[proofplan] The essential input is the closed-subgroup theorem for matrix groups: a closed subgroup of $GL(n,\mathbb C)$ has a smooth embedded manifold structure whose tangent space at the identity is exactly the infinitesimal subgroup defined by one-parameter subgroups. We then choose a sufficiently small exponential neighbourhood in $\mathfrak g$, translate its logarithm chart by left multiplication, and use the smoothness of logarithm-chart transition maps to obtain a smooth atlas compatible with the [subspace topology](/page/Subspace%20Topology). Finally, multiplication and inversion are smooth because locally they are expressed through matrix multiplication, inversion, the exponential map, and the logarithm chart. [/proofplan]

[step:Invoke the closed subgroup theorem to obtain the embedded smooth structure]Since $G\le GL(n,\mathbb C)$ is closed as a subset of $M(n,\mathbb C)$, it is in particular a closed subgroup of $GL(n,\mathbb C)$ in the subspace topology. By [Closed Subgroups Of General Linear Groups][citetheorem:8814], $G$ is a matrix Lie group. Applying the [Closed Subgroup Theorem For Matrix Groups][citetheorem:8774], $G$ carries a smooth embedded submanifold structure inside the open manifold $GL(n,\mathbb C)\subset M(n,\mathbb C)$, and the restrictions of matrix multiplication and matrix inversion to $G$ are smooth. Let $\mathfrak g_G:=T_I G\subset T_I M(n,\mathbb C)$. Under the standard affine identification \begin{align*} T_I M(n,\mathbb C)\cong M(n,\mathbb C), \end{align*} the [Lie algebra](/page/Lie%20Algebra) of this matrix Lie group is the tangent space at the identity. By [Lie Algebra as Tangent Space][citetheorem:8785], this tangent space agrees with the one-parameter-subgroup definition: \begin{align*} \mathfrak g_G=\{X\in M(n,\mathbb C):\exp(tX)\in G\text{ for every }t\in\mathbb R\}=\mathfrak g. \end{align*} Therefore the real dimension of the embedded manifold $G$ is $\dim_{\mathbb R}\mathfrak g$.[/step]

[guided]The point of this step is to separate the deep theorem from the coordinate construction. The hypothesis says that $G$ is a subgroup of $GL(n,\mathbb C)$ and that, as a subset of the ambient finite-dimensional [vector space](/page/Vector%20Space) $M(n,\mathbb C)$, it is closed. This is exactly the hypothesis needed to apply [Closed Subgroups Of General Linear Groups][citetheorem:8814], which promotes $G$ to a matrix Lie group. Once $G$ is known to be a matrix Lie group, the [Closed Subgroup Theorem For Matrix Groups][citetheorem:8774] gives the actual smooth structure: $G$ is a smooth embedded submanifold of $GL(n,\mathbb C)$, and since $GL(n,\mathbb C)$ is open in $M(n,\mathbb C)$, also an embedded submanifold of $M(n,\mathbb C)$. The same theorem also gives smoothness of the group operations inherited from matrices. It remains to connect this smooth structure with the specific vector space $\mathfrak g$ appearing in the statement. Let \begin{align*} \mathfrak g_G:=T_I G\subset T_I M(n,\mathbb C) \end{align*} be the tangent space at the identity. The ambient space $M(n,\mathbb C)$ is a finite-dimensional real vector space, so translation identifies \begin{align*} T_I M(n,\mathbb C)\cong M(n,\mathbb C). \end{align*} Under this identification, [Lie Algebra as Tangent Space][citetheorem:8785] says that the tangent-space Lie algebra equals the infinitesimal subgroup defined by one-parameter subgroups: \begin{align*} \mathfrak g_G=\{X\in M(n,\mathbb C):\exp(tX)\in G\text{ for every }t\in\mathbb R\}. \end{align*} The right-hand side is exactly the set denoted $\mathfrak g$ in the theorem statement. Hence the manifold dimension is \begin{align*} \dim_{\mathbb R}G=\dim_{\mathbb R}T_I G=\dim_{\mathbb R}\mathfrak g. \end{align*}[/guided]

[step:Choose an exponential neighbourhood at the identity] By [Exponential Is A Local Diffeomorphism][citetheorem:8791], there exist open neighbourhoods $U_0\subset\mathfrak g$ of $0$ and $V_0\subset G$ of $I$, where $V_0$ is open in the subspace topology on $G$, such that \begin{align*} \exp|_{U_0}:U_0\to V_0 \end{align*} is bijective and has a smooth inverse. Denote this inverse by \begin{align*} \log_{U_0}:V_0&\to U_0 \end{align*} \begin{align*} c&\mapsto (\exp|_{U_0})^{-1}(c). \end{align*} Let $U\subset U_0$ be an open neighbourhood of $0$ such that $\exp|_U$ is injective and $V:=\exp(U)$ is a neighbourhood of $I$ in $G$. Then \begin{align*} \log_U:V&\to U \end{align*} \begin{align*} c&\mapsto (\exp|_U)^{-1}(c) \end{align*} is the restriction of $\log_{U_0}$ to $V$, and is therefore smooth. Since $\exp|_{U_0}$ is a homeomorphism onto $V_0$, the set $V=\exp(U)$ is open in $G$ whenever $U$ is chosen inside $U_0$ with $V$ an identity neighbourhood. Thus the exponential coordinates on $V$ are compatible with the subspace topology. [/step]

