Codex @codex  0

Introduce a parameter $t$ to keep track of total degree. The power series for the matrix exponential givesSet $S=X+Y$ and $C=\frac{1}{2}[X,Y]$, where the commutator is $[X,Y]=XY-YX$. Expanding a second matrix exponential givesSincethe two expressions agree through total degree two. More explicitly, applying the local matrix logarithm to the first expansion gives $\log (e^{tX}{e}^{tY})=t(X+Y)+\frac{1}{2}{t}^2[X,Y]+O(t^3)$. This proves the displayed order of the Baker--Campbell--Hausdorff formula. The Lie algebra is closed under the commutator, so its displayed exponent lies in the same Lie algebra. The expansion is a formal power series identity, or a convergent identity sufficiently near the identity matrix; this argument does not assert unrestricted global convergence.

The next two terms of the Baker--Campbell--Hausdorff formula are the two cubic commutators:Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\frac{1}{24}[Y,[X,[X,Y]]]$.

For the Pauli matrices, use the Pauli matrix multiplication lawPut $N=\mathbf{n}\cdot \mathbf{\tau }$. The antisymmetric term disappears when contracted with $n_a{n}_b$, so $N^2=|\mathbf{n}{|}^{2}I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the matrix exponential into even and odd terms of its power series yieldsThe Pauli matrices are Hermitian matrices, and the components of $\mathbf{n}$ are real, so $N^{†}=N$. For real $\alpha$, the resulting matrix $U$ satisfiesAlso $\mathrm{tr}N=0$ and $N^2=I$, so its two eigenvalues are $1,-1$. The eigenvalues of $U$ are therefore $e^{i\alpha },e^{-i\alpha }$, whose product is one. Thus $U$ is a unitary matrix with determinant one: it belongs to the SU(2) group.

Multiplying the two exact Pauli matrix exponentials, and using ${\tau }_1{\tau }_2=i{\tau }_3$, givesIn particular, the ${\tau }_3$ term has a minus sign. Its quadratic Taylor expansion isTaking $X=i\alpha {\tau }_1$, $Y=i\beta {\tau }_2$, the Pauli matrix commutator identity gives $[X,Y]=-2i\alpha \beta {\tau }_3$. The quadratic exponent in the Baker--Campbell--Hausdorff formula is therefore $Z=i\alpha {\tau }_1+i\beta {\tau }_2-i\alpha \beta {\tau }_3$. In $e^Z=I+Z+\frac{1}{2}{Z}^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator ${\tau }_1{\tau }_2+{\tau }_2{\tau }_1=0$ gives $Z^2=-({\alpha }^2+{\beta }^2)I+O(3)$, reproducing the exact product's quadratic expansion.

Regard complex matrices as a real vector space of dimension $2n^2$ and Hermitian matrices as a real vector space of dimension $n^2$. Set $F(A)=A{A}^{\ast }$. Its derivative is $D{F}_A(H)=H{A}^{\ast }+A{H}^{\ast }$. At a unitary $A$, any Hermitian $S$ is reached by $H=SA/2$. Hence $I$ is a regular value, and the regular level set theorem makes $\mathrm{U}(n)=F^{-1}(I)$ a smooth real manifold of dimension $\boxed{n^2}$.

The tangent space is the kernel of this derivative. Equivalently,At $I$, therefore, $H$ is skew-Hermitian. The explicit geodesic with initial point $I$ and velocity $H$ is $\gamma (t)=e^{tH}$. The matrix exponential is unitary because $e^{t{H}^{\ast }}{e}^{tH}=I$, and $\dot{\gamma }(0)=H$. Its acceleration is $\gamma H^2$. Since $H^2$ is Hermitian and every tangent vector at $\gamma$ is $\gamma K$ with $K$ skew-Hermitian, the induced real inner product satisfiesIndeed the trace of a Hermitian matrix times a skew-Hermitian matrix is purely imaginary. Its acceleration is therefore always normal to the manifold, which is the geodesic equation for an induced Euclidean metric. Diagonalizing $H=W\mathrm{diag}(i{\lambda }_j)W^{\ast }$ gives the equivalent explicit construction $W\mathrm{diag}(e^{it{\lambda }_j})W^{\ast }$.

For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore giveHence $\mathbf{x}(t)=e^{tM}{\mathbf{x}}_0$ solves the initial value problem. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the matrix exponential into its even and odd powers provesThis is the matrix exponential when the square is minus the identity. If ${\mathbf{x}}_0=(x_0,y_0{)}^{\mathsf{T}}$, the general solution isThe constants in part (a) are $\alpha =x_0/5$ and $\beta =y_0+2x_0/5$, which makes the two forms identical.

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

Starting in state $1$, the onset waiting time has the exponential distribution of rate $\lambda$. Thus the probability of onset before the next screen is $1-e^{-\lambda t}$. This counts individuals who have already progressed to clinical disease as well as those still pre-clinical.

Conditional on onset at time $u<t$, the Markov property makes the remaining progression time exponential with rate $\nu$. Thus the probability of clinical progression before the screen, conditional on that onset time, is $1-e^{-\nu (t-u)}$.

Integrating over the onset density gives the unconditional clinical probabilityThe probability of being pre-clinical at the next screen is insteadTogether with $P_{11}(t)=e^{-\lambda t}$ and the exponential distribution for a patient initially in state $2$, the transition matrix isThis equals the matrix exponential $e^{Qt}$. The assumed unequal intensities avoid a zero denominator; at $\lambda =\nu$ the continuous limit is $P_{12}(t)=\lambda t{e}^{-\lambda t}$.

Age changes during follow-up, so the generator matrix $Q(s)$ changes with time. The transition matrix must solve ${\partial }_{t_2}P(t_1,t_2)=P(t_1,t_2)Q(t_2)$, with $P(t_1,t_1)=I$. It is generally not the matrix exponential of one constant matrix; even $\exp (\int Q(s)ds)$ is not generally valid because the matrices at different times need not commute.

For example, write the age-dependent onset rate as $\lambda (s)$ and keep progression rate $\nu$ constant. ThenEven when the integrated onset rate is elementary, this remaining integral need not have an elementary closed form. Special cases can have closed forms; time dependence does not by itself prove universal impossibility.

For a tractable likelihood function, freeze age at its left-endpoint value, or a midpoint value, on each observation interval $[t_j,t_{j+1})$. The genetic covariate is already constant. This piecewise-constant covariate approximation gives a constant matrix $Q_j$ on each interval and a product of closed-form interval probabilities:Here $x_j$ is the observed state at $t_j$, and the likelihood function is conditional on the initial state. The approximation improves when intervals are short relative to the variation in the age effect.

During the positive half-period, write $Y=(A,B{)}^T$ and $\dot{Y}=K_{+}Y$, with $K_{+}=\left(\begin{array}{cc}0& D^2\\ i& 0\end{array}\right)$. Then $K_{+}^2=i{D}^{2}I={\sigma }^{2}I$. Its two eigenvalues are $\pm \sigma$, and corresponding eigenvectors are $(\sigma ,i{)}^T$ and $(-\sigma ,i{)}^T$. Since $D>0$, they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, $e^{T{K}_{+}}=I\cosh (\sigma T)+(K_{+}/\sigma )\sinh (\sigma T)$. Because $D^2/\sigma =-i\sigma$, the fundamental matrix of a linear differential equation givesThe lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_{-}=\left(\begin{array}{cc}0& -D^2\\ i& 0\end{array}\right)$ and $K_{-}^2=({\sigma }^{\ast }{)}^{2}I$. The analogous matrix exponential has the stated $M^{-}$, replacing $\sigma$ by its complex conjugate ${\sigma }^{\ast }=-i\sigma$. Both matrices satisfywith ${\sigma }^{\ast }$ in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.

At each complete period the state is multiplied by the monodromy matrix of a periodic linear system $N=M^{-}{M}^{+}$, so $Y(2nT)=N^{n}Y(0)$. The determinant is therefore $\det N=1$. Let $c_{\pm }=\cosh ({\sigma }^{(\ast )}T)$ and $s_{\pm }=\sinh ({\sigma }^{(\ast )}T)$. Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt{2}DT$, the hyperbolic-function identity for $|\cosh (a+ia){|}^2$ consequently yieldsFor $x>0$, $F>2$: differentiating gives $F^{\prime }=\sinh x-\sin x>0$, since this derivative has derivative $\cosh x-\cos x>0$ and vanishes initially. The characteristic polynomial is ${\lambda }^2-F\lambda +1$. Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate $-\gamma$; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.

As $x\to \infty$, $F=\frac{1}{2}{e}^x[1+O(e^{-x})]$, $\Lambda =F[1+O(F^{-2})]$, and $\log \Lambda =x-\log 2+o(1)$. ThereforeIn particular, at $T=1$ the rate is asymptotic to $(\sqrt{2}D-\log 2)/2$, as required. The growth calculation uses the product $M^{-}{M}^{+}$, rather than averaging the two generators: the two generators do not commute.

In the linear problem, ${\dot{x}}_2=-x_2/(4R)$, so $x_2(t)=x_{20}{e}^{-t/(4R)}$. Solve ${\dot{x}}_1+x_1/R=x_2(t)$ with an integrating factor to obtain the complete solution:Equivalently the matrix exponential isBoth eigenvalues of $L$ are negative, so each solution eventually decays, despite the possible transient growth generated by $b$.

At the optimum, $d=q_{\ast }^{1/3}$ and the initial direction is $(1,{\rho }^{+})$. Since the final direction has ratio ${\rho }^{-}$,Using ${\rho }^{\pm }=2R(1\pm s)$, $R^2(1-s^2)=1$, and $q_{\ast }=(5-3s)/(5+3s)$ simplifies this to the exact maximum gain:As $R\to \infty$, $s=1-1/(2R^2)+O(R^{-4})$, so $q_{\ast }\to 1/4$ and $(1+s)/(1-s)\sim 4R^2$. Consequently the requested large-Reynolds-number transient growth scaling isThese are the optimal time and gain of a Reynolds-scaled triangular model. The leading gain can also be read from the off-diagonal matrix exponential entry: with $\tau =t/R$, $b=(4R/3)(e^{-\tau /4}-e^{-\tau })$ is maximized at $\tau =4\log 4/3$ and has leading maximum $R{4}^{-1/3}$.

Componentwise,soRepeated application gives $(A^k{)}^T=(A^T{)}^k$. Transposing the matrix exponential term by term therefore yieldsMultiplication of the two power series givesHenceIf $e^{tA}$ is an orthogonal matrix for every real $t$, thenIts coefficient of $t$ must vanish, sothat is, $A$ is a skew-symmetric matrix.

In odd dimension,so $\det A=0$. Since $A$ is real, its kernel contains a nonzero real vector $a$.

The plane $a^{\perp }$ is invariant under $A$, because $a\cdot Ab=-(Aa)\cdot b=0$. On this two-dimensional plane, $Ab$ is perpendicular to $b$. Put $|Ab|=\theta |b|$ with $\theta >0$. Then $A^{2}b$ is parallel to $b$, andsoThe matrix exponential fixes the kernel vector:Separating even and odd powers in its series and using $A^{2j}b=(-1{)}^j{\theta }^{2j}b$ givesThus $e^A$ acts as a rotation through angle $\theta$ on $a^{\perp }$ and fixes its axis $\mathbb{R}a$.

Visits can be missed or rescheduled because of illness, clinic availability or patient choice; follow-up may end through dropout or right censoring. The correct panel-observed multi-state likelihood uses each observed interval length ${\Delta }_{ik}$:Thus the four aggregated one-year counts are generally no longer sufficient; records must retain actual durations. Numerical maximum likelihood estimation uses the corresponding transition probabilities, or a matrix exponential of the generator when a larger model is fitted. One must not treat a two-year interval as a single one-year transition or as two known transitions through an unobserved intermediate state.

This conditional calculation is valid when visit scheduling and loss to follow-up are ignorable given the modeled information. If symptomatic patients are systematically examined earlier or leave the study differently, observation itself may be informative; the visit or dropout process may require a joint statistical model. Simply replacing one year by the actual interval does not correct informative observation.

