The standard or unit second-order cone of dimension n+1 is defined as :\mathcal{C}_{n+1}=\left\{ \begin{bmatrix} x \\ t \end{bmatrix} \Bigg| x \in \mathbb{R}^n, t\in \mathbb{R}, \|x\|_2\leq t \right\}. The second-order cone is also known by the names
quadratic cone,
ice-cream cone, or
Lorentz cone. For example, the standard second-order cone in \mathbb{R}^3 is :\left\{(x,y,z) \Big| \sqrt{x^2 + y^2} \leq z \right\}. The set of points satisfying a second-order cone constraint is the inverse image of the unit second-order cone under an affine mapping: :\lVert A_i x + b_i \rVert_2 \leq c_i^T x + d_i \Leftrightarrow \begin{bmatrix} A_i \\ c_i^T \end{bmatrix} x + \begin{bmatrix} b_i \\ d_i \end{bmatrix} \in \mathcal{C}_{n_i+1} and hence is convex. The second-order cone can be embedded in the cone of the
positive semidefinite matrices since :||x||\leq t \Leftrightarrow \begin{bmatrix} tI & x \\ x^T & t \end{bmatrix} \succcurlyeq 0, i.e., a second-order cone constraint is equivalent to a
linear matrix inequality. The nomenclature here can be confusing; here M\succcurlyeq 0 means M is a semidefinite matrix: that is to say :x^T M x \geq 0 \text{ for all } x \in \mathbb{R}^n which is not a linear inequality in the conventional sense. Similarly, we also have, :\lVert A_i x + b_i \rVert_2 \leq c_i^T x + d_i \Leftrightarrow \begin{bmatrix} (c_i^T x+d_i)I & A_i x+b_i \\ (A_i x + b_i)^T & c_i^T x + d_i \end{bmatrix} \succcurlyeq 0. == Relation with other optimization problems ==