Why isn't the lie algebra of $GL(n, \mathbb{F})$ abelian? From what I understand in this answer, derivations of a manifold (in my case, I'm intrested in the lie group $GL(n,\mathbb{F})$) are a linear combinations of $\frac{\partial}{\partial{x_{i}}}$.
To be more formal, we can look at a derivation (at $e$)
$X:GL(n,\mathbb{F})\to \mathbb{F}$ of $GL(n,\mathbb{F})$ and transfrom it to a derivation of $\mathbb{F}^{n^2}$ using the chart of $GL(n, \mathbb{F})$ that assigns each matrix to the vector of the flattened matrix. Such a derivation is a sum:
$$\sum a_{i,j} \frac{\partial}{\partial x_{i,j}}$$
But from what I know:
$$[\frac{\partial}{\partial x_{i,j}}, \frac{\partial}{\partial x_{p,q}}] = 0$$
Implying that the lie algebra of $GL(n, \mathbb{F})$ is abelian. Of course, this is false. Can someone please clarify what is my misunderstanding?
 A: It's true that $\left(\frac{\partial}{\partial x^i{}_j}\right)$ is a global frame of commuting vector fields on $GL(n, \Bbb F)$, but the Lie algebra of $GL(n, \Bbb F)$ consists precisely of the left-invariant vector fields on $GL(n, \Bbb F)$, which (except for the zero vector field) do not have constant coefficients with respect to that frame. Under the usual identification of each tangent space $T_B GL(n, \Bbb F)$, $B \in GL(n, \Bbb F)$, with the space $\Bbb F^{n \times n}$ of $n \times n$ matrices, the left-invariant vector field whose value at $I$ is $X$ is $\hat X$, where $$\hat X_A := T_I L_A \cdot X ,$$ and where $L_A$ denotes the map that multiplies by $A$ on the left. Since $L_A$ is the restriction of a linear map, under the usual identifications with concrete matrices we have $\hat{X}_A = AX$, whose components with respect to the coordinate frame $\left(\frac{\partial}{\partial x^i{}_j}\right)$ are not constant (again except when $X = 0$).
Exercise Computing the components of the left-invariant vector field determined by the coordinate vector $\left.\frac{\partial}{\partial x^i{}_j}\right\vert_I \in T_I GL(n, \Bbb F)$.
Unwinding definitions shows that $[\hat X, \hat Y]_I = X Y - Y X$, so we typically identify $\mathfrak{gl}(n, \Bbb F)$ (with its Lie bracket) with the space $M(n, \Bbb F)$ of $n \times n$ matrices endowed with the matrix commutator $$(X, Y) \mapsto X Y - Y X .$$ (For $n > 1$) not all $n \times n$ matrices commute, so $\mathfrak{gl}(n, \Bbb F)$ is nonabelian. (In fact, its center is the space of scalar matrices $\lambda I$ and so has dimension $1$.)
