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  "body": "\n## Definition\n\nLet $G$ be a [[Lie group|7JBRGCVJRB]] with identity $e$ and Lie algebra $\\mathfrak{g} = T_eG$, the [[tangent space|E7A5DHFT01]] at $e$.\nFor $X \\in \\mathfrak{g}$, let $\\gamma_X \\colon \\R \\to G$ be the unique [[one-parameter subgroup|YHA64AP1HN]] with $\\gamma_X'(0) = X$.\nThe *exponential map* of $G$ is\n$$\\exp \\colon \\mathfrak{g} \\to G, \\qquad \\exp(X) = \\gamma_X(1).$$\n\n## Remarks\n\nEquivalently, $\\gamma_X$ is the integral curve through $e$ of the [[left-invariant vector field|SNA3DVPK0C]] with value $X$ at $e$.\nBy rescaling, $\\exp(tX) = \\gamma_X(t)$ for all $t \\in \\R$.\n\nExample: for $G = GL_n(\\R)$, $\\mathfrak{g} = M_n(\\R)$ and $\\gamma_A(t) = e^{tA}$ is a one-parameter subgroup with $\\gamma_A'(0) = A$.\nSo $\\exp(A) = e^A$ is the [[matrix exponential|020M04FN1H]].\n\nNon-example: in $G = SL_2(\\R)$, the matrix $\\begin{pmatrix} -1 \u0026 1 \\\\ 0 \u0026 -1 \\end{pmatrix}$ is not of the form $\\exp(X)$ for any $X$ in the Lie algebra $\\mathfrak{sl}_2(\\R)$.\nSo $\\exp$ need not be surjective, even for a connected group.\n\nThe map $\\exp$ is smooth and its differential at $0$ is the identity of $\\mathfrak{g}$.\nBy the inverse function theorem, it restricts to a diffeomorphism from a neighbourhood of $0$ in $\\mathfrak{g}$ onto a neighbourhood of $e$ in $G$.\nFor a compact connected Lie group, $\\exp$ is surjective.\n",
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