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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:

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

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  • $\begingroup$ 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. $\endgroup$
    – user347489
    Nov 15, 2017 at 10:00
  • $\begingroup$ Is the matrix you refer to the one whose columns are the vectors? $\endgroup$
    – leastaction
    Nov 15, 2017 at 13:19
  • $\begingroup$ Yes, it's the change of coordinates matrix. $\endgroup$
    – user347489
    Nov 15, 2017 at 18:04
  • $\begingroup$ 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. $\endgroup$
    – leastaction
    Nov 15, 2017 at 18:06
  • $\begingroup$ 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. $\endgroup$
    – user347489
    Nov 15, 2017 at 18:23

1 Answer 1

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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.

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