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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 $SO(d) \cong SO(d-1,1)$?

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  • $\begingroup$ 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). $\endgroup$ – Cosmas Zachos Jan 11 at 22:54
  • $\begingroup$ ...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 $X^0,X^1,X^2$... $\endgroup$ – Cosmas Zachos Jan 11 at 23:12
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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.

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  • $\begingroup$ 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? $\endgroup$ – MBolin Jan 13 at 1:09
  • $\begingroup$ Hi MBolin, I add a proof. Please let me know if there is any mistake in the proof. $\endgroup$ – Hao Zhang Jan 13 at 11:46

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