I think we're just working with slightly different definitions of "works fine". Isometry is a pretty strong definition of "works fine", but R^n with 1-norm is i

I think we're just working with slightly different definitions of "works fine". Isometry is a pretty strong definition of "works fine", but R^n with 1-norm is isomorphic and quasi-isometric to R^n with 2-norm which is still fairly strong.