Orientation preserving isometries on a hyperboloid 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 $$$$\begin{pmatrix}A & \beta\\\ \beta^T & 0\end{pmatrix}$$$$, where $A$ is a $d\times d$  skew symmetric matrice and $\beta\in\Bbb R^d$. 
Question
I don't understand why the elements of Lie algebra should take this shape. What kind of transformation does it generate? Is $A$ the rotation around $d$ axis and $\beta$ kind of translation of the $d+1$ axis? What kind of transformation does the matrix $$$$\begin{pmatrix}0 & \beta\\\ \beta^T & 0\end{pmatrix}$$$$ generates?
 A: That the Lie algebra has the given explicit form is worked out in this answer (for the case $d = 2$, but the solution works just as well for general $d$).
As discussed in the comments, $\mathfrak{so}(d, 1)$ is the Lie algebra of the (oriented) Lorenz group $\operatorname{SO}(d, 1)$, which in turn consists of the (oriented) linear transformations that preserve the given bilinear form $I$.
The elements of $\mathfrak{so}(d, 1)$ of the form $\pmatrix{0&\beta\\ \beta^{\top}&0}$ do not generate translations: Fix such an element; by making an appropriate orthogonal change of basis (so that the form of $I$ is the same w.r.t. both the old and new bases) we can assume that $\beta$ has the form $\pmatrix{0&\cdots&0&t}^{\top}$, so that the element has the form
$${\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.
