Consider the algebraic curve $\mathcal{C}$ given by $f(x,y)=0$, where $(x,y)\in\mathbb{C}^2$.

Suppose that the singular point of $f$ is $p=(x,y)=(0,0)$.

The blow-up of $p$ is given by $\{((x,y),[x_{1}:y_{1}])\in\mathbb{C}^{2}\times\mathbb{P}^1\}\rightarrow\mathbb{C}^2$. This produces the following two charts.

Chart 1

The blow-up of $p$ has the new coordinates $(x_{1},y_{1})$ given by $x_{1}=x$ and $y_{1}=\frac{y}{x}$.

The exceptional divisor, denoted as $E$, is given by $x_{1}=0$ and let $\widetilde{\mathcal{C}}$ denote the strict transform of $\mathcal{C}$.

Suppose that $E$ and $\widetilde{\mathcal{C}}$ intersect at $(x_{1},y_{1})=(0,\alpha)$, where $\alpha\neq 0$.

Chart 2

The blow-up of $p$ has the new coordinates $(x_{1}',y_{1}')$ given by $x_{1}'=\frac{x}{y}$ and $y_{1}'=y$.

The exceptional divisor, denoted as $E'$, is given by $y_{1}'=0$ and let $\widetilde{\mathcal{C}}'$ denote the strict transform of $\mathcal{C}$.

The result of Chart 1 implies that $E'$ and $\widetilde{\mathcal{C}}'$ should intersect at $(x_{1}',y_{1}')=(1/\alpha,0)$.


Why are the intersection points $(x_{1},y_{1})=(0,\alpha)$ and $(x_{1}',y_{1}')=(1/\alpha,0)$ equivalent? What is the geometric meaning of this?

Thanks, Jack.


2 Answers 2


Let me give an explicit example. Let $\mathcal{C}$ be a curve in $\mathbb{C}^2$ given by the vanishing of equation $f(x,y)=y^2-x^2-x^3$, a nodal cubic with a singularity at $(0,0)$. The blow up of $\mathcal{C}$ at point $(0,0)$ is defined as follows:

$Bl_{(0,0)} \mathbb{C}^2:= \{ ((x,y),[u:v]) \in \mathbb{C}^2 \times \mathbb{P}^1 \text{ } \vert \text{ } xv-yu=0 \} $, where $[u:v]$ are coordinates of $\mathbb{P}^1$ and $(x,y)$ are coordinates of $\mathbb{C}^2$.

Chart I. $u \neq 0$ $\implies u=1 \implies y=xv$

$\tilde{\mathcal{C}} \cup E$ in this chart is then given by equation $(xv)^2-x^2-x^3=0$ in $\mathbb{C}_{xv}^2$ where $x=0$ is the equation of the exceptional divisor $E$ in this chart and $v^2-1-x=0$ is the equation of $\tilde{\mathcal{C}}$ in this chart. Draw a picture - get a parabola intersecting a line. I.e. $\tilde{\mathcal{C}} \cap E = \{(0,1) , (0,-1) \}$. Emphasizing again: these are the intersection points in Chart I.

Chart II. $v \neq 0$ $\implies v=1 \implies x=yu$. In this chart, $\tilde{\mathcal{C}}$ is given by $1-u^2-yu^3=0$ in $\mathbb{C}_{uy}^2$ and the exceptional divisor is given by $y=0$. Get intersection points in Chart II: $\tilde{\mathcal{C}} \cap E = \{(1,0) , (-1,0) \}$.

Take, for example, the point $(0,1)$ in chart I and $(1,0)$ in chart II. They satisfy the criterion in your question. These 2 intersection points in different charts are "equivalent" or to be precise, they both represent the same point in the blow-up. Let's see that explicitly:

If we carefully go through what we did to the variables in Chart I, we see this: in Chart I, $u=1$, the point $(0,1)$ tells us $x=0, v=1$. Finally, $y=xv=0$. So, the point $(0,1)$ "back in the blow-up" is a point $((0,0),[1:1]) \in Bl_{(0,0)} \mathbb{C}^2$.

In Chart II, do the same. We get that $v=1$, $u=1, y=0$ and $x=0$. So again we get the point $((0,0),[1:1])$.

In this case, we can "see" the strict transform $\tilde{\mathcal{C}}$ in both charts. From my experience, we usually get something only in 1 chart. For example, if one blows up $V(x^2-y^3)$.

The simple thing is, both charts "see" a lot of the blow-up and sometimes they "see" overlapping things, that's normal, except very tedious to keep the track of.

  • $\begingroup$ Thank you. I overlooked the part where $[u,v]$ are the homogeneous in $((x,y),[u,v])$ so that $[u,v]=[1,a]=[1/a,1]$. $\endgroup$
    – Jack
    Mar 13, 2015 at 2:55

The blow-up of $\mathbb C^2$ in $(0,0)$ is not the entire set $\mathbb C^2\times \mathbb P^1$ but very imporantly, it is the incidence variety

$$\operatorname{Bl}_{(0,0)}(\mathbb C^2) = \{ (p,[q]) \mid p\in\mathbb C q \}\subseteq \mathbb C^2\times\mathbb P^1$$

The blow-up is then the projection $\pi:\operatorname{Bl}_{(0,0)}(\mathbb C^2)\to\mathbb C^2$.

The exceptional divisor of this blow-up is the projective line $E=\{(0,0)\}\times\mathbb P^1$.

So, you can consider those points $((x,y),[\tilde x:\tilde y])$ with $\tilde x\ne0$ and those with $\tilde y\ne0$. If you have $\tilde x\ne0$, you may scale to achieve $\tilde x=1$, then $(x,y)\in \mathbb C\cdot (1,\tilde y)$ so $x\tilde y = y$. For $x=0$ this implies $y=0$, so in this case your point is in $E$. If the point is not in $E$, then $\tilde y=\frac yx$ is defined by $(x,y)$, so the only point above $(x,y)$ in this case is $((x,y),[x:y])$.

This works analogously for the chart where $\tilde y\ne0$. If you have a point with $\tilde y\ne0$ and $\tilde x\ne0$, then still the only point above $(x,y)$ is the point $((x,y),[x:y])$.

I hope this helped? The geometric meaning is that you have $$\pi^{-1}(x,y) = \left\{\begin{array}{lcl}\{(0,0\}\times\mathbb P^1&;& (x,y)=(0,0)\\\{((x,y),[x:y]))\}&;&\text{otherwise}\end{array} \right.$$

so above every nonzero point of $\mathbb C^2$ there is a unique preimage - except for $(0,0)$ which has an entire copy of a $\mathbb P^1$ in its preimage. You may want to think of this $\mathbb P^1$ as parametrizing all the different possible tangents to a curve in the point $(0,0)$.

This way, if $\mathcal C$ has a singularity at $(0,0)$ which is due to the fact that it approaches the point with two distinct tangents, this blow-up is a strategy to separate these two approaches, pulling the curve slightly apart to remove the singularity. Mind you, though, a single blow-up will not remove every possible singularity.

  • $\begingroup$ Thank you very much for your help. $\endgroup$
    – Jack
    Mar 13, 2015 at 2:57

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