# Does a strongly convex cone have a unique set of minimal generators?

I am studying toric varieties from the book "Toric Varieties" by Cox, Little and Schenck. They define rays as the edges of a strongly convex cone, $\sigma$. Each ray, $\rho$ has a unique ray generator $u_\rho$ which generates the semigroup $\rho\cap N$ where $N$ is the underlying lattice.

Lemma 1.2.15 says that $\sigma$ is generated by it's ray generators but I am wondering if the converse is true.

If $\sigma$ has a set of minimal generators (over $\mathbb{R}$), $x_1,...,x_n$, do the $x_i$ have to lie in the rays of $\sigma$?.

I am having trouble proving this because the definition of an edge is a 1 -dimensional intersection $H_m\cap \sigma$ where $H_m$ is a halfplane. This seems very different from being a generator to me.

I have tried showing that if some $u_\rho$ can be generated (over $\mathbb{R}$) by another sets of point in the cone, then the cone is not strongly convex, but I am not sure if this is actually true.

• Hint: First show that if $v$ is a vertex of a convex polytope $P$, then $v$ cannot lie in the interior of a segment contained in $P$. Jul 30, 2017 at 17:20

I spoke to my supervisor who helped me find an answer to this question, so I will share for any future readers who have the same question.

Assume for the sake of contradiction that there are two minimal generating sets for the cone: $x_1,...,x_n$ and $v_1,...,v_m$, and that $x_1$ does not lie in the same line as any $v_j$.

Since each $x_i$ is in the cone, it can be generated by $v_1,...,v_n$. Therefore we can write $x_i=\sum a_{ij}vj$ for some $a_{ij}\in \mathbb{R}_{\ge 0}$. Similarly each $v_j$ can be written as $\sum b_{ij} x_i$ for some $b_{ij}\in \mathbb{R}_{\ge 0}$.

Combining these we get $x_i=\sum_{i=1}^n a_{ij}b_{ij}x_i$.

Therefore, renaming our constants and collecting like terms, there exists $c_i\in \mathbb{R}$ such that $c_1x_1=\sum_{i=2}^n c_ix_i$. Here $c_1=1-a_{11}b_{11}\in \mathbb{R}$ and $c_i=a_{1i}b_{1i}\in \mathbb{R}_{\ge 0}$. We can assume that $c_2,...,c_n$ are not all zero as this would only be possible if $x_1$ was in the same line as some $v_j$.

There are three cases: $c_i>0$, $c_i=0$ and $c_i<0$.

If $c_i>0$, then $x_1=\sum_{i=2}^n\frac{c_i}{c_1}x_i$. Since $\frac{c_i}{c_1}\in \mathbb{R}_{\ge 0}$, this implies $x_1\in \text{Cone}(x_2,...,x_n)$ which contradicts the fact that $x_1,...,x_n$ is a minimal generating set.

If $c_i=0$, then $0=\sum_{i=2}^nc_ix_i$. Not all of the $c_i$ are zero, so without loss of generality we can assume $c_2\ne 0$. Then $-c_2=\sum_{i=3}^n\frac{c_i}{c_2}x_i\in \text{Cone}(x_1,...,x_n)$, which contradicts the strong convexity of the cone.

Finally if $c_i<0$ then $x_1=\sum_{i=2}^n\frac{c_i}{c_1}x_i$, which again contradicts the strong convexity of the cone.

Therefore, up to scalar multiples, $v_1,...,v_n$ is the only minimal generating set.

• Isn't the final case impossible, since each $a_{ij}$ and $b_{ij}$ are non-negative? Jul 27, 2021 at 4:23
• You are very right. Especially considering that I literally wrote that the c_i were in the positive reals in the previous paragraph, and then promtly forgot. Silly me from 2017!
– CEH
Jul 28, 2021 at 13:14