $n=0$ all matrices and maps are uniquely zero; for $X=0$ or $Y=0$ the commutator vanishes as the formula says. No invertibility is required of $X$ or $Y$, there is no metric or endpoint condition, and no iff is asserted. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used exactly through [F2] and [F3]; differentiating the fixed curves adds no choice.

Direct calculation using [F3] gives $\det A=1$, so $A\in {\mathrm{GL}}_2^{+}(\mathbb{R})$.

By [F1], the positive-determinant open subgroup is a Lie group; it is path connected. Indeed, for any $B=(b_1{b}_2)$ in it, put $u=b_1\mathrm{/}\parallel b_1\parallel$, let $v$ be the positive quarter-turn of $u$, and set $Q=(uv)\in \mathrm{SO}(2)$. Then $Q^{T}B=R=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ with $r=\parallel b_1\parallel >0$ and $t=\det (B)\mathrm{/}r>0$. The path $R_{\lambda }=\left(\begin{array}{cc}(1-\lambda )r+\lambda & (1-\lambda )s\\ 0& (1-\lambda )t+\lambda \end{array}\right)$ joins $R$ to $I$ through positive-determinant matrices, while writing the fixed $Q$ as a rotation through some angle $\theta$ gives the path of rotations from $Q$ to $I$. Concatenating $B=QR$ first to $Q$ and then to $I$ proves path connectedness.

Assume for contradiction that a real matrix $X$ satisfies $e^X=A$. The defining power series commutes with $X$, so $XA=AX$. Since $A$ has the two distinct eigenspaces $\mathbb{R}{e}_1$ and $\mathbb{R}{e}_2$, commutation makes each of them $X$-invariant. Hence $X{e}_1=x{e}_1$ for some real $x$, and the power series gives $e^X{e}_1=e^x{e}_1$ with $e^x>0$, whereas $A{e}_1=-2e_1$. This is impossible.

Thus $A$ has no real matrix logarithm. By [F2], it is not in the image of the Lie exponential of the connected group established in step 1.2, disproving surjectivity. The witness is nonsingular and two-dimensional; no claim is made in dimensions zero or one. There is no boundary, metric, interval endpoint, or iff issue. The logarithm obstruction and path construction are choice-free; ${\mathrm{A}\mathrm{C}}_{\omega }$ is present only because the current Lie-exponential interface [F2] carries it.

By [F1], $[X,Y]=E_{13}$, $[X,[X,Y]]=E_{03}$, and $[Y,[X,Y]]=0$. Every product of four strictly upper-triangular $4\times 4$ matrices is zero. The coefficient of the surviving cubic commutator will be determined directly below, without applying a local BCH theorem outside its neighbourhood.

The failure can be checked without relying on formal uniqueness. Since $X^2=E_{02}$, $X^3=Y^2=0$, direct multiplication gives $e^X{e}^Y=I+E_{01}+E_{12}+E_{23}+\frac{1}{2}{E}_{02}+E_{13}+\frac{1}{2}{E}_{03}$.

For $Z_0=E_{01}+E_{12}+E_{23}+\frac{1}{2}{E}_{13}$, [F1] gives $Z_0^2=E_{02}+E_{13}+\frac{1}{2}{E}_{03}$, $Z_0^3=E_{03}$, and $Z_0^4=0$. Hence $e^{Z_0}=I+E_{01}+E_{12}+E_{23}+\frac{1}{2}{E}_{02}+E_{13}+\frac{5}{12}{E}_{03}$.

The $E_{03}$ coefficients in steps 1.2 and 1.3 are respectively $1\mathrm{/}2$ and $5\mathrm{/}12$, so $e^{Z_0}\mathrm{\ne }{e}^X{e}^Y$. Moreover $E_{03}$ annihilates every strictly upper-triangular matrix on either side, so it commutes with $Z_0$ and has square zero. Hence $e^{Z_0+E_{03}\mathrm{/}12}=e^{Z_0}(I+E_{03}\mathrm{/}12)=e^X{e}^Y\mathrm{.}$ The exponential is injective on strictly upper-triangular $4\times 4$ matrices: for $N^4=0$, its polynomial inverse is $\log (I+K)=K-K^2\mathrm{/}2+K^3\mathrm{/}3$, and direct finite expansion gives $\log (e^N)=N$. Thus $Z_0+E_{03}\mathrm{/}12$ is the exact logarithm and the omitted term is precisely $[X,[X,Y]]\mathrm{/}12$. The matrix calculation is choice-free; ${\mathrm{A}\mathrm{C}}_{\omega }$ is stated only for [F2]. No endpoint, metric, or biconditional occurs.