# Distinguished points of a cone

Sorry, as this is a rather trivial question that I am misunderstanding, but I do not understand how the distinguished point is defined. We define it as a homomorphism from some semigroup $S_{\sigma}$ to either 1 or 0 depending on whether the point is contained in $S_{\sigma} \cap \sigma^{\perp}$.

Now, consider the fan $\Sigma$ with $P^{2} \simeq X_{\Sigma}$ with the usual fan as depicted here: https://rigtriv.files.wordpress.com/2008/10/p2.jpg

The limit point is supposed to coincide with the distinguished point (cf Cox, Little, Schenck, Prop 3.2.2). So for example, the limit point for the first quadrant is (1,0,0). If we take some point contained in the first quadrant, say (1,1,1), this point should be taken to 0 by the distinguished point since it is clearly not contained in $\sigma^{\perp}$. How does (1,0,0) take this point to 0 then? Similarly, I'm having trouble seeing this with the limit points for the other 6 cones.

Thanks

There are points of the variety, and there are lattice points in the characters $M$, and there are lattice points in the dual lattice $N$. What you have in the picture you link are the points in $N$. What the homomorphism is defined on is the points of $S$, which is a subset of $M$. Finally, the special point is in $V$. Your special point is $[1:0:0]$ in $P^2$. The "first quadrant" cone $\sigma$ in the fan is dual to the "first quadrant" cone $\sigma^\vee$ in M (a coincidence). That cone has a semigroup which is given by all $x^a y^b$ with $a, b \geq 0$. The point $[1:0:0]$ has coordinates $(0,0)$ in the affine chart $U_\sigma$, and thus corresponds to the homomorphism $x^a y^b \to 0^a 0^b$, which is indeed zero on all points of $S$ except the vertex/zero in $S$, where it is 1.