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Puzzled
Puzzled
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The relative Picard group of a minimal resolution of a Du Val singularity is well understodunderstood.

Let $(X,p)$ be a normal surface singularity and $\epsilon:\widetilde{X}\rightarrow X$ be a minimal resolution. Let $Pic(\widetilde{X}/X) = Pic(\widetilde{X})/\epsilon^{*}Pic(X)$ be the relative Picard group.

If $p\in X$ is a canonical singularity (Du Val), and $E$ is the exceptional divisor, then $$Pic(\widetilde{X})\cong \epsilon^*Pic(X)\oplus H^{2}(E,\mathbb{Z}).$$

Furthermore $E = \bigcup_{i=1}^{n}E_i$ is a connected union of $(-2)$-curves intersecting transversally in at most one point. Therefore we have

$$Pic(\widetilde{X}/X)\cong H^{2}(E,\mathbb{Z}) = \bigoplus_{i=1}^n H^{2}(E_i,\mathbb{Z}) = \mathbb{Z}^n.$$

For instance:

  • If $p\in X$ is an $A_n$ singularity $(xy-z^{n+1} = 0)$ then $Pic(\widetilde{X}/X)\cong \mathbb{Z}^n$. Furthermore $Cl(X)\cong\mathbb{Z}/(n+1)\mathbb{Z}$. If $X$ is a quadric cone then the Picard number of $\widetilde{X}$ is two. The Picard group if generated by the strict transform a line through the vertex and the exceptional divisor.

You could take a look to these papers:

The relative Picard group of a minimal resolution of a Du Val singularity is well understod.

Let $(X,p)$ be a normal surface singularity and $\epsilon:\widetilde{X}\rightarrow X$ be a minimal resolution. Let $Pic(\widetilde{X}/X) = Pic(\widetilde{X})/\epsilon^{*}Pic(X)$ be the relative Picard group.

If $p\in X$ is a canonical singularity (Du Val), and $E$ is the exceptional divisor, then $$Pic(\widetilde{X})\cong \epsilon^*Pic(X)\oplus H^{2}(E,\mathbb{Z}).$$

Furthermore $E = \bigcup_{i=1}^{n}E_i$ is a connected union of $(-2)$-curves intersecting transversally in at most one point. Therefore we have

$$Pic(\widetilde{X}/X)\cong H^{2}(E,\mathbb{Z}) = \bigoplus_{i=1}^n H^{2}(E_i,\mathbb{Z}) = \mathbb{Z}^n.$$

For instance:

  • If $p\in X$ is an $A_n$ singularity $(xy-z^{n+1} = 0)$ then $Pic(\widetilde{X}/X)\cong \mathbb{Z}^n$. Furthermore $Cl(X)\cong\mathbb{Z}/(n+1)\mathbb{Z}$. If $X$ is a quadric cone then the Picard number of $\widetilde{X}$ is two. The Picard group if generated by the strict transform a line through the vertex and the exceptional divisor.

You could take a look to these papers:

The relative Picard group of a minimal resolution of a Du Val singularity is well understood.

Let $(X,p)$ be a normal surface singularity and $\epsilon:\widetilde{X}\rightarrow X$ be a minimal resolution. Let $Pic(\widetilde{X}/X) = Pic(\widetilde{X})/\epsilon^{*}Pic(X)$ be the relative Picard group.

If $p\in X$ is a canonical singularity (Du Val), and $E$ is the exceptional divisor, then $$Pic(\widetilde{X})\cong \epsilon^*Pic(X)\oplus H^{2}(E,\mathbb{Z}).$$

Furthermore $E = \bigcup_{i=1}^{n}E_i$ is a connected union of $(-2)$-curves intersecting transversally in at most one point. Therefore we have

$$Pic(\widetilde{X}/X)\cong H^{2}(E,\mathbb{Z}) = \bigoplus_{i=1}^n H^{2}(E_i,\mathbb{Z}) = \mathbb{Z}^n.$$

For instance:

  • If $p\in X$ is an $A_n$ singularity $(xy-z^{n+1} = 0)$ then $Pic(\widetilde{X}/X)\cong \mathbb{Z}^n$. Furthermore $Cl(X)\cong\mathbb{Z}/(n+1)\mathbb{Z}$. If $X$ is a quadric cone then the Picard number of $\widetilde{X}$ is two. The Picard group if generated by the strict transform a line through the vertex and the exceptional divisor.

You could take a look to these papers:

Source Link
Puzzled
Puzzled
  • 9k
  • 1
  • 38
  • 65

The relative Picard group of a minimal resolution of a Du Val singularity is well understod.

Let $(X,p)$ be a normal surface singularity and $\epsilon:\widetilde{X}\rightarrow X$ be a minimal resolution. Let $Pic(\widetilde{X}/X) = Pic(\widetilde{X})/\epsilon^{*}Pic(X)$ be the relative Picard group.

If $p\in X$ is a canonical singularity (Du Val), and $E$ is the exceptional divisor, then $$Pic(\widetilde{X})\cong \epsilon^*Pic(X)\oplus H^{2}(E,\mathbb{Z}).$$

Furthermore $E = \bigcup_{i=1}^{n}E_i$ is a connected union of $(-2)$-curves intersecting transversally in at most one point. Therefore we have

$$Pic(\widetilde{X}/X)\cong H^{2}(E,\mathbb{Z}) = \bigoplus_{i=1}^n H^{2}(E_i,\mathbb{Z}) = \mathbb{Z}^n.$$

For instance:

  • If $p\in X$ is an $A_n$ singularity $(xy-z^{n+1} = 0)$ then $Pic(\widetilde{X}/X)\cong \mathbb{Z}^n$. Furthermore $Cl(X)\cong\mathbb{Z}/(n+1)\mathbb{Z}$. If $X$ is a quadric cone then the Picard number of $\widetilde{X}$ is two. The Picard group if generated by the strict transform a line through the vertex and the exceptional divisor.

You could take a look to these papers: