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We say a cone at the origin in $R^n$ means that it is an intersection of finitely many halfspaces, i.e. $$C=\bigcap_{i\in I}H_i,\text{ where }|I|<\infty.$$

A cone is strongly convex if $C\cap -C=\{0\}$. Then we can define the linear dimension of $C$ as $$ldim(C):=dim(C\cap -C).$$

Here we assume the dimension of $C$ is always $n$. (NOTE! not $ldim(C)=n$ but $dim(C)=n$.) If $ldim(C)=k>0$, then can we use only $n-k$ halfspaces to represent the original cone $C$ as $$C=\bigcap_{i\in [n-k]}H_i$$

Of course for $(n,k)=(a,a)$ or $(a,a-1)$ or $(a,a-2)$ such cases can be reduced into $R^2$ or $R^3$ is obvious. Here $(n,k)$ means it is a $C$ in $R^n$ with whose $dim(C)=n,ldim(C)=k$. But I don't know how to prove it for general situations.

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This is false in the other cases.

Let $\alpha=2\pi/n$ and observe that the interior of the polygon with $n$ given by $(cos(i\alpha),sin(i\alpha))_{i=0,\dots,n}$ is given by $n$ affine inequalities (of degree $1$ but with constants), but not less.

Hence, the cone in $\mathbb{R}^3$ generated by $(cos(i\alpha),sin(i\alpha),1)_{i=0,\dots,n}$ is not given by less than $n$ linear inequalities, and is thus not the intersection of less than $n$ halfspaces (just consider the intersection of your halfspaces with the plane where the third coordinate is $1$).

The same argument works for the cone in $\mathbb{R}^4$ generated by $(cos(i\alpha),sin(i\alpha),1,0)_{i=0,\dots,n}$ and $(0,0,0,1)$ and $(0,0,0,-1)$ (intersect with the plane where the third coordinate is $1$ and the last is $0$).

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  • $\begingroup$ Thank you! It is really different for $R^n$ if $n\geq 4$. So in the case above, $n$ halfspaces are necessarily needed as $\{v_i, v_{i+1}, e\}_{i\in [n]}$ form their supporting hyperplanes, where $v_i$ is $(cos(i\alpha),sin(i\alpha),1,0)$ and $e=(0,0,0,\pm 1)$? If it is correct I think I understand the anti-example you give. $\endgroup$
    – A.T.Saaki
    Mar 19, 2014 at 8:02
  • $\begingroup$ Sorry i edit this comment too many times... $\endgroup$
    – A.T.Saaki
    Mar 19, 2014 at 8:05
  • $\begingroup$ Yes, this is it. If you look for any convex cone, you can see with a picture that dimension $1$ and $2$ are always generated respectively $1$ or $2$ linear inequations (look at the extremal rays). But in dimension $3$ it does not work anymore, with the example I gave. As you wanted something which is not strongly convex, you need one line contained in your cone, so this shifts the dimension by $1$, and in your case the dimension $n\ge 4$ is really different. $\endgroup$ Mar 19, 2014 at 9:08

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