Lie-algebraic elements as derivations.

  • Thread starter Kreizhn
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  • #1
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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.
 

Answers and Replies

  • #2
743
1
Nothing? Okay, let me say something else that will maybe be answerable instead.

Let M be an [itex] n^2 [/itex] dimensional matrix Lie group, and specify a local coordinate system [itex] (x^{ij})_{i,j=1}^n [/itex]. Let I be the identity element, and take [itex] A \in T_I M \cong \text{Lie}(M) [/itex]. If we write

[tex] A = \sum_{i,j} a^{ij} \left.\frac{\partial}{\partial x^{ij}} \right|_I [/tex]

do the [itex] a^{ij} [/itex] correspond to the matrix elements in this representation?
 

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