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I have a question regarding the logarithmic map $log: G\mapsto \mathfrak{g}$ and exponential map $exp: \mathfrak{g}\mapsto G$ between the Galilei group and its Lie algebra. The Galilei group of two dimensions can be represented by the matrix Lie group $$Gal(2)=\begin{bmatrix}R&v&a\\0&1&s\\0&0&1\end{bmatrix}$$ and its Lie algebra by $$\mathfrak{gal(2)}=\begin{bmatrix}X&w&b\\0&0&t\\0&0&0\end{bmatrix}.$$ Because the Galilei group $Gal(2)\cong \mathbb{R}^3\rtimes SE(2)\cong \mathbb{R}^3\rtimes (\mathbb{R}^2\rtimes SO(2))$ and there is a closed-form for the exponential and logarithmic maps for the special Euclidean group $SE(2)$ I wonder, if such a closed form also exists for the Galilei group $Gal(2)$? If yes, would that also apply to $Gal(3)$?

Lino

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  • $\begingroup$ I took the liberty of correcting a crucial mistake in your generators. See this. You can isolate them by setting, successively, all parameters =0, except for X; or w ; or b, or t, respectively. All in all, 6 parameters, of course. $\endgroup$ Jul 12 '20 at 22:46
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Yes, of course. You may intuit the name of the game from the simplest ever Affine Lie Group, and verify the 2×2 matrix expressions. I'll get you started on Gal(2), but of course this also works for Gal(3), etc.

In your case, you are working with upper triangular 4×4 matrices. I'll use vectors and 2×2 matrices for the upper two rows, which, somewhat confusingly, you conflated with scalar symbols.

The generic group element is $$G=\begin{bmatrix}{\mathbf R}(\theta')&\vec v&\vec a\\0&1&s\\0&0&1\end{bmatrix}$$
and the generic Lie algebra element is $$ g=\begin{bmatrix}{\mathbf X}&\vec w&\vec b\\0&0&t\\0&0&0\end{bmatrix}.$$

First, recall the SO(2) rotation is generated by $$ {\mathbf X}=i\theta \sigma_2= \theta \begin{bmatrix} 0&1\\-1&0\end{bmatrix} \qquad \implies \\ \exp {\mathbf X}= \begin{bmatrix} \cos\theta &\sin\theta \\-\sin\theta &\cos\theta \end{bmatrix} = {\mathbf R}(\theta), $$ with ${\mathbf R}( 0)={\mathbf 1} $.

So go to g which is a linear combination of 6 generators/parameters, $\theta; t; \vec w; \vec b$. Evaluate the group elements, exponentials of each of them separately: $$ \exp \begin{bmatrix} {\mathbf X} & 0&0\\0&0&0\\0&0&0\end{bmatrix} = \begin{bmatrix} {\mathbf R}(\theta) & 0& 0\\0&1&0\\0&0&1\end{bmatrix} ; $$ $$ \exp \begin{bmatrix} 0& 0&0\\0&0&t\\0&0&0\end{bmatrix} = \begin{bmatrix}{\mathbf 1} & 0& 0\\0&1&t\\0&0&1\end{bmatrix} ; $$ $$ \exp \begin{bmatrix} 0& \vec w&0\\0&0&0\\0&0&0\end{bmatrix} = \begin{bmatrix}{\mathbf 1} & \vec w&0\\0&1&0\\0&0&1\end{bmatrix} ; $$$$ \exp \begin{bmatrix} 0& 0&\vec b\\0&0&0\\0&0&0\end{bmatrix} = \begin{bmatrix}{\mathbf 1}& 0&\vec b\\0&1&0\\0&0&1\end{bmatrix} . $$

The product of all 6 (4) of them, in any order, will be an upper triangular matrix of the generic G type above; for instance , with the first one on the right, you get $$\begin{bmatrix}{\mathbf R}(\theta)&\vec w&\vec b + \vec w t\\0&1&t\\0&0&1\end{bmatrix}.$$

But also, by adroit orderings and CBH compositions, for example the trivial $$ \exp \begin{bmatrix} 0& \vec w&0\\0&0&0\\0&0&0\end{bmatrix} \exp \begin{bmatrix} 0& 0&\vec b\\0&0&0\\0&0&0\end{bmatrix}=\exp \begin{bmatrix} 0 & \vec w&\vec b\\0&0&0\\0&0&0\end{bmatrix} , $$ it will be the single exponential of an expression like g, $$\begin{bmatrix}\tilde{\mathbf X}&\tilde{\vec{w}}&\tilde{ \vec{b}}\\0&0&\tilde t\\0&0&0\end{bmatrix}.$$

but with its parameters being functions of the 6 parameters of your input. You compare the two expressions, and you get a relatively simple expression of group parameters as a function of algebra parameters. The exercise is not as trivial as the 2×2 matrix affine group expression I linked at the beginning, but it is likely to be tractable. (I wouldn't be shocked if you found it in some book, but I have not seen it.)

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  • $\begingroup$ Thank you very much for your answer! I am still confused about the approach to find the closed-form matrix exponential. The product of the 6(4) generators evaluates to $$ G=\begin{bmatrix} {\mathbf R}(\theta) & {\mathbf R}(\theta)\vec{w}&{\mathbf R}(\theta)\vec{b}+t{\mathbf R}(\theta)\vec{w}\\0&1&t\\0&0&1\end{bmatrix}, $$ but this does not give me the same result as $$ \exp \begin{bmatrix} {\mathbf X} & \vec{w}&\vec{b}\\0&0&t\\0&0&0\end{bmatrix}. $$ In my understanding, the group elements of $G$ are expressed as a function of the algebra parameters already, but I am confused. $\endgroup$
    – Antoni G.
    Jul 16 '20 at 5:39
  • $\begingroup$ Well, you multiplied them in the order written, whereas I chose to put the leftmost on the right. My last paragraph tells you of the last, crucial step: You have to BCH your way to a joint exponential of a messy linear combination of the generators, which I did not do for you, with different parameters, not the ones you wrote!, and compare answers to reparameterize , to find the map you are seeking. That's why I sent you to practice with the simpler affine case whose answer you know. $\endgroup$ Jul 16 '20 at 10:12

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