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I have come across two different definitions of the second-order/Lorentz cone. The first is the standard form where $t$ is a scalar and $\mathbf{y} \in \mathbb{R}^n$. $$ \mathcal{C}_1 = \bigg\{ \begin{bmatrix}\mathbf{y} \\ t\end{bmatrix} \in \mathbb{R}^{n+1}: t \ge \lVert \mathbf{y}\rVert_2\bigg\} $$

The second definition is from this paper. Here the cone is defined for $\mathbf{x} \in \mathbb{R}^{n+1}$. $$ \mathcal{C}_2 = \big\{\mathbf{x} : \mathbf{x}^{\sf T} \mathbf{1} \ge \sqrt{n}\lVert \mathbf{x}\rVert_2 \big\} $$

Are $\mathcal{C}_1$ and $\mathcal{C}_2$ the same?

Srinivas Eswar
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1 Answers1

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The axis of the first cone is in the direction $(0,0,\ldots,0,1)$, while the axis of the second cone is in the direction $(1, 1, \ldots, 1)$. There is also some scaling difference in the sharpness of the cone (the factor $\sqrt{n}$).

angryavian
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  • Thanks, that makes it clearer. I'm going to try and "rotate" $\mathcal{C}_2$ to be in the canonical form. – Srinivas Eswar Nov 03 '20 at 23:18
  • Is $\mathcal{C}_3 = {\mathbf{x}: \mathbf{x}^{\sf T}\mathbf{1} \ge \lVert\mathbf{x}\rVert_2}$ the appropriate "rotated" version of $\mathcal{C}_1$? – Srinivas Eswar Nov 15 '20 at 07:13
  • @SrinivasEswar Actually, now that I think about it, there is no scaling difference. The condition is $x^\top u \ge |x|_2$ where $u$ is a unit vector. In the first cone, $u=(0,\ldots,0,1)$ and in the second cone $u=(1,\ldots,1)/\sqrt{n}$. – angryavian Nov 15 '20 at 14:59
  • Isn't that slightly different then since the variable $t$ in $\mathcal{C}_1$ should appear in the norm on the RHS as well? – Srinivas Eswar Nov 16 '20 at 17:37
  • @SrinivasEswar If $t$ is the last element of a vector $x$ ,then $t=x^\top (0,\ldots, 0, 1)$. – angryavian Nov 16 '20 at 20:20
  • Yes but the RHS is no longer $\lVert x \rVert$ but rather $\sqrt{x_1^2 + x_2^2 + \ldots + x_{n-1}^2}$, right? – Srinivas Eswar Nov 17 '20 at 21:14
  • @SrinivasEswar Ah, you are correct. So they are not quite the same thing. Apologies for my mistakes. – angryavian Nov 17 '20 at 23:29
  • So does the rotated cone looks correct? – Srinivas Eswar Nov 21 '20 at 05:16