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Let $a,b,c$ be mutually relatively prime positive integers. The weighted projective plane $X:=\Bbb CP^2(a,b,c)$ is the quotient space $\Bbb C^3-\{0\}/(z_1,z_2,z_3)\sim (\lambda^a z_1,\lambda^b z_2,\lambda^c z_3)$ for $\lambda\in \Bbb C-\{0\}$. $X$ is a singular 4-manifold with three singular points $[1,0,0], [0,1,0], [0,0,1]$, and near these singular points, $X$ is a cone on lens space. Let $Y$ be the complement of small open neighborhoods of the singular points. Then $Y$ is a smooth 4-manifold with boundary. How can we compute the intersection form of $Y$?

Since $H_2(Y)=\Bbb Z$, it suffices to compute $\alpha^2$ for a generator $\alpha \in H_2(Y)$. Such a generator $\alpha$ may be represented by an embedded surface in $H_2(Y)$ which I can't find.

According to Weighted projective plane as a quotient of $\Bbb CP^2$, $X$ can be regarded as a quotient space of $\Bbb CP^2$. So we have a quotient map $f:\Bbb CP^2\to X$ and a covering map $f:f^{-1}(Y)\to Y$. A generator $H$ of $H_2(\Bbb CP^2)=\Bbb Z$ is represented by a line in $\Bbb CP^2$ and its self-intersection is $1$. Using these, I tried to compute the self-intersection of $f($line$)$ but I got stuck in the simple case $(a,b,c)=(2,3,5)$ (Intersection of two lines in weighted projective plane).

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  • $\begingroup$ @hm2020 Can we get the information of cohomology ring of $X/G$ from its Chow group? $\endgroup$
    – blancket
    Commented Dec 4, 2021 at 19:07

1 Answer 1

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Question: "How can we compute the intersection form of Y?"

Answer: There is another construction of the weighted projective plane $X:=\mathbb{P}(a,b,c):=\mathbb{P}^2/G:=X/G$ where $G:=\mathbb{Z}/a×\mathbb{Z}/b×\mathbb{Z}/c$ which is a finite group (Harris "Algebraic geometry - a first course"). There is an explicit formula for its Chow group with rational coefficients:

$$CH^∗(X/G)≅CH^∗(X)^G.$$

@hm2020 Can we get the information of cohomology ring of $X/G$ from its Chow group?

Response: I do not have a reference but I believe there are similar formulas on the form

$$\text{ I1. }H^∗(X/G,\mathbb{Q})≅H^∗(X,\mathbb{Q})^G$$

when $H^∗(−)$ is singular cohomology. Since there is an isomorphism

$$CH^*(\mathbb{P}^2)_{\mathbb{Q}} \cong H^*(\mathbb{P}^2,\mathbb{Q})$$

it may be this induce an isomorphism

$$CH^*(\mathbb{P}(a,b,c))_{\mathbb{Q}}\cong CH^*(\mathbb{P}^2/G)_{\mathbb{Q}} \cong H^*(\mathbb{P}^2/G, \mathbb{Q}) \cong H^*(\mathbb{P}(a,b,c) ,\mathbb{Q})$$

but this must be checked. If this is the case it may help in the calculation of the intersection form of $Y$.

Question: "So we have a quotient map $f:\Bbb CP^2\to X$ and a covering map $f:f^{-1}(Y)\to Y$. A generator $H$ of $H_2(\Bbb CP^2)=\Bbb Z$ is represented by a line in $\Bbb CP^2$ and its self-intersection is $1$. Using these, I tried to compute the self-intersection of $f($line$)$ but I got stuck in the simple case $(a,b,c)=(2,3,5)$".

Whenever there is a quotient map $ \pi: Y \rightarrow Y/G$ (this holds if $Y$ is quasi-proejctive of finite type over a field) this induce a sequence (with rational coefficients $k$)

$$ CH^*(Y/G)_k \rightarrow^{\pi^*} CH^*(Y)_k^G \rightarrow CH^*(Y)_k$$

and an isomorphism $CH^*(Y/G)_k \cong CH^*(Y)_k^G$ of abelian groups, induced by $\pi^*$.

Note: The Chow group of the projective plane is $CH^*(\mathbb{P}^2) \cong \mathbb{Z}[t]/(t^3)$ by the "projective bundle formula". From this you get a formula for the Chow group of the weighted projective plane with rational coefficients. There are similar "projective bundle formulas" for singular cohomology, hence it should be possible to calculate

$$\phi:H^*(\mathbb{P}(a,b,c),\mathbb{Q}) \cong H^*(\mathbb{P}^2,\mathbb{Q})^G$$

if $I1$ holds. With such an isomorphism $\phi$ it is easy to calculate any product in $H^*(\mathbb{P}^2,\mathbb{Q})^G$ or $CH^*(\mathbb{P}^2)_\mathbb{Q}^G$. Take any pair of cycles

$$\alpha, \beta \in CH^*(Y/G)_k \cong CH^*(Y)_k^G$$

and calculate the product $\alpha \beta$ in $CH^*(Y)_k$ which by the above construction is well known. A similar construction holds for singular cohomology.

Note: The intersection product is defined geometrically for smooth schemes and the weighted projective plane is not smooth. Hence care must be taken when defining the "product": It is not completely clear how to define the product on $CH^*(X/G)_k$. You may for any subvarieties $V,W \subseteq X/G$ define

$$ [V]\bullet[W]:=\frac{1}{\# G} \eta_*(\pi^*[V]\pi^*[W])$$

where $\eta:\pi^{-1}(V \cap W) \rightarrow V \cap W$

is the canonical map. Here $\pi^*[V]\pi^*[W]$ is the product in $CH^*(X)_k$ and this product is well known by the above calculation. This product is "intrinsic" in a certain sense.

If $G$ acts on $H^*(X/G, \mathbb{Q})$ as ring-automorphisms, it follows the invariant ring has canonically a ring structure.

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