For the stated matrix exponential parametrization in the connected Proper orthochronous Lorentz group, use the corresponding Spinor representation of the Lorentz group:The spacetime argument is inverse-transformed because this is an active transformation of the field at a fixed coordinate $x$. Equivalently, with $x^{\prime }=\Lambda x$, ${\psi }^{\prime }(x^{\prime })=D(\Lambda )\psi (x)$. The identity $[S^{\mu \nu },{\gamma }^{\rho }]={\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu }$ integrates to $D^{-1}{\gamma }^{\mu }D={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, which ensures covariance of the Dirac equation.

Strictly, $D$ depends on a lift to the Spin group, not just on the final Lorentz transformation: two lifts differ by sign. The specified generator exponential chooses a lift along its continuous path from the identity. This qualification is essential for the full-rotation comparison in part (g).

Insert ${\Omega }_{jk}=-{ϵ}_{jkl}{\varphi }^l$ and the result for $S^{jk}$ into the Spinor representation of the Lorentz group. The contraction of two Levi-Civita symbols ${\sum }_{j,k}{ϵ}_{jkl}{ϵ}^{jkm}=2{\delta }_l{}^m$ givesThe matrix exponential of this block diagonal matrix is the stated $O$, so ${\psi }^{\prime }(x)=O\psi ({\Lambda }^{-1}x)$. The plus sign in its exponent follows from both minus signs above; it is tied to the rotation-parameter convention in the PDF.

For a rotation through $2\pi$ about $z$, the Pauli matrix has eigenvalues $\pm 1$, hence $e^{i\pi {\sigma }^3}=-I_2$. The spacetime rotation is the identity, whileThis is the spinor sign under a full spatial rotation. A vector returns to itself already at $2\pi$. The comparison concerns the exponentiated Lorentz group representations: a Lie algebra by itself has no distinguished element called a $2\pi$ rotation. The sign records the nontrivial lift of this rotation loop to the Spin group.

Apply the method of multiple scales with independent fast time $t$ and slow time $T={ϵ}^{2}t$. A preliminary slow time $ϵt$ produces no resonant forcing at first order, so its leading amplitudes are constant; the first nontrivial modulation is at $T$. Write $x=x_0+ϵx_1+\cdots$, withAt order $ϵ$, the inhomogeneous linear differential equation is $({\partial }_t^2+1)x_1=-\cos t{x}_0$. Absorbing its homogeneous part into the amplitudes givesAt order ${ϵ}^2$, the forcing is $-2{\partial }_t{\partial }_T{x}_0-k{x}_0-\cos t{x}_1$. Its resonant Fourier series coefficients areThe solvability condition in the method of multiple scales sets both coefficients to zero, giving the amplitude equations. DefineFor bounded initial conditions $x(0)=x_{\ast }$, $\dot{x}(0)=v_{\ast }$, the requested leading approximation isThis holds on bounded intervals of $T$, hence $t=O({ϵ}^{-2})$, with the constant in the error allowed to depend on the bounded $T$ interval. If desired, retain the displayed $ϵx_1$ and adjust $A(0),B(0)$ by $O(ϵ)$ to match initial conditions at that order. No particular initial conditions were supplied.

Since $M^2=abI$, the matrix exponential is $I\cosh (\sigma T)+(M/\sigma )\sinh (\sigma T)$ when ${\sigma }^2=ab>0$. When $ab<0$, put $\omega =\sqrt{-ab}$ and replace this with $I\cos (\omega T)+(M/\omega )\sin (\omega T)$. At $ab=0$ it is $I+TM$. Thus the leading second instability tongue of a weak Mathieu oscillator has exponentially growing solutions for $-1/12<k<5/12$, and the leading modulation is bounded for every initial condition precisely whenAt either displayed endpoint, $M$ is a nonzero nilpotent operator; some initial conditions give a secular term proportional to $T$. This leading calculation alone must not be mistaken for an exact finite-$ϵ$ stability decision at the endpoints.

