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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.

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.