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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?

Arctic Char
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Alexey
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    Commutators of derivations (even at a point) are not $C^\infty$-linear, so $[L,L']= 0$ doesn't imply $[fL,f'L'] = 0$ for functions $f,f'$ and derivations $L,L'$. – Indraneel Tambe 2 Jul 10 '22 at 16:58
  • @IndraneelTambe2 Isn't the commutator linear? as in $[L + L', K] = [L, K] + [L', K]$ where $L,L',K$ are derivations? – Alexey Jul 10 '22 at 17:01
  • It's linear over "constants" ($\mathbb{F}$-linear) but not over "functions" ($C^\infty$-linear). (Note: if using derivations at a point, replace "functions" by "germs of functions at the point".) For example, $\partial_y$ commutes with $\partial_x$, but not with $y\partial_x$. – Indraneel Tambe 2 Jul 10 '22 at 17:02
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    The Lie algebra of a Lie group $G$ is the space of left-invariant vector fields on $G$. Indeed, each $\frac{\partial}{\partial x_{i,j}}$ is a vector field on $G$, but it's not left invariant. – Arctic Char Jul 10 '22 at 17:06
  • @IndraneelTambe2 I'm re-reading the question I linked, and in the answer they write: Our goal is to show that $D$ can be written in these coordinates as: $$D = \sum c_i \frac{\partial}{\partial x_i}$$ for some constants $c_i$. Is this wrong (those constant actually being functions maybe)? – Alexey Jul 10 '22 at 17:17
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    A tangent vector at a point can be expressed as a constant linear combination of the coordinate tangent vectors, at that point. However, to specify a vector field, you cannot just use constant linear combinations of the coordinate vector fields. And to compute a Lie bracket, you need the full left-invariant vector field, not just the tangent vector at a point. (The Lie bracket of vector fields depends on their values in a neighborhood, not just at a point! So the Lie bracket of two single vectors makes no sense; you can only take Lie brackets of vector fields, defined on open neighborhoods.) – Indraneel Tambe 2 Jul 10 '22 at 17:21
  • @IndraneelTambe2 I see, and when extending the derivation at a point into a $G$-invariant derivation of the whole space those constants will be functions? – Alexey Jul 10 '22 at 17:23
  • Yes, and the "constants" will be the values of those functions, at the identity (but to reiterate, you need the full functions). – Indraneel Tambe 2 Jul 10 '22 at 17:27
  • @IndraneelTambe2 Got it, thanks! If you wish to collect your comments into an answer I'll gladly accept it. If you can explicitly write into what functions those constant turn I would be also thankful, but I'll try to understand it myself as well. – Alexey Jul 10 '22 at 17:29
  • Unfortunately I don't have the time to write a full answer; but hopefully the comments above should be enough to get you started. – Indraneel Tambe 2 Jul 10 '22 at 17:34

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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 $1$.)

Travis Willse
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