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For matrix Lie groups, the exponential map is usually defined as a mapping from the set of $n \times n$ matrices onto itself.

However, sometimes it is useful to have a minimal parametrization of our Lie algebra elements. Therefore, there is the hat operator $\hat{\cdot}$ which maps a $m$-vector onto the corresponding Lie algebra element

$$ \hat{\cdot}: \mathbb R^m \rightarrow g, \quad \hat{\mathbf x} = \sum_{i=0}^m x_i \mathtt G_i, $$

where $\mathtt G_i$ is called the $i$th generator of the matrix Lie algebra $g\subset \mathtt R^{n\times n}$ (especially by physicists). I am very much interested in the properties of this function.

Example for SO3: $\hat{\mathbf x} = \begin{bmatrix} 0&-x_3& x_2\\ x_3&0,&-x_1\\-x_2&x_1&0\end{bmatrix}$

Question 1: Does the function $\hat{\cdot}$ has a common name? Would "change of basis" transformation (from the standard basis to the Lie algebra basis) be the right name?

I guess in this context it make sense to assume that the family $\mathtt G_1,...,\mathtt G_m$ is linear independent. From the linear independence it follows that $\hat{\cdot}$ is bijective, right? Thus, there is an inverse function $v$ which maps a matrix Lie algebra onto the corresponding Lie algebra:

$$ v(\cdot): g \rightarrow \mathbb R^m$$ Example for SO3: $v(\mathtt R)= \begin{bmatrix}R_{3,2}\\R_{1,3}\\R_{2,1}\end{bmatrix}=\frac{1}{2} \begin{bmatrix}R_{3,2}-R_{2,3}\\R_{1,3}-R_{3,1}\\R_{2,1}-R_{1,2}\end{bmatrix} = -\begin{bmatrix}R_{2,3}\\R_{3,1}\\R_{1,2}\end{bmatrix} $

Question 2: Is there a closed form solution to write down/calculate the inverse of $\hat{\cdot}$?

I find it a bit confusing that the standard basis vectors are element of $\mathbb R^m$ while $\mathtt G_i$ are element of $\mathbb R^{n^2}$.

(Edit: Heavy changes after I developed a better understanding of the problem.)

(Edit 2: Replaced $[\cdot]$ by the common hat notation $\hat{\cdot}$.)

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1  
I think you mean physicists. –  joriki Sep 19 '11 at 10:52
    
Yes :-) Corrected! –  Hauke Strasdat Sep 19 '11 at 10:56
    
$[\cdot]$ is almost never going to be bijective. for $so(3)$ for example, its image is the set of all skew-symmetric matrices, which is not all matrices. –  Eric O. Korman Sep 19 '11 at 12:34
    
Thanks, I was sloppy in my definition, not making sure that $[\cdot]$ is surjective. I corrected it. –  Hauke Strasdat Sep 19 '11 at 12:46

2 Answers 2

up vote 1 down vote accepted

As to question 1, I'm not sure I'd qualify it as a transform, it seems purely notational to me. Consider the vector space $\mathbb{R}^n$, then the vector $\vec{x} = (x_1\ x_2\ \ldots \ x_n)^T \in \mathbb{R}^n$ is shorthand for

$$\vec{x} = \sum_i x_i \hat{e}_i$$

where $\{\hat{e}_i\}$ is a basis of $\mathbb{R}^n$. Your function, $\hat{\cdot}$, operates in the same manner. In other words, what you think of as your vector is just shorthand.

As to question 2, most likely yes, but it will be basis and Lie algebra dependent.

The confusion you mentioned is really a restatement that unless $m = n^2$, the basis does not span $\mathcal{M}^{n\times n}(\mathbb{R})$ which is okay as $SO(3) \subset \mathcal{M}^{3\times 3}(\mathbb{R})$. Although, if I remember correctly, the basis for $SU(2)$ spans $\mathcal{M}^{2\times 2}(\mathbb{C})$.

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Thanks! If $[\cdot]$ is purely notational, maybe you could comment on my original question (math.stackexchange.com/questions/63551/…) too. ;-) –  Hauke Strasdat Oct 6 '11 at 11:25
    
give me a little time, it is a lot to digest. –  rcollyer Oct 6 '11 at 13:33
    
Not that I fully understand it, but more and more I got used to the view that $\hat{\cdot}$ is purely notational. So, I guess one should completely avoid to introduce explicitly a function $v(\cdot)$. –  Hauke Strasdat Oct 19 '11 at 15:53

The function $\hat{\cdot}$ seems simply to be called hat-operator.

A google search revealed that some authors call the inverse of $\hat{\cdot}$ the vee-operator $(\cdot)^{\vee}$ and it can be defined as

$$( \hat{\mathbf x} )^\vee := \mathbf x$$

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