# Lie derivative on Lie group in the direction of an element of Lie algebra

I want a reference to the definition of the Lie derivative of a smooth function $$f:G \to \mathbb R$$ on a Lie group $$G$$ in the direction of an element $$\theta$$ of the Lie algebra $$\mathfrak G$$.

I can see in many places the definition of the Lie derivative with respect to a vector field, and I also understand that an element of a Lie algebra $$\theta$$ can be considered as a left (or right) invariant vector field whose value at the identity is $$\theta$$.

But I would really like a sweet, short reference that combines both of these into one definition, or at least states this explicitly that this can be done.

What I really want is something accessible to engineers with a weak math background. As an example, think of $$SO(n)$$ as embedded in $$\mathbb R^{n^2}$$, and suppose $$f:SO(n) \to \mathbb R$$. Then the 'Lie derivative' of $$f$$ in the direction of a an anti-symmetric matrix $$A$$ at $$Q$$ is $$\mathcal L_A(Q) = \frac d{dt} f\big(Q(I+tA)\big) \Bigg|_{t=0} .$$

• Lie derivative along a vector field is by definition a map from tensor fields to tensor fields, and so in particular for scalar fields (as an argument) is determined by a vector field (as a parameter). If you are interested in the derivative in a given direction at a single point, you should take a look at any textbook on differential geometry, for this is not specific to Lie groups and algebras. Jan 17 at 15:11
• @AndreiSmolensky I added more to my question clarifying what I really want. (I do know that for a scalar field that the Lie derivative is equivalent to a directional derivative, as you say.) Jan 17 at 18:27

What about the formula $$(D_X\varphi)(g)=\frac{\rm d}{{\rm d}t}\Bigl\vert_{t=0}\varphi(g\exp_G(tX))$$ for $$\varphi\in C^\infty(G)$$, $$g\in G$$, and $$X\in \mathfrak g$$, where $$G$$ is a Lie group with its Lie algebra $$\mathfrak g$$? See for instance Eq. (5) in Ch. II of the book by S.Helgason, "Differential geometry, Lie groups, and symmetric spaces".