Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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$

share|improve this question
1  
I wanted to understand the origin of linear momenta as translation generators in the $ E(2) $ (Euclidean) group. –  user35952 Feb 7 at 2:08
    
Did you make any notes of the calculation of the generator of translation group? If yes, then I would like to see it. –  Ome Feb 21 at 18:18
add comment

migrated from physics.stackexchange.com Feb 7 at 16:00

This question came from our site for active researchers, academics and students of physics.

3 Answers

up vote 7 down vote accepted

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: $$ P_x\sim \begin{pmatrix} 0 & 0 &1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\quad,\quad P_y\sim \begin{pmatrix} 0 & 0 &0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\quad\textrm{and} \quad M \sim \begin{pmatrix} 0 & 1 &0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$

share|improve this answer
    
Thanks. I checked this already. But this is not a unitary representation. Is it ? –  user35952 Feb 7 at 3:03
1  
+1: I had forgotten about this way of packaging things; very slick. –  joshphysics Feb 7 at 3:14
5  
@user35952 You are right. $E(2)$ is non-compact and has no finite-dimensional unitary representations. I think from the view of pedagogy, unitarity is a secondary issue especially when is trying to understand the Lie Algebra and Lie Group connection. –  suresh Feb 7 at 3:25
    
The finite-dimensional unitary representation is not possible because of the commutation relations of these operators ? –  user35952 Feb 7 at 3:39
    
@user35952 I don't think so. I think you may be thinking of a converse of the Stone-von Neumann theorem, which converse does not hold. –  WetSavannaAnimal aka Rod Vance Feb 10 at 0:29
show 5 more comments

For one thing, it's impossible to write a nonzero translation on $\mathbb R^2$ as a linear transformation (matrix). I encourage you to try to prove this. Therefore, since exponentiating a matrix gives another matrix, given a nonzero translation $T(\mathbf x) = \mathbf x + \mathbf a$, there does not exist a matrix $M$ for which $T(\mathbf x) = e^M\mathbf x$ for all $\mathbf x\in\mathbb R^2$.

However, let $f$ denote a smooth real-valued function on the real line, then translations act on such a function as follows: \begin{align} (T_af)(x) = f(x+a). \end{align} Now, notice that Taylor expansion gives \begin{align} f(x+a) &= \left(1 +a\frac{d}{dx} + \frac{a^2}{2!}\frac{d^2}{dx^2} + \cdots + \frac{a^n}{n!}\frac{d^n}{dx^n} + \cdots\right)f(x) \\ &= \exp\left(a\frac{d}{dx}\right) f(x) \end{align} to that the translation operator can be written as the exponential of the derivative; \begin{align} T_a = \exp\left(a\frac{d}{dx}\right) \end{align} But recall that, up to normalization, the derivative is precisely the momentum operator in the position space representation of a particle moving on the real line in quantum mechanics. This is one way of seeing that momentum generates translations.

This can easily be generalized to higher dimensions where the generator of translations in the direction of the standard ordered basis vector $\mathbf e_1$ is $\partial/\partial x^i$.

share|improve this answer
    
Thanks !! Linear transformations are something that will preserve the form of an equation in that space. In that case translations do preserve the form of equation right ? I might be wrong with definition though !! –  user35952 Feb 7 at 3:11
    
@user35952 I don't quite understand what you mean. Could you be more specific about what sorts of "equations" you're referring to? –  joshphysics Feb 7 at 3:13
    
Am sorry, I meant Equation of a curve in that space. For instance, equation of circle, straight line and so on. –  user35952 Feb 7 at 3:14
    
@user35952 Well let's see. A circle is described by the equation $x^2 + y^2 = r^2$, and if we translate $x\mapsto x+a$, $y\mapsto y+b$, then we get $(x+a)^2 + (y+b)^2 = r^2$ which is certainly not an equation of the same form. In fact, the only such equation that will have the same form is the equation for any line in the direction of the translation $(a,b)$ itself. –  joshphysics Feb 7 at 3:21
    
But is that not equation circle too ? Am sorry, I am not able to understand preserval of the form of the equation. –  user35952 Feb 7 at 3:24
show 3 more comments

As suresh mentioned if the vector is just a two component object then you can't translate it without expanding the vector. However, if you consider the vector to be variable (which are essentially infinite vectors) then it can be translated.

To find the differential form of a translation, start with the translation of a 1D dimensional vector, $x$: \begin{align} e ^{ i\epsilon {\cal P} } x & = x + \epsilon \\ \left( 1 + i \epsilon {\cal P} \right) x &= x + \epsilon \\ {\cal P} x & = - i \end{align} Thus we must have $ {\cal P} = - i \frac{ \partial }{ \partial x } $.

Now it is easy to extend this to two dimensions:

\begin{align} e ^{ i\epsilon _x {\cal P} _x + i \epsilon _y {\cal P} _y } \left( \begin{array}{c} x \\ y \end{array} \right) & = \left( \begin{array}{c} x + \epsilon _x \\ y + \epsilon _y \end{array} \right) \\ i \left( \epsilon _x {\cal P} _x + \epsilon _y {\cal P} _y \right) \left( \begin{array}{c} x \\ y \end{array} \right) &= \left( \begin{array}{c} x + \epsilon _x \\ y + \epsilon _y \end{array} \right) \end{align} where we have two different generators since you have two degrees of freedom in the transformation you gave in your question. This expression requires, \begin{align} & {\cal P} _x = \left( \begin{array}{cc} - i \frac{ \partial }{ \partial x } & 0 \\ 0 & 0 \end{array} \right) \\ & {\cal P} _y = \left( \begin{array}{cc} 0 & 0 \\ 0 & - i \frac{ \partial }{ \partial y } \end{array} \right) \end{align}

share|improve this answer
    
Thanks. In the $E(2)$ space, is not the vector definitely a variable. It is a space with continuous counting. –  user35952 Feb 7 at 3:07
    
@Ome: I'm not sure what you mean? –  JeffDror Feb 19 at 21:05
    
@JeffDror: We know that any element of a matrix lie group, $D(\alpha_m) = e^{\alpha_iX_i}$; $\alpha_i$ are the parameters and $X_i$ are the generators of the group. For translation group in $R^n$, what are the parameters, $\alpha_i$ ? While the $X_i$s can be constructed as the $P_x$ and $P_y$. –  Ome Feb 19 at 21:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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