- #1
Kreizhn
- 743
- 1
Hey,
So I'm trying to figure out how the matrix representatives of Lie-algebras can act as derivations. In particular, let [itex] N \in \mathbb N [/itex] and consider the Lie group of special unitary matrices [itex] \mathfrak{SU}(N)[/itex]. Now we know that the Lie-algebra is the set of skew-Hermitian matrices [itex] \mathfrak{su}(N) [/itex], so let us choose an element [itex] X \in \mathfrak{su}(N) [/itex].
Since we can identify the Lie-algebra with the tangent space at the group identity [itex] T_{\text{id}} \mathfrak{SU}(N) \cong \mathfrak{su}(N) [/itex] we can view X as a tangent vector to identity. Furthermore, given a function [itex] f: \mathfrak{SU}(N) \to \mathbb R [/itex] we know that X acts on f to give a real value; namely, [itex] Xf \in \mathbb R[/itex].
Now let's say we're working in the standard matrix representation of [itex] \mathfrak{su}(N) [/itex], 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.
So I'm trying to figure out how the matrix representatives of Lie-algebras can act as derivations. In particular, let [itex] N \in \mathbb N [/itex] and consider the Lie group of special unitary matrices [itex] \mathfrak{SU}(N)[/itex]. Now we know that the Lie-algebra is the set of skew-Hermitian matrices [itex] \mathfrak{su}(N) [/itex], so let us choose an element [itex] X \in \mathfrak{su}(N) [/itex].
Since we can identify the Lie-algebra with the tangent space at the group identity [itex] T_{\text{id}} \mathfrak{SU}(N) \cong \mathfrak{su}(N) [/itex] we can view X as a tangent vector to identity. Furthermore, given a function [itex] f: \mathfrak{SU}(N) \to \mathbb R [/itex] we know that X acts on f to give a real value; namely, [itex] Xf \in \mathbb R[/itex].
Now let's say we're working in the standard matrix representation of [itex] \mathfrak{su}(N) [/itex], 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.