[proofplan] The proof first chooses a real vector-space complement to the [Lie algebra](/page/Lie%20Algebra) inside the ambient [matrix space](/page/Matrix%20Space) and uses products of exponentials to build local coordinates near the identity matrix. The derivative of this coordinate map at the origin is the direct-sum isomorphism, so the finite-dimensional [inverse function theorem](/theorems/51) gives ambient coordinates. A no-small-transverse-elements result then identifies the nearby points of $G$ with the Lie-algebra slice in those coordinates. This gives the local exponential chart on $G$, and the final differential statement follows directly from the matrix exponential series. [/proofplan] [step:Split the ambient matrix space into tangent and transverse directions] Regard $M(n,\mathbb C)$ as a real [vector space](/page/Vector%20Space) of dimension $2n^2$. Since $\mathfrak g$ is a real vector subspace of $M(n,\mathbb C)$, choose a real vector subspace $\mathfrak h\subset M(n,\mathbb C)$ such that \begin{align*} M(n,\mathbb C)=\mathfrak g\oplus\mathfrak h. \end{align*} Let $\pi_{\mathfrak g}:M(n,\mathbb C)\to\mathfrak g$ and $\pi_{\mathfrak h}:M(n,\mathbb C)\to\mathfrak h$ denote the linear projections associated to this direct-sum decomposition. [/step] [step:Construct ambient coordinates using two exponentials] Define the real-smooth map $F:\mathfrak g\oplus\mathfrak h\to M(n,\mathbb C)$ by \begin{align*} F(X,Y)=\exp(X)\exp(Y). \end{align*} Here $\exp Z=\sum_{m=0}^{\infty}Z^m/m!$ is the matrix exponential, which is real analytic on the finite-dimensional real vector space $M(n,\mathbb C)$. For $(A,B)\in\mathfrak g\oplus\mathfrak h$, the differential of $F$ at $(0,0)$ is \begin{align*} dF_{(0,0)}(A,B)=A+B. \end{align*} Indeed, the [differential of the matrix exponential](/theorems/8834) at $0$ is $Z\mapsto Z$, and multiplication by $\exp(0)=I$ on either side does not change the first-order term. [guided] We introduce the second variable $Y\in\mathfrak h$ because $\exp:\mathfrak g\to G$ alone is the map we want to prove is locally invertible, while the ordinary inverse function theorem applies most directly to maps between open subsets of vector spaces of the same dimension. The complement $\mathfrak h$ supplies the missing transverse directions. Define $F:\mathfrak g\oplus\mathfrak h\to M(n,\mathbb C)$ by \begin{align*} F(X,Y)=\exp(X)\exp(Y). \end{align*} The domain and codomain are both finite-dimensional real vector spaces, and $F$ is real analytic because the matrix exponential is given by an absolutely convergent [power series](/page/Power%20Series) and matrix multiplication is polynomial in the entries. We compute the derivative at $(0,0)$. For real $t\ne 0$ and fixed $(A,B)\in\mathfrak g\oplus\mathfrak h$, the exponential series gives \begin{align*} \lim_{t\to 0}\frac{\exp(tA)-I-tA}{t}=0 \end{align*} and \begin{align*} \lim_{t\to 0}\frac{\exp(tB)-I-tB}{t}=0 \end{align*} in the real vector space $M(n,\mathbb C)$. Multiplying the first-order expansions and using bilinearity of matrix multiplication gives \begin{align*} \lim_{t\to 0}\frac{F(tA,tB)-I-t(A+B)}{t}=0. \end{align*} Therefore the first-order part of $F$ at $(0,0)$ is the [linear map](/page/Linear%20Map) \begin{align*} dF_{(0,0)}:\mathfrak