Embed the intersection of an n-dimensional unit $L_1$ sphere and a hyperplane  into an (n-1)-dimensional unit $L_1$ sphere. In $\mathbb{R}^n$, given an unit $L_1$ sphere $\mathcal{B}_n: |x_1|+|x_2|+\ldots+|x_n|\leq 1$ and a hyperplane $\mathcal{P}: a_1x_1+a_2x_2+\ldots+a_nx_n=0$. Does there always exist a rotation such that  $\mathcal{B}_n\cap\mathcal{P}$ is embedded into the $(n-1)$ dimensional unit $L_1$ sphere: $\mathcal{B}_{n-1}: |x_1|+|x_2|+\ldots+|x_{n-1}|\leq 1, x_n=0$ after the rotation?
 A: Thanks for the comments above. I have just proofed this problem is only true when $n\leq 4$.
When $n\leq 4$, without loss of generality, assume $|a_n|\geq |a_1|, \ldots, |a_{n-1}|$. Let $Q$ be the rotation on the plane spanned by $(a_1, \ldots, a_n)$ and $(0,\ldots, 0, 1)$ such that applying $Q$ to $(a_1, \ldots, a_n)$ rotates it to  $(0,\ldots, 0, 1)$. Apply the rotation to $\mathcal{B}\cap\mathcal{P}$ will inscribe it into $\mathcal{B}_{n-1}$.
For $n=5$, the intersection between $\mathcal{B}_n$ and $\mathcal{P}: a_1+a_2+a_3+a_4+a_5=0$ can not be inscribed in to the 4-dimension $L_1$ sphere. The intuition of the proof is that the $\mathcal{B}_n\cap\mathcal{P}$ contains the intersection between the 3-dimension $L_1$ sphere and hyperplane $a_1+a_2+a_3+a_4=0$. This intersection can only be inscribed into the 4-dimensional $L_1$ sphere under two rotations (with symmetry cases ignored). One can verified that non of those two rotations can fit $\mathcal{B}_n\cap\mathcal{P}$ into a 4-dimensional $L_1$ sphere.
