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?



Embed the intersection of an n-dimensional unit $L_1$ sphere and a hyperplane  into an (n-1)-dimensional unit $L_1$ sphere.
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.


