MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $L$ is a lattice (free abelian group) and $\sigma$ is a (pointed) spanning rational cone in $L\otimes\mathbb Q$. Then $M=L\cap \sigma$ is a monoid with $M^{gp}=L$. A monoid of this form is called a toric monoid. Toric monoids are are precisely the finitely generated, commutative, integral (cancellative), sharp (unit-free), saturated monoids. Let $TMon$ be the category of toric monoids.

A face of a monoid $M$ is a submonoid $F\subseteq M$ so that $a+b\in F$ implies $a,b\in F$. The faces of a toric monoid $\sigma \cap L$ are exactly submonoids of the form $\tau \cap L$, where $\tau$ is a face of the cone $\sigma$.

Suppose $D$ is a diagram of toric monoids in which every morphism is the inclusion of a face, and so that for every pair of monoids $D_i$ and $D_j$ in the diagram, there is a unique maximal common face $D_i\cap D_j$ in the diagram. Let $M$ be the colimit of the diagram in $TMon$. Are the maps $D_i\to M$ inclusions of faces?

My feeling is that this problem should be straightforward, but I've done a good job getting myself confused.

Problem 1: What is $M$? Colimits exist in the category of finitely generated commutative monoids, but they're unwieldy. See, for example, the first chapter of William Gillam's notes on log geometry. The inclusion of integral saturated monoids into all commutative monoids has a left adjoint, so the colimit $M$ can be formed by taking the colimit in commutative monoids, and then "integral-and-saturifying" it. In particular, $M$ exists. This description makes it impossible to deal with $M$.

Here is another description which is a bit better, but which I'm still not sure how to handle. Let $L$ be the colimit of the induced diagram of free abelian groups $D^{gp}$. This $L$ is probably free even if you take the limit in the category of abelian groups (rather than free abelian groups). Then $M$ is the image of $\bigoplus D_i\to L$. The problem with this description is that it's hard to keep track of faces once you turn everything into groups.

Problem 2: The fact that every pair of monoids in the diagram has a unique maximal common face in the diagram is necessary, but I'm having trouble making use of it. Here is a counterexample where this condition fails.

Let $\def\N{\mathbb N}M_n\subseteq \N^2$ be the submonoid generated by $\{(1,0), (1,1),\dots, (1,n)\}$

Let $f_1,g_1:\N\to M_2$ be given by $f_1(1)=(1,0)$ and $g_1(1)=(1,2)$. Let $f_2,g_2:\N\to M_3$ be given by $f_2(1)=(1,0)$ and $g_2(1)=(1,3)$. Consider the diagram

$$\begin{array}{ccc} & \N & \\ {}^{f_1}\swarrow & & \searrow^{f_2}\\ M_2 & & M_3\\ {}_{g_1}\nwarrow & & \nearrow_{g_2}\\ & \N \end{array}$$

In this case, I'm pretty sure the colimit is $M_6$ with the maps $\begin{pmatrix}1&0\\ 0&3\end{pmatrix}:M_2\to M_6$ and $\begin{pmatrix}1&0\\ 0&2\end{pmatrix}:M_3\to M_6$. These maps are not inclusions of faces.

share|cite|improve this question
up vote 3 down vote accepted

Yes, assuming that $D$ includes all faces (i.e. if $D_i$ is in $D$, the so are all of its faces). This is Corollary 2.12 of Toric Stacks II: Intrinsic Characterization of Toric Stacks.

The basic solution to both problems is to note that a toric monoid $M$ is determined by its dual $M^\vee = \{\chi\in Hom(M^{gp},\mathbb Z)| \chi$ is non-negative on $M\}$, and that the faces of $M$ are precisely the vanishing sets of elements of $M^\vee$. The advantage of this observation is that by the universal property of the colimit, functions on the colimit are equivalent to compatible systems of functions on the diagram.

Given an element $D_i$ of the diagram, one then inductively shows that

  1. It is possible to extend any linear function on $D_i$ to a compatible system of linear functions on $D$. This shows that the restriction map $Hom(colim(D)^{gp},\mathbb Z)\to Hom(D_i^{gp},\mathbb Z)$ is surjective, from which it follows that $D_i\to colim(D)$ is injective.
  2. The zero function on $D_i$ can be extended to a compatible system of functions on $D$ which are strictly positive away from $D_i$ and its faces. This shows that $D_i$ is in fact a face of $colim(D)$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.