# Lie-algebraic elements as derivations.

Hey,

So I'm trying to figure out how the matrix representatives of Lie-algebras can act as derivations. In particular, let $N \in \mathbb N$ and consider the Lie group of special unitary matrices $\mathfrak{SU}(N)$. Now we know that the Lie-algebra is the set of skew-Hermitian matrices $\mathfrak{su}(N)$, so let us choose an element $X \in \mathfrak{su}(N)$.

Since we can identify the Lie-algebra with the tangent space at the group identity $T_{\text{id}} \mathfrak{SU}(N) \cong \mathfrak{su}(N)$ we can view X as a tangent vector to identity. Furthermore, given a function $f: \mathfrak{SU}(N) \to \mathbb R$ we know that X acts on f to give a real value; namely, $Xf \in \mathbb R$.

Now let's say we're working in the standard matrix representation of $\mathfrak{su}(N)$, and fix the elements X and f. How can we compute Xf? I'm not certain what to do here and would appreciate any help.

Let M be an $n^2$ dimensional matrix Lie group, and specify a local coordinate system $(x^{ij})_{i,j=1}^n$. Let I be the identity element, and take $A \in T_I M \cong \text{Lie}(M)$. If we write
$$A = \sum_{i,j} a^{ij} \left.\frac{\partial}{\partial x^{ij}} \right|_I$$
do the $a^{ij}$ correspond to the matrix elements in this representation?