# Number of the Generators of Poincare Group

It is said that the Poincare group, $P(3,1)$ has $10$ generators. $6$ of them are the generators of the Lorentz group, $O(3,1)$ and the other $4$ generators are the generators of $4D$ translational group.

But Brian C Hall in his book named Lie Groups, Lie Algebras,and Representations puts that the Lie algebra of $P(n;1)$ is the space of all $(n+2) \times (n+2)$ real matrices of the form

\begin{align} \begin{pmatrix} Y & y \\ 0 & 0 \\ \end{pmatrix} \end{align} with $Y \in$ so $(n,1)$ and $y \in \mathbb{R}^{n+1}$ is given by \begin{align} y = \begin{pmatrix} y_1 \\ y_2 \\ \vdots\\ y_{n+1} \end{pmatrix} \end{align}

If I use $n=3$ then $Y \in$ so $(3,1)$. Now there are $6$ generators in so $(3,1)$. For them I will get in total $6$ generators of the Poincare group, $P(3,1)$.

But from where I will get the other $4$ generators according to the scheme given in Hall's book?

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Physically, the other four generators come from translations in the spatial (3) and time (1) directions. –  Danu Feb 22 '14 at 8:19
@Danu: How can I get the other 4 generators according to the scheme given in Hall's book? –  Ome Feb 22 '14 at 9:17
The components of the vector $y$ are generators too and there are $n+1=4$ of them making $6+4=10$ in total. It is as simple as that. –  Philip Gibbs - inactive Feb 22 '14 at 12:13
@Philip Gibbs: Then for each $Y$ there are four $y$s. And therefore we will get $6 \times 4 = 24$ generators. Where are my mistakes? –  Ome Feb 22 '14 at 15:49
There is no "for each" relationship here. There are just 6 generators in Y and 4 in y. –  Philip Gibbs - inactive Feb 22 '14 at 16:00

Preliminary remarks.

As Danu writes in his comment, the physics of the other four generators has to do with spacetime translations, one for each spatial direction, and one for time. But how do we see this explicitly in the math behind the somewhat odd-looking presentation of the Poincare group and its Lie algebra that Hall discusses.

First, recall that any $d+1$-dimensional Lorentz transformation is a Linear transformation on $\mathbb R^{d+1}$, so it can be representation by multiplication by a $(d+1)\times(d+1)$ matrix $\Lambda$.

Second, and most crucially, recall that translations of $\mathbb R^{d+1}$ are not linear transformations; there is no way to write spacetime translation as multiplication by a matrix.

However, here's the really cool thing. If we embed a copy of $(d+1)$-dimensional spacetime into the vector space $\mathbb R^{d+2}$, namely into a space with one higher dimension, then we can implement translations as linear transformations. Here's how it works.

The main construction.

For each $x\in \mathbb R^{d+1}$, we associate an element of $\mathbb R^{d+2}$ as follows: \begin{align} x \mapsto \begin{pmatrix} x \\ 1 \\ \end{pmatrix} \end{align} Now, for each Lorentz transformation $\Lambda\in \mathrm O(d,1)$, and for each spacetime translation characterized by a vector $a\in\mathbb R^{d+1}$, we form the matrix \begin{align} \begin{pmatrix} \Lambda & a \\ 0 & 1 \\ \end{pmatrix} \tag{$\star$} \end{align} and we notice what this matrix does to the embedded copies of points in $\mathbb R^{d+1}$; \begin{align} \begin{pmatrix} \Lambda & a \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ 1 \\ \end{pmatrix} = \begin{pmatrix} \Lambda x+a \\ 1 \\ \end{pmatrix} \end{align} Whoah! That's really cool! What has happened here is that when we augment the dimension of spacetime by one with the embedding give above, and when we correspondingly embed Lorentz transformations and translations appropriately into square matrices of dimension $d+2$, then we actually do get a way of representing both Lorentz transformations and translations as linear transformations on $\mathbb R^{d+2}$ that act in precisely the correct way on the copy of Minkowski space embedded in $\mathbb R^{d+2}$!

In other words, the Poincare group in $d+1$ dimensions can be thought of as the set of all $(d+2)\times(d+2)$ matrices of the form $(\star)$ where $\Lambda\in \mathrm O(d,1)$ and $a\in\mathbb R^{d+1}$.

A natural question then arises: what do the Lie algebra elements look like as matrices when we represent the Lie group elements this way? Well, I quick standard computation will show you that the Lie algebra of the Poincare group can, in this representation, be regarded as all matrices of the form \begin{align} \begin{pmatrix} X & \epsilon \\ 0 & 0 \\ \end{pmatrix} \tag{$\star$} \end{align} where $X\in\mathfrak{so}(d,1)$ and $\epsilon\in \mathbb R^{d+1}$, precisely as Hall indicates. But from the remarks above, we see clearly that the parameter $\epsilon$ precisely corresponds to the generators that span the subspace of the Poincare algebra that yield spacetime translations.