g\oplus\mathfrak h\to M(n,\mathbb C),\qquad dF_{(0,0)}(A,B)=A+B. \end{align*} Because $M(n,\mathbb C)=\mathfrak g\oplus\mathfrak h$, this linear map is an isomorphism. [/guided] [/step] [step:Apply the inverse function theorem to obtain ambient product coordinates] Since $dF_{(0,0)}:\mathfrak g\oplus\mathfrak h\to M(n,\mathbb C)$ is the direct-sum isomorphism $(A,B)\mapsto A+B$, the ordinary inverse function theorem for finite-dimensional real vector spaces gives open neighbourhoods $U_0\subset\mathfrak g$ of $0$ in the real vector space $\mathfrak g$, $W_0\subset\mathfrak h$ of $0$ in the real vector space $\mathfrak h$, and $\Omega\subset M(n,\mathbb C)$ of $I$ in the real vector space $M(n,\mathbb C)$ such that \begin{align*} F:U_0\times W_0\to\Omega \end{align*} is a real-smooth diffeomorphism. Let $F^{-1}:\Omega\to U_0\times W_0$ denote its inverse, and define coordinate maps $a:\Omega\to U_0$ and $b:\Omega\to W_0$ by \begin{align*} F^{-1}(Z)=(a(Z),b(Z)). \end{align*} Both $a$ and $b$ are real-smooth. [/step] [step:Shrink the ambient chart so that $G$ is the slice with transverse coordinate zero] We apply [citetheorem:8812] with ambient matrix Lie group $GL(n,\mathbb C)$, closed subgroup $G\le GL(n,\mathbb C)$, ambient Lie algebra $M(n,\mathbb C)$, infinitesimal subgroup $\mathfrak g$, and vector-space complement $\mathfrak h$ satisfying $M(n,\mathbb C)=\mathfrak g\oplus\mathfrak h$. The hypotheses hold because the theorem statement assumes that $G$ is closed in $GL(n,\mathbb C)$, because $\mathfrak g$ is the Lie algebra of $G$ by hypothesis, and because $\mathfrak h$ was chosen as a complement to $\mathfrak g$ in the ambient Lie algebra. Therefore, after intersecting the neighbourhood supplied by [citetheorem:8812] with $W_0$ and then choosing a smaller open neighbourhood of $0$ if necessary, there is an open neighbourhood $W_1\subset W_0$ of $0$ in $\mathfrak h$ such that, with \begin{align*} \exp(W_1):=\{\exp Y:Y\in W_1\}, \end{align*} one has \begin{align*} G\cap\exp(W_1)=\{I\}. \end{align*} Since $U_0\times W_1$ is open in $\mathfrak g\oplus\mathfrak h$ and $F$ is a diffeomorphism on $U_0\times W_0$, replace $\Omega$ by the open neighbourhood $F(U_0\times W_1)$ of $I$. We claim that \begin{align*} G\cap\Omega=F(U_0\times\{0\}). \end{align*} The inclusion $F(U_0\times\{0\})\subset G\cap\Omega$ holds because $X\in\mathfrak g$ implies $\exp(X)\in G$. Conversely, let $Z\in G\cap\Omega$. Write $F^{-1}(Z)=(X,Y)$ with $X\in U_0$ and $Y\in W_1$. Since $X\in\mathfrak g$, $\exp(X)\in G$, and since $Z=\exp(X)\exp(Y)\in G$, we get \begin{align*} \exp(Y)=\exp(-X)Z\in G. \end{align*} Thus $\exp(Y)\in G\cap\exp(W_1)$, so $\exp(Y)=I$. Since $F$ is injective on $U_0\times W_1$ and $F(0,0)=I=F(0,Y)$, we obtain $Y=0$. Hence $Z=F(X,0)$, proving the claim. [guided] The point of this step is to prove that the ambient product coordinates separate the directions tangent to $G$ from directions transverse to $G$. The theorem [citetheorem:8812] says that, after shrinking the transverse neighbourhood, no non-identity element of the subgroup $G$ can be written as $\exp(Y)$ with $Y\in\mathfrak h$. Its hypotheses are satisfied here: the ambient matrix Lie group is $GL(n,\mathbb C)$, the subgroup $G\le