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 :
$$ \left( \begin{array}{ccc}
x+a \\
y+a \end{array} \right) =
\left( \begin{array}{ccc}
x \\
y \end{array} \right) +
\left( \begin{array}{ccc}
a \\
b \end{array} \right)$$
Now the set of all translations $ T = \left( \begin{array}{ccc}
a \\
b \end{array} \right) $ form a two parameter 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.
I know the generators of translation are linear momenta in the corresponding directions. But I am not able to see this here.
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, $\operatorname{SO}(2)$ 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.
EDIT :This edit is to summarise and get a view of the answers obtained.
Firstly, the vector representation of the translation group (for 2D) would in general have the form :
$$ \begin{pmatrix} 1 & 0 & a_x\\ 0 & 1 & a_y \\ 0 & 0 & 1 \end{pmatrix}\ $$
with generators (elements of Lie algebra)
$$ T_x =\begin{pmatrix} 0 & 0 & i\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ , \;\;
T_y = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & i \\ 0 & 0 & 0 \end{pmatrix}\ $$
Secondly, the scalar-field representation of the same is given by the differential operators
$$ exp^{ i(a_x\frac{\partial}{\partial x}+ a_y\frac{\partial}{\partial y} )} $$
with generators
$$ T_x^s = i\frac{\partial}{\partial x},\;\;T_y^s = i\frac{\partial}{\partial y} $$
The Lie algebra is two-dimensional and abelian : $ [T_x,T_y] = 0$
Best Answer
The issue is that translations add an inhomogeneous piece and so there is no matrix associated with it. Change it to the following so that we can associate a matrix: $$ \begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix}= \begin{pmatrix} 1 & 0 & a\\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\ \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}\ . $$ Note that $(x,y,1)$ has the same data as $(x,y)$ and thus both are realizations of Euclidean space. Now you can write the $3\times 3$ matrix as the exponential of (the nilpotent matrix) $$ \begin{pmatrix} 0 & 0 &a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}\ . $$ This construction is related to the fact that the two-dimensional Euclidean group can be obtained as a (Inonu-Wigner) contraction of $SO(3)$ (but don't worry if this statement doesn't make sense right away). So you now obtain three generators for the Lie algebra for the Euclidean group: $$ P_x\sim \begin{pmatrix} 0 & 0 &1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\quad,\quad P_y\sim \begin{pmatrix} 0 & 0 &0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\quad\textrm{and} \quad M \sim \begin{pmatrix} 0 & 1 &0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$