If I constrain a n-dimensional vector $v_n$ to have a 2-norm of $a$, it can be interpreted geometrically to limiting $v_n$ to the space of vectors that touch the shell of a radius $a$ sphere.
$$ v_n \in \mathbb{R}^n \,\,\,\,;\,\,\, ||v_n||_2 = a $$
If I instead specify that $v_n$ must have a 1-norm of $b$, does this imply the vector $v_n$ must lie on the surface of some other geometric object?
$$ v_n \in \mathbb{R}^n \,\,\,\,;\,\,\, ||v_n||_1 = b $$
Yes, $1$-norm sphere is a diamond.
For $2$ dimension, try to connect $(0,b), (-b,0), (0,-b), (b,0)$ using line segments to see the diamond.
when $b=1$.