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I am reading Miles Reid's notes on weighted projective spaces, and I'm a little confused about a particular paragraph (notes here, page 8):

A famous case is the $E_8$ singularity $X: (x^2+y^3+z^5=0)$, which is naturally weighted homogeneous with weights 15,10,6. The $\mathbb{G}_m$ quotient morphism $X \to \mathbb{P}^1$ defined by the ratio $x^2:y^3:z^5$ has stabiliser of order 2, 3, and 5. The weighted blowup $Y \to X$ (the graph of the quotient morphism $X \to \mathbb{P}^1$) is a surface having cyclic quotient singularities of order 2,3,5 at the 3 points, giving rise to the Dynkin diagram of $E_8$.

I'd like to see this very explicitly. I agree that $\mathbb{P}(15,10,6) \cong \mathbb{P}^2$, and I can see that the equation $x^2 + y^3 + z^5$ becomes $u+v+w$ in $\mathbb{P}^2$, with coordinates $(u,v,w)$, so I agree that $X \to \mathbb{P}^1$. However, I am having trouble writing the equations for the graph and observing the singularities Reid describes.

Here is what I can do: The map $\mathbb{A}^3 \setminus \{0\} \to \mathbb{P}(15,10,6)$ is given by $(x,y,z) \mapsto [x:y:z]$, and the isomorphism $\mathbb{P}(15,10,6) \to \mathbb{P}^2$ is $[x:y:z] \mapsto [x^2:y^3:z^5]$. The graph of this map is $$ \Gamma = \{(x,y,z) \times [u:v:w] \,|\, uy^3=vx^2, uz^5=wx^2,wy^3=vz^5\} $$ Restricting to $w=-u-v$ I get the equations $(uy^3=vx^2, uz^5=(-u-v)x^2,(-u-v)y^3=vz^5)$. Taking partial derivatives, this appears to be singular everywhere. What have I done wrong?

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  • $\begingroup$ In the usual blowup I know of the $E_8$ surface singularity, the fiber over the singular point has $8$ components (all $\mathbb P^1$s), bearing the multiplicities 2,4,6,5,4,3,2 and 3. In particular, none of them is generically reduced, i.e. the fiber is singular everywhere. I'm not sure about the relation between that and this weighted blowup but don't find the everywhere-singularness surprising. $\endgroup$ Mar 3, 2016 at 19:30
  • $\begingroup$ Being a graph, the projection to $X$ had better be birational. To get the (weighted) blowup, I think you should impose the equation $z^5 = x^2 + y^3$ on the first set of variables, not on $u,v,w$. When $(x,y)$ is not $(0,0)$, you should be able to solve for $[u,v,w]$. And indeed, it is determined by $[u,v,w] = [x^2,y^3,-x^2-y^3]$. $\endgroup$ Mar 3, 2016 at 20:49
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    $\begingroup$ Great question! Check out this similar question by @kopper on StackExchange. $\endgroup$ Mar 3, 2016 at 22:10

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Edit: Your computation is correct. The weighted blowup $Y \to X$ as defined in Example 3.7 of Reid's notes (i.e. the graph of the quotient morphism $X \to \mathbb{P^1}$) is singular at all the points of the exceptional line. However, the statement about $Y$ having 3 cyclic singularities becomes true when you replace $Y$ by its normalization $Y'$.

As you write, $\Gamma$ is defined in $X \times \mathbb{P}^2$ by the equations $$uy^3 = vx^2,\ uz^5 = wx^2,\ wy^3 = vz^5,\ w+u+v = 0$$ Now $w+u+v = 0$ defines a hypersurface in $X \times \mathbb{P}^2$ isomorphic to $ X \times \mathbb{P}^1$, with coordinates $((x,y,z), [u:v])$. Now $\Gamma$ is defined in this $ X \times \mathbb{P}^1$ by your equations $$uy^3 = vx^2,\ uz^5 = -(u+v)x^2,\ -(u+v)y^3 = vz^5$$ It follows that on $ X \times \mathbb{A}^1$ where $v \neq 0$, $\Gamma$ is defined by $$u'y^3 = x^2$$ where $u' := u/v$. Let $U := \Gamma \cap \{v \neq 0\}$. As you noted, $U$ is singular at all the points on $u'$-axis. In particular, $U$ is not normal. Indeed, it is straightforward to note that $x/y, z^2/y$ and $xz/y^2$ are integral over the coordinate ring of $U$. It turns out that you need to only adjoin these elements to get the normalization $U'$ of $U$; i.e. $U'$ is the closure in $U \times \mathbb{A}^3$ of the graph of the map $U\setminus\{y=0\} \to \mathbb{A}^3$ given by $(u',x,y,z) \mapsto (x/y, z^2/y, xz/y^2)$. $U'$ has $2$ singular points: $(0,\ldots,0) $ and $(-1,0,\ldots, 0) $ (where the first coordinate corresponds to $u'$) - corresponding respectively to the stabilizers of $x=0$ and $z=0$ on $X$. The other singular point (corresponding to the stabilizer of $y=0$ on $X$) on $Y'$ is on (the normalization of) the chart $\Gamma \cap \{u \neq 0\}$.

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  • $\begingroup$ This might me a stupid question, but what is the relationship between the weight blowup and regular blowup (i.e., blowup $\mathbb C^3$ at the origin and look at the proper transform of $x^2+y^3+z^5=0$)? I know the minimal resolution of such singularity is obtained by successive regular blowups at isolated singular points. $\endgroup$
    – AG learner
    Mar 9, 2020 at 3:37
  • $\begingroup$ @AGlearner: what do you mean by "relationship"? If you choose a system of local coordinates $(x_1, \ldots, x_n)$ centered at a nonsingular point of an $n$-dimensional variety, then a weighted blow-up (with respect to these coordinates) at this point would be the graph of the map $x \to [x_1^{w_1}: \cdots: x_n^{w_n}]$, where the square brackets denote the homogeneous coordinate on $\mathbb{P}^n$. If $w_1 = \cdots = w_n = 1$, then you have the regular blowup. Unlike the regular one, the weighted blow-up depends on the system of coordinates. But I suspect you already know all this. $\endgroup$
    – pinaki
    Mar 10, 2020 at 15:00

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