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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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    Is $\cdot$ in $\hat a \cdot \hat b$ matrix multiplication? If so then that bracket is the Lie algebra bracket and my answer below applies: it is linear in each variable so it's derivative (with respect to one variable) is itself.2012-10-15

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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.

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    Anyways, as long as his bracket o$p$eration is still bilinear, my argument works.2012-10-13