When is a projective space smooth from the toric perspective?

I am studying the lecture notes on toric geometry (for physicists) by Bouchard (arxiv:hep-th/0702063) and Skarke (arXiv:hep-th/9806059). In these lecture notes, the authors give the following criterion for smoothness of a toric variety $\mathcal{M}_{\Sigma}$ ($\Sigma$ is the associated fan):

If every $n$-dimensional cone of $\Sigma$ is generated by vectors that generate the whole lattice $N$, then $\mathcal{M}_{\Sigma}$ is smooth.

Bouchard considers the examples of $\mathbb{CP}^2$ and $\mathbb{CP}^{(2,3,1)}$: the former is smooth whereas the latter is not. He says that for $\mathbb{CP}^2$, every two-dimensional cone in the fan is generated by vectors that generate $N$; hence $\mathbb{CP}^2$ is smooth. He adds that two of the three two-dimensional cones of $\mathbb{CP}^{(2,3,1)}$ are not generated by vectors generating $N$; hence $\mathbb{CP}^{(2,3,1)}$ has orbifold singularities.

But I am a bit confused by the statement of this criterion: below I describe the one- and two-dimensional cones of these two spaces.

Is it being claimed that for $\mathbb{CP}^{(2,3,1)}$ if one chooses the basis $v_1, v_3$ or $v_2, v_3$ then not every point of the lattice $N$ can be reached by a linear combination of these basis eleemnts? This is confusing because any point in the lattice $N \cong \mathbb{Z}^2$ is of the form

$$a e_1 + b e_2$$

and I can always write this as $\alpha v_1 + \beta v_3$ or $\alpha v_2 + \beta v_3$ for suitable $\alpha, \beta$ determined in terms of $a$ and $b$. This fact holds for $\mathbb{CP}^2$ and $\mathbb{CP}^{(2,3,1)}$. So where is the distinction?

EDIT: Is this just because of the simple fact that for $\mathbb{CP}^{(2,3,1)}$, the condition

$$a e_1 + b e_2 = \alpha v_2 + \beta v_3$$

for integers $a$ and $b$ leads to $\alpha = (b-3a/2)$ and $\beta = -a/2$ which need not have to be integers, whereas for $\mathbb{CP}^{2}$, such a condition would leads to $\alpha, \beta \in \mathbb{Z}$?

Without doing this expansion, could I have seen this?

Case of ${\mathbb{CP}^2}$

Lattice $N$ generated by $e_1 = v_1 = \{1, 0\}$ and $e_2 = v_2 = \{0, 1\}$

One-dimensional cones generated by:

$v_1 = e_1 = (1, 0) \\v_2 = e_2 = (0, 1) \\v_3 = -e_1-e_2 = (-1,-1)$

Two-dimensional cones generated by:

$\{(1,0), (0,1)\} = \{v_1, v_2\}\\ \{(-1,-1), (0,1)\} = \{v_3, v_2\}\\ \{(-1,-1), (1,0)\} = \{v_3, v_1\}$

Case of ${\mathbb{CP}^{(2,3,1)}}$

Lattice $N$ generated by $e_1 = v_1 = \{1, 0\}$ and $e_2 = v_2 = \{0, 1\}$

One-dimensional cones generated by:

$v_1 = e_1 = (1, 0) \\v_2 = e_2 = (0, 1)\\ \\v_3 = -2e_1-3e_2 = (-2,-3)$

Two-dimensional cones generated by:

$\{(1,0), (0,1)\} = \{v_1, v_2\}\\ \{(1,0), (-2,-3)\} = \{v_1, v_3\}\\ \{(0,1), (-2,-3)\} = \{v_2, v_3\}$

• Your question has already been answered below, but here's something that might be useful for you in the future: Consider $n$ integral vectors $v_1,\dots,v_n\in\mathbb{Z}^n$. Then, they span the lattice if and only if the determinant of the $n\times n$ matrix they form is a unit in $\mathbb{Z}$, i.e., $\pm 1$. In this case you see neither $\{v_1,v_3\}$ or $\{v_2,v_3\}$ form a basis. Nov 15, 2017 at 10:00
• Is the matrix you refer to the one whose columns are the vectors? Nov 15, 2017 at 13:19
• Yes, it's the change of coordinates matrix. Nov 15, 2017 at 18:04
• Just to confirm..there are 3 two-dimensional vectors (v1, v2, v3)...what is the "change of coordinates matrix"? You need a square matrix to take a determinant. Nov 15, 2017 at 18:06
• You want to see which pairs of these vectors form a basis, so you need to compute $\det(v_i|v_j)$ for $i\neq j$. For example, for $\mathbb{P}(3,2,1)$ the set $\{(1,0),(0,1)\}$ clearly forms a basis for $N$, so the affine chart it defines is smooth. On the other hand, $\det((1,0),(-2,-3))=-3$, so these two vectors don't form a basis of the lattice and the chart they define is not smooth. A geometrical way of seeing this is that the (unsigned) area of the parallelogram spanned by these two vectors is greater than one, so there are lattice points inside the parallelogram. Nov 15, 2017 at 18:23

The sets $\{v_1,v_3\}$ and $\{v_2,v_3\}$ in the second example do not generate $\Bbb Z^2$ as a $\Bbb Z$-module. This can be seen by noting that the $y$-coordinate in any $\Bbb Z$-linear combination of $v_1,v_3$ is always a multiple of 3, while the $x$-coordinate of any $\Bbb Z$-linear combination of $v_2,v_3$ is always a multiple of 2. This is what the text is referring to when it talks about the lattice not being generated by the vectors generating the cone.