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 $Gal(3)$?


  • $\begingroup$ 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

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.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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