(0.6,2.5) [[$u$]{}]{}; at (2.9,1.1) [[$X$]{}]{}; at (-2.4,4.5) [[$\scriptstyle(x,u)=1$]{}]{};

The $n$-simplex $\Delta^n$ is of course a particular pyramid. It is obtained using the vector $u=(1,\ldots,1)$.

The Minkowski functional of the pyramid $X$ associated with $u$ is $${\ensuremath{\mu_{X}}}(x)=
        \begin{cases}
            (x,u) &\hspace*{-3mm},\quad x\geq 0,
            \\
            \infty &\hspace*{-3mm},\quad \text{otherwise}.
        \end{cases}$$ Write $u=\sum_{j=1}^n\alpha_je_j$, where $e_j$ is the $j$-th canonical basis vector, and set $y_j=\frac 1{\alpha_j}e_j$. Clearly, $\{y_1,\ldots,y_n\}$ is linearly independent. Each vector $x=\sum_{j=1}^n\xi_je_j$ can be written as $x=\sum_{j=1}^n(\xi_j\alpha_j)y_j$, and this is a subconvex combination if and only if $\xi_j\geq 0$ and $\sum_{j=1}^n\xi_j\alpha_j\leq 1$, i.e., if and only if $x\in X$. Thus $X$ is generated by $\{y_1,\ldots,y_n\}$ as a .

The linear map given by the diagonal matrix made up of the $\alpha_j$’s induces a bijection of $X$ onto $\Delta^n$, and maps the $y_j$’s to the corner points of $\Delta^n$. Hence, $X$ is free with basis $\{y_1,\ldots,y_n\}$.

If $u$ is not strictly positive, the situation changes drastically. Then $X$ is not finitely generated as a , because it is unbounded whereas the subconvex hull of a finite set is certainly bounded.

(0,0) – (0,2) – (16,4) – (16,0) – (0,0); (0,0) – (12,0); (0,0) – (0,5); (-1,2-1/8) – (16,4); (0,0) – (-3/8,3); at (-0.8,3.2) [[$u$]{}]{}; at 