Why $dS^d \cong SO(d,1)/SO(d-1,1)$?

I have found a similar question, but there they give a seemingly rigorous proof, and what I am looking for is just an intuition.

I understand that $$S^2 \cong SO(3)/SO(2)$$: for every point in $$S^2$$ there is a continuous set of elements of $$SO(3)$$ (3D rotations) that leave it invariant. It is intuitive that this set is isomorphic to $$SO(2)$$, so $$SO(3)/SO(2)$$ gives us the points in the sphere. Having this, I am okay with saying that, in general, $$S^d \cong SO(d+1)/SO(d).$$

Now take de Sitter space $$dS^d$$, defined by the embedding equation

$$-(X^0)^2 + (X^1)^2 + ... + (X^d)^2 = L^2,$$

so it is invariant under the rotations $$SO(d, 1)$$. Defining $$\textbf{e}_0 = (1,0,...,0)$$, $$\textbf{e}_1 = (0,1,...,0)$$, ... $$\textbf{e}_d = (0,0,...,1)$$, we see that any point $$\textbf{e}_i$$ with $$i>0$$ is left invariant by rotations $$SO(d-1,1)$$, so it seems OK to write $$dS^d \cong SO(d,1)/SO(d-1,1).$$ However, if we take $$\textbf{e}_0$$ in particular, we see it is not invariant under $$SO(d-1,1)$$, but under $$SO(d)$$. Why is this not important? Is $$SO(d) \cong SO(d-1,1)$$?

• Related. No, SO(d)≠ SO(d−1,1). The respective coset spaces are very different manifolds. Take d=3 for specificity, and contrast the two resulting 3-dim manifolds. Remember, of the 6 isometries, 3 are realized linearly and 3 nonlinearly (~spontaneously broken: this is a physics SE). Commented Jan 11, 2019 at 22:54
• ...moding out SO(3) by solving for $X^0$ in terms of $\vec X$ ensures its 3 rotations leave it invariant, while the 3 "s.broken" boosts mix it up with $\vec X$. While modding out SO(2,1) by solving, for, e.g., $X^3$, ensures its 2 boosts and one rotation leave it invariant, while the "s.broken" 2 rotations and one boost mix it up with $X^0,X^1,X^2$... Commented Jan 11, 2019 at 23:12

