Is the constraint $w^2 = 1$ a quadratic constraint or a combination of two affine constraints?

Question

Suppose I need to minimize the least squares function $\|Aw - B\|^2$ wrt this constraint $w^2 \leq 1$ and $w^2 = 1$ where $w$ is a weight vector and the norm is $\ell_2$ norm. In the first case since the function is convex, the inequality need not be affine but in the second case when the equality is present, it must be affine. Now it seems to be a quadratic constraint but in actual, can it be expressed as two affine constraints $w = 1$ or $w = -1$? In the first case, is the reasoning correct? Also does $w^2$ mean $w^Tw$ or not?

Answer 1

When $w \in \mathbb{R}$, you get $w=1$ or $w=−1$, which is not affine (it is binary). When $w \in \mathbb{R}^n$ ($n\geq2$), $w^Tw=1$ means that $w$ is on a circle (or sphere or hypersphere). Either way, the constraint is nonconvex quadratic.