What does the “standard basis” of $O(1,n)$ mean?

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?

• 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 $O(1,n)$. – W. mu Mar 4 at 1:37
• 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. – Ben Mar 4 at 5:33