Why is there no isometric isomorphism between $\ell^{1}(\mathbb{R}^{3})$ and $\ell^{\infty}(\mathbb{R}^{3})$?
I know that there is such an isomorphism if $\mathbb{R}^{3}$ is replaced with $\mathbb{R}^{2}$ (consider $\psi: \ell^{\infty}(\mathbb{R}^{2}) \rightarrow \ell^{1}(\mathbb{R}^{2})$ defined by $\psi(a, b) = ((a + b)/2, (a - b)/2)$). But why is this not true for $\mathbb{R}^{3}$?
The (closed) unit ball of $\ell^\infty(\mathbb{R}^3)$ is a cube ($[-1,1]^3$). It has six faces and eight vertices.
The (closed) unit ball of $\ell^1(\mathbb{R}^3)$ is an octahedron, it has eight faces and six vertices.
Since an isometric isomorphism must map vertices to vertices - as these are precisely the extreme points of the closed unit ball - it is necessary for the existence of such an isomorphism that the closed unit balls have the same number of vertices.