# Verifying that the weighted projective space $\Bbb{P}_{\Bbb{Q}}(1,1,2,3)$ is singular.

I while ago I attended a talk that was somewhat over my head, and the speaker mentioned in passing that the weighted projective space $\Bbb{P}:=\Bbb{P}_{\Bbb{Q}}(1,1,2,3)$ is singular. I suppose this means that there is a point $P\in\Bbb{P}$ for which $$\dim\mathcal{O}_{\Bbb{P},P}\neq\dim_{k(P)}\mathfrak{m}_P/\mathfrak{m}_P^2,$$ where $\mathfrak{m}_P$ is the maximal ideal of $\mathcal{O}_{\Bbb{P},P}$. I believe $P$ need not be a $\Bbb{Q}$-point, but please correct me if I'm wrong.

Unfortunately I have no clue how to check this claim; how would I find such a singular point? Checking this (in)equality for a point $P\in\Bbb{P}$ is very cumbersome (I have tried for a handful of points, to no avail), and it is obviously impossible to check all points $P\in\Bbb{P}$. Is there an easy way to check the (in)equality above? Or some way to narrow down the number of candidates for singular points (drastically)?

Thanks in advance for any answers or pointers as to where to look for them.

## 1 Answer

We can explicitly describe the singular locus of $\Bbb P(q_0,\dotsc,q_n)$. Recall that $\Bbb P(q_0,\dotsc,q_n)$ has homogeneous coordinates $(x_0:\dotsb:x_n)$ satisfying \begin{align*} (x_0:\dotsb:x_n)&=(\lambda^{q_0} x_0:\dotsb:\lambda^{q_n}x_n)& \lambda\in\Bbb C^* \end{align*} For every prime $p$ define $$\DeclareMathOperator{Sing}{Sing} \Sing_p\Bbb P(q_0,\dotsc,q_n)=\{x\in\Bbb P(q_0,\dotsc,q_n): x_i\neq0\Rightarrow p\mid q_i\}$$

One may then prove that the singular locus of $\Bbb P(q_0,\dotsc,q_n)$ is $$\Sing \Bbb P(q_0,\dotsc,q_n)=\bigcup_{p\text{ prime}}\Sing_p\Bbb P(q_0,\dotsc,q_n)$$ In our case we have $$\Sing_p\Bbb P_{1123}= \begin{cases} \{(0:0:1:0)\} & p=2 \\ \{(0:0:0:1)\} & p=3 \\ \varnothing & p\neq 2,3 \end{cases}$$ so that $$\Sing\Bbb P_{1123}=\{(0:0:1:0),(0:0:0:1)\}$$ Hence we have two singular points $P=(0:0:1:0)$ and $Q=(0:0:0:1)$.

I'll leave it to you to prove that \begin{align*} \dim\mathfrak m_P/\mathfrak m_P^2 &=1 & \dim\mathfrak m_Q/\mathfrak m_Q^2 &=1 \end{align*}

If you're familiar with toric varieties, then proving that $\Bbb P_{1123}$ is not smooth is quite simple.

Let \begin{align*} \rho_0 &= \langle-3,\,-2,\,-1\rangle & \rho_1 &= \langle0,\,0,\,1\rangle & \rho_2 &= \langle0,\,1,\,0\rangle & \rho_3 &=\langle1,\,0,\,0\rangle \end{align*}

The space $\Bbb P_{1123}\DeclareMathOperator{Cone}{Cone}$ is the toric variety $X_\Sigma$ corresponding to the fan $\Sigma$ generated by the cones \begin{align*} \sigma_0 &= \Cone(\rho_1,\rho_2,\rho_3) & \sigma_1 &= \Cone(\rho_0,\rho_2,\rho_3) & \sigma_2 &= \Cone(\rho_0,\rho_1,\rho_3) & \sigma_3 &= \Cone(\rho_0,\rho_1,\rho_2) \end{align*} A cone is called smooth if its generators are a subset of a $\Bbb Z$-basis of the ambient lattice. The toric variety $X_\Sigma$ is smooth if and only if each $\sigma_i$ is smooth. However, $\sigma_2$ and $\sigma_3$ are not smooth. Hence $X_\Sigma$ is not smooth.

In fact, under the orbit-cone-correspondence, our two singular points correspond to the cones $\sigma_2$ and $\sigma_3$.

• Thank you for your answer. I'm not familiar with toric varieties beyond having heard the term before, but I'm curious to see a proof of the fact that $$\bigcup_{p\text{ prime}}\Sing_p\Bbb P(q_0,\dotsc,q_n),$$ is the singular locus of $\Bbb{P}(q_0,\dotsc,q_n)$. Do you have a reference (or perhaps a very short proof)? – Servaes Jun 13 '16 at 12:28
• Also, I'm confused by the claim that $\dim\mathfrak{m}_P/\mathfrak{m}_P^2=1$. I thought that because $\mathcal{O}_{\Bbb{P},P}$ is a Noetherian local ring we should have $$\dim\mathfrak{m}_P/\mathfrak{m}_P^2\geq\dim\mathcal{O}_{\Bbb{P},P}=3.$$ – Servaes Jun 13 '16 at 12:37