# Finding properties of Poincare Transformation

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?

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 $$(\Lambda,a)_{\text{inv}}=(\Lambda^{-1},-\Lambda^{-1} a).$$

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.