----- > [!proposition] Proposition. ([[general linear group is a Lie group]]) > The [[general linear group|general linear groups]] $\text{GL}_{n}(\mathbb{R})$ and $\text{GL}_{n}(\mathbb{C})$ are open subsets of $\mathbb{R}^{n^{2}}$ resp. $\mathbb{R}^{2n ^{2}}$ and are thus [[Lie group|Lie groups]] of dimension $n^{2}$ resp. $2n^{2}$ with [[Lie algebra|Lie algebras]] $\mathfrak{gl}_{n}(\mathbb{R}) \cong\mathbb{R}^{n \times n}$ resp. $\mathfrak{gl}_{n}(\mathbb{C}) \cong \mathbb{C}^{n \times n }$. > Moreover, the [[Lie algebra|Lie bracket]] on $\mathfrak{gl}_{n}$ is given by the standard matrix commutator $[X,Y]=XY-YX \in \mathbb{F}^{n \times n}$. Moreover, each of these groups has two [[connected component|connected components]] $\operatorname{GL}_{n}^{+}$, $\operatorname{GL}_{n}^{-}$ corresponding to [[orientation-preserving and orientation-reversing operators|orientation-preserving vs. orientation-reversing]] maps. (We prove this by constructing a path first to $\operatorname{SO}(n)$ and then using the analogous result there.) First note that $\operatorname{GL}_{n}^{+}=\det ^{-1}(+1)$ and $\operatorname{GL}_{n}^{-}=\det ^{-1}(-1)$ are disjoint open subsets of $\operatorname{GL}_{n}$, by [[continuous|continuity]] of $\det$. So all we have to show is that each is [[connected]]; equivalently [[path-connected]] since $\operatorname{GL}_{n}(\mathbb{R})$ and $\operatorname{GL}_{n}(\mathbb{C})$ are [[smooth manifold|smooth manifolds]]. We will focus on the real case. Starting with $\operatorname{GL}_{n}^{+}$, it suffices construct a [[parameterized curve|path]] $P(t)$ in $\operatorname{GL}_{n}^{+}$ from any $A \in \operatorname{GL}_{n}^{+}$ to $I$. The straight-line path $P(t)=(1-t)A + tI$ does not work, but it is basically all we can try so let's see if we can modify it. [[QR decomposition|Factorize]] $A=QR$ for $Q \in \operatorname{SO}(n)$ [[upper-triangular matrix|and]] $R \in \mathfrak{b}_{n}$ such that $R$ has positive [[diagonal]] entires. Define $R(t)=(1-t)R + t I \in \operatorname{GL}_{n}^{+}, t \in I$ Then define $M(t)=QR(t), t \in I.$ Note that $\det M(t)=\underbrace{ \det(Q) }_{ =1 }\det R(t)$, so $M$ is a path in $\operatorname{GL}_{n}^{+}$ from $A$ to $Q$. Now, $Q \in \operatorname{SO}(n)$ [[special orthogonal group|and]] $\operatorname{SO}(n)$ [[identity component of a Lie group|is path-connected]]; obtain a path $\gamma$ from $Q$ to $I$ in $\operatorname{SO}(n) \subset \operatorname{GL}^{+}_{n}$. Then the composition $P:=\gamma \circ M$ is our desired path. $\operatorname{GL}^{-}_{n}$ is now immediate, because it is [[diffeomorphism|diffeomorphic]] to $\operatorname{GL}_{n}^{+}$ via left-translation $L_{B}$ with any $B \in \operatorname{GL}_{n}$ with $\det B<0$. > [!proof]+ Proof. ([[general linear group is a Lie group]]) > First assume $\mathbb{R}$ is the base [[field|field]]. We know that $\text{GL}_{n}(\mathbb{R}) = \{ \boldsymbol A \in \mathbb{R}^{n^{2}} : \det A \neq 0 \}.