Second order integer cone and polar

Question

Consider the two sets $C_1$ and $C_2$ that are defined as follows: $$ \left\{ \begin{array}{ll} C_1=\{\,(x_1,x_2,x_3)\in \Bbb{R}^3 \mid x_3\ge 0\,,\,x_3^2 \ge x_1^2+x_2^2\,\} &,\\ \\ C_2=\{\,y\in \Bbb{R}^3 \mid \forall x \in C_1 \,,\, y^t\cdot x \le 0\,\} &. \end{array} \right. $$ Suppose that $$ C_3 =\{\,y\in \Bbb{R}^3 \mid \forall x \in C_1\cap \Bbb{Z}^3 \,,\,y^t\cdot x \le 1 \, \} $$

My question: How to prove that $C_2=C_3$?

Thanks for any suggestions.

I just realized that changing it to $\le 0$ does not change $C_3$. Indeed, for each $x \in C_1 \cap \mathbb Z^3$, you have $n , x \in C_1 \cap \mathbb Z^3$ for all $n \in \mathbb N$. Hence, $y^\top x \le \frac1n$ which goes to $0$ as $n \to \infty$. — gerw, Jun 07 '17 at 13:24
By showing that $C_2^\circ = C_3^\circ$. Since both $C_2$ and $C_3$ are closed, convex cones, this yields $C_2 = C_3$ by the bipolar theorem. — gerw, Jun 08 '17 at 06:59

score 1 · Answer 1 · answered Jun 07 '17 at 13:19

1

For an arbitrary set $C \subset \mathbb R^3$, we define the polar cone via $$C^\circ = \{y \in \mathbb R^3 \mid \forall x \in C : y^\top x \le 0\}.$$

Then, we have $$C_2^\circ = C_1$$ from the bipolar theorem and, similarly, $$C_3^\circ = \operatorname{clcone}( C_1 \cap \mathbb Z^3 ),$$ where $\operatorname{clcone}(B)$ is the closed, convex, conical hull, i.e., the smallest closed convex cone containing $B$.

Finally, one can check $$C_1 = \operatorname{clcone}( C_1 \cap \mathbb Z^3 ).$$

answered Jun 07 '17 at 13:19

gerw

31,359

yes - how will u do that? – famfjord Jun 07 '17 at 13:25
Actually you didn't do the $99 %$ % of problem. – Red shoes Jun 09 '17 at 04:37
@Ashkan: Actually, I did not intend to offer a full solution. My answer give some hints to attack the problem, only some (rather straightforward) calculations in between are missing. Somebody who want to understand the topic should fill the details him/herself. – gerw Jun 09 '17 at 06:24

Second order integer cone and polar

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