Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: $$\frac{\partial [\mathbf{a},\mathbf{b}]}{\partial \mathbf{a}}$$?

share|improve this question
    
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
show 1 more comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.