If the "$$(1,0,\cdots,0)$$" you used to present $${\bf{e}}_0$$ is its components in the coordinate basis in a Lorentz coordinate system $$\displaystyle{\left(\frac{\partial}{\partial x^\mu}\right)^a}$$, then the $${\bf{e}}_0$$ is unfortunately not on the $$dS^d$$. You have to replace it with some point (with non-zero "$$0$$"-component you want, I guess) like $$(1,\sqrt{1+L^2},0,\cdots,0)$$. It is obviously that this point is $$SO(d-1,1)$$ invariant.
The proof is not too hard. First, we notice that the hypersurface $$dS^d:=\left\{\left(X^0,X^1,\cdots,X^d\right)\in{\bf R}^{d+1}\bigl|~~-\left(X^0\right)^2+\left(X^1\right)^2+\cdots+\left(X^d\right)^2=L^2,~L>0\right\}$$ is the invariant subspace of the Lorentz transformation $$SO(d,1)$$. This is because the equation can be written as $$g_{\mu\nu}X^\mu X^\nu=L^2$$ (let us use the "$$-+\cdots+$$" signature in this discussion). On the other hand, there is Lorentz transformation $$\Lambda(p)$$ which transforms the point $$p$$ on the $$dS^d$$ to the standard point (you may choose it as you want, I would like to choose it to be) $${\bf e}_1=(0,L,0,\cdots,0)$$. For example, the Lorentz transformation $$\Lambda_0=\left(\begin{matrix} \frac{\sqrt{1+L^2}}{L} & -\frac{1}{L} & 0 & \cdots & 0 \\ -\frac{1}{L} & \frac{\sqrt{1+L^2}}{L} & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & 0 & \ddots & 0 \\ 0 & 0 & 0 & \cdots & 1 \\ \end{matrix}\right)$$ maps $${\bf e}_0$$ to $${\bf e}_1$$ ($${\bf e}_1^T=\Lambda_0{\bf e}_0^T$$). It is certainly not the unique one. Now, for any element $$g\in\left\{\Lambda\in SO(d,1)~\bigl|~\Lambda {\bf e}_1^T={\bf e}_1^T\right\}$$ in the $${\bf e}_1$$-invariant subgroup, it is obviously that $$\Lambda_1^{-1}g\Lambda_1{\bf e}_0=\Lambda_1^{-1}g{\bf e}_1=\Lambda_1^{-1}{\bf e}_1= {\bf e}_0.$$ So $$\Lambda_1^{-1}g\Lambda_1$$ is an element in the $${\bf e}_0$$-invariant subgroup. Since $$\Lambda_1$$ is invertible, it offers a natural isomorphism between the invariant subgroup on different points. Thus the $${\bf e}_0$$-invariant subgroup, which must be isomorphic to the $${\bf e}_1$$-invariant subgroup, is also $$SO(d-1,1)$$.
To make it intuitive, let us take a boost as an example. Take the boost along the $${\bf e}_2$$ direction $$\Lambda_1=\left(\begin{matrix} \cosh\zeta & 0 & \sinh\zeta & 0 & \cdots & 0 \\ 0 & 1 & 0 & 0 & \cdots & 0 \\ \sinh\zeta & 0 & \cosh\zeta & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & 0 & \ddots & 0 \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ \end{matrix}\right)$$ which keeps $${\bf e}_1$$ invariant, we have $$\Lambda_0^{-1}\Lambda_1\Lambda_0=\left(\begin{matrix} \frac{\left(L^2+1\right)\cosh\zeta}{L^2} & -\frac{\left(\cosh\zeta-1\right)\sqrt{1+L^2}}{L^2} & \frac{\sqrt{1+L^2}\sinh\zeta}{L} & 0 & \cdots & 0 \\ \frac{\left(\cosh\zeta-1\right)\sqrt{1+L^2}}{L^2} & \frac{L^2-\cosh\zeta+1}{L^2} & \frac{\sinh\zeta}{L} & 0 & \cdots & 0 \\ \frac{\sqrt{1+L^2}\sinh\zeta}{L} & -\frac{\sinh\zeta}{L} & \cosh\zeta & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & 0 & \ddots & 0 \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ \end{matrix}\right).$$ You may check that it is an $${\bf e}_0$$-invariant boost by direct calculation. To understand this result, we can study the infinitesimal transformation by set $$\zeta\to0$$, then it becomes $$\Lambda_0^{-1}\Lambda_1\Lambda_0\to\left(\begin{matrix} 1 & 0 & \frac{\sqrt{1+L^2}}{L}\zeta & 0 & \cdots & 0 \\ 0 & 1 & \frac{\zeta}{L} & 0 & \cdots & 0 \\ \frac{\sqrt{1+L^2}}{L}\zeta & -\frac{\zeta}{L} & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & 0 & \ddots & 0 \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ \end{matrix}\right)$$ which is a combination of rotation in the $${\bf e}_1-{\bf e}_2$$ plane and boost along the $${\bf e}_1$$ direction.
BTW, this result also tells us that the little group of tachyon in the $$d+1$$-Dim flat spacetime is $$SO(d-1,1)$$. Because its orbit under the $$SO(d,1)$$ group is the $$dS^d$$ by the mass-shell equation.
• Ok, of course, you’re right. Just a further doubt: I think it makes sense that the non zero 0th component point you wrote is $SO(d-1,1)$ invariant, but I don’t find it obvious. How would you see that? Commented Jan 13, 2019 at 1:09