Subjects: Group Theory, Differential Geometry
Links: Lie Groups, Matrix Representation of Linear Transformations, Automorphism Group, Trace of Matrix, Special Linear Group, Topological Connectedness

If $V$ is a vector space over the field $F$, the general linear group of $V$ is defined as $\text{GL}(V):=\text{Aut}(V)$. . This related to the Automorphism Group in any category

Let $F$ being an field, and $\text{M}(n,F)$ be the set of all $n\times n$ matrices over the field $F$. Then $\text{GL}(n,F):=\{A\in \text{M}(n,F)\mid detA\ne 0\}$.

If $V$ has finite dimension $n$, then $\text{GL}(V)$ and $\text{GL}(n,F)$ are isomorphic. This is because we have the matrix representation of linear transformations.

Lemma: Let $A$ be a $2\times 2$ matrix over a field $k$. If $A$ is not a scalar multiple of the identity matrix, then $A$ is similar to the matrix $$\begin{pmatrix}0 & -\det(A) \ 1 & \text{tr}(A)\end{pmatrix} $$
Cor: Two non scalar $2\times 2$ matrices over $k$ are similar iff they have the same eigenvalues.

General Linear Group over $\mathbb{R}$

We have that $\text{GL}(n,\mathbb{R})$ has extra properties. In particular we have that $\text{GL}(n,\mathbb{R})=\stackrel{-1}{det}[\mathbb{R}\setminus \{0\}]$, meaning is an open set of ${\mathbb{R}}^{n\times n}$. Then we have that as an open subset of ${\mathbb{R}}^{n\times n}$, it is a manifold. It is fairly easy to prove that $\mu$ (matrix multiplication) and $\iota$ (matrix inverse) are ${\mathcal{C}}^{\mathrm{\infty }}$. Meaning that $\text{GL}(n,\mathbb{R})$ is an Lie group.

Let $det:\text{GL}(n,\mathbb{R})\to \mathbb{R}$ be the determinant map. The tangent space $T_I(\text{GL}(n,\mathbb{R}))$ to $\text{GL}(n,\mathbb{R})$ at the identity $I$ is the vector space ${\mathcal{M}}_n(\mathbb{R})$.

Prop: If $V$ is a finite-dimensional real or complex vector space then there's a Lie group isomorphism $\text{GL}(V)\cong \text{GL}(n,\mathbb{R})$ or $\text{GL}(n,\mathbb{C})$.

Prop: For any $X\in {\mathcal{M}}_n(\mathbb{R})$, $d(det{)}_I:{\mathcal{M}}_n(\mathbb{R})\to \mathbb{R}$, $d(det{)}_I(X)=\text{tr}(X)$.

Prop: For any $X\in {\mathcal{M}}_n(\mathbb{R})$, $d(det{)}_A:{\mathcal{M}}_n(\mathbb{R})\to \mathbb{R}$ ,$d(det{)}_A(X)=(detA)(\text{tr}(X))$.

Prop: $Z(\text{GL}(n,\mathbb{R}))=\{\lambda I\mid \lambda \in {\mathbb{R}}^{\times }\}\cong {\mathbb{R}}^{\times }$.

Prop: Let $\varphi :\text{GL}(2n+1,\mathbb{R})\to \text{SL}(2n+1,\mathbb{R})\times {\mathbb{R}}^{\times }$, defined as $\varphi (A)=((detA{)}^{-1/(2n+1)}A,detA)$. Then $\varphi$ is a Lie group isomorphism. Meaning that $\text{GL}(2n+1,\mathbb{R})\cong \text{SL}(2n+1,\mathbb{R})\times {\mathbb{R}}^{\times }$.

Prop: $Z(\text{GL}(2n,\mathbb{R}))\ncong Z(\text{SL}(2n,\mathbb{R})\times {\mathbb{R}}^{\times })$. Meaning that $\text{GL}(2n,\mathbb{R}))\ncong \text{SL}(2n,\mathbb{R})\times {\mathbb{R}}^{\times }$.

Prop: $\text{GL}(n,\mathbb{R})\cong \text{SL}(n,\mathbb{R})⋊{\mathbb{R}}^{\times }$, we can think of this as the Semidirect Product of Lie Groups.

