Lorentz cone: $$L_{n+1} := \{(x_1,\cdots,x_{n+1})\in \Bbb R^{n+1} : x_1^2 - x_2^2-\cdots-x_{n+1}^2\geq 0, \text{and} \ x_1 \geq 0 \}$$
I have verified that Lorentz Cone is a full cone. But stuck with showing that it is not-polyhedral.
We know a cone is polyhedral if it can be written as intersection of finitely many closed half-spaces.
We have a hint: Since Lorentz Cone has infinitely many extreme rays so it is not polyhedral.
Can one please elaborate on that reason or give some other reason?
Thank You.
A polyhedral is formed by intersecting finitely many halfspace, hence it can only have finitely many extreme ray. A non-zero element of a cone is an extreme ray if there are $n-1$ linearly independent active constraint at $x$. An upper bound for the number of extreme ray if you have $m$ constraints would be $\binom{m}{n-1}$ , which is finite.
Hence if a cone has infinitely many extreme ray, it can't be a polyhedral.