# Derivative wrt. to Lie bracket.

Let $\mathbf{G}$ be a matrix Lie group, $\frak{g}$ the corresponding Lie algebra, $\widehat{\mathbf{x}} = \sum_i^m x_i G_i$ the corresponding hat-operator ($G_i$ the $i$th basis vector of the tangent space/Lie algebra $\frak{g}$) and $(\cdot)^\vee$ the inverse of $\widehat{\cdot}$: $$(X)^\vee := \text{ that } \mathbf{x} \text{ such that } \widehat{\mathbf{x}} = X.$$

Let us define the Lie bracket over $m$-vectors as: $[\mathbf{a},\mathbf{b}] = \left(\widehat{\mathbf{a}}\cdot\widehat{\mathbf{b}}-\widehat{\mathbf{b}}\cdot\widehat{\mathbf{a}}\right)^\vee$.

(Example: For $\frak{so}(3)$, $[\mathbf{a},\mathbf{b}] = \mathbf{a}\times \mathbf{b}$ with $\mathbf{a},\mathbf{b} \in \mathbb{R}^3$.)

Is there a common name of the derivative: $$\frac{\partial [\mathbf{a},\mathbf{b}]}{\partial \mathbf{a}}$$?

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 How are the generators defined? As images under a basis of the $\exp$ map? And why is it that $\hat a \hat b - \hat b \hat a$ is necessarily a combination of generators of $G$? I'm guessing though that this is just the bracket on $\mathfrak g$ (since at least what you have agrees up to highest order if you're using the notion of generators I mentioned). – Eric O. Korman Oct 13 '12 at 15:24 The generators are simply a set of Cartesian basis vectors which span the tangent space. – Hauke Strasdat Oct 14 '12 at 23:06 I think you are conflating Lie groups and Lie algebras... – Eric O. Korman Oct 14 '12 at 23:32 Because I called $G_i$ the generator of the "Lie group?" This is a common convention in physics, but I am happy to call $G_i$ just the basis vector of the tangent space/Lie algebra. – Hauke Strasdat Oct 14 '12 at 23:39 Eric, thanks for your comments. I understand that it can be very misleading to call $G_i$ generators of the Lie group, so I edited the OP. – Hauke Strasdat Oct 15 '12 at 0:06
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I may be misunderstanding your question, but it seems like you are asking for the derivative of the map $$\mathfrak g \to \mathfrak g, ~~ a \mapsto [a, b]$$ where $b \in \mathfrak g$ is fixed. Since $\mathfrak g$ is a vector space, the derivative at a point can be viewed as a map $\mathfrak g \to \mathfrak g$. But the above map is linear so it is it's own derivative at any point.
 So the point is, you implicitly say, that $[{\bf a},{\bf b}]$ as defined by the hat operator above just coincides with the original Lie bracket $[a,b]$ in a way, am I right? – Berci Oct 13 '12 at 15:16 Ah ok... actually I'm not sure-- it should be. This notion is non-standard and I don't even see why it's well defined (see my comment to OP). – Eric O. Korman Oct 13 '12 at 15:22 Anyways, as long as his bracket operation is still bilinear, my argument works. – Eric O. Korman Oct 13 '12 at 15:25