The subset ${\text{GL}}^{+}(n,\mathbb{R})\subseteq \text{GL}(n,\mathbb{R})$ consisting of real $n\times n$ matrices with positive determinants is a subgroup. It is an open subset of $\text{GL}(n,\mathbb{R})$ by continuity of the determinant, thus it is an embedded Lie subgroup of dimension $n^2$.

We would like to consider the Lie algebra of $\text{GL}(n,\mathbb{R})$. Since $T_{I_n}\text{GL}(n,\mathbb{R})$ can be identifies with $\text{GL}(n,\mathbb{R})$, then the Lie algebra of $\text{GL}(n,\mathbb{R})$ is itself. To make the distinction that is equipped with the bracket commutator of matrices we denote it as $\mathfrak{g}\mathfrak{l}(n,\mathbb{R})$.

Prop: Since $det:\text{GL}(n,\mathbb{R})\to {\mathbb{R}}^{\times }$ is a Lie group homomorphism, then its induced Lie algebra homomorphism is $\text{tr}:\mathfrak{g}\mathfrak{l}(n.\mathbb{R})\to \mathbb{R}$.

Define a map $\beta :\text{GL}(n,\mathbb{C})\to \text{GL}(2n,\mathbb{R})$ by replacing each complex matrix entry $a+bi$ with the $2\times 2$ block $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$: $$\beta \begin{pmatrix}
a_1^1+ ib_1^1 & \dots &a_1^n +ib^n_1 \
\vdots &&\vdots \
a_n^1+ ib_n^1 & \dots &a_n^n +ib^n_n
\end{pmatrix} = \begin{pmatrix}
a_1^1 & -b_1^1 & \dots &a_1^n & -b^n_1 \
b_1^1& a_1^1 & \dots & b_1^n & a^n_1 \
\vdots &\vdots&&\vdots&\vdots \
a_n^1 & -b_n^1 & \dots &a_n^n & -b^n_n \
b_n^1& a_n^1 & \dots & b_n^n & a^n_n \
\end{pmatrix}$$We can see that $\beta$ is an injective Lie group homomorphism whose image is a closed Lie subgroup of $\text{GL}(2n,\mathbb{R})$. Thus $\text{GL}(n,\mathbb{C})$ is isomorphic to this Lie subgroup of $\text{GL}(2n,\mathbb{R})$.

Prop: Let us consider the evaluation map $\epsilon :\text{Lie}(\text{GL}(n,\mathbb{C}))\to T_{I_n}\text{GL}(n,\mathbb{C})$, and the usual identification between tangent spaces to an open subset of a vector space and the vector space itself $\phi :T_{I_n}\text{GL}(n,\mathbb{C})\to \mathfrak{g}\mathfrak{l}(n,\mathbb{C})$. The composition of these maps yield a Lie algebra isomorphism between $\text{Lie}(\text{GL}(n,\mathbb{C}))$ and the matrix algebra $\mathfrak{g}\mathfrak{l}(n,\mathbb{C})$, i.e., $\phi \circ \epsilon :\text{Lie}(\text{GL}(n,\mathbb{C}))\to \mathfrak{g}\mathfrak{l}(n,\mathbb{C})$ is a Lie algebra isomorphism.

Prop: The connected components of $\text{GL}(n,\mathbb{R})$ are ${\text{GL}}^{+}(n,\mathbb{R})$ and ${\text{GL}}^{-}(n,\mathbb{R})$.

Prop: We can ge that $\text{GL}(n,\mathbb{R})$ is diffeomorphic to $\text{O}(n)\times {\text{T}}^{+}(n,\mathbb{R})$, where ${\text{T}}^{+}(n,\mathbb{R})$ is the Lie group of $n\times n$ upper triangular real matrices with positive diagonal entries. In particular, we get the diffeomorphism $\text{GL}(n,\mathbb{R})\cong \text{SO}(n)\times {\text{T}}^{+}(n,\mathbb{R})$.

Prop: $\text{GL}(n,\mathbb{C})\cong \text{SL}(n,\mathbb{C})⋊{\mathbb{C}}^{\times }$, we can think of this as the Semidirect Product of Lie Groups.