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Given the Minkowski metric on $\Bbb R^d$ with $I(x,y)=\sum_{j=1}^d x_jy_j-x_{d+1}y_{d+1}$ and the hypberboloid $$\Bbb H^d=\{x\in\Bbb R^{d+1}:I(x,x)=1,x_{d+1}>0\}$$. Let's denote the group of orientation preserving isometries on $\Bbb H^d$ as $SO(d,1)$. There is a claim that the Lie-algebra $\mathfrak{so}(d,1)$ is the set of $(d+1)\times(d+1)$ matrices of the form $$$$\begin{pmatrix}A & \beta\\\ \beta^T & 0\end{pmatrix}$$$$, where $A$ is a $d\times d$ skew symmetric matrice and $\beta\in\Bbb R^d$.

Question I don't understand why the elements of Lie algebra should take this shape. What kind of transformation does it generate? Is $A$ the rotation around $d$ axis and $\beta$ kind of translation of the $d+1$ axis? What kind of transformation does the matrix $$$$\begin{pmatrix}0 & \beta\\\ \beta^T & 0\end{pmatrix}$$$$ generates?

quallenjäger
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  • Typically you'd see this Lie algebra written as $\mathfrak{so}(d, 1)$---it's the Lie algebra preserving a symmetric, nondegenerate binlinear form of signature $(d, 1)$. – Travis Willse Nov 25 '17 at 17:30
  • @Travis Sorry, edited. Why do the element of the Lie algebra take this shape? – quallenjäger Nov 25 '17 at 17:34
  • It depends on what you mean by "why". One option is to take the matrix representation $[I] = \operatorname{diag}(1, \ldots, 1, -1)$ of $I$ and just compute the subalgebra of $\mathfrak{gl}(d + 1, \Bbb R)$ that fixes $[I]$ under in standard action. Do you know how to carry out this computation? – Travis Willse Nov 27 '17 at 14:47
  • @Travis Is it not just the Poincare-Group preserving the minkowski metric? – quallenjäger Nov 27 '17 at 16:24
  • The group $\operatorname{SO}(d, 1)$ is the Lorenz group---the Poincare group is the corresponding affine group, a semidirect product of $\operatorname{SO}(d, 1)$ and its standard representation: $\operatorname{SO}(d, 1) \ltimes \Bbb R^{d, 1}$. – Travis Willse Nov 28 '17 at 11:09

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That the Lie algebra has the given explicit form is worked out in this answer (for the case $d = 2$, but the solution works just as well for general $d$).

As discussed in the comments, $\mathfrak{so}(d, 1)$ is the Lie algebra of the (oriented) Lorenz group $\operatorname{SO}(d, 1)$, which in turn consists of the (oriented) linear transformations that preserve the given bilinear form $I$.

The elements of $\mathfrak{so}(d, 1)$ of the form $\pmatrix{0&\beta\\ \beta^{\top}&0}$ do not generate translations: Fix such an element; by making an appropriate orthogonal change of basis (so that the form of $I$ is the same w.r.t. both the old and new bases) we can assume that $\beta$ has the form $\pmatrix{0&\cdots&0&t}^{\top}$, so that the element has the form $${\bf 0}_{d-1} \oplus \pmatrix{0&t\\t&0\\}$$ The transformation it generates is $$\exp\left[{\bf 0}_{d-1} \oplus \pmatrix{0&t\\t&0\\}\right] = \Bbb I_{d - 1} \oplus\pmatrix{\cosh t&\sinh t\\ \sinh t&\cosh t} .$$ Such transformations are called (inverse) Lorentz boosts. See the Wikipedia article on Lorentz transformations for more.

Travis Willse
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  • Thank you. This is very clear. So it’s a Lorentz boost in the d-direction? – quallenjäger Nov 28 '17 at 12:06
  • And maybe one more question, how can I interprete the rapidity parameter $t$? – quallenjäger Nov 28 '17 at 12:11
  • You're welcome. Yes, it's a Lorentz boost in the $x_d$-direction, but of course this statement is dependent on a choice of coordinate system. Since (only) for $t = 0$ the transformation is the identity, we can interpret $t$ as a measure of how far from the identity the boost is. Geometrically, it is the hyperbolic analogue of the angle of rotation (cf. the usual matrix representation of a rotation in $\Bbb R^2$). In terms of physical interpretation in the setting of special relativity, you might fight this helpful: https://en.wikipedia.org/wiki/Minkowski_diagram – Travis Willse Nov 28 '17 at 13:17