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Translations in two dimensions - Group theory

+ 7 like - 0 dislike
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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 :

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

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$

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
asked Feb 7, 2014 in Mathematics by user35952 (155 points) [ no revision ]
I wanted to understand the origin of linear momenta as translation generators in the $ E(2) $ (Euclidean) group.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
Did you make any notes of the calculation of the generator of translation group? If yes, then I would like to see it.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user Ome

3 Answers

+ 6 like - 0 dislike

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

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user suresh
answered Feb 7, 2014 by suresh (1,535 points) [ no revision ]
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Thanks. I checked this already. But this is not a unitary representation. Is it ?

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
+1: I had forgotten about this way of packaging things; very slick.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user joshphysics
@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.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user suresh
The finite-dimensional unitary representation is not possible because of the commutation relations of these operators ?

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
@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.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
Most recent comments show all comments
$n$ from translations and $n(n-1)/2$ from rotations giving a total of $n(n+1)/2$.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user suresh
@suresh: can you please suggest me a reference where the generators of the matrix lie groups are discussed clearly? I would be very pleased. :)

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user Ome
+ 5 like - 0 dislike

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 $\partial/\partial x^i$.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user joshphysics
answered Feb 7, 2014 by joshphysics (830 points) [ no revision ]
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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 !!

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
@user35952 I don't quite understand what you mean. Could you be more specific about what sorts of "equations" you're referring to?

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user joshphysics
Am sorry, I meant Equation of a curve in that space. For instance, equation of circle, straight line and so on.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
@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.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user joshphysics
But is that not equation circle too ? Am sorry, I am not able to understand preserval of the form of the equation.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
@user35952 Yes it's a circle too. It's a circle with the same radius but whose center is at $(-a,-b)$. We could only have said that the equation for the circle was preserved in form (at least if we're using standard terminology) if there were some $R$, say, for which the transformed equation were $x^2+y^2 = R^2$; that equation has the same form. So, for example, a scaling $x\mapsto \alpha x, y\mapsto\alpha y$ would preserve the form of the equation for a circle at the origin because the new equation would be $x^2 + y^2 = (r/\alpha)^2$. Are thinking of a different notion of "preserve."

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user joshphysics
Thanks. Makes sense ! So is that itself a proof that it is impossible to write translation as a linear transformation ?

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
+ 2 like - 0 dislike

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}

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user JeffDror
answered Feb 7, 2014 by JeffDror (650 points) [ no revision ]
Thanks. In the $E(2)$ space, is not the vector definitely a variable. It is a space with continuous counting.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user user35952
@Ome: I'm not sure what you mean?

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user JeffDror
@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 $P_y$.

This post imported from StackExchange Mathematics at 2014-07-04 11:56 (UCT), posted by SE-user Ome

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