0
$\begingroup$

I have started studying the Poincare group for the first time, in preparation for my first QFT course, and I wish to be able to solve the following problem:

A Poincare transformation ($\Lambda,a)$ can be written as: $$x'^{\mu}=\Lambda_{\nu}^{\mu}x^{\nu}+a^{\mu}$$ Determine the multiplication rule, ($\Lambda_1,a_1)$($\Lambda_2,a_2)$, as well as the inverse and unity element in this group.

I know that this is very basic, but it is probably because of it that I have not been able to find an explanation for beginners as to how to do it.

I know that the multiplication rule means that the multiplication of two elements of the group must still be a member of the group, but I don't know how to start proving it.

Similarly, I know the definitions of unity element and inverse, but I am lost as to how to begin working on this problem. Could you please point the way?

$\endgroup$
1
$\begingroup$

In order to not get lost with too many indices let's define $x'$, $x$ and $a$ as column vectors (with 4 elements) and $\Lambda$ as a matrix (with 4x4 elements).

Then the Poincaré transformation $$ x'^{\mu}=\Lambda^{\mu}{}_{\nu}x^{\nu}+a^{\mu} $$ can be written more concisely (by using matrix multiplication and vector addition) as $$x'=\Lambda x +a.$$

I know that the multiplication rule means that the multiplication of two elements of the group must still be a member of the group, but I don't know how to start proving it.

Consider a first Poincaré transformation $(\Lambda_1,a_1)$ as $$x'=\Lambda_1 x + a_1, \tag{1}$$ and a second Poincaré transformation $(\Lambda_2,a_2)$ as $$x''=\Lambda_2 x' + a_2. \tag{2}$$ You get the composition by inserting (1) into (2): $$\begin{align} x''&=\Lambda_2 (\Lambda_1 x + a_1) + a_2 \\ &= \Lambda_2 \Lambda_1 x + \Lambda_2 a_1 + a_2 \end{align}$$ Now it has the form $x''= \Lambda x + a$ with $$(\Lambda,a)_{\text{composed}} = (\Lambda_2\Lambda_1,\Lambda_2 a_1 + a_2).$$

Similarly, I know the definitions of unity element and inverse, but I am lost as to how to begin working on this problem. Could you please point the way?

The unity transformation is simply $$x'=x.$$ You can rewrite this as $$x'=\mathbf{I} x + \mathbf{0},$$ where $\mathbf{I}$ is the unity matrix ($\Lambda^{\mu}{}_{\nu}=\delta^\mu_\nu$) and $\mathbf{0}$ is the null vector ($a^\nu=0$).
Now it has the form $x'= \Lambda x + a$ with $$(\Lambda,a)_{\text{unity}}=(\mathbf{I}, \mathbf{0}).$$


You can find the inverse transformation of $$ x' = \Lambda x + a$$ by resolving this equation for $x$: $$\begin{align} x &= \Lambda^{-1}(x'-a) \\ &= \Lambda^{-1}x'-\Lambda^{-1}a \end{align}$$ where $\Lambda^{-1}$ is the inverse matrix of $\Lambda$. Now it has the form $x= \Lambda x' + a$ with $$(\Lambda,a)_{\text{inv}}=(\Lambda^{-1},-\Lambda^{-1} a).$$

$\endgroup$
1
$\begingroup$

The element $(\Lambda, a)$ can be seen as a function, let's call it $g_{\Lambda,a}$ $$ \begin{aligned} g_{\Lambda,a} : \;\;\mathbb{R}^{1,3} &\to \mathbb{R}^{1,3}\\ \quad x^\mu &\to\Lambda^\mu_{\phantom{\mu}\nu}\,x^\nu + a^\mu \,. \end{aligned} $$ This is a representation of the group.${}^1$ Representations are useful because we can do explicit computations with them and prove abstract properties of the group. A representation is fully specified only if we also say what does the group product $(\Lambda_1,a_1)\cdot (\Lambda_2,a_2)$ map to in this space. Obviously the answer is: function composition $g_{\Lambda_1,a_1} \circ g_{\Lambda_2,a_2}$.

We need to compute the following composition: $g_{\Lambda_1,a_1} \circ g_{\Lambda_2,a_2}$ $$ \begin{aligned} x^\mu \;\;&\overset{g_{\Lambda_2,a_2}}{\longrightarrow}\;\; (\Lambda_2)^\mu_{\phantom{\mu}\nu}\,x^\nu + a_2^\mu \equiv y^\mu\\ \;\;&\overset{g_{\Lambda_1,a_1}}{\longrightarrow}\;\; (\Lambda_1)^\mu_{\phantom{\mu}\nu}\,y^\nu + a_1^\mu = \\ &\quad\;\, = (\Lambda_1)^\mu_{\phantom{\mu}\nu}\,((\Lambda_2)^\nu_{\phantom{\nu}\rho}\,x^\rho + a_2^\nu) + a_1^\mu = \\ &\quad\;\, = (\Lambda_1\cdot \Lambda_2)^\mu_{\phantom{\mu}\rho}\,x^\rho + (\Lambda_1)^\mu_{\phantom{\mu}\nu}\,a_2^\nu + a_1^\mu\,. \end{aligned} $$ Now with $\cdot$ I mean plain simple matrix multiplication. I could also rename $\rho\to\nu$ in the last line. The task now is finding an element of the group $(\Lambda_?,a_?) \leftrightarrow g_{\Lambda_?,a_?}$ that does the exact same job $$ \begin{aligned} x^\mu \;\;&\overset{g_{\Lambda_?,a_?}}{\longrightarrow}\;\;(\Lambda_?)^\mu_{\phantom{\mu}\nu}\,x^\nu + a_?^\mu\\ &\quad\;\, = (\Lambda_1\cdot \Lambda_2)^\mu_{\phantom{\mu}\nu}\,x^\nu + (\Lambda_1)^\mu_{\phantom{\mu}\nu}\,a_2^\nu + a_1^\mu\,. \end{aligned} $$ This equation is easily solved to $$ (\Lambda_?)^\mu_{\phantom{\mu}\nu} = (\Lambda_1 \cdot \Lambda_2)^\mu_{\phantom{\mu}\nu}\,,\qquad a_?^\mu = (\Lambda_1)^\mu_{\phantom{\mu}\nu}\,a_2^\nu + a_1^\mu\,. $$ And this will be true in the abstract sense, not just in this particular representation, therefore we can say $$ (\Lambda_1,a_1)\cdot (\Lambda_2,a_2) = (\Lambda_1\cdot \Lambda_2, \,\Lambda_1 \cdot a_2 + a_1)\,. $$


$\quad{}^1$ Representations are actually linear transformations (endomorphisms) on a vector space. What we have is an affine transformation because of the translation piece $a^\mu$. But it's well known that we can still represent this as a linear transformation on the projective space.

$\endgroup$

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.