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Hello,

I'm trying to understand the relation between the points of view of log geometry (monoids) and toric geometry (fans).

Suppose that $k$ is a field and $P$ is a finitely generated monoid. Then $k[P]$ has a natural log structure and furthermore, any choice of generators $\mathbf N^r\to P$ induces a closed embedding $Spec(k[P])\subset\mathbf A^r$.

On the other hand, starting from a cone $\sigma$ satisfying some properties in a lattice $N\otimes\mathbf R$, where $N = \mathbf Z^r$, we obtain a monoid $P' = \sigma^\vee\cap M$, where $M = Hom(N,\mathbf Z)$ and $\sigma^\vee$ is the set of all $x\in M\otimes\mathbf R$ such that $x(\sigma) \geq 0$.

Question: if we start with $P$ (and a choice of generators as above), can one write a corresponding cone so as to recover $P$ by the construction in the previous paragraph?

Thanks!

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Just to clarify about generators: you don't want to use the same $r$ for the number of generators of $P$ and the rank of $N$. For example, take the monoid generated by $2$ and $3$ inside $M={\Bbb Z}$. Then $N={\Bbb Z}$ also, but you need two generators for $P$. –  Dave Anderson Nov 18 '10 at 5:05
    
You have to assume that $P$ is saturated, cancellative and with the group completion torsion free in order to be able to recover it from the cone. "Saturated" means that if $p$ belongs to the group completion and $np\in P$ for some $n>0$ then $p\in P$. –  Torsten Ekedahl Nov 18 '10 at 5:56

3 Answers 3

up vote 4 down vote accepted

Not all finitely generated monoids $P$ will come from the cone construction. You need to assume that:

  1. $P^{gp}$ is torsion-free: If $x \in P^{gp}$ and $n\cdot x = 0$ then $x = 0$.
  2. $P$ is cancellative: If $x + y = x + y'$, then $y = y'$. This is equivalent to saying that the map $P \rightarrow P^{gp}$ is injective.
  3. $P$ is saturated: If $x \in P^{gp}$ and $x^n \in P$, then $x \in P$. Assuming the previous two proporties, this is equivalent to $k[P]$ being normal.

Here, $P^{gp}$ refers to the group formed by inverting all the elements of $P$.

If $P$ is finitely generated and satisfies 1, then $P^{gp}$ is a lattice, i.e. isomorphic to $\mathbb Z^r$ for some $r$, and this is the lattice $M$ from the cone construction. The dual lattice $N$ is $\textrm{Hom}(M, \mathbb Z)$, and $\sigma$ can be taken to be those $\lambda$ in $N \otimes_{\mathbb Z} \mathbb R = \textrm{Hom}(M, \mathbb R)$ such that $\lambda(x) \geq 0$ for all $x \in P$.

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In Condition (3), I think you mean to say $k[P]$ is a normal (i.e., integrally closed) domain. To me, at least, "integral domain" just means no zerodivisors, which is implied by (1) and (2). –  Dave Anderson Nov 18 '10 at 6:31
    
@Dave, yes, (1) and (2) are exactly what you need for k[P] to be an integral domain, and then having (3) in addition says that it is a normal domain. –  Dustin Cartwright Nov 18 '10 at 16:18

Sloppy stab at a reformulation:

I think we are assuming P is commutative? (\sigma^\vee \cap M certainly is.) If so, then isn't P spanned by all linear combinations of the generators? But then there are relations, too, etc. So the question might be, does P embed as a submonoid of a free abelian group over Z? (I don't know.) If so, convexity of the corresponding domain is clear, and finite generatedness means it's cut out by a finite number of conditions. Your lattice is the Z-span of P and your cone is then the (dual of the) R-convex hull.

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Picking up some of Eric Zaslow's reformulation: Assume $P$ is commutative, saturated, and cancellative, as well as finitely generated. The answer to your question is affirmative if and only if the "groupification" $P^{gp}$ of $P$ is torsion-free. (As mentioned in Dustin's answer, saturated means that for all $p$ in $P^{gp}$, $np \in P$ implies $p\in P$. Cancellative means $p_1+q=p_2+q$ implies $p_1=p_2$.)

All this amounts to $P$ being embeddable as a sub-monoid of ${\Bbb Z}^n$ for some $n$. Then take the subgroup of ${\Bbb Z}^n$ spanned by $P$. This is isomorphic to some ${\Bbb Z}^m$; take the dual of the convex hull of $P$ in ${\Bbb R}^m$ and you've got your cone $\sigma$, just as Eric says. When $P$ is saturated, it is equal to $\sigma^\vee \cap M$; otherwise, this gives the saturation of $P$, corresponding to the integral closure of $k[P]$.

Depending on what references you use, when $P^{gp}$ is torsion-free, $P$ is called either integral or toric. (See, e.g., the toric variety notes on M. Mustata's webpage versus the log geometry notes on Danny Gillam's webpage; both sources are worth looking at.) It seems the latter terminology is more standard in the log geometry world, where "integral" sometimes just means "cancellative".

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There is an extension of toric geometry which starts with a fan in an abelian group which is not assumed to be torsion free. Then one obtains a toric stack. –  Sasha Nov 18 '10 at 5:12

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