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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 $g \\in G$, let $C_g \\colon G \\to G$, $C_g(h) = g h g^{-1}$, be the [[inner automorphism|0RN00NTM3S]] given by $g$.\nSince $C_g(e) = e$, its [[differential|6P9PXQDSEN]] at $e$ is a linear map $\\operatorname{Ad}(g) = (dC_g)_e \\colon \\mathfrak{g} \\to \\mathfrak{g}$.\nThe *adjoint representation* of $G$ is the map\n$$\\operatorname{Ad} \\colon G \\to GL(\\mathfrak{g}), \\qquad g \\mapsto \\operatorname{Ad}(g).$$\n\n## Remarks\n\nFrom $C_{gh} = C_g \\circ C_h$ and the chain rule, $\\operatorname{Ad}(gh) = \\operatorname{Ad}(g)\\operatorname{Ad}(h)$, so $\\operatorname{Ad}$ is a [[representation|WYGRXCG2RN]]; it is also smooth.\n\nExample: for $G = GL_n(\\R)$ and $\\mathfrak{g} = M_n(\\R)$, the map $C_g$ is the restriction of the linear map $X \\mapsto g X g^{-1}$ of $M_n(\\R)$.\nSo $\\operatorname{Ad}(g)X = g X g^{-1}$.\n\nNon-example: the map $g \\mapsto (dL_g)_e$, with $L_g(h) = gh$, does not define a representation on $\\mathfrak{g}$.\nFor $g \\ne e$, $L_g$ does not fix $e$, so $(dL_g)_e$ sends $T_eG$ to $T_gG$, not to $\\mathfrak{g}$.\n\nThe differential of $\\operatorname{Ad}$ at $e$ is the adjoint representation $\\operatorname{ad}$ of the Lie algebra $\\mathfrak{g}$, given by $\\operatorname{ad}(X)(Y) = [X, Y]$.\nThe two are related by $\\operatorname{Ad}(\\exp X) = e^{\\operatorname{ad} X}$, and $g \\exp(X) g^{-1} = \\exp(\\operatorname{Ad}(g)X)$.\n",
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