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I have just started learning Lie Groups and Algebra. Considering a flat 2-d plane if we want to translate a point from $(x,y)$ to $(x+a,y+b)$ then can we write it as :

$$ \left( \begin{array}{ccc} x+a \\ y+b \end{array} \right) = \left( \begin{array}{ccc} x \\ y \end{array} \right) + \left( \begin{array}{ccc} a \\ b \end{array} \right)$$

Now the set of all translations $ T = \left( \begin{array}{ccc} a \\ b \end{array} \right) $ form a two paramater lie group (I presume) with addition of column as the composition rule.

If that is so, how do I go about finding the generators of this transformation.

PS: In my course I have been taught that the generators are found by calculating the taylor expansion of the group element about the Identity of the group. For instance, $SO(2)$ group $$ M = \left( \begin{array}{cc} cos \:\phi & -sin \:\phi \\ sin \:\phi & cos \:\phi \end{array} \right) $$ I obtain the generator by taking $$ \frac{\partial M}{\partial \phi}\Bigg|_{\phi=0} = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) $$

Now if I exponentiate this, I can obtain back the group element. My question how do I do this for Translation group.

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marked as duplicate by user35952, Sami Ben Romdhane, Michael Hoppe, Claude Leibovici, mau Feb 21 '14 at 8:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you calculated the generators? – omehoque Feb 10 '14 at 18:24
@Ome : I posted this question in Physics.SE (then got migrated to Math.Se). Have a look at this – user35952 Feb 11 '14 at 5:49
Thank you very much! – omehoque Feb 11 '14 at 7:14

Your Lie group in this case is just $\mathbb R^2$. One way to think of the infinitesimal generators is as the set of tangent directions at the origin. I'm not sure how you've defined the Lie bracket in your studies, but one way to do it is to extend a vector at the origin to a vector field by left multiplication. In this case, that amounts to translation, giving rise to a constant vector field on $\mathbb R^2$. The Lie bracket of any two constant vector fields is trivial, so the infinitesimal generators (the Lie algebra) are given by $\mathbb R^2$ with a zero bracket.

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+1. Thanks for your reply. I got the question resolved, I guess. Have a look at [this] (…) – user35952 Feb 21 '14 at 6:39

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