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

1. Torsion-free: If $x \in P$ and $n\cdot x = 0$ then $x = 0$.
2. Cancellative: If $x + y = x + y'$, then $y = y'$. This is equivalent to saying that the map $P \rightarrow P^{gp}$ is injective.
3. 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.

In conditions 2 and 3, $P^{gp}$ 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 $x \in P$.

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 $x \in P$.

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

1. Torsion-free: If $x \in P$ and $n\cdot x = 0$ then $x = 0$.
2. Cancellative: If $x + y = x + y'$, then $y = y'$. This is equivalent to saying that the map $P \rightarrow P^{gp}$ is injective.
3. 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 an integral domain.

In conditions 2 and 3, $P^{gp}$ 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 $x \in P$.

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

1. Torsion-free: If $x \in P$ and $n\cdot x = 0$ then $x = 0$.
2. Cancellative: If $x + y = x + y'$, then $y = y'$. This is equivalent to saying that the map $P \rightarrow P^{gp}$ is injective.
3. 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.

In conditions 2 and 3, $P^{gp}$ 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 $x \in P$.