Let $O(1,n)$ be the orthogonal group of the quadratic form $b(x)=-x_0^2+\sum\limits_{i=1}^n x_i^2$. In other words, $$O(1,n)=\{T\in GL(n+1,\Bbb{R}|b(Tx)=b(x),\forall x\in R^{n+1}\}$$ Elements in $O(1,n)$ will be written in the block form with respect to the natural basis of $\Bbb{R}^{n+1}$. The standard basis $\{e_i\}_0^n$ corresponds to block matrices of the form $\begin{pmatrix}1\times 1 & 1\times n \\n\times 1 & n\times n \end{pmatrix}$

What is this "standard basis" $\{e_i\}_0^n$ that is referred to here? $O(1,n)$ is clearly not a vector space. So does $\{e_i\}_0^n$ generate the whole of $O(1,n)$ as a group?

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    $\begingroup$ I think $e_i$ represent a $n+1$ dimensional unit column vector, s.t. the $i+1$-th element is 1. $\{e_i\}_0^n$ is a set of all these vectors, and they constitute a basis of $\mathbb{R}^{n+1}$, not for $O(1,n)$. $\endgroup$ – W. mu Mar 4 at 1:37
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    $\begingroup$ I agree - it sounds like saying “a standard basis for the quadratic form $O(1,n)$”, While $O(1,n)$ is not a quadratic form, it corresponds to one, so we can guess what it means. $\endgroup$ – Ben Mar 4 at 5:33

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