Pome

Example 21.13.15B328D39711DF

Example 21.13.15.B328D39711DF Let $\Sigma $ be the fan shown in 21.13.16.

Mathematical diagram from the source
C619293CCE8A21.13.16 The fan of $\mathrm{Bl}_0\mathbb {P}^2$. Adding the ray $\sigma _7$ generated by $(1,1)$ subdivides the cone $\sigma _1$ into two full-dimensional cones $\sigma _1',\sigma _1''$.

Comparing with our previous example 21.13.13, we have divided $\sigma _1$ from the middle, giving rise to two additional $2$-dimensional cones $\sigma _1'$ and $\sigma _1''$, and one additional $1$-dimensional cone $\sigma _7$.

The corresponding $X=\mathrm{Bl}_0 \mathbb {P}^2$ is just the blow-up of $\mathbb {P}^2$ at the origin $0=[1:0:0]$ and hence $L=\pi ^*\mathcal{O}_{\mathbb {P}^2}(1)$. Let $\pi \colon X\rightarrow \mathbb {P}^2$ denote the blow-up morphism. Let

\[ D=D_{\sigma _4}. \]

Then $D$ is the pull-back of the divisor $D$ in example 21.13.12. Note that $D$ is not ample, since it has degree $0$ on the exceptional divisor.

In this case, we have

\[ \Sigma (1)=\{ \sigma _4,\sigma _5,\sigma _6,\sigma _7\} , \]

and $D_{\sigma _7}$ is just the exceptional divisor.

The corresponding ray generators are

\[ u_{\sigma _4}=(-1,-1),\quad u_{\sigma _5}=(1,0),\quad u_{\sigma _6}=(0,1),\quad u_{\sigma _7}=(1,1), \]

while

\[ m_{\sigma _1'}=m_{\sigma _1''}=(0,0),\quad m_{\sigma _2}=(1,0),\quad m_{\sigma _3}=(0,1). \]

Therefore, $P_ D$ is the same as in 21.13.14.

We also observe that

\[ \begin{aligned} Q_{\sigma _1'}=& \{ (0,0)\} ,\quad Q_{\sigma _1''}=\{ (0,0)\} ,\quad Q_{\sigma _7}=\{ (0,0)\} ,\\ Q_{\sigma _2}=& \{ (1,0)\} ,\quad Q_{\sigma _3}=\{ (0,1)\} ,\\ Q_{\sigma _4}=& \{ (m_1,m_2):m_1\geq 0,m_2\geq 0,m_1+m_2=1\} ,\\ Q_{\sigma _5}=& \{ 0\} \times [0,1],\quad Q_{\sigma _6}= [0,1]\times \{ 0\} ,\\ Q_{\sigma _0}=& P_ D. \end{aligned} \]
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