Dimension of the space of lines in the light cone in $\Bbb{C}^{n+1,1}$ Consider the Lorentzian space $\Bbb{C}^{n+1,1}$. This link says that the space of lines in the light cone can be identified with the sphere $S^{2n+1}$. 
Shouldn't this be $S^{2n+2}$? The reason is the following: $\Bbb{C}^{n+1,1}$ has dimension $2n+4$. The light cone would then have codimension $1$, and hence would have one less dimension. Consequently, if we identify each of these lines again as one point, we would go still one dimension lower, and the space of lines would have dimension $(2n+4)-2=2n+2$. 
Why is this not true?

What you want to look at is the space of complex lines in the light cone of $\mathbb C^{n+1,1}$, so you should subtract two dimensions in the last step.  
The best way to see which sphere you get is to view $\mathbb C^{n+1,1}$ as $\mathbb C^{n+1}\times\mathbb C$ and correspondingly write vectors as $(z,\lambda)$. Denote by $\langle\ ,\ \rangle$ the standard (positive definite) Hermitian form on $\mathbb C^n$ and write the Hermitian form on $ \mathbb C^{n+1,1}$ as $h((z,\lambda),(w,\mu))=\langle z,w\rangle-\lambda\bar\mu$. Now for a non-zero isotropic vector, the second component has to be non-zero and this readily implies that any isotropic complex line contains a unique element for which the second component equals $1$. But then $(z,1)$ is isotropic if and only if $\langle z,z\rangle=1$. This establishes the bijection between the space of complex isotropic lines and the unit sphere in $\mathbb C^{n+1}$, which is $S^{2n+1}$. 