$The [[determinant]] map $\mathbb{R}^{n^{2}} \to \mathbb{R}$ is [[continuous]] ([[determinant of a matrix|it is a polynomial]]). Since $\text{GL}_{n}(\mathbb{R})=\det ^{-1}(\{ \mathbb{R} \setminus 0 \})$, it is an open subset of $\mathbb{R}^{n^{2}}$ as the inverse image of the open set $\mathbb{R} \setminus \{ 0 \}$ under $\det$. We therefore can cover it with a single $n^{2}$-dimensional coordinate chart: the [[identity map]] on $\mathbb{R}^{n^{2}}$ restricted to this open subset. Multiplication and inversion are [[smooth maps between manifolds|smooth]] because both are [[polynomial 4|polynomials]] or [[matrix inverse is adjugate over determinant|ratios thereof]]. As an open subset of $\mathbb{R}^{n \times n}$ $\operatorname{GL}(n, \mathbb{R})$ inherits its [[tangent space at a point of a smooth manifold|tangent spaces]] to just be $\mathbb{R}^{n \times n}$, including $\mathfrak{gl}_{n}(\mathbb{R})\cong T_{I}\operatorname{GL}(n, \mathbb{R}) \cong \mathbb{R}^{n \times n}$. It remains to justify that the Lie bracket of $X,Y \in \mathfrak{gl}_{n}(\mathbb{R}) \cong \mathbb{R}^{n \times n}$ on $\operatorname{GL}(n,\mathbb{R})$ equals their matrix commutator $XY-YX$. This can be done by viewing $[X,Y]$ as the [[Lie derivative]] of [[Lie algebra of a Lie group|left-invariant vector fields]] First it is helpful to note the general fact that $dL_{A} |_{A'} X_{A'}=AX$ for $\operatorname{GL}(n, \mathbb{R})$. [[derivative|Indeed]] we see that for $A'\in \operatorname{GL}_{n}(\mathbb{R})$, $X_{A'} \in T_{A'}\operatorname{GL}_{n}(\mathbb{R})$, $L_{A}(A' + X_{A'})=\underbrace{ A A' }_{ L_{A} (A') } + \underbrace{ AX_{A'} }_{ \text{must be derivative} }$and conclude in general that $dL _{A} |_{A'}(X_{A'})=AX_{A'}$. In particular, $dL_{\alpha_{A}(t)} |_{I} X_{I}=\alpha_{A}(t)X_{I}$. Since $X$ is [[Lie algebra of a Lie group|left-invariant]] $dL_{A} |_{A'} X_{A'}=X_{A A'}$, in particular $dL_{\alpha_{A}(t)} |_{I} X_{I}=X_{\alpha_{A}(t)}$. Put together, have $X_{\alpha_{A}(t)}=\alpha_{A}(t) X_{I}.$ Thus the defining [[ODE]] for $\alpha_{A}(t)$, $\dot{\alpha}_{A}(t)=X_{\alpha_{A}(t)}$, can be formulated as $\dot{\alpha}_{A}(t)= \alpha_{A}(t) X_{I}.$ This is an [[solutions to linear homogenous system of ODEs with constant coefficients|LCCDE]] with initial condition ${\alpha}_{t}(0)=A$, [[solutions to linear homogenous system of ODEs with constant coefficients|with solution]] $\alpha_{A}(t)=Ae^{tX_{I}}$. So the flow of left-invariant $X$ is $\Phi_{t}(A)=Ae^{t X_{I}}$ and since $\Phi_{t}(A+ H)=\underbrace{ Ae ^{t X_{I} } }_{ \Phi_{t}(A) } + \underbrace{ He^{t X_{I}} }_{ \text{must be derivative} }$ (IOW $\Phi_{t}$ is linear in $A$) we have $d\Phi_{-t} |_{A} H_{A} =H_{A} e^{-t X_{I}}$ hence $(\Phi_{-t})_{*} Y_{\Phi_{t}(I)}=Y_{e^{t X_{I}}}e^{-tX_{I}}=e^{tX_{I}}Y_{I}e^{-t X_{I}}$ where we used left-invariance of $Y$ to get $Y_{ e ^{ tX_{I}}}=dL_{e^{t X_{I}}} |_{I}Y_{I}=e^{tX_{I}}Y_{I}$. Expanding the [[matrix exponential]] we have that for all $X,Y \in \operatorname{Lie}(\operatorname{GL})$, $\begin{align} [X,Y]=\mathcal{L}_{X}Y&= (I + tX_{I} + O(t^{2})) Y_{I} (I - tX_{I} + O(t^{2})) \\ &= (I + tX_{I} + O(t^{2})) (Y_{I} - tY_{I}X_{I} + O(t^{2})) \\ &= Y_{I} - tY_{I} X_{I} + tX_{I} Y_{I} + O(t^{2}) \\ &= Y_{I} + t [X_{I} , Y_{I}] + O(t^{2}) \end{align}$ from which we conclude the result (note that $[-,-]$ is pushforward-invariant in general, in patricular left-invariant, which is why computing at $I$ was enough.) ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```