May 31st: A Comparison of Two Curves (Deciphering the Meaning of Normalization)

In this post, we compare the process of resolving singularities for the two curves $y^2 = x^2(x-1)$ and $y^2 = x^2(x+1)$ .

Curve 1: $y^2 = x^2(x-1)$ .

Notice that the singularity lies at the origin.

Key step: we put this curve into the blow up space of $\mathbb{R}^2$ at the origin (i.e. find a correspondence between the lines passing through the origin and the points on the curve). By this process, we find a correspondence between this curve (say C) and the real line $\mathbb{R}$ , it follows that the polynomial functions restricted to C has an (almost) one-to-one correspondence to the polynomial functions restricted to $\mathbb{R}$ . We can then find the function missing in $\mathbb{R}[x, y]/<y^2 - x^2(x-1)>$ but that is in $\mathbb{R}[t]$ . This function corresponds to our singular point.

Putting C into $Bl_0 \mathbb{A}_{\mathbb{R}^2}^2$ , every point on the curve is now expressed as $((x, y), [X, Y])$ , where $[X, Y]$ is the line passing through the origin and the pointe $(x, y)$ .

When $x \neq 0$ , the points can be expressed as $((x, y), [1, t])$ . The corresponding point in our original space $\mathbb{R}^2$ is given by $(x, tx)$ .

Substituting this into the equation for our curve gives:

$(tx)^2 = x^2(x-1)$ .

This simplifies to:

$x = t^2 + 1$

and so $y = t^2 + t$ .

Claim: $t \Rightarrow (t^2 +1, t^3 + t)$ is a bijection from $\mathbb{R}$ to C\{(0, 0)}. (I will not verify this in the post.)

What about the point at $(0, 0)$ ? What value of t does it correspond to? This will be $t = \infty$ .

Physical interpretation: The parameterization describes our movement from line to line in our space of lines. Adding the vertical line allows us to move in a ‘complete’ and ‘smooth’ loop. As we move along our curve by going from line to line, we can now travel from the ‘top’ of our curve to the ‘bottom’ by transition through that vertical line.

It follows that we have a bijection from $\mathbb{RP}^1 = \mathbb{R} \cup \infty \Rightarrow C$ .

This parameterization also gives $\mathbb{R}[x, y]/<y^2 - x^2(x-1)> \cong \mathbb{R}[t^2 + 1, t^3 + t]$ .

My observation: $t \notin \mathbb{R}[t^2 + 1, t^3 + t]$ ; $t$ satisfies $\frac{t^3 + t}{t^2 + 1} = t$ . Adding $t$ to $\mathbb{R}[t^2 + 1, t^3 + t]$ which gives us $\mathbb{R}[t]$ is equivalent to adding $\frac{y}{x}$ to $\mathbb{R}[x, y]/<y^2 - x^2(x-1)>$ . Notice that $y^2 = x^2(x-1)$ gives us $(\frac{y}{x})^2 - x + 1 = 0$ .

Key (don’t know if this is relevant):

$\frac{t^3 + t}{t^2 + 1} = t \Leftrightarrow (\frac{y}{x})^2 - x + 1 = 0$ .

Thus, adding $\latex t$ to $\latex \mathbb{R}[t^2 + 1, t^3 + t]$ is equivalent to adding $\frac{y}{x}$ to $\mathbb{R}[x, y]/<y^2 - x^2(x-1)>$ . This is equivalent to saying $\frac{y}{x}$ is well-defined near the origin, i.e. $t = \infty$ .

We can also geometrically see why the curve C is in bijection with a line through the following picture:

Now, let’s look at $y^2 - x^3 - x^2 = 0$ .

We again study the correspondence between points on the curve and the space of lines by putting this curve into $Bl_0 \mathbb{A}_{\mathbb{R}^2}^2$ .

When $x \neq 0$ , the points can be expressed as $((x, y), [1, t])$ . The corresponding point in our original space $\mathbb{R}^2$ is given by $(x, tx)$ .

Substituting this into the equation for our curve gives:

$(tx)^2 = x^2(x+1)$

We can simplify this to obtain:

$x = t^2 - 1$

and it follows that $y = t^3 - t$ .

Claim: the map $t \rightarrow (t^2 - 1, t^3 - t)$ is a bijection from $\mathbb{R}$ \{-1, 1} to C\{(0, 0)}.

The claim is easy to verify, so I will give a proof by picture:

Notice that the origin corresponds to two points (1, 1) and (1, -1) on the line.

Thus, we have a birational map from the curve (removed one point) to the line $\mathbb{R}$ (removed 2 points). The new curve is this line (or would it be $\mathbb{R}$ ?). The process of normalization is identifying the points -1 and 1 as the same point on $\mathbb{R}$ .

May 31st: A Comparison of Two Curves (Deciphering the Meaning of Normalization)

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