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As the translation group is a subgroup of the Euclidean group, I can compute the 4 generators of the $4D$ translation group. I see that in physics literatures they call these 4 generators as momentum operators. Why this is so? However, a slip of the pen: "Well, I quick standard computation will show you..." should be "Well, a quick standard computation will show you". –  Ome Feb 22 '14 at 9:41
@Ome If you consider the representations of the translation subgroup on functions, then you find that finite translations are obtained by exponentiating scalar multiples derivative operators $\partial/\partial x^i$, but now notice that such derivative operators are precisely what we call momentum operators in quantum mechanics in the position space representation of a massive particle moving in space. –  joshphysics Feb 22 '14 at 9:51
So, in the context of physics, in the lie algebra of the Poincare group, the $P^\mu$s are matrices of momentum operators? –  Ome Feb 22 '14 at 9:55
@Ome It depends on what you mean when you write the symbol $P^\mu$. That symbol is usually used to refer to generators of the translation group in a given representation. In the matrix representation of the translation subgroup of the Poincare group given above, the $P^\mu$ would be matrices, in quantum mechanics, however, the $P^\mu$ would be operators (such as differential operators in the position space representation), and in that case, the operators are interpreted as momentum operators. –  joshphysics Feb 22 '14 at 9:58
got it. alhamdulillah! thanks a lot. –  Ome Feb 22 '14 at 10:00

Joshphysics beat the answer I was writing and which I left off to eat my tea. Just a couple of further points:

The vector $y$ is an arbitrary vector in $\mathbb{R}^4$. Hall's rendering of a general Lie algebra member:

$$\left(\begin{array}{cc}Y&y\\0&0\end{array}\right)$$

exponentiates to:

$$\left(\begin{array}{cc}e^Y&\frac{e^Y-1}{Y}y\\0&1\end{array}\right)$$

Note that "$\frac{e^Y-1}{Y}$" (in quotes) is really a mnemonic for $I + \frac{1}{2!} Y + \frac{1}{3!} Y^2 + \cdots$: the expression in quotes is in general not literally meaningful since $Y$ is in general not invertible, although the entire (universally convergent) matrix function is always defined, always invertible and is equal to the identity at $Y=0$. You can work the exponentiation out readily from first principles from the matrix exponential Taylor series.

Now, as in JoshPhysic's answer, recall that these matrices can be thought of as acting on $(n+2)\times 1$ column vectors to represent the inhomogeneous translation as a homogeneous linear transformation in the double dimension space. What probably needs a few more comments is the extraction of the Lie algebra member, because $\frac{e^Y-1}{Y}\,y$ would seem to include parameters $Y$ from the Lorentz group's Lie algebra as well as the translation components $y$. However, this matrix block in the tangent space member when the group member shifts is:

$$\left. \lim\limits_{\epsilon_1\to0}\epsilon_1^{-1}\left(\left(\frac{e^{Y+\epsilon_1 \Lambda}-1}{Y+\epsilon_1 \Lambda}\right)-I\right)\,y+ \lim\limits_{\epsilon_2\to0}\frac{e^Y-1}{Y}\,(y+ \epsilon_2\,\Delta - y)\epsilon_2^{-1}\right|_{Y=0,\,y=0} = \Delta$$

where $\Lambda \in so(n,1)$ and $\Delta \in \mathbb{R}^4$ are the contributions to the tangent from the "building block" Lie algebras, so indeed the $y$ appears unscaled and is, notwithstanding appearances from the $\frac{e^Y-1}{Y}\,y$ term, purely the four components of the translation. So, as in Joshphysics's answer, you count the six components of $Y$ and the four components of $y$ to get ten as the dimension of the Lie algebra of the Poincaré group.

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Excellent answer. Thank you very much. From Hall's presentation it becomes clear that how do we get the 6 generators of Poincare group, but it is not clear how we get the other 4 generators. You calculation of $\Delta$ simply fills up the gap. Superb answer. Thanks again. However, in the physics literature these topics are not clearly discussed. The discussion in the physics literature have seemed very messy to me. –  Ome Feb 22 '14 at 10:30
Did the tea turn out well ;) ? –  joshphysics Feb 22 '14 at 10:44
@WetSavannaAnimal aka Rod Vance: Can you please explain this line a bit: "recall that these matrices can be thought of as acting on $2(n+1) \times 1$ column vectors..."? –  Ome Feb 22 '14 at 11:12
@Ome I got my dimensions wrong. Should have been $(n+2)\times 1$. Does that make more sense? –  WetSavannaAnimal aka Rod Vance Feb 22 '14 at 11:55
@WetSavannaAnimal aka Rod Vance: yah. :D Now it is clear to me. –  Ome Feb 22 '14 at 12:07