[step:Translate the logarithm chart to every point of the group] For each $a\in G$, define the left translation map \begin{align*} L_a:G&\to G \end{align*} \begin{align*} b&\mapsto ab. \end{align*} Since multiplication on $G$ is smooth, $L_a$ is a smooth diffeomorphism with inverse $L_{a^{-1}}$. Define \begin{align*} aV:=L_a(V)=\{av:v\in V\}. \end{align*} Because $V$ is open in $G$ and $L_a$ is a homeomorphism, $aV$ is open in $G$. Now define \begin{align*} \varphi_a:aV&\to U \end{align*} \begin{align*} b&\mapsto \log_U(a^{-1}b). \end{align*} Its inverse is \begin{align*} \varphi_a^{-1}:U&\to aV \end{align*} \begin{align*} X&\mapsto a\exp X. \end{align*} The equality $\varphi_a^{-1}(\varphi_a(b))=b$ follows from $a^{-1}b\in V$ and the definition of $\log_U$, while $\varphi_a(\varphi_a^{-1}(X))=X$ follows from $X\in U$ and injectivity of $\exp|_U$. Since \begin{align*} G=\bigcup_{a\in G}aV \end{align*} because $a\in aV$ for every $a\in G$, the charts $\{(aV,\varphi_a):a\in G\}$ cover $G$. [/step]

[step:Verify smooth compatibility of translated logarithm charts] Let $a,b\in G$ be such that $aV\cap bV\neq\varnothing$. On the overlap, the transition map is \begin{align*} \varphi_b\circ\varphi_a^{-1}:\varphi_a(aV\cap bV)&\to \varphi_b(aV\cap bV) \end{align*} \begin{align*} X&\mapsto \log_U(b^{-1}a\exp X). \end{align*} Indeed, for $X\in\varphi_a(aV\cap bV)$, the point $a\exp X$ lies in $aV\cap bV$, so $b^{-1}a\exp X\in V$, and the displayed formula is defined. By [Smooth Transition Maps For Logarithm Charts][citetheorem:8793], this transition map is smooth on its domain. Interchanging $a$ and $b$ gives smoothness of the inverse transition map. Therefore the translated logarithm charts form a smooth atlas on $G$. Because every chart domain $aV$ is open in the subspace topology and each chart map is a homeomorphism onto an open subset of $\mathfrak g$, this atlas induces the same topology as the subspace topology on $G$. [/step]

[step:Express the group operations in logarithm coordinates] Let $a,b\in G$. To check multiplication near $(a,b)$, choose chart domains $aV$, $bV$, and $abV$. In these charts the multiplication map has the local expression \begin{align*} M_{a,b}:\Omega_{a,b}&\to U \end{align*} \begin{align*} (X,Y)&\mapsto \log_U\bigl(b^{-1}\exp X\, b\, \exp Y\bigr), \end{align*} where \begin{align*} \Omega_{a,b}:=\{(X,Y)\in U\times U:b^{-1}\exp X\, b\, \exp Y\in V\}. \end{align*} This formula follows from \begin{align*} (ab)^{-1}(a\exp X)(b\exp Y)=b^{-1}\exp X\, b\, \exp Y. \end{align*} The map $(X,Y)\mapsto b^{-1}\exp X\,b\,\exp Y$ is smooth as a map from the open subset $\Omega_{a,b}\subset\mathfrak g\times\mathfrak g$ into $G$, because matrix multiplication is smooth, conjugation by the fixed matrix $b$ is linear, and the matrix exponential is smooth. Composing with the smooth logarithm chart $\log_U:V\to U$ shows that multiplication is smooth in these coordinates. For inversion near $a$, use the charts $aV$ and $a^{-1}V$. Its local expression is \begin{align*} J_a:\Omega_a&\to U \end{align*} \begin{align*} X&\mapsto \log_U(a\exp(-X)a^{-1}), \end{align*} where \begin{align*} \Omega_a:=\{X\in U:a\exp(-X)a^{-1}\in V\}. \end{align*} This formula follows from \begin{align*} (a^{-1})^{-1}(a\exp X)^{-1}=a\exp(-X)a^{-1}. \end{align*} The map $X\mapsto a\exp(-X)a^{-1}$ is smooth by smoothness of scalar multiplication on $\mathfrak g$, the matrix exponential, and matrix multiplication. Composing with $\log_U$ proves that inversion is smooth. Thus the smooth atlas constructed above makes $G$ a Lie group. [/step]

[step:Identify the tangent space at the identity with the matrix Lie algebra] In the identity chart \begin{align*} \varphi_I:V&\to U \end{align*} \begin{align*} c&\mapsto \log_U(c), \end{align*} the curve associated to $X\in\mathfrak g$ is \begin{align*} \gamma_X:\mathbb R&\to G \end{align*} \begin{align*} t&\mapsto \exp(tX). \end{align*} For $t$ sufficiently close to $0$, $\gamma_X(t)\in V$, and \begin{align*} \varphi_I(\gamma_X(t))=tX. \end{align*} Therefore the tangent vector $\gamma_X'(0)$ has coordinate representative $X$ in $T_0\mathfrak g\cong\mathfrak g$. Conversely, if $\xi\in T_I G$, then in the chart $\varphi_I$ it corresponds to a unique vector $X\in T_0\mathfrak g\cong\mathfrak g$. The curve \begin{align*} t\mapsto \exp(tX) \end{align*} has tangent vector $\xi$ at $I$ in these coordinates. Hence the chart differential identifies $T_I G$ bijectively and naturally with $\mathfrak g$. This completes the proof of the stated smooth manifold structure, its translated logarithm charts, smooth group operations, and the tangent-space identification. [/step]