Differential Operators

Question

I was given some lecture notes on invariant differential operators. In the notes, there was something that I did not quite understand (link to original image)

and recall the matrix exponential function $$ \exp(X) = \sum\limits_{k=0}^\infty \frac{X^k}{k!}, $$ which satisfies $\det\exp(X) = e^{\text{trace}(X)}$. So, if $\text{trace}(X)=0$, we can define a differential operator on the manifold $G = \text{SL}(2,\Bbb R)$ via $$ \underline X.f(g) := \frac{d}{dt}\biggr\rvert_{t=0}f(g\exp(tX)) $$

I don't understand the correlation of having a matrix $X$ having a zero trace and being able to define a differential operator on $SL_2(\mathbb{R})$. Here $f$ is any differentiable function on $SL_2(\mathbb{R})$.

Answer 1

Closed subgroups $G$ of $Gl(n,k)$ ($k=\mathbb{R}$ or $\mathbb{C}$) are defined by a system of equations (they are called matrix Lie groups). A tangent vector $X$ at $1_G$ (the unit of $G$) is described by the one-parameter group with $X$ as infinitesimal generator (in fact, any path $\gamma$ such that $\gamma(0)=1$ and $\gamma'(0)=X$), this group is $\{exp(tX)\}_{t\in \mathbb{R}}$ in general. Here for all $X\in \frak{sl}(2,\mathbb{R})$ (i.e. with $tr(X)=0$) the path $\{exp(tX)\}_{t\in \mathbb{R}}$ is drawn on $SL(2,\mathbb{R})$ and satisfies the two preceding conditions ($\gamma(0)=1$ and $\gamma'(0)=X$), whence this description.

Answer 2

Recall that $\text{SL}_2(\Bbb R)$ is the group of $2\times2$ matrices with determinant $1$ (with the group operation of matrix multiplication). This means that any $g\in \text{SL}_2(\Bbb R)$ must have $\det g = 1$.

Next, recall that $\det ab = \det a \cdot \det b$ for any $n\times n$ matrices $a,b$.

Now, if $f: \text{SL}_2(\Bbb R) \to M$ is a function between manifolds, the expression $f(g\cdot\exp(tX))$ is only defined if

$$1 = \det(g\cdot\exp(tX)) = \det g \cdot \det \exp(tX) = \det\exp(tX) = \exp\text{tr}(tX)$$

and so we need $\text{tr}X = 0$.