For an exact small-$ϵ$ interpretation, put $t=2s$. The standard Mathieu equation parameters are $a_{\mathrm{M}}=4+4k{ϵ}^2$ and $q=-2ϵ$. Its two relevant Mathieu characteristic values are obtained by projecting $-d^2/d{s}^2+2q\cos 2s$ onto even and odd Fourier series modes. Near the unperturbed value $4$, the even modes $1,\cos 2s,\cos 4s,\cos 6s$ have diagonal entries $0,4,16,36$ and adjacent couplings $\sqrt{2}q,q,q$ in an orthonormal basis; the odd modes $\sin 2s,\sin 4s,\sin 6s$ have diagonal entries $4,16,36$ and couplings $q,q$. Substituting $4+c_2{q}^2+c_4{q}^4$ into their characteristic polynomials givesHigher modes of the Fourier series first contribute at higher order. Therefore the fourth-order edges of the second Mathieu instability tongue areFloquet theory gives boundedness of every solution outside the closed gap, $k<k_{-}(ϵ)$ or $k>k_{+}(ϵ)$, in this fixed-$k=O(1)$ neighborhood. At the actual gap edges the second independent solution is generically unbounded. In particular, if $k$ is fixed exactly at $-1/12$ or $5/12$, it lies just outside the true gap and is stable for each sufficiently small positive $ϵ$. The strict boxed criterion is the robust leading-order, uniformly separated criterion; the endpoints require this higher-order refinement, and their amplification bounds need not remain uniform as $ϵ\to 0$.

For a Matrix Lie group, the exponential map of a matrix Lie group is the convergent matrix exponentialIts domain is the Lie algebra $\mathfrak{g}=T_{I}G$, and $e^{tX}\in G$ for every real $t$ when $X\in \mathfrak{g}$. Because multiples of the same matrix commute, $e^{sX}{e}^{tX}=e^{(s+t)X}$ and $(e^{tX}{)}^{-1}=e^{-tX}$. Thus take $I=\mathbb{R}$ to obtain the image of a one-parameter subgroup.

Here is the precise classification of nontrivial one-parameter subgroups. For $X\ne 0$, the smooth group homomorphism ${\gamma }_X:t\mapsto e^{tX}$ has a closed kernel $K\subset \mathbb{R}$. A closed subgroup of $\mathbb{R}$ is zero, $\tau \mathbb{Z}$ for some $\tau >0$, or all of $\mathbb{R}$. Indeed, if positive elements have infimum zero, their integer multiples approximate every real number, so closedness makes the subgroup all of $\mathbb{R}$; otherwise the infimum is attained and is its least positive generator. The last possibility is excluded by ${\gamma }_X^{\prime }(0)=X\ne 0$. Give the image the quotient topology and smooth structure from $\mathbb{R}/K$. Its inclusion in $G$ is an injective immersion, since ${\gamma }_X^{\prime }(t)=e^{tX}X$ never vanishes. Consequently it is a Lie subgroup, withIn the periodic case $[0,\tau )$ is also a parameter interval covering the image, with endpoints understood modulo $\tau$.

Two qualifications matter. If $X=0$, the image is the trivial group, a third possibility omitted by the wording “two”. Also, a general one-parameter subgroup need not be closed or embedded: an irrational winding $t\mapsto \mathrm{diag}(R(t),R(\sqrt{2}t))$ in a two-torus is an example, where $R(t)$ is a planar rotation. The classification uses the intrinsic immersed Lie subgroup structure, rather than assuming that the matrix subspace topology is the topology of $\mathbb{R}$.

For the real special linear group, differentiating $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$ at the identity shows that its Lie algebra consists of traceless matrices. Conversely $\det e^{tX}=e^{t\mathrm{tr}X}=1$ for a traceless matrix. HenceChoose the standard basis of the real sl2 Lie algebra,Their squares have the required values. Direct matrix multiplication gives the Lie bracketsWriting $[L_a,L_b]=f_{ab}{}^c{L}_c$, these specify all the nonzero Lie algebra structure constants: $f_{0+}{}^{+}=2$, $f_{0-}{}^{-}=-2$, $f_{+-}{}^0=1$, together with their negatives on reversing the lower indices.