GL(n,\mathbb C)$ is closed by hypothesis, the infinitesimal subgroup is the Lie algebra $\mathfrak g$, and the chosen vector space $\mathfrak h$ satisfies $M(n,\mathbb C)=\mathfrak g\oplus\mathfrak h$. Therefore, after intersecting the neighbourhood supplied by [citetheorem:8812] with $W_0$ and then choosing a smaller open neighbourhood of $0$ if necessary, there is an open neighbourhood $W_1\subset W_0$ of $0$ in $\mathfrak h$ such that, with \begin{align*} \exp(W_1):=\{\exp Y:Y\in W_1\}, \end{align*} one has \begin{align*} G\cap\exp(W_1)=\{I\}. \end{align*} We now shrink the ambient inverse-function-theorem chart so that only transverse vectors in $W_1$ occur. Since $U_0\times W_1$ is open in $\mathfrak g\oplus\mathfrak h$ and $F$ is a diffeomorphism on $U_0\times W_0$, replace $\Omega$ by the open neighbourhood $F(U_0\times W_1)$ of $I$. We prove \begin{align*} G\cap\Omega=F(U_0\times\{0\}). \end{align*} First, if $X\in U_0\subset\mathfrak g$, then the defining property of the Lie algebra gives $\exp(X)\in G$, so $F(X,0)=\exp(X)\in G\cap\Omega$. This proves $F(U_0\times\{0\})\subset G\cap\Omega$. Conversely, take $Z\in G\cap\Omega$. Since $F:U_0\times W_1\to\Omega$ is a diffeomorphism, there are unique $X\in U_0$ and $Y\in W_1$ such that \begin{align*} Z=F(X,Y)=\exp(X)\exp(Y). \end{align*} Because $X\in\mathfrak g$, we have $\exp(X)\in G$ and hence $\exp(-X)\in G$. Since also $Z\in G$, closure of $G$ under multiplication gives \begin{align*} \exp(Y)=\exp(-X)Z\in G. \end{align*} But $Y\in W_1$, so $\exp(Y)\in\exp(W_1)$. Hence $\exp(Y)\in G\cap\exp(W_1)=\{I\}$, and therefore $\exp(Y)=I$. Finally, injectivity of $F$ on $U_0\times W_1$ gives $Y=0$ from $F(0,Y)=I=F(0,0)$. Thus $Z=F(X,0)$, proving the reverse inclusion and the slice identity. [/guided] [/step] [step:Identify the exponential chart on $G$] Set \begin{align*} U:=U_0 \end{align*} and \begin{align*} V:=G\cap\Omega. \end{align*} The set $V$ is open in the [subspace topology](/page/Subspace%20Topology) on $G$ because $\Omega$ is open in $M(n,\mathbb C)$. By the slice identity from the previous step, \begin{align*} V=\{\exp X:X\in U\}. \end{align*} The map $\exp|_U:U\to V$ is surjective by this equality. It is injective because if $\exp X_1=\exp X_2$ for $X_1,X_2\in U$, then \begin{align*} F(X_1,0)=F(X_2,0), \end{align*} and injectivity of $F$ on $U\times W_1$ gives $X_1=X_2$. The inverse map $\log_G:V\to U$ is \begin{align*} \log_G(Z)=a(Z), \end{align*} where $a:\Omega\to U$ is the first component of $F^{-1}$. Since $a$ is real-smooth on $\Omega$, its restriction to $V$ is smooth in these exponential coordinates. [/step] [step:Compute the differential of the exponential at the identity of the Lie algebra] Finally, define $E:\mathfrak g\to M(n,\mathbb C)$ by \begin{align*} E(X)=\exp X. \end{align*} For $A\in\mathfrak g$ and real $t\ne 0$, the exponential series gives \begin{align*} \lim_{t\to 0}\frac{E(tA)-I-tA}{t}=0 \end{align*} in $M(n,\mathbb C)$. Therefore \begin{align*} dE_0(A)=A. \end{align*} Thus the differential of the ambient exponential map at $0$ is the inclusion $\mathfrak g\hookrightarrow M(n,\mathbb C)$, and after identifying $\mathfrak g$ with its image this is the identity map on $\mathfrak g$. This completes the